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in this lesson we're going to talk about the real number system and specifically we're going to learn about classifying numbers within our real number system so basically we're going to discuss the definition of the natural numbers the whole numbers the integers the rational numbers and the irrational numbers so we're going to start out today by talking about the natural numbers these are also known as the counting numbers so i've kind of set this up to where we're going to look at everything in set notation if you don't know what a set is essentially a set is just a collection of stuff so we've named our set here we have an n which just represents the natural numbers and inside of these curly braces these are known as set braces so inside these braces we just put the elements or members of the set so what we have is elements or members of the set of natural numbers we start with the number one that's the smallest natural number and then we increase in increments of one indefinitely so obviously if we're starting with one and we increase by one we get to two then if we do that again we get to three then four then five so on and so forth now because we can't obviously list all of the natural numbers because it goes on forever and ever and ever we put these three dots here right the three dots this is an ellipsis so it just says that the pattern is going to continue forever we're going to run into this a lot in our study of mathematics so again after 4 would come 5 then 6 and 7 so on and so forth now another thing we can do with our set of numbers here we can draw a visual representation of the set of natural numbers so this right here is a horizontal number line so we're going to encounter a lot of different number lines throughout our study of mathematics this is probably the simplest one you're going to work with just a horizontal number line you'll see at the left most point here we have a notch here which represents the smallest natural number which is one as we move to the right we have additional notches that represent a distance of one unit more or an increase of one so in other words if i move to the right from one i get to two then three then four then five you know so on and so forth so this is another important concept in math as we move to the right our numbers are increasing as we move to the left our numbers are decreasing so this is just a way to visually represent this set and i just want to call your attention to one last thing here you don't have an arrow on the left because the smallest natural number is listed you have an arrow on the right because we don't list the largest natural number because there is none right it's an infinite set it just continues forever and ever and ever if i thought about the largest natural number i could let's say it's just one trillion well i can just add one of that one trillion and one so there is no largest natural number so that's why we have that arrow there and that's why we have the ellipsis here all right now let's talk about the whole number so the whole numbers essentially are just the natural numbers with the inclusion of the number zero so this allows us to count nothing right so we have w here as the set name and then the set contains the elements 0 1 2 3 and then again we have the ellipses here to show that hey this pattern is going to continue forever now again we have a number line that shows this our smallest number is now zero it's not one and again there's no arrow on the left because we have the smallest number and this continues forever in the right direction right because numbers increase moving to the right so again we have our arrow here on the right because there will be no largest whole number so one more thing i want to bring your attention to you can see that the number line for the whole numbers and the number line for the natural numbers are basically identical with the exception of having a zero so this tells us that the natural numbers are all whole numbers but the whole numbers are not all natural numbers so in other words if you've talked about sets before you'll understand this as the natural numbers are a subset of the whole numbers meaning every natural number can be thought of as a whole number but it's not the same vice versa right not every whole number meaning 0 can be thought of as a natural number so next we want to think about the integers so in our everyday life we want to be able to think about negative values so suppose you have 20 in your bank account and you take your debit card and you go out and you spend 30. well now your bank balance is going to be negative 10. it'll actually be a little bit less than that because your bank's probably going to charge you you know some sort of fee for going overdraft but if it wasn't an overdraft fee let's just say your bank balance is negative 10 now the integers allow us to count negative values like that it also would allow us to think about negative temperatures or you know going below c level there's all kinds of applications for the set of numbers known as the integers so we're going to call our set z and we have an ellipsis here and here which tells us that i'm going to kind of start on the middle at 0 here if i decrease by 1 i'm going to go to negative 1 the negative 2 the negative 3. so after negative 3 i'd have negative 4 the negative 5 the negative 6. so this guy right here is telling me i'm decreasing by 1 indefinitely over on the right if i start from 0 again and i go to the right well i'm increasing by 1 so i'm going to 1 then 2 then 3 then 4. you know so on and so forth now when i think about the integers again all of the whole numbers so all of the sets that i've worked with before the smaller sets so the whole numbers is going to be a smaller set than the integers right because they don't have any negative numbers and then the natural numbers are going to be a smaller set than the whole numbers because it doesn't have 0. so every time i kind of work myself backwards i can say okay well every integer is not going to be a whole number but every whole number is going to be an integer and every natural number is going to be an integer right if i look at the integers i can see that i have 1 2 3 and then you know i have that same pattern so i have my natural numbers there or i can say okay well i have 0 1 2 3 and then again my ellipsis i have my whole numbers there so every natural number is an integer and every whole number is an integer but not every integer is a whole number or a natural number all right so we can again visually represent this set using a number line if we look at our number line here it's drawn a little bit differently you'll see you have an arrow at the left and an arrow at the right that's because there is no smallest integer and there is no largest integer so we have numbers continuing in both directions forever you can kind of think about the center point of this as 0. if i move to the right i'm increasing right so i go by 1. so i go to 1 then 2 then 3 then 4 you know so on and so forth if i start at 0 and i go to the left well i'm decreasing by 1. so i go to negative 1 negative 2 negative 3 negative 4 negative 5 negative 6. again so on and so forth now let's think about the rational numbers so for this guy we've written it in what's known as set builder notation in case you're unfamiliar with this notation i'll walk you through it it's pretty easy so we have a capital letter q that represents the rational numbers and i'm just going to read this and then i'll explain it so it's the set of all elements p over q such that p and q are elements of z so this set z we talked about above this is the set of integers and we have another condition that q is not equal to zero so for set builder notation it always starts out by saying the set of all elements and then whatever this is so in this case it's lowercase p over lowercase q and let me make that p to where it looks a little more lower case and then we have this vertical bar here this means such that so this is where we're going to list a condition so this right here is the condition this is the condition and our condition is that p is an element of the set z q is an element of the set z so in other words p and q are both integers and then the last condition is that q because it's in the denominator of this fraction is not going to be equal to 0 because again we cannot divide by 0. so p and q are integers and q is not equal to 0. now we can take any integer we want and write it as a rational number so again as we move up we think about the numbers that were covered before they're a smaller set so we can take every integer and write it as a rational number we can take every whole number right as a rational number we can take every natural number and write as a rational number all right so what i'm going to say here is that every integer is a rational number every whole number is a rational number and every natural number is a rational number but not every rational number is an integer not every rational number is a whole number not every rational number is a natural number and you can think about how you could take an integer like let's just say the number six how can i write this as a rational number well from the definition it's the numerator and denominator of this guy have to be integers well i can take any integer i want and just write it over the number one so six over one is the same thing as six right so this is how i would write an integer or a whole number or a natural number as a rational number so let's think about some rational numbers that are not integers so one would be something like you have an integer four over an integer five right so this fits the definition we have up here we have an integer four of an integer five this is not an integer this is not a whole number this is not a natural number if i do the division here and write this in decimal form this is going to be 0.8 now this is an important point about rational numbers rational numbers in decimal form will either terminate like this one terminated there's nothing after the eight or will repeat the same pattern of digits forever okay so a good example of this let's say i had 2 9 so 2 9 and we all know how to do basic division right if i divide 2 by 9 i know i'm going to put a decimal point there bring this up here put a 0 here so i can do my division 9 would go into 20 twice so put a 2 here 2 times 9 is 18. subtract and i would get 2. so to continue this division i would put a 0 here and bring this down and i would have 20 again well i started out with that so let's do this again so 9 goes into 20 how many times twice 2 times 9 is 18. subtract then i get 2. put a 0 bring this down i got 20 again so you can see that this would continue forever and ever and ever so in other words this would be 0.2222 i could put as many 2s as i'd like so to represent this concept what we'll do is we'll put a bar over the two to say that it continues forever or another thing you can do we can use the ellipsis here but we've got to make it completely crystal clear i'm going to put 0.222 [Music] so it's crystal clear that that 2 is going to continue forever and then put my three dots there so that's two different ways you can kind of notate a repeating decimal but the important thing here is to understand that if you have a rational number in decimal form it's either going to repeat the same pattern forever or it's going to terminate so now let's talk about the irrational numbers so the irrational numbers are basically real numbers that are not rational numbers so we have this listed as p okay so p is standing for the irrational numbers and again i have it in set builder notation so the set of all elements x so in other words any real number that you can think of such that x is real but not rational so in the real number system any real number that's not rational is by definition irrational so even if you've never heard of an irrational number before you've probably heard of pi right it's very famous it's the ratio in geometry of a circle circumference to its diameter right so the symbol for it looks like this and basically it's approximately 3.14 but after that four they have an infinite number of digits that continues and there's no pattern to it okay so it just continues forever and ever and ever and you can list as many digits of pi as you want you'll never finish right another one that's pretty famous is the number e and e is used in a lot of applications we'll see it later on in this course and it deals with problems that involve growth decay things like that you might also see something like the square root of two this is something you can punch in on your calculator right now and your calculator depending on how it works will give you a series of digits it's going to go 1.41 and then after the 1 it's going to give you some digits but at some point it's going to stop it's either going to truncate mean it's just going to cut it off or it's going to do some type of rounding procedure but there is no exact decimal answer that you can give for the square root of 2 because again the decimal continues forever and ever and ever without a pattern so these are three examples of irrational numbers all right so then for the real numbers a real number is any number that is either rational or irrational so if we look at the number line we use for the integers here it doesn't have to just represent the integers this can represent the set of all real numbers although our notches are in increments of 1 i can go in here and put a notch anywhere i want so between 0 and 1 i can put a notch there for the number one half right that's a rational number and i could put a notch there somewhere for the square root of 2. let's just say that's about right here i don't know exactly where it would be but let's just say this is the square root of 2. so with the real numbers you can basically say that it represents a point on the number line where the number line is just all the real numbers together and when we think about an official definition for the real numbers basically a real number is any number that is either rational which means again we can write it as the quotient of two integers where the denominator is not 0 or it's irrational right it's something like pi or e or square root of 2 or square root of 5 something like that so the set of all x such that x is an element of the set of q which we said were the rational numbers where x is an element of the set p which we said were the irrational numbers so we have a little diagram here to hopefully make this clear for you this can be helpful to understand that no rational number can be an irrational number also so you're either an irrational number or a rational number you can't be both if we're thinking about this in terms of sets these two sets the irrational numbers and the rational numbers are disjoint right they have no elements in common so irrational numbers we know already this is something like pi or e or square root of 2 or square root of 5. but with the rational numbers we can think about this if i start at kind of the smallest set here well for the natural numbers i know this could be something like the number 1 or 2 or something like that well this is also a whole number it's also an integer and it's also a rational number okay so we kind of work our way up with whole numbers we just take the natural numbers and we include 0. so you could say 0 1 2 you know with integers we have all the whole numbers but we also have the negatives of those so you could say you know negative 1 0 1 you know something like that and then with the rational numbers we have all the integers all the whole numbers all the natural numbers we just include all the you know quotients that we can make with two integers so we would have our four-fifths we would have our two-ninths we would have our negative 1 8 you know so on and so forth so let's wrap up our lesson by just looking at a quick sample problem so we have set a which is equal to we have the elements which are negative 3 14 negative 1 7 0 square root of 5 and 2 9. so what elements of set a are natural numbers so we know the set of natural numbers starts with 1 increases by 1 indefinitely so we know negative 3 is not right because it's less than 1 14 would be negative 1 7 is not 0 is not because that's a whole number square root of 5 is not and 2 9 is not so what elements of set a are natural numbers just 14. all right so same said now we're asking what elements of set a are whole numbers so again we know all natural numbers are whole numbers so we know 14 would go in there and we also know zero is a whole number so that would be in there as well so we would have 0 and 14 only all right so what elements of set a are integers again i'm going to always list what was listed before so whole numbers we had 0 and 14 and now we want to think about negative whole numbers so basically any whole number with a negative sign in front of it so we have negative 3 that we can add to this list so we put negative 3 over here so we have negative 3 0 and 14. all right so what elements of set a are rational numbers so again i'm going to take what i just listed so i have negative 3 0 and 14. these can all be written as rational numbers because i could put them each over 1. so we would be good there so what else do we have in here now so i can kind of not think about this or this or this because they're already listed negative one over seven is that the quotient of two integers with a non-zero denominator yes it is so i can kind of throw that in there and that's gone what about 2 9. well yeah i can throw that in there as well that's the quotient of two integers with a non-zero denominator what about the square root of five if you think about the square root of five this is an irrational number one way you can tell is just punching that up on a calculator you're going to get a nasty nasty decimal with no pattern that is going to continue forever or anything like that it's just a non-terminating non-repeating decimal so we know this is an irrational number so for the rational numbers we have negative 1 7 negative 3 0 14 and 2 9 and then lastly what elements of set a are irrational numbers we just have the square root of 5. in this lesson we want to talk about exponents and the order of operations at this point you should be fairly comfortable working with exponents and simplifying any type of numerical or algebraic expression that you come across it's really important that we take a moment and review this type of material so that you don't get stuck on this later on in the course when we're working with more advanced topics so we're going to start out today by just talking about what is an exponent and why are exponents used in the study of mathematics so most of us start learning about exponents when we're in pre-algebra some of us wait until algebra to learn about exponents and others will be exposed to exponents in grammar school no matter what your experience is with exponents you start out by learning about exponents that are whole number exponents larger than the number one so in other words you start with exponents of two then three then four you know so on and so forth so these type of exponents are used to write a repeated multiplication of the same number in a more compact form so let me give you an example if we had something like 9 times 9 times 9 times 9 times 9 times 9 this takes up a lot of room on my screen what i can do i can write this in exponential form or exponential notation what i would want to figure out here is how many factors of this number 9 do i have so i can count i have 1 2 3 4 5 and then finally 6. so i have 6 factors of 9. so to write this using an exponent i would say this is equal to the number that's being multiplied by itself in this case that's 9 this is referred to as the base this is the larger number this is the first number you're going to write so let me just label this as the base again this is the number that's multiplied by itself and we have 6 factors of this 9. so what we're going to do is we're going to write an exponent of 6. so this is going to go at the top right of the number and this tells me again how many factors of 9 that we have so this 6 is my exponent so when you work with a whole number exponent that's larger than 1 it always follows this format so as another example let's say i had something like 4 times 4 times 4 times 4 times 4 times 4 times 4 times 4. if i count this up i have 1 2 3 4 five six seven eight so eight factors eight factors of four so that means i can write this as four right that's the number that's being multiplied by itself that's going to be my base raised to the eighth power right because i have eight factors of four so four raised to the eighth power so eight is my exponent all right so let's look at a very simple example we're going to write each repeated multiplication in exponent form we're starting out with 2 times 2 times 2 times 2 so we have 4 factors of 2 so this means we would write this as 2 raised to the 4th power again this 2 is the base it's the number that's multiplied by itself the 4 is the exponent it's the number of factors of 2 that we have all right then we have 5 times 5 times 5 we have 3 factors of the number five so we'll have five to the third power otherwise known as five cubed right if you have something that is raised to the second power you say it's squared something raised to the third power you say it's cubed all right as another example we're going to evaluate each which means we're going to find the value 4 and then we're going to identify the base and the exponent so we have 7 to the second power again if something raised to the second power we say it's squared so we'd say 7 squared so to evaluate this i would write it as a repeated multiplication this 2 tells me i have 2 factors of the number 7. so 7 times 7 and now i can just multiply 7 times 7 is 49. guy also wants me to identify the base and the exponent so the base is 7 and let me make that a little better and the exponent is 2. as another example we have four to the fifth power so to evaluate this we would have four times four times four times four times four right five factors of the base 4. so again this is the base this is the exponent and we can do this without a calculator although this is kind of a big number 4 times 4 is 16 16 times 464. 64 times 4 is 256 and then 256 times 4 is 1024 and kind of an easy way to do that in your head if you think about 250 times four that's easy it's a thousand and then you could take the sixth part multiply that by four and that would give you the 24. so you can just add those two numbers together and you get a thousand twenty-four all right now let's talk about exponents with a negative base so this is not negative exponents please don't be confused here we're talking about exponents that have a negative base so when working with a number in exponential form where we have a negative base we have to be extra careful the negative is not part of the base unless it is wrapped inside of a set of parentheses so let me show you the difference between these two if we have negative 3 inside of parentheses squared this is equal to negative 3 multiplied by negative 3 which is positive 9. if we have negative 3 squared like this by rule we are to treat this as the negative negative one multiplied by three squared the base here is three it is not negative three this is a very common source of confusion so by the rules of the order of operations we know we need to apply the exponent before we multiply so this would be negative 1 times 9 which is negative 9. so that's why you get two different answers here and if you don't believe me punch up negative 3 and then squared on your calculator it will give you negative 9 as a result then take negative 3 and wrap it in parentheses and square that on your calculator and you will get positive 9 as a result now something we want to talk about before we kind of move further if we think about exponents with a negative base it's only going to have an impact if the exponent is even okay so if i saw something like i don't know let's say negative 4 and this is cubed or i had negative 4 like this cubed i'm going to end up with the same answer right although it's evaluated different this is negative 1 times 4 cubed which is negative 1 times 4 cubed is going to be 4 times 4 which is 16 times 4 which is 64. so negative 1 times 64 which is negative 64. the process won't be the same here but i'll get the same result so this is negative 4 times negative 4 times negative 4 because again the parentheses surround the negative and the 4. so this becomes what negative 4 times negative 4 is positive 16 positive 16 times negative 4 is negative 64. although again we had different methods here we get the same answer and the reason for that is because we have an odd exponent in this case an odd number of negative factors will always give you a negative result so it doesn't matter in the case we have an odd exponent with an even exponent an even number of negative factors will give you a positive result so it will definitely change the answer whether you have parentheses around the base or not when you have a negative number involved all right let's look at a quick example here so we want to evaluate each and then identify the base and the exponent so we have negative 5 where negative 5 is wrapped inside of parentheses and this is cubed so i'm just going to identify the base as negative five so the base is negative five the exponent is three to evaluate this i would have negative five times negative five times negative five and that would give me negative 5 times negative 5 is 25 25 times negative 5 is negative 125. and again it didn't matter that i had parentheses around the base here we have an odd number for our exponent right 3 is an odd number so we ended up with an odd number of negative factors right you have 1 2 3 negatives there so you're going to end up with a negative product right so you get negative 125. all right what about negative 8 squared with no parentheses around the base so here we're going to say that our base is 8 it is not negative 8 it is 8. because really i could write this as negative 1 times 8 squared so the base is 8 and the exponent the exponent is 2. all right so we have negative 1 times 8 squared which is basically negative 1 times 64 right 8 times 8 is 64. so this is just equal to negative 64. again if i had set this up as inside of parentheses negative 8 squared well then the negative and the 8 are both squared so this is negative 8 times negative 8 which is 64. so please don't make this mistake when you're evaluating expressions or if you're working on something like that make sure you understand the difference between when you have parentheses around a negative and when you don't because it can definitely change your answer all right so let's talk about the order of operations most of us will remember this as pemdas pam or please excuse my dear aunt sally right so it stands for parentheses exponents multiplication and division addition and subtraction so this is basically the method that you're going to use whenever you evaluate any type of numerical expression or algebraic expression and you have multiple operations involved right so you need to know the order to perform the operations in so that you get the correct answer so the first one here isn't really a step it's just a general guideline a lot of people make this mistake you need to work separately above and below any fraction bar so that's very important you want to separate the two so work the numerator completely work the denominator completely so make sure those are completely simplified and then you can perform the main division all right so our first real step here is to work inside of any grouping symbols so this is where the p comes in pemdas right parentheses where really it doesn't have to be parentheses it can be brackets it can be absolute value bars it could be any type of grouping symbols that are present these take the absolute priority over everything and if you have multiple grouping symbols you basically start at the innermost set on work outward it's kind of like when you get inside of a set of parentheses you have to reapply the order of operations the next thing we want to do the e in pemdas stands for exponents but this means we will simplify powers and roots so you could also have a square root a cube root something like that so these have kind of equal priority levels and we're going to evaluate these working from left to right so if i had a power on the left i would do that one first versus something that i saw later on on the right now the next thing the m and the d in pemdas and this is a source of confusion stands for perform all the multiplication and division because m comes before d and pemdas students always assume that multiplication is more important than division it's not they're the same priority level we work these left to right so i have working left to right so if you had something like 3 times 5 divided by let's say 5 you would make sure to multiply first before you divide because in this particular case the multiplication occurs to the left of the division but if i had something like 20 divided by 4 times 3 i would divide before i multiply because the division is to the left of the multiplication all right the last step in our order of operations it's the a and the s in pemdas it's perform all addition and subtraction again these have the same priority level it's the same thing as we just saw with multiplication division it's working left to right so again don't assume because the a comes before the s that addition is more important it's worked left to right all right let's take a look at an example so we have a fraction here so the first thing i would note before i do anything is that if you didn't write this as a fraction what you would do is you would put parentheses around the numerator so 3 squared minus 5 times 17 plus 81 divided by 9. so parentheses around that and then divided by put this in parentheses so you have the absolute value of negative 10 squared plus 15 and then minus 6 times the square root of 9. this is separate from this so i would simplify in here simplify in here and then do that division that's what we need to do when we have a fraction bar involved so very very important to understand that so let me erase this we don't even need to write it like this we're just going to work with it in fractional form but if you didn't want to that's how you would set it up all right so the first thing i'm just going to work in my numerator i'm going to simplify that completely and then i'll look at the denominator so what do i have i have an exponent i have subtraction multiplication addition and division so no parentheses or grouping symbols in the numerator so i'm going to start with my highest priority which is to apply the exponents so 3 squared is 9. so i'm just going to replace 3 squared with 9 and then i'll copy so minus 5 times 17 plus 81 divided by 9 over we have the absolute value of negative 10 squared plus 15 and then minus 6 times the square root of 9. okay so the next thing if we look we have subtraction multiplication addition and division multiplication is the highest priority here so we want to work on 5 times 17 next so what is 5 times 17 well 5 times 10 is 50 5 times 7 is 35 if you add those two together 50 plus 35 is 85. so i would have 85 so 9 minus 85 and then plus 81 divided by 9. again this is over we have the absolute value of negative 10 squared plus 15 and then minus 6 times the square root of 9. okay so now we have subtraction addition and division division is the highest priority so we're going to work on 81 divided by 9. 81 divided by 9 is 9 and just to save kind of some space here let's just erase this and put a 9 and now we have subtraction and addition subtraction occurs to the left of addition so it's a higher priority so 9 minus 85 is going to give me negative 76. so let's erase this and put negative 76 and then negative 76 plus 9 would give me negative 67. so negative 67. okay so the numerator is simplified i don't need to worry about it anymore i can just work in my denominator so in the denominator we do have grouping symbols we have the absolute value operation so we start inside of there first so once we get inside of there we want to reapply the order of operations we have negative 10 squared plus 15. now this negative here is not inside of a set of parentheses so basically i have negative 1 times 10 squared or negative 1 times 100 so let's put this is equal to negative 67 over the absolute value of this right here will be negative 100 and then plus 15 and then we have minus 6 times the square root of 9. kind of finishing this up negative 100 plus 15 is negative 85 and in other words if we had negative 85 inside of absolute value bars we take the absolute value we get 85. so we would have negative 67 over 85 minus 6 times the square root of 9. okay let me scroll down get a little room going so now in my denominator i have subtraction i have multiplication 6 times the square root of 9 and i have a root right i have the square root of 9. so square root of 9 will have the highest priority square root of 9 is 3. so we can really write this as 6 times 3 here so now we have subtraction and multiplication multiplication is a higher priority 6 times 3 here would be 18. so we'll end up with 85 minus 18 which is 67 so this will end up giving me negative 67 over 67 or in other words negative 67 divided by 67 which we know is negative one right this would cancel with this and just leave me with negative 1. in this lesson we want to review ratios rates and proportions so for any of you that took a pre-algebra or an elementary algebra course you covered ratios rates and proportions pretty thoroughly but again it's just something we want to review so you don't get caught up on something very trivial later on in the course so let's begin our lesson today by thinking about ratios so what is a ratio in math it's basically just a comparison of two quantities so how much of one thing there is compared to how much of another thing there is so if we had something like pineapples and oranges we could compare the number of pineapples we have to the number of oranges that we have okay we can use a ratio to do that so if i said what is the ratio of pineapples to oranges what would be your answer so what is the ratio of pineapples to oranges okay and when we work with ratios you're going to see this keyword 2 a lot so essentially what i want to do i want to count the number of pineapples that i have so i have one two three four five six i have six pineapples and then i want to count the number of oranges i have i have 1 2 two three four five six i have two rows of six so that would be twelve so i have twelve oranges okay so six pineapples and twelve oranges so if i want the ratio of pineapples to oranges all i need to do since i said pineapples first i'm just going to basically put the number six which is the number of pineapples i have and then the word two okay that's a key word when you're working with ratios and then since i have oranges last i'm just going to put the number of oranges which is 12. so the ratio of pineapples to oranges is 6 to 12. okay now i can simplify that it means something to say i have 6 to 12 but it means even more if i simplify that and say okay well between 6 and 12 the greatest common factor there or the greatest common divisor you could say would be 6. so if i divide 6 by 6 i would get 1. if i divide 12 by 6 i would get 2. so now i have something that has some meaning if i said the ratio of pineapples to oranges was 1 to 2 then really you could start to think in your head if you didn't have a picture that i have one pineapple for every two oranges okay so it becomes kind of clear what the relationship is between pineapples and oranges now if i reverse the wording here and i said i wanted the ratio of oranges to pineapples what i would do is i would just reverse the numbers so instead of one to two i would put two the number of oranges two the key word and then one the amount of pineapples i have per two oranges okay so now i'm saying that i have two oranges for every one pineapple now we don't have to write a ratio like this there's other ways so instead of using the keyword two we can also write a colon between the numbers okay and then also which is most common we can use a fraction so we can say two to one like this okay now you might be tempted to kind of write this as just the number two right if we see a fraction two over one or if we see a fraction four over one or five over one or a million over 1 we're tempted to kind of get rid of the denominator and just write the numerator part most of us would just say okay this is 2. but with a ratio it represents a relationship it's the amount of oranges to the amount of pineapples so i need that one there to express the second part of the relationship so i would leave this as two to one in fractional form like that all right let's take a look at a very simple example so on a local church trip there are 18 boys and 27 girls what is the ratio of boys to girls and then what is the ratio of girls to boys so the first thing is what is the ratio of boys to girls so boys to girls okay well it's very easy i just list the number of boys first which in this case there's 18 boys so just put 18. again the wording matters because boys came first the number of boys came first then you could put 2 you could put a colon you could write it as a fraction whatever you want to do and then you put the number of girls because that comes second so in this case that's 27 and you can reduce this a lot of people like to write this as a fraction because they're used to reducing fractions and it makes it mentally easier for them if you want to do it that way that's fine if you want to put the word 2 it's fine if you want to put a colon it's fine it's all the same so 18 over 27 the greatest common factor of 18 and 27 would be 9. if i divide 18 by 9 i get 2 if i divide 27 by 9 i get 3. so essentially the ratio of boys to girls is 2 to 3 okay so that's telling me that i have two boys for every three girls now if it asks for what is the ratio of girls to boys well then girls two boys i would just flip this around right if i have two boys for every three girls then i have three girls for every two boys right it's the ratio of three to two all right let's look at a little application problem so a local bar is having a promotion party for 98 customers okay so remember this number 98 customers each customer will receive a free vodka bottle okay so they're giving out presents there are two different bottle types one is pink and one is black now the pink vodka bottles will be given to the female customers and the black vodka bottles will be given to the male customers if the ratio of males to females is five to two how many of each bottle type should be prepared so in other words the bar owners want to know how many black vodka bottles they need to get ready and how many pink vodka bottles need to be ready given that again the ratio of males to females is five to two and they're going to have 98 customers okay so how do you solve something like this we don't need a variable we don't need to get into algebra how do we solve this with just some plain math well the first thing is to realize that if i have the ratio of males to females and let me just write this here we have males to females and that's five to two right so what does that tell me that tells me in a group of seven people that i've got five males and two females so in a group of seven you've got five males and you've got two females okay so that means if i had 98 total customers how many groups of seven could i make well 98 divided by 7 is 14. so i'd have 14 groups where each group had five males and two females okay well it's pretty simple from here if i want to know how many males are going to go i just multiply 14 the number of groups of seven that i can make by five the number of males in each group so 14 times 5 is going to give me 70. right so there would be 70 males let me make that better and then if i wanted to know how many females again i would multiply 14 times 2 so 14 times 2 is 28 so there would be 28 females does that make mathematical sense yes it does 70 plus 28 is 98 that's the total number of customers in the bar right for that party and then if you look at 70 to 28 if i erase all this it should still be a ratio of five to two right if i said i had 70 males in a bar and 28 females in a bar i should get the ratio of 5 to 2. if you divide 70 by 14 you get 5. if you divide 28 by 14 you get 2. so you get 5 to 2 there so to answer our question here to answer our question here again it says how many of each bottle type should be prepared we would say that he's going to need 70 black bottles black bottles then he's going to need 28 pink bottles okay pretty simple overall all right let's talk a little bit about rates so a rate is a special type of ratio where the units are different generally speaking when we work with rates we're going to be talking about unit rates so how much of some quantity relates to one unit of another quantity a very easy example to understand is miles per gallon right if you're driving in your car so how many miles do you get to go per one gallon of gas and the way you find this is you basically fill your tank up and you drive around right you you might go a certain amount of miles and then you divide it by the number of gallons that were in your tank when it gets to completely empty you can say i went this many miles per each gallon of gasoline right so for this blue car it goes 700 miles per 10 gallons and for this red car it goes 600 miles per 6 gallons so to kind of compare them right now it's a little bit up in the air you might be able to do the math in your head and say well i know this car has a better miles per gallon but it's a little bit easier if we just divide the numerator by denominator so 700 divided by 10 we know that we would just basically cancel this with one of the zeros so we could say that this is 70 you could say miles over one gallon mostly we write this as 70 mpg right miles per gallon but this is the official way to write it and then this guy i know that 600 divided by six would be a hundred so this one would be a hundred miles per one gallon okay so which car gets the better gas mileage of course this one does right it goes 100 miles per every one gallon whereas this one only goes 70 miles per 1 gallon all right let's take a look at a quick problem so jeremy works at a local bank and earns 760 for a 40 hour work week his sister beth works at a local school and only earns 735 dollars for a 35 hour work week who makes more on a per hour basis so very easy problem to solve i would just take jeremy's information so he makes seven hundred sixty dollars for a forty hour work week and you can put seven or sixty dollars over forty hours like this and then we would compare this to for beth she makes 735 dollars for a 35 hour work week so for 35 hours just divide the number in the numerator over the number in the denominator don't really worry about the units because what's going to happen is you're going to keep the units in the numerator the unit in the denominator is going to be a single unit right so 760 divided by 40 is going to give us 19. so really i could say this is 19 per 1 hour right or we could say 19 per hour for this guy 735 divided by 35 is 21. so this would be 21 over one hour or again you could say 21 dollars per hour so obviously beth makes more money per hour right she doesn't make more money in total because she didn't work enough hours but on a per hour basis she actually makes more right she makes 21 dollars per hour so to kind of answer this question where it says who makes more on a per hour basis we can say beth right beth makes more on a per hour basis right she makes 21 dollars an hour versus her brother jeremy who's only making 19 per hour okay let's wrap up the lesson by talking about proportions so a proportion basically states that two ratios or two rates are equal we'll know if we have a proportion by checking to see if the cross products are equal right so we have here that we can check to see if two ratios or rates are equal by using the equality test for fractions okay if you don't remember that it basically tells you that two fractions are equal if the cross products are equal so as an example if we have 2 over 9 and we say is this equal to 10 over 45 we can check to see if we have a proportion here by just cross multiplying right take the denominator of 1 multiply by the numerator of the other so 9 times 10 is 90 and then the denominator of this one times the numerator of this one 45 times 2 is also 90. so the cross products are equal so we could say yes this is a proportion right these two are equal all right let's check this one so again we want to do the cross product so what is 5 times 28 that's going to be 140. what's 9 times 14 that's going to be 126. so those two are not equal so no this is not a proportion all right so again for this guy we're just going to cross multiply so 92 times 57 is 5 244 and then 276 times 19 is again 5 244. so the cross products here are equal so yes this is a proportion all right so this is typically what you would see in a ratio and proportion section a problem like this where we have seven males to 28 females we're saying is this equal or do we have a proportion here for 28 males to 88 females as long as the units match up so in other words in the numerator you have males and the numerator you have males in the denominator you have females in the denominator you have females once you've checked that you're just going to multiply the numbers right you're going to cross multiply so you have 28 times 28 which is 784 so if i multiply across here this is 784 and then what's 88 times 7 that's 616. so because the cross products there are not equal we don't have a proportion so this guy is a no right these two are not equal all right what about this one we have nine cups of flour to 19 cups of sugar we're saying is this equal to we have 190 cups of flour to 81 cups of sugar so again just check your units you have cups of flour and cups of flour in the numerator you have cups of sugar and cups of sugar in the denominator so you can just really cross multiply so you have 19 times 190 that would be 3610 let me make that a little clearer and then you would have 81 times 9 which is 729 so those two amounts are not equal so we can say no this is not a proportion in this lesson we're going to look at some properties of real numbers so all the properties that we're going to look at today these are things you would have learned in pre-algebra or algebra 1. they're very very simple just things that we need to know to be successful in our course so we're going to start by just talking about the commutative property of addition and multiplication and we'll begin by focusing on just the commutative property of addition so this rule basically tells us that we can add in any order so for example if we see something like 5 plus 9 we could rearrange the add-ins remember add-ins are what we refer to as the numbers being added together so we could rearrange the add-ins so we could say 9 plus 5 and the result will be the same 5 plus 9 is 14 also 9 plus 5 is 14. so changing the order did not change the result right 5 plus 9 equals 14 and also 9 plus 5 equals 14. now this extends to all real numbers so if i'm working with negatives decimals fractions whatever it is the same rule applies so as an example if i had something like negative 7 plus negative 3 this is equal to or the same as if i reordered the add-ins and said negative 3 plus negative 7. either way i would get negative 10 as a result right so negative 7 plus negative 3 is negative 10 and of course negative 3 plus negative 7 is also negative 10 right same thing either way this property also extends to more than just two numbers if i'm only performing addition then i can add in any order that i'd like i could have 4 trillion numbers i could change the order up as much as i want i will never change the result so as a quick example of this suppose we had something like four plus two plus seven so we can say this is equal to what four plus two is six six plus seven is thirteen let's just say i changed it to four plus seven plus two what do you expect again you'd expect an answer of 13. 4 plus 7 is 11 11 plus 2 is 13. and let's just do one more so let's just say we put 2 first so 2 plus let's say 4 plus 7. again we'd still get 13. 2 plus 4 6 6 plus 7 is 13. so just to kind of recap this property the commutative property of addition just tells us that we can add in any order i will never change the result so when we think about the commutative property of multiplication it's the same thing that we saw with addition we can multiply in any order we won't change the product so if we had something like 3 times 9 we could rewrite this or we could change the order of the factors right the numbers being multiplied together known as factors so we could say 9 times 3 and 3 times 9 is 27 and 9 times 3 is of course also 27. again this property extends to all real numbers so if you're working with decimals if you're working with fractions you know negatives positives whatever you're working with it will not change because you change the order of the multiplication so if i had something like negative one-fifth multiplied by let's say two-sevenths this would be equal to two-sevenths times negative one-fifth in each case if i have negative one fifth times two sevenths negative one times two is negative two five times seven is thirty five if i did it this way two sevenths times negative one fifth again two times negative one is negative 2 7 times 5 is 35 either way you're getting negative 2 35 as a result all right so the next property we want to cover is known as the associative property of addition and the associative property of multiplication so the associative properties allows us to group the addition or multiplication of three or more real numbers in any order so if i'm only performing addition i can group the add-ins however i want if i'm only performing multiplication i can group the factors however i want so let me give you an example so for the associative property of addition let's say i had something like negative 9 plus let's say the quantity five plus seven so i can say this is equal to the quantity negative nine plus five then plus seven we know from the order of operations that whatever is inside of parenthesis gets evaluated first so on the left here this would be evaluated first on the right this would be evaluated first what our associative property tells us that it doesn't matter you can end up with the same answer either way on the left if i have negative 9 plus the quantity 5 plus 7 5 plus 7 done first is 12 so you'd have negative 9 plus 12 which gives us positive 3. if i look at the one on the right here i have the quantity negative nine plus five this is done first and then plus seven so the quantity negative nine plus five so negative nine plus five is going to give me negative four so this is negative 4 plus 7 and negative 4 plus 7 is also 3. so you can see that changing the grouping or changing which add-ins we had parentheses around did not change the result all right let me give you another example of this so for addition let's do the quantity 14 plus 2 and then plus 11. again we could rewrite this by shifting the parentheses so i can put 14 plus the quantity 2 plus 11. you may see some practice exercises like this if you're taking a more advanced course in algebra or pre-calculus course typically in the beginning of the course in the review section you will see this property addressed so if you saw something like this and they said hey use the associative property of addition to rewrite this it's just what i did so you would take 14 plus two plus eleven you just shift where the parentheses are so if it's around fourteen plus two you now put it around two plus eleven okay that's all they're looking for and again to prove that this is a valid property if i evaluate the left side what do i get 14 plus 2 done first is 16 16 plus 11 is 27 on the right side 2 plus 11 done first is 13 13 plus 14 is also 27 so no matter what you put the parentheses around you get the same result again this is what the associative property of addition is telling us now for the associative property of multiplication it's essentially the same principle so we can group the multiplication of three or more real numbers in any order and not change the product so something like 4 times 3 times 7 where we have parentheses around 4 times 3 would be equal to or the same as 4 times 3 times 7 where parenthesis around 3 times 7. again with the order of operations we know on the left we do 4 times 3 first because it's inside of parentheses 4 times 3 is 12. so i would have 12 times 7 and 12 times 7 is going to give us 84. on the right hand side i do 3 times 7 first so that's 21 so i'd have 4 times 21 and 4 times 21 is also 84. so again it doesn't matter what we put parentheses around or how we group this multiplication the associative property multiplication tells us that we're going to get the same answer either way all right let's look at a quick exercise here so the problem tells us to use the commutative property to simplify each so we have negative 10 multiplied by x multiplied by four-fifths so the idea here is that if i want to simplify this i want to multiply this number here by this number here right i can't really do anything with that x there to simplify other than put the x next to the number so what i'm going to do is just reorder the multiplication so i'm going to write this as negative 10 multiplied by i'll just switch the position of this and this so four fifths then multiplied by x so now that they're next to each other i can multiply so negative 10 divided by 5 would give me negative 2 and then of course negative 2 times 4 would be negative 8 so this would end up being negative 8 x so this is something we do without even thinking about it right if we saw this problem and you didn't even know about the commutative property you would know automatically that you can just straight up multiply negative 10 by four fifths take that result which is negative eight and just multiply it by x to simplify the problem all right for the next problem we want to use the associative property to simplify each so we have 5 plus the quantity 3 plus x so the idea here is that i can add 5 and 3 but i can't add x to any of these numbers right it's not going to combine they're not really like terms so i can switch the parentheses here and put the parentheses around five plus three and then plus x i can now combine five and three five plus three is eight so i can write this as eight plus x or we can just switch that around and put x plus eight so next let's talk about the identity properties so the identity property states special properties of the numbers zero and one and we're going to use these when we're solving equations so we first have 0 is the additive identity okay 0 is the additive identity so adding 0 to a number leaves the number unchanged now you might think this is very obvious but when we're solving equations a lot of times we end up saying x plus 0 equals what and students get confused x plus 0 just equals x right adding 0 does not change the value of this no matter what x represents it's just like if i had five million three hundred and i added zero what's the result it's just five million three hundred [Music] right adding 0 to something does not change it that's why it's known as the additive identity all right next we have 1 is the multiplicative identity so multiplying 1 by a number leaves the number unchanged so if we looked at an example of that let's say we had something like negative 558 times one what's the result it's negative 558. again when we're solving equations a lot of times we'll end up with 1 times x what is this 1 does not change x when it multiplies it 1 times x is just x and you also need to think about this because a lot of times if you see a variable by itself you need to know that it has an implied coefficient of 1 right because i can always write x as 1 times x all right so the inverse properties tell us that each real number has an opposite or some people call it a negative or an additive inverse there's a few different ways you can kind of describe that in each non-zero okay non-zero real number has a reciprocal so when we add a real number and it's opposite together the result is zero so first off you know a number is an opposite of the other if the absolute value of the number is the same they just have different signs so in other words three and negative three or two and negative 2 or 117 and negative 117. so the number part is the same it's just the the sign is different you have a positive and a negative when you sum two opposites together you're going to get zero so if i add 3 plus negative 3 what do i get i get 0. if i add 2 plus negative 2 what do i get i get 0. if i add 117 plus negative 117 you guessed it we get 0. so why is this useful well if we're solving equations and let's just say we have something like x plus 4 equals 2. i know that i want to get x by itself and i know i need to get rid of this 4 here so if i need to get rid of the 4 what can i add to 4 to make it go away well i can add the opposite of 4 or negative 4. so i would say x plus 4 plus negative 4 is equal to we'll talk more about this rule later on we're going to review solving equations but essentially whatever i do to the left i've got to do to the right so i'm going to add negative 4 to the right side as well and so what do i have here negative 4 plus 4 is 0. so x plus 0 equals 2 plus negative 4 2 plus negative 4 is negative 2. now we just talked about the fact that 0 is the additive identity so x plus 0 is just x so this gives me x equals negative 2 and that's the solution for my problem right if i had negative 2 plus 4 i would in fact get a result of 2. so next when we multiply any non-zero real number and its reciprocal together the result is 1. so hopefully at this point you know what a reciprocal is and basically to find the reciprocal of a number if it's a fraction you can just flip the fraction if it's not a fraction you would write it as a fraction and then flip it so in other words if i had the number 5 i could write this as five over one i would flip it so in other words the numerator would go into the denominator the denominator would go into the numerator and if i multiply these two guys together i should get one right a number of times this reciprocal is always 1. so 5 over 1 times 1 over 5. what's going to happen is this guy will cancel with this guy and i'm going to be left with just 1 right that's going to be my answer and this is true anytime you multiply a non-zero real number by its reciprocal so let's just say we had something like negative 3 7 and i multiplied this by 7 over negative 3. you can see that this 7 would cancel with this 7 this negative 3 would cancel with this negative 3 you basically have 1 times 1 or just 1 as a result so again we're going to use this when we're solving equations and let's just look at an example real quick suppose we had something like 1 4 times x is equal to let's just say 12. well if i want x by itself what can i do to get rid of this 1 4. i'm not in a situation where i can really add things because this is multiplication here so what can i multiply by 1 4 to make it not 0 but 1 right because if this over here was 1x it would just be x right we talked about that earlier so if i multiplied 1 4 by its reciprocal which is 4 over 1 i would accomplish that because this would cancel with this and i'd basically have 1x on the left hand side to make this legal i'd have to also multiply by 4 over 1 over here 12 times 4 is 48 of course so we would get 1x is 48 or just x is 48 and again if you wanted to check that you could say well 1 4 times 48 48 over 4 is 12 so yes this is true 48 over 4 does equal 12. all right let's look at some examples so we're going to use the inverse and identity properties to solve the following equation so we have x plus 15 equals 12. so what can we do here we have this 15 that is being added to x what can i add to 15 to make it go away what can i add to 15 to make it into a zero well i can add the opposite of 15 or negative 15. again remember the additive inverse property so we would have x plus 15 plus negative 15. whatever i do to the left i also do to the right so 12 plus negative 15 and 15 plus negative 15 is 0. so i would have x plus 0 is equal to 12 plus negative 15 is negative 3. we know that x plus 0 is just x so i have x is equal to negative 3 as my answer does this work as a solution negative 3 plus 15 is 12. all right let's look at the next one so we have use the inverse and identity properties to solve the following equation we have 8x equals 40. so what can we do here we have multiplication so 8 multiplied by x equals 40. again in this situation i don't want i don't want this to end up being 0 because 0 times x would be 0 my variable will be gone what i want when i'm thinking about multiplication i want the left side to be 1x because 1x is just x so what can i do here if i want this guy to turn out to be 1 i can multiply it by its reciprocal which is 1 8. to make it legal i'd have to do that on this side as well and what would happen is this would cancel with this and give me a 1x and over here this would cancel with this and give me a 5. so i'd have 1x equals 5 or just x equals 5. and if you check this if i had 8 times 5 that would be 40 right so 8 times 5 would give me 40. all right so lastly let's talk about the distributive property which is a very important property in mathematics something we use all the time to simplify and solve equations and just everything so the distributive property allows us to distribute multiplication over addition and subtraction we will use this property when we need to change a product into a sum so right now if i look at this guy i have 3 multiplied by the quantity 4 plus 5. so this is a product i can change this into a sum by distributing the 3 to each number inside of the parentheses so 3 multiplied by 4 plus 3 multiplied by 5. so 3 times 4 is 12 plus 3 times 5 is 15 this would give me 27. you might say well why in the world would you do that you could just add inside the parentheses first 4 plus 5 is 9 then multiply by 3 and get 27 much quicker that's true but in the case where you have something like a variable involved you won't be able to do that so let's say i had 3 times the quantity 4x plus 5. well now i can't sum 4x and 5. if i want to change this guy from a product to a sum my only option here is to use my distributive property i would take 3 and multiply it by 4x and get 12x and then plus 3 would be multiplied by 5 to give me 15. so i've changed it from 3 multiplied by the quantity 4x plus 5 into a sum where i have 12x plus 15. this is going to be something we do very often in algebra we're going to kind of move things back and forth between one form to another this will allow us to solve equations and allow us to simplify or just do a variety of things that we need to do all right so lastly let's talk a little bit about factoring so with our distributive property you saw that we can write a product as a sum but there's an equality involved there so we can also use the distributive property to write a sum as a product so when we go in reverse we refer to this as factoring so what we would do here let's say we have this problem 8x plus 24. we would look for what's common to each term so between these two we know that they have a common factor of 8. so i would start by just writing this as 8 times x like this and then plus 8 times 3 right 8 times 3 is 24 and let me write this in a different color so it's really clear what's going on so 8 and 8. now what i can do is i can pull the 8 out from each and place it outside of a set of parentheses and inside the parentheses i'll have what's left so i pull the 8 out here i'd have an x so i'll have an x and then plus if i pulled the 8 out from here i'd have a 3. so 3 and then close my parentheses and the great thing about this is you can check it really easily if i went back a times x is 8x and then plus 8 times 3 is 24. so this is basically factoring or reversing the distributive property if i saw 8 times the quantity x plus 3 i know i could write it like that if i saw 8x plus 24 i know i could write it like that so essentially because of the equality i can write a sum as a product and a product as a sum all right so let's look at a sample problem so we want to use the distributive property to rewrite each product as a sum we have negative 2x multiplied by the quantity 4 minus y all we're going to do is just distribute so this guy is going to get distributed to each term inside the parentheses so negative 2x multiplied by 4 is negative 8x and then negative 2x multiplied by negative y would be positive 2xy so this turns out to be negative 8x plus 2xy and again if i saw just this and if i was asked to factor i could rewrite it as this right i can take my sum and write it as a product or i can start out with a product and write it as a sum all right last problem we want to use the distributive property to rewrite each sum as a product so we have 25x minus 10 and the first thing we want to do if we want to factor this is think about what's going to be common here i only have an x in one of the terms so that's not going to be common so just think about the number parts 25 is 5 times 5 10 is 5 times 2. so let's write this to where we have 5 multiplied by 5 minus we're going to have 5 times 2 right the 5 is what's going to be common to each and let's not forget our x right here so let's put is equal to i'm going to pull out this 5 from each term and i'm going to place that outside of a set of parentheses so my 5 will be outside and inside the parentheses i have what's left so here it would be 5x right so i'm going to write 5x here and then minus here i would just have 2. so put the 2 there so we have 5 multiplied by the quantity 5x minus 2 that is the factored form of 25x minus 10. again if you were presented with this which is a product 5 multiplied by the quantity 5x minus 2 you could write it as a sum as 25x minus 10 and then of course if you have 25x minus 10 like we saw in this problem you could write that as a product as 5 times the quantity 5x minus 2. in this lesson we're going to review inequalities the number line in interval notation so we're going to start by just talking about the basic definition of an inequality an inequality is a statement that two algebraic expressions are not equal in value so generally when we're talking about an inequality we're talking about a less than a greater than a less than or equal to or a greater than or equal to so let's start out by talking about the less than so when we look at a less than you notice that the symbol is pointing to the left okay it's pointing to the left and this tells us that the number on the left is smaller or less than the number on the right so if you wanted to write something like 5 is less than 6 that would be a true statement if you wanted to write 3 is less than 7 that would be a true statement if you wanted to write negative 10 is less than let's say negative 1 that would also be a true statement so the thing to remember here is that the smaller number is always going to have the symbol pointing towards it okay so see how 5 is the smaller number and the symbol is pointing towards it 3 is the smaller number the symbol is pointing towards it negative 10 is the smaller number and the symbol again is pointing towards it so even if you can't remember the difference between a less than and a greater than symbol if you just remember that when we're dealing with inequalities this symbol will always point to the smaller number you're going to be good to go all right now the other symbol you deal with is known as a greater than so this guy is going to point to the right so kind of the open side here is facing the larger number okay so the larger number is now going to be on the left so if we had something like 17 is greater than let's say 4 okay the larger number's on the left if we had something like 3 is greater than negative 1 and let's just say we had something like negative 20 is greater than negative 50. so again always always always the smaller number has the inequality symbol pointing towards it so see this guy is pointing towards the 4 this guy is pointed towards the negative 1 and this guy is pointing towards the negative 50. and you can always tell again if you have a less than the symbol points to the left if you have a greater than the symbol points to the right all right so let's talk a little bit about the difference between strict inequalities and non-strict inequalities so a strict inequality is something that you just saw so you saw strictly less than and strictly greater than so if i said 3 is less than 10 right that's a strict inequality but if i said something like 3 is less than or equal to right you have that little bar there underneath the inequality symbol well now i'm allowing myself the opportunity for inequality so i could say 3 is less than or equal to 3 and this would be a true statement although the first part 3 is less than 3 is false 3 is equal to 3. so when you look at something like this you basically have to ask yourself two questions to see if it's true so you would first ask the question is three less than three so the answer to that is no right so you can write no but that's okay you're not done only one of these needs to be true the next question you ask yourself is three equal to three and that guy is yes that guy is yes so this is a true statement this is true so only one of those needs to evaluate to true now if i said something like 3 is less than or equal to let's say 1 well that wouldn't be true is 3 less than 1 no is 3 equal to 1 no so both of those are no so in this case this guy is going to evaluate to false okay so that would not be a true statement now you also have a greater than or equal to and basically it's the same principle so if i had something like 6 is greater than let's say negative 1 this is true of course but i could also have something like 6 is greater than or equal to let's say 5. is 6 greater than 5 yes it is so you can stop there if that's a true statement if i had 6 is greater than or equal to 6 well first we'd say is 6 greater than 6 no it isn't right 6 and 6 are the same number then you'd ask the question is 6 equal to 6 yes it is so this is also a true statement right so this guy is true so essentially when you have a non-strict inequality like less than or equal to or greater than or equal to you're allowing for the possibility of equality right it can be the strict part can be true or the equality can be true you'll get a true statement if you have a strict inequality like less than or greater than well then it's got to be strictly less than or strictly greater than to get a true statement another thing we need to talk about here is the concept of the number line so we've looked at the number line a few times now and we want to just refresh our memory on how the number line works i want you to recall that numbers increase as we move to the right and decrease as we move to the left so as we're moving this way the values are increasing so if you start at any point so let's just say you start at the point negative three as you move to the right each number to the right of negative three is larger than negative three so i can pick any number on the number line that's to the right of negative 3 let's just say it's 5 and i know based on what i said that 5 would be greater than negative 3 because 5 lies to the right of negative 3 on the number line similarly if i chose 2 random points let's say this is 6 and let's just say negative 2. well 6 lies to the right of negative 2. you could also say negative 2 lies to the left of 6 so negative 2 is therefore a smaller number right if a number lies to the left of another number on the number line it's a smaller number so i can say negative 2 is less than 6 since negative 2 again lies to the left of 6 on the number line so something you might see in your textbook in a review section we would say for any three real numbers a b and c so again when we say real numbers we mean anything you can think of so negatives decimals fractions you know square root cube root whatever you want these three real numbers you can say the following things so a is less than b if a lies to the left of b on the number line a is greater than b if a lies to the right of b on the number line and then lastly a is greater than b and b is greater than c if here a needs to be to the right of b and c on the number line right it's got to be to the right of both of them b has to be to the right of c and to the left of a and c has to be to the left of b and a okay now typically we would not write this situation like this so let's just put a scenario on the board here let's just say we had something like 6 is greater than 3 and 3 is greater than let's say 1. typically the way you want to write this is in the order of the number line so you would want to put 1 is less than 3 which is less than 6 because the number line goes like this right if you had a number line let's just say this is 1 this is 3 and this is 6. this would be the way it's displayed and this is how you want to write it you want the smallest number all the way to the left just like it is on the number line then the next smallest just like it is on the number line then you know your largest number which in this case is 6 and that's how it would be on the number line so this is typically how we're going to write this this is not wrong but it's just not how we typically would do it all right let's take a look at a quick example we want to replace each question mark with the correct inequality symbol so we have four question mark two so again obviously we know that four is a larger number than two if we had four dollars in our wallet we obviously have more money than if we had two dollars but to prove this we can look at our number line and say okay 4 lies to the right of 2 on the number line so therefore 4 is greater than 2. again the symbol is always going to point to the smaller number so notice i use the greater than symbol there the symbol is going to point to the right it's pointing to that 2 which is the smaller value now you could technically also use a greater than or equal to here this is still true because 4 is greater than 2 it doesn't matter that 4 is not equal to 2 because only one of those needs to be correct but we typically only use these when we're working with variables although this is technically correct we wouldn't say this is a good answer we'd say this is a more precise answer right 4 is just greater than 2. now you can also flip this around and say well since 2 lies to the left of four on the number line two is less than four right so two different ways to kind of display the same information all right what about negative seven question mark negative three now this type of problem trips people up because negative seven seems like or it might seem like a bigger number to you right because you're used to dealing with seven being larger than three well what happens is a bigger negative is further to the left on the number line and it actually represents a smaller number if you look at negative 7 here it is to the left of negative 3 on the number line so by definition it is smaller so we can say that negative seven is less than negative three okay and the way i always remember this is i think okay a bigger negative is a smaller number okay and again i can flip this around and i can say that negative 3 is greater than negative 7. okay so if you had something like let's say negative 50 and negative 200 would be the relationship there negative 200 is a larger negative and therefore it's a smaller number so in this situation we would say negative 50 is greater than negative 200 or we could say negative 200 is less than negative 50. all right let's take a look at another example so display the relationship between the numbers using inequality symbols all right so if we look at the numbers here we have negative 11 we have 7 we have negative 5 and we have 1. so the smallest number here would be the biggest negative number and so negative 11 is a larger negative than negative 5 right it's further to the left on the number line so negative 11 is going to be the smaller number and so i'm going to list that first and i'm going to say this is less than my next negative is negative 5. then if i look at the positive numbers i know 1 is smaller than 7 so i'm going to put is less than 1 and then is less than 7. so this is again in the order of the number line negative 11 will be the furthest to the left then negative 5 then 1 then finally 7. so negative 11 is less than negative 5 which is less than 1 which is less than 7. all right so let's wrap up the lesson by talking about interval notation so this is a very important way to kind of notate inequality relationships and we'll see this a lot throughout our course so if we had something like let's say x is greater than 5 of course this way displays a lot of information we know x is just any value larger than 5 but interval notation is another way to display this information so typically with interval notation you want to find a way to list the smallest number that x can take on and you want to find a way to list the largest number okay the largest number that x can take on so i want to just think about this for a second let's just stop and think about this what is the smallest possible value that x can take on it's not 5 because x is anything larger than 5. and in terms of the largest number there is none right it would just go on for infinity but to focus on the smallest number right now if we think about that what would i put here for the smallest number would it be accurate to put 5.1 can i think of a number that's smaller than that that is larger than 5 well of course i could put 5.01 and then i could put 5.001 i can keep doing this forever i can keep getting closer and closer to 5 without actually touching it and i can do that for infinity i can just keep going so to solve this problem what we're going to do in interval notation is we're going to list the 5. we're just going to put that down and we're going to put a parenthesis next to it to say 5 is not included so this means not included and so this tells me that x can be anything that is larger than this number five okay so five isn't the actual smallest number the smallest number would be anything that's larger than that then we put a comma and we list the largest number since x is anything that's larger than 5 technically x can go out to infinity right so we would just put the infinity symbol and it looks like a sideways 8 and then we always use a parenthesis next to infinity because infinity is not a number it is just a concept right it just keeps going on forever and ever now let's think about this using a non-strict inequality let's say i had x is greater than or equal to 5. well now the 5 would need to be included so i would need a different notation instead of using a parenthesis i would want to use a bracket okay this bracket tells me that the 5 is included in my solution so the smallest number is actually 5 now okay so if i saw something like x is greater than or equal to let's say negative 1 i would put a bracket and a negative 1 there okay that's the only thing that would change so very very easy the other scenario would be dealing with a less than so let's say x is less than let's say 7. well this has to change itself up since x is less than 7 what is the largest number that x can take on well it's any number that is smaller than seven again i can play that game that i just played i could list seven then i could list six point nine nine nine nine i could keep going and going and going get as close to seven as i'd like i can keep doing that forever and ever and ever so again to solve that problem i just put 7 and i put a parenthesis next to it to say hey 7 is not included not included and i'll put a comma and for the smallest number because x is anything less than 7 we go to the left right forever so we would say this goes out to negative infinity so we'll put a negative and then the infinity symbol and let me kind of scooch this down and i'll put a parenthesis there so we have from negative infinity out to and not including 7. now again if i had a less than or equal to here the only change i would make is instead of a parenthesis i would just use a bracket okay so very very easy to use interval notation the other scenario you would run across is a three part inequality so let's say x was greater than three and less than seven well three would not be included and seven would not be included but it could be any value in between right so again we'd list three we'd list seven and we would just put a parenthesis next to each to say three is not included seven is not included but any number in between would be and again if i change this to a non-strict inequality these guys right here would just change to a bracket in each case okay so very very simple all right so let's look at x is greater than negative two so how would i write this in interval notation so i have a strict inequality here so i know that negative 2 is going to be excluded right so anything larger than negative 2 works but negative 2 itself does not work so for my smallest value i'll put negative 2 and then i'll put a parenthesis next to that to say negative 2 is not included then comma and then i'll have out to infinity always use a parenthesis with infinity now one thing i want to show you is how to graph this and graphing it is the same notation as the interval notation so what i would do is at negative 2 let me kind of change my marker up a little bit so i'll go to 3.5 so at negative 2 i would put a parenthesis there facing to the right and i would just shade everything this way including the arrow so all this is telling me is that negative 2 is not included that's what the parenthesis is telling me i've shaded all this area going to the right including the arrow to say hey every number works as a solution for x with the exception of negative 2 or anything to the left of that but anything to the right of negative 2 works now you might see in your textbook a different scenario you might see them put an open circle here let me kind of change this marker a little bit down so you might see them do this that's perfectly valid it's the same thing it's just an open circle to say negative 2 is not included and they're shading everything to the right this is perfectly acceptable as an answer and it's just another way to kind of notate the same thing all right what if we saw x is less than negative 3. if we think about the largest number i'm going to put negative 3 there it's a strict inequality it's not included so i put a parenthesis and then over here for the smallest value i put negative infinity so you have a parenthesis negative infinity comma negative 3 parenthesis right so anything smaller than negative 3 negative 3 is not included works as a solution for x and again graphically we can show that by putting a parenthesis facing to the left at negative 3 and shading everything to the left so negative 3 doesn't work but anything to the left works and again if you want you can do an open circle and shade everything to the left like that all right what about x is less than or equal to the number one so let's start out with the graphical display here and i would find one and i would put a bracket there facing to the left so the bracket again tells me that one is included and i would just shade everything to the left and again you saw the open circle in the other scenario here instead of the open circle we would have a filled in circle right so it would look like that to say hey 1 is included anything to the left is also included all right so in interval notation we would have from negative infinity always use a parenthesis with either negative infinity or infinity up to and including the number one so because we have the non-strict inequality here 1 is included so i'm going to use a bracket all right so we have x is greater than or equal to 7. so graphically if we look at 7 again we can either put a bracket facing to the right and shade everything to the right or we can put a filled in circle at 7 and shade everything to the right again either one is acceptable when we look at the interval notation we would put a bracket on the left okay and then put seven so seven is included seven is the smallest value that x can take on and then comma x can be anything going out to positive infinity again we always use a parenthesis with infinity so we have our bracket next to 7 showing that 7 is included and then comma we have infinity and then our parenthesis next to infinity all right for the last one we see that x is greater than negative four and less than seven so graphically we would look at negative four which is right here we'd put a parenthesis there facing to the right to show again that negative 4 is not included and we would find 7 and we put a parenthesis facing to the left again to show that 7 is not included so x can basically be any value between negative 4 and 7 where negative 4 and 7 are not included so we show that graphically that way again if you wanted to you could put an open circle here and an open circle there again it means the same thing right so i'll stick to the parenthesis throughout this course so i'm just going to display it that way but again if your book shows you the open circle and you feel more comfortable with that you're good to go with that method alright so interval notation here we want to list a parenthesis then a negative 4 then comma a 7 and then a parenthesis in this lesson we want to review the concept of absolute value so let's just start out with the basic definition the absolute value of a number is the distance between the number and 0 on the number line so if i was to ask you for the absolute value of let's just say the number 6 you could find the number 6 on the number line and you could count the number of units away it is from zero so if we take this here from six and we go one two three four five and then you'll see that 6 is 6 units away from 0 on the number line so the absolute value of 6 is just 6. now similarly if we looked at negative 6 you'd find that it has the exact same absolute value if we start here we go one two three four five six units to the right to get to zero so this leads us to an important concept two numbers that lie on the opposite side of the number line with the same absolute value are known as opposites so we've talked about opposites before so 6 and negative 6 would be opposites 3 and negative 3 would be opposites 4 and negative 4 122 and negative 122 you know so on and so forth one important property here and we talked about this in a previous review session if you add two opposites together you get a result of zero think about adding negative six and six well if i started at zero and i went six units to the left and then i went six units to the right and end up back at zero it's because their weight or absolute value is going to cancel each other out right one is negative one is positive they have the same absolute value so they cancel and they end up in the middle which is at zero so in general we don't need to pull out a number line every time we want to find the absolute value we can follow a very very simple rule the absolute value of some real number a is just a if a is non-negative meaning a is greater than or equal to 0. so if you have 0 or some positive number you take the absolute value it's just the number so in our example we took the absolute value of 6 it was just 6. right you just used the number the absolute value of some real number a is the opposite or the negative of a if a is less than zero so if i had something like the absolute value of negative six this would be equal to the opposite of negative six which is just six right you don't even need to go through all this just know that if you take the absolute value of something that's negative you just make it positive right so if i had the absolute value of negative 122 i would just make the number positive i know this is 122. similarly if i had the absolute value of let's say negative one million three hundred twenty two thousand five hundred fifteen what would i do i would just make the number positive let me kind of erase this to get a little room so i would just take this number and say okay it's one million three hundred twenty two thousand five hundred fifteen right just make it positive so if it's zero or a positive number it's just a number if it's a negative number just make it positive okay that's all you really need to remember about finding the absolute value of a number it's very very simple all right so now let's think a little bit deeper about the concept of absolute value especially in terms of what we're going to see in this course we think about a statement such as the absolute value of x is greater than 5. so what does this mean let's break this down so that you can understand it the absolute value of a number again is the distance from the number to zero on the number line so what this is saying right here is any number here that has a distance from zero that is larger or greater than five okay so x can be any number whose distance from 0 is larger than 5. so if i start at 0 and i go in the right direction 5 units and kind of circle this here at 5. if i start at 0 and i go left 5 units and i circle this here at negative 5. so what does that tell me well x could be anything larger than 5 we know from interval notation that we can kind of mark this this way i could put a parenthesis here to say that 5 is not included and i could shade everything to the right and x could be anything less than negative 5. so i could put a parenthesis at negative 5 facing to the left and i could shade everything to the left like this right so anything in between here so from 5 to negative 5 with 5 and negative 5 included would not work because in those situations the values would not have a distance from 0 that is larger than 5. so as an example let's say i took negative 1. if i plug that in for x what would i get so the absolute value of negative one is that greater than five no the absolute value of negative one is what i just told you if you have a negative number just make it positive so this evaluates to one is greater than five which is false one is not greater than 5. if i plug 5 in there the absolute value of 5 is 5 5 is not greater than 5. that doesn't work if i plug in negative 5. the absolute value of negative 5 is 5. 5 is not greater than 5 that doesn't work so all these numbers in between here don't work but if i chose something that is to the right of 5 or to the left of negative 5 that would work if i chose 7 for example let's see what happens there the absolute value of 7 is 7 7 is greater than five so we're good if i chose let's say negative nine okay let's say negative nine the absolute value of negative nine is nine nine is greater than five so that works so if we wanted to write our solution we could say that x could be anything larger than five or okay you use the or statement here x could be anything that is less than negative five either situation would work in interval notation we would use our union symbol so we would have the interval from negative infinity let me make that symbol better up 2 but not including negative 5. and it's this set and we're going to have a union with this set so anything larger than 5 again out to positive infinity so this is how you would write it kind of using the standard notation where you know you're using inequality symbols and this is how you would write it using interval notation and this is how you could display this graphically now let's think about the other scenario we have here this is kind of a trick question so before i even read it why don't you see if you can think about this one on your own so i assume you tried this and hopefully you figured out that this is again a trap if you see this on a test you should be able to answer right away all real numbers why is that the case well the absolute value of x is greater than negative four okay let's think about that again give me a number here that you take the absolute value of it and it's larger than negative 4 well it doesn't matter what i plug in there if i plugged in a negative number the absolute value would make it positive okay if i plugged in 0 the absolute value would make it 0. if i plugged in a positive number it would keep it positive so anything i plug in is always going to be greater than negative four because the smallest this can be is if x is zero so the absolute value of zero is zero so zero is automatically greater than negative 4 and that's as small as this can be using the absolute value operation okay again if i plug in something negative let's say i plugged in negative 6 there the absolute value of negative 6 is 6 6 is larger than negative 4. so you see that no matter what i plug in for x it would work so we say all real numbers okay you could say it like that you could do all kinds of different ways to notate this you could say you know the interval from negative infinity to positive infinity you could shade the entire number line graphically right you could take and just shade this entire thing to show that any real number on the real number line would be a solution for your inequality here all right so let's look at the other scenario here so we start out with the absolute value of x is less than 2. so again when i think about x it's it's going to be any value whose absolute value or distance from 0 is smaller or less than 2. so if i started at 0 and i went 1 and then 2. so it couldn't be negative 2 right it couldn't be negative 2 because it can't actually go 2 units got to be less than that so let's kind of notate this by putting a parenthesis at negative 2 facing to the right and i could do the same thing i could travel 1 2 it can't quite be 2 so i'm going to put a parenthesis here facing to the left i could just shade everything in between so essentially what i have here is a scenario where x can be any number that's between negative 2 and 2 with negative 2 and 2 not included right so in interval notation that's the interval from negative 2 to 2 again neither is included so i use a parenthesis at each end and you can try this out with let's say the number negative one so you plug in a negative one there the absolute value of negative one is one so you would have that one is less than two which is true if you tried something outside of this range here let's say you tried 3 the absolute value of 3 is 3 3 is not less than 2 so that doesn't work if you tried let's say negative 5 the absolute value of negative 5 is 5 5 is not less than 2. so again you've got to be within this range here to have an absolute value or a distance from 0 that is smaller than 2. so that's your solution there and you could also write this as x is greater than negative 2 and less than two okay so kind of three different ways to show the solution there all right let's think about another one and again i advise you to kind of stop the video here and think about this what's going on here because i'm going to tell you in advance this is a trick one all right so i assume you tried this and you would see that there's no solution here and why is there no solution again x represents a number whose absolute value is less than negative 3. can absolute value be negative just think about that for a second can it be negative no it has to be either 0 if you're plugging in 0 or positive if i plug in a negative i get a positive i plug in a 0 i get a 0 if i plug in a positive i get a positive so there's no way that the absolute value of something can be less than negative three it's not possible it has to be zero or larger so there's no solution here there's no solution okay you can write no solution or a lot of teachers will say make the symbol for the null or empty set because the solution set here has no elements and again if i'm going over your head there just write no solution that's all you really need for the majority of the classes so the last thing we want to think about in this lesson on an absolute value is how to find the distance between two points on a number line so this is a very very easy thing to do and we have a very simple formula that we'll see in a second so the distance between two points on a number line is the absolute value of the difference between their coordinates in any order so in other words if i wanted to find the distance between let's say 6 and let's say negative 1. well what this is telling me is that i can say the absolute value of 6 minus a negative 1 would give me the distance between the two so 6 minus a negative 1 would be six plus one that's seven so i want the absolute value of seven which is just seven i could also because this says in any order i could flip this around so i could say the absolute value of negative one minus six and the absolute value of negative 1 minus 6 is the absolute value of negative 7 which again is also 7. so kind of the order doesn't matter here because of that absolute value operation it's going to end up taking the sign and disregarding it right if you get a positive 7 it's still 7 if you get a negative 7 it makes it positive 7 so the order isn't going to matter here because of the absolute value operation so officially you'll see this in your textbook if p and q are points and don't get confused by p and q p and q are just something we can use to label the points so in this example here where i said we want to find the distance between 6 and negative 1 on the number line i could label this guy as p i could label this guy as q right or vice versa it doesn't really matter so if p and q are points on the number line with coordinates a and b so when we say coordinates that refers to the number on the number line itself and then we'll say the distance between the two points is given by this formula here so the distance between p and q is equal to the absolute value of b minus a or the distance between p and q is equal to the absolute value of a minus b so again we can change the order around and get the same result because of this absolute value operation so let's look at a sample problem here we want to find the distance between negative 3 and 6. so here's negative three and here's six if i was to just count the units out i could start at negative three and go one two three four five six seven eight nine right so they're nine units apart if i started at negative three and went to six it takes me nine units to get there if i started at six and went back to negative three it would also take me nine units to get there but again we don't wanna do that each time we don't want to have to pull out a number line we want a formula and that formula tells us to subtract right one coordinate from the other it doesn't matter the order because we're going to do this inside of the absolute value operation so i can say the absolute value of negative 3 minus 6 which would be equal to the absolute value of negative 9 which would be 9. again that's what we found by just counting we could also flip this around we could say the absolute value of 6 minus a negative three so six minus a negative three is the same as six plus three so this would be the absolute value of nine which is nine so either way we do it if we do negative three minus six that gives us negative nine when we take the absolute value we get nine if we do six minus a negative three that gives us six plus three which is nine we take the absolute value it's still nine so again because of the absolute value operation the order here doesn't matter all right let's look at one more so we want to find the distance between negative four and negative seven so here's negative four here's negative seven you can eyeball that and see it's three units right one two three they're three units away from each other and again to use our formula we could do negative seven minus a negative four inside of absolute value bars so this would be negative seven minus a negative 4 is negative 7 plus 4 so that would be the absolute value of negative 3 which is 3. or again we could change the order around right inside of absolute value bars we could say negative 4 minus a negative 7. negative four minus a negative seven is negative four plus seven that would be positive three so the absolute value of three which is three again because of the absolute value operation the order here doesn't matter once we evaluate if we get negative three the absolute value of that is three if we get positive three the absolute value of that is three so either way you do it you end up with three as the distance between negative four and negative seven on the number line in this lesson we want to review the product and power rules for exponents we also want to review having an exponent of 0. so this is something that you probably already know you're probably very familiar with exponents in the rules of exponents but again to be successful in this course and specifically to be successful in the next section where we review polynomials we need to really understand how to work with exponents so let's start by talking about the product rule for exponents so basically when we multiply powers with like bases we keep the base the same and we just add exponents so if i have some real number a raised to the power of m multiplied by the same real number a raised to the power of n this is equal to a raised to the power of m plus n so keep the base the same right so a is the base in each case so we keep it the same and then add the exponents you have an exponent of m plus an exponent of n okay so to give you a few examples let's just start out with 2 as a base because it's pretty simple so let's say we had 2 cubed multiplied by let's say 2 squared we would keep the base the same so we'd write 2 and we just add the exponents so i have an exponent of 3 and an exponent of 2 so you could say 3 plus 2 up here so this would be 2 to the fifth power and it's pretty easy to prove this to yourself that it works if you think about what 2 cubed means it's 3 factors of 2. if you think about what 2 squared means it's two factors of 2. so essentially if i just count the number of factors i have here it's 3 and if i count the number of factors i have here it's 2. so if i add those together i get the total number of factors that i'm going to have 3 plus 2 is 5 so i get 2 to the fifth power so really really simple really really easy to prove that to yourself now another example could be something like let's say 5 to the 4th power multiplied by let's say 5 to the 11th power something like that so let's say this is equal to the base 5 stays the same and we just add the exponents 4 plus 11 is 15 so this would be 5 to the 15th power this also works with variables if we have something like x times y to the third power and let's say we multiply this by x squared y to the fourth power we would keep the base the same so we think about x and y separately for right now so x times x squared would be x keep the base the same this has an implied exponent of one so anytime you see something hanging out by itself with no visible exponent it has an implied exponent of one so this would be x raised to the power of one the exponent here plus 2 the exponent theta all right so 1 plus 2 and then for y we have y cubed and we have y to the 4th power so we'd have y raised to the power of 3 plus 4. so this would be equal to x raised to the power of 1 plus 2 which is 3 so we have x cubed times y raised to the power of 3 plus 4 which is 7. so x cubed y to the 7th and we can mix this up and have some numbers and variables involved so let's say we had 2 squared x to the fifth power multiplied by let's say 2 to the fourth power x cubed and then just y so what do we do here so if we think about 2 2 squared times 2 to the 4th power would be 2 raised to the power of 2 plus 4 so this would be 2 to the sixth power if we think about x we'd have x to the fifth and x cubed so if we multiply those together x stays the same you add 5 and 3 that's 8. and then the y here i don't really have anything that i can combine here so i just write y so this would simplify to two to the sixth power x to the eighth power y right and of course i can put a visible multiplication symbol there but again when we're working with algebra when we're putting things next to each other so if i put a 2 next to an x next to a y next to a z all those things are multiplied together so putting a number next to a variable implies multiplication okay but if it makes you feel more comfortable it is completely valid to put a multiplication symbol there if you would like okay now some teachers might actually ask you if you have a number raised to a power to evaluate that so what is 2 to the 6th power we know that's 64. so we could really write this as 64 x to the 8th power y all right so let's talk about the power to power rule now and this is another rule that's going to come up quite often so let's say we had some real number a and it's raised to the power of m so we already have something going on here with an exponent now i'm going to take this guy and we're going to raise it to the power of n so with this guy we're going to keep the base a the same and we're going to multiply the exponents m times n so to give you an easy example suppose we had 2 squared and this guy was raised to the 3rd power or cubed well what i'm going to do is i'm going to keep 2 the same and i'm going to multiply 2 times 3 and get 6. and we know that's 64 we don't need to write that but we'll just keep it as 2 to the 6th power now how can i prove this to myself well if i cube something it means i have 3 of them right so i could really say 2 squared cubed is 2 squared times 2 squared times 2 squared which is what using the product rule for exponents keep the base the same add the exponents i would get 2 plus 2 plus 2 which is 6. right so it's the same either way right either using this method or this method this is just really a shortcut a way to kind of get the answer in a quicker fashion so let's think about another example so let's suppose we had something like 8 to the 4th power raised to the 7th power so i would keep the base 8 the same and i would multiply 4 times 7 and get 28 so that would be 8 to the 28th power again we can involve variables here let's suppose i had x squared and y to the fourth power and this guy was raised to the fifth power what would i do here well i would raise each one of these individually to the fifth power so x squared raised to the fifth power x would stay the same i would multiply exponents 2 times 5 would give me 10 then for the y to the fourth power part y stays the same multiply the exponents 4 times 5 is 20. so we get x to the 10th power y to the 20th power all right the next rule is product to a power rule so this will help when simplifying essentially this rule tells us that if i'm multiplying here if i have a product a times b is a product if i'm raising this to a power what i can do is i can say this is equal to each factor raised to that power so a b raised to the power of m is equal to a to the power of m times b to the power of m so as an example let's just keep using 2. so let's say we had 2 times 3 raised to the second power what this is saying is that it's equal to 2 squared so that's one factor squared times the other factor squared and you can prove this to yourself 2 times 3 is 6. so 6 squared would be 36 2 squared is 4 3 squared is 9 we know 4 times 9 is 36 so it's the same either way you might say well why would you want to go through and do this there's going to be situations where you might need to do that in order to simplify or it might make it easier to solve something you just you need to know all the rules so that when you get in the situation you can manipulate things and get the answer that you're looking for let's take a look at another example of this let's suppose we had 4 squared x to the fourth power y cubed and this guy is going to be raised to the third power so we could simplify this by raising each power here using the power power rule each power to the power and we're going to do it individually right so we're going to take each factor and raise to the power so 4 squared cubed is 4 raised to the power of 2 times 3 or 6. then times we'd have x to the fourth power raised to the third power so this is x raised to the power of 4 times 3 or 12 then we have y raised to the power of 3 raised to the power of 3. so y cubed cubed so y stays the same multiply the exponents 3 times 3 is 9. so this is one way that you might want to simplify again you see something like this you can remove the parentheses and write it as 4 to the 6th power x to the 12th power y to the ninth power let's talk a little bit about the quotient to a power rule so this is similar to what we just saw essentially if i have a quotient so a over b or a divided by b this is raised to the power of m i can say this is equal to a raised to the power of m over b raised to the power of m so if you think about this as a division problem your dividend would be raised to the power of m and your divisor would be raised to the power of m if you think about it as a fraction your numerator is raised to the power of m over a denominator which raises the power of m however you want to think about that and as an example let's suppose we had something like 15 over 3 and this guy squared well what i can do here is i can raise 15 to the power of 2 over i can raise 3 to the power of 2 and this should be the same either way if we divide 15 by 3 we know we get 5 5 squared is 25. if i did 15 squared that's 225 if i did 3 squared that's 9 225 divided by 9 is also 25 right so it's the same thing either way in this particular case it would be a lot simpler to deal with this right because you're taking a smaller number and squaring it here you're squaring and you're squaring then you have to do an extra division problem so again if i saw something like this i'd probably want to put it in this form to simplify because it's quicker right especially if you don't have a calculator but again you can go back and forth because of the equality so whatever is easier for you so to look at another example of this let's suppose we had something like 1000 and this is over 25 and this is raised to the fourth power now 1000 divided by 25 is going to give us 40. so this would be 40 to the fourth power but again if you wanted to do this in another way you could say this is 1000 to the fourth power over 25 to the fourth power okay it's the same either way so lastly let's talk about an exponent of zero so the power of zero we have a some real number raised to the power of zero is equal to one and we have a is not equal to zero okay so let's let's think about this a lot of you know where this comes from but in case you don't i'm just going to show you a quick pattern and then it should make sense so we start with the number 2 2 is an easy number to work with if i have 2 cubed we know this is 8. if i have 2 squared we know this is what it's 4. so notice how if i go from 8 to 4 i divide by 2. so i divide by 2 when i reduce the exponent by 1 if i increase the exponent by 1 i multiply by 2. so 4 times 2 would get me back to 8. if i go down to 2 to the first power i can take 4 and divide by 2 that gives me 2. so again i'm dividing by 2. so following that same pattern if i want 2 to the power of 0 then i would take 2 and divide it by 2 and i would get 1. right it's just we can teach this with a pattern and we'll talk about negative exponents later but most of you know that if i had 2 to the power of negative 1 i would divide 1 by 2 and i would get 1 half right you just keep dividing by 2. and again it works the same way if i go up i can multiply by 2. so 1 half times 2 is 1. if i go up 1 times 2 is 2 if i go up 2 times 2 is 4 if i go up 4 times 2 is 8. again coming back down you're dividing by 2 going up you're multiplying by 2. so this works for any real number so if you take some real number and you raise it to the power of 0 it's equal to 1 and that's because you're dividing the number by itself this is why we can't have 0 raised to the power of 0 because you would end up by definition with 0 over 0 which is undefined okay we can't divide by zero so let's think about an example so we know any number raised to the power of zero other than zero is one so at 155 raised to the power of zero this would be one if i had negative 212 raised to the power of zero notice how it's in parentheses this is equal to one if i had negative eight without the negative inside of parentheses raised to the power of zero this is negative 1 right because you have a to the power of 0 which is 1 and you want the negative of that that's how you get negative 1 there so kind of a trap question with this let's say you had something like 215 x squared y cubed z and this guy is raised to the power of 0. do i need to go through and use the power of the power rule here no everything inside of here is going to be raised to the power of 0 and everything inside of there is going to be 1. so the result of raising something to the power of 0 is 1 1 times itself is always 1. so you could just stop you don't need to go through here the long way this is just 1. okay doing this the slow way 215 is really 215 raised to the power of 1. if i use the power of the power rule here 1 times 0 is just 0. so this becomes a 0 here x squared raised to the power of 0 is x to the power of 0 y cubed raised to the power of 0 is y to the power of 0 z to the first power raised to the power of 0 is going to be z to the power of 0. so all of these here are going to equal 1. right this is going to be 1 times 1 times 1 times 1 which is just 1. so if you see something like this save yourself the time and just write 1. all right so to kind of wrap up the lesson let's think about something like this so let's say we had 2x to the fifth power and this is squared and then this is over let's say we had 3 x y z this is raised to the power of 0 and this is multiplied by we would have 3 squared and this is times y squared and this is raised to the power of 4. okay so what can we do here so in my numerator i would use my power to power rule 2 basically to the first power raised to the power of 2 keep 2 the same multiply the exponents 1 times 2 is 2 so this is 2 squared then x to the fifth power squared keep x the same multiply exponents 5 times 2 is 10. in the denominator again if you see something like this this is a typical trap question you think about this as just 1 right 3xyz this is all raised to the power of 0 this is going to be 1. then times this guy right here 3 squared y squared raised to the power 4. keep three the same multiply the exponents two times four is eight so this would be three to the eighth power y squared to the fourth power keep y the same multiply the exponents two times 4 is 8. so this would give me as a final answer we would have 2 squared which you could write as 4 if you want x to the 10th power over 3 to the 8th power y to the eighth power now if you wanted to you could write 3y inside of parentheses like this raised to the 8th power okay you could do that but to make it simpler i'd rather put it like this we have 3 to the 8th power y to the 8th power again you could write that either way you want it means the same thing but i like this answer a little bit better so 2 squared x to the 10th power over 3 to the 8th power y to the eighth power in this lesson we want to review the concept of a polynomial so before we kind of jump into the definition of a polynomial we want to start with a definition of a term so this is something that is fundamental to the study of math in case you've forgotten the definition or you never heard it basically a term is a number okay so it could be a single number like let's just say 13 or the product of a number and one or more variables raised to powers so you could also have something like 2x or you could have something like negative 4x squared y or we could have something like 5 x to the fourth power y squared z cubed you know so on and so forth now one thing to point out here is that if you have a number that is not multiplying a variable so like in this case we have 13 is just hanging out by itself it's referred to as a constant you could say it's a constant term or just a constant its value is not going to change it's always just going to be 13. if you have a number that's multiplying a variable or variables let's just take this 2 here it's referred to as a coefficient so here 2 is the coefficient of the variable x all right now when we work with terms you're going to come across this definition called like terms and like terms are terms with the same variable parts so this is very very simple when you only have one variable when you have more than one variable it can get a little bit more complex so with one variable let's just say we had 3x squared and 4x squared we have the same variable you have x and you have x and you have the same exponential power here right you have 2 and 2. so these are like terms these are like terms but if i had something like let's say 3x to the fourth power and 4x cubed these are not like terms these are not like terms and why is that the case well they're not the same right you have an x and an x but you have a four and a three so you have to have the same variable parts we could have like terms if we change this guy into a four right because you'd have a four and a four but that's not what we were given right we're given an exponent of a 3 there so these are not like terms now if you have more than one variable it gets a little bit more complicated basically for that situation each variable has to be present and each of those variables each copy of that has to have the same power so in other words let me show you with an example if i had 3 x squared y z and 5 x squared y z these are like terms why because i have x squared and x squared y to the first power and y to the first power z the first power and z the first power if i had y to the fourth power and y to the fourth power still got like terms if i had z to the fifth power and z to the fifth power still like terms if i change this to z cubed i don't have like terms because in this case i have z to the fifth and z cubed so the whole thing fails although everything else is good right all it takes is one to fail so you have to have the same variables so i have an x an x a y and a y a z and a z and they each have to be raised each each copy of that variable has to be raised to the same power so my x has to be raised to the second power in each case my y has to be raised to the fourth power in each case my z has to be raised to the fifth power in each case okay that's what gives me like terms if i saw something like let's say two x to the fourth power y squared and let's say negative seven x to the fifth power y cubed these are not like terms okay not like terms again i have x and x but i have the fourth power and the fifth power and all it takes is one of them to fail and you could just stop and say they're not like terms but then you have a squared here on y and a cubed here on y again doesn't work not like terms so when we place our terms together using either a plus operation or minus operation we end up with an algebraic expression so an algebraic expression could be one single term so let's say you had the number 14 that's an algebraic expression 14x squared would be an algebraic expression you know so on and so forth it can be a single term but it can also be multiple terms that are separated by plus or minus operations so 14x squared plus 5x plus you could do square root of x minus 5 something like this in each case you have a term a term a term and a term and they're separated by plus or minus operations so this is an algebraic expression okay now the simplest type of algebraic expression that we're going to come across is known as a polynomial so a polynomial is defined as a single term okay so it could be one single term or the sum of a finite number of terms so it just means it has an end where each variable has only non-negative integer exponents so let's make this a little easier when we say non-negative integer exponents we really mean whole number exponents so let's write that whole number exponents okay so when you're asked if something is a polynomial you want to look at the variables and you want to look for whole number exponents so let's look at some examples so if i saw something like 5x squared minus 1 would this be a polynomial yes it would be you have a term here that consists of a variable part that is x squared the exponent there is a 2 that's a whole number exponent you're good to go and then you just have minus 1 here you don't need to worry about any kind of constant term you're good to go there you're only focusing on the variables that you're going to encounter and you need to make sure the variable has a whole number exponent so as another example let's say i had seven x squared plus the square root of x plus one is this a polynomial the answer to that is no and you might say well i just see a 2 here as an exponent you might not remember this but remember the square root of x can be written as x to the power of one-half okay so you got to watch out for that if you're taking the square root or the cube root or something like that that's not going to be a polynomial in most cases right there are some cases where it might be if you did something tricky like said i had the square root of x squared okay well in that case it'll pass but you just got to kind of look through things in most cases when i see the square root of x i know i can count that as x to the power of one-half and so this is not a polynomial right because this guy right here this exponent is not a whole number as another example of one that's not a polynomial let's say we had something like 11x cubed plus let's say we had 15 over x squared this is not a polynomial you might say well why i have x cubed and i have x squared it's a polynomial it's not and the reason is is because you have a variable in the denominator watch out for a variable that's in a radicand and watch out for a variable that's in a denominator because remember i can write this as x to the power of negative 2. this exponent is not okay is not a whole number exponent it's an integer but it's not a non-negative integer it is a negative integer so we have a problem there so that's not a polynomial all right so once you kind of understand the basic definition of a polynomial you might get some questions in a review section on this asking you is this a polynomial is this not a polynomial pretty easy again all you're looking for is are the exponents on the variables whole number exponents if you have a variable that's in a radicand you need to watch out for that if you have a variable that's in the denominator you need to watch out for that if you're given fractional exponents or negative exponents you need to watch out for that okay so now let's move on and talk about the degree okay so the degree of a term and then we're also going to talk about the degree of the polynomial so let's start off with the degree of a term so the degree of a term is the sum of the exponents on all variables of the term so as an example if i have something like 14 x cubed i've only got one variable there the degree is 3 right i just look at the exponent there if i had something like let's say 55 x to the seventh power y squared i want to sum the exponents 7 plus 2 is 9 so my degree is 9. if i had something like let's say 27 x to the fourth power y to the 13th power z to the fifth power again just sum the exponents 4 plus 13 is 17 17 plus 5 is 22 so the degree there is 22. now what if i put something out here like let's just say the number 5. what's the degree there okay so we're going to use a little trick for this remember that raising a number or specifically raising a non-zero number to the power of 0 gives me 1. so if i multiplied 5 times 1 it would just be 5. well i can replace this with x to the power of 0. as long as x is not zero okay as long as x is not zero x to the power of zero would be one and this would just be five okay so i'm not doing anything illegal here but i can look at this guy right here and say okay well my exponent is zero so five the constant term 5 has by definition a degree of 0. so anytime you run into a constant you can just say it has a degree of 0. all right so the degree of a polynomial is the largest degree of any term in the polynomial so let me give you an example let's say that we had 27 x to the ninth power y to the 11th z then plus let me kind of make that a little better 15 x y z then minus 2. so what's the degree of the polynomial so you look term by terms you say okay this one's 9 plus 11 plus 1. remember if a variable does not have a visible exponent it's implied to be 1. so 9 plus 11 is going to be 20 and then 20 plus 1 is 21. this guy right here you'd have a 1 a 1 and a 1 1 plus 1 plus 1 is 3. and we just said that a constant term has a degree of 0. so you have 21 3 and 0. we just said the degree of a polynomial is the largest degree the largest degree of any term in the polynomial so this guy right here has the largest degree it's 21 so 21 is the degree of the polynomial let's take a look at another one let's say we had negative 5 x to the fourth power y squared minus 2 x to the fifth power y and then plus 7 z what's the degree of the polynomial well you sum the exponents so you say 4 plus 2 is 6. so this one's 6 this one's 5 plus 1 this is six and this guy right here is just a one so you have a tie for first place basically but that doesn't change the degree the degree of the polynomial is still going to be six all right so let's talk a little bit about standard form when we write a polynomial it is expected that we place the polynomial in standard form this means our powers are in descending order from left to right so this is basically something you want to do especially when you're first working with polynomials if you get polynomials with a lot of variables involved it's typically not as important to write something in standard form because it gets a little messy but when you work with one single variable there's kind of no excuse not to because it's very straightforward so if i'm working with something like let's say six x plus nine x to the fifth power plus seven x squared minus three x to the fourth power plus two i just look at my exponents and i say okay this one has the highest then the next highest then the next highest then this would be the next highest and then lastly this guy doesn't have a variable at all the constant term will go last so i could just reorder this and write it in standard form and say okay well 9x to the fifth power goes first then negative 3x to the fourth power or minus 3x to the fourth power is next and then you could say plus 7x squared and then plus 6x and then plus 2. so this would be standard form here so that's very very simple very very easy to wrap your head around that all right so when we encounter a polynomial that has more than one variable and we want to write it in standard form we basically want to go by degree first okay so let's say i gave you an example where we had 2 y squared plus negative 3x squared y squared and then let's say we had plus 6x to the seventh power okay so if i look at the degree of each term this guy is a two this guy is a four this guy is a seven so if i go an order of degree i would put 6x to the 7th power first i would put minus 3x squared y squared you put plus negative it doesn't matter and then i would put plus 2 y squared okay so this would be standard form now let's say we had a tie okay we had a tie so let me give you an example like that so let's say we had something like 9 x to the fourth power y plus 7 x squared y cubed minus 15 then plus 12 x cubed y cubed so what's going to go first what's going to go second you know so on and so forth this guy would have a degree of five four plus one is five this guy would have a degree of five two plus three is five this guy would have a degree of zero right a constant term has a degree of zero this guy would have a degree of six so we could write this guy first it has a degree of 6 but what do we do here we have two terms with the same degree so when this happens most people will tell you to go in alphabetical order so in other words we have the variable x and y x comes before y so we would look for the term that has the highest exponent on x so this guy has an x to the fourth power this guy has x squared so this is the term that's going to go before this term because in alphabetical order x comes before y and this guy has an x to the fourth power so we would put plus nine x to the fourth power y and then plus 7x squared y cubed and then minus 15. so hopefully that gives you a little bit of clarification on the topic you might see some computer programs or some people out there that show a different method to do this that's okay you know generally speaking when you think about standard form you're talking about a polynomial with one single variable it's agreed upon that that goes in descending order of the powers and essentially you don't ever have any problems with that when you run into a situation with two or more variables then you kind of get a little bit of disagreement out there you might punch this into a calculator and say simplify and it might give you a different order than this at the end of the day you're going to have the same answer and that's what matters all right so lastly we have special names for certain polynomials a single term polynomial is known as a monomial so that would be something like the number 3 or it could be something like 4x squared or it could be something like negative 7x cubed these are each monomials a binomial okay think about bi that means 2 a binomial means we have a 2 term polynomial so something like 4x minus 1 or 12x squared plus 3 or let's say 15 x to the seventh power minus 2x something like that and then a trinomial has three terms right think about tri tri means three right you think about tricycle it has three wheels so we would have something like let's say five x squared minus x minus three or as another example 90x to the fourth power plus x squared minus 1. right so two examples of a trinomial in this lesson we want to review adding and subtracting polynomials so before we look at our sample problems we want to review the concept of like terms so recall when you're adding and subtracting polynomials you need to be able to combine like terms so first and foremost what are like terms well if i have only one variable involved like terms have the same variable raised to the same power so something like 2x squared and 5x squared we have the same variable we have an x and we have an x and we also have the same exponent so we have a 2 and we have a 2. so these are like terms these are like terms now if i saw something else let me kind of scroll down a little bit we'll get back to that picture in a minute we saw something else like let's say 3x squared and 5 y squared these are not like terms you have different variables you have a variable of x and you have a variable of y the exponent is the same but that doesn't matter because the variables are different so those are not like terms so we can say not like terms and to give you another example where we don't have like terms suppose we had 5 x to the fifth power and let's say 7 x to the fourth power we have the same variable you have x and you have x but the exponents different you have an exponent of 5 and an exponent of 4 so these are not like terms okay so let's erase this real quick and i'm going to leave this example up here because we're going to work with it in a second and i want to talk a little bit about like terms with more than one variable so this can be a little bit challenging to describe with words so let me kind of give it my best shot so essentially what you're looking for is the exact same variable parts okay plural so each variable that you have has to be present in each term so if you have an x and a y and one term you have to have an x and a y in the other term if you have x y z in one term you got to have x y z in the other term then each variable okay that's similar has to be raised to the same power so what i mean by that is if you have x that's squared in 1 you've got to have x squared in the other okay so you've got to have the same variables and they have to be raised to the same power not all of them raised to the same power but the x has to be raised to the same power as the other x the y has to be raised to the same power as the other y okay so as an example suppose we had 2 x y z squared and we had 5 x y z squared now i have an x i have an x i have a y i have a y i have a z i have a z remember when we don't have a visible exponent we can say it's an exponent of 1. so i'm just going to fill these in real quick for reference sake so we have x to the first power and x to the first power right same exponent y to the first power y to the first power same exponent z squared and z squared same exponent so i'm not saying the exponent has to be the same on all the variables what i'm saying is that if you have a z squared here you've got to have a z squared over here to have like terms okay if you had let's say a y cubed here this guy over here would have to be y cubed for you to have like terms if you had an x to the let's say seventh power then this guy over here would have to be x to the seventh power to have like terms the coefficients or the numbers that are multiplying the variables or variable never matter okay it doesn't affect anything when you're talking about having like terms only the variable parts that matter so these would be like terms so as an example where you don't have like terms suppose you had x to the fourth power y squared and then something like 9 x cubed y squared so you have an x and you have an x you have a y and you have a y so the variables are the same you have x and y but the exponents are not the same right so here you have x to the fourth power here you have x cubed so right there you can stop it fails you don't have like terms although you have y squared here and y squared here this guy right here is what stops you okay that 4 and that 3. if i change this into a 4 then i have like terms if this gets changed back into a 3 i don't have like terms okay so once you understand the concept of like terms and what you're looking for there you're ready to move on and think about how to combine like terms okay so let's start out with just addition and i'm going to kind of scroll up just a little bit and i'm going to show you a few different things and i'm going to show you with a little illustration that often helps students that struggle with this concept so if we have 2x squared and 5x squared we can sum these using the distributive property so before i even do that let me just tell you it's very very easy to find the sum you just add the coefficients if i have 2 of something plus 5 of the same thing it's 7 of that thing so i have 7x squared but if you can't wrap your head around that at first use the distributive property so if i had 2 x squared plus 5 x squared we know at this point that we can use our distributive property and we can factor out the x squared so we're going to pull this out in front of a set of parentheses and inside i'd have the 2 plus the 5. so if i did 2 plus 5 that's 7. so this ends up giving me 7x squared so that's kind of two different ways you can think about this another way to think about this and kind of erase everything if you think about 2x squared plus 5x squared like this 2x squared is 1. it's x squared plus x squared okay this is 2x squared then plus 5x squared is x squared plus x squared plus x squared plus x squared plus x squared one two three four five of those guys so if i have again 2 of something plus 5 of the same thing i know i'm going to have 7 of that right if i just went through and dropped the parenthesis and said x squared plus x squared plus x squared plus x squared plus x squared plus x squared plus x squared i'm going to have seven x squared okay so there's a few different ways you can kind of think about this now i want to turn to this kind of image i have up here and this is something that really gets students thinking if you ever can't figure out why you can't combine like terms you can go to this simple example which is taught in pretty much every algebra one course so you think about having apples and oranges the apples you can think of as being like terms with each other and the oranges you can think of being like terms with each other so in other words if i have four apples here and two apples here i can combine these two and say i have six apples okay very easy if i have two oranges here and one orange here i can combine these and say this is three oranges now what do i do here i have an apple and an orange plus another orange can i combine everything here no all i can do is add the orange and the orange i would have to say this is one apple and or you could say plus if you wanted to two oranges the orange and the apple are different you can't combine them right i mean you can put them together but you don't combine them like you would if you had all apples or all oranges so you have to say you have one apple and two oranges it's the same thing if you had something like 3x plus 5x squared plus 2x squared we know 5x squared plus 2x squared is 7x squared so we would have 7x squared plus 3x this right here i can't combine this any further because they're not like terms i have 7 of some quantity x squared plus 3 times a different quantity so i basically have seven apples and three oranges here okay so i just gotta list it as seven apples and three oranges i can't really make it any simpler so that's why we write seven x squared plus three x all right so now that we understand the concept of like terms and how to combine like terms let's talk about how we're going to add polynomials so adding polynomials is very very simple you're just going to drop the parentheses okay you can add in any order because of the commutative property the easiest thing to do is just to put your like terms next to each other and then you're going to combine all the like terms you can and then you want to place your answer in standard form okay so for this one i'm going to just write 3x squared okay that's this one right here plus do i have any other terms with x squared i have this guy right here so 19x squared and plus i have 2x and 7x so 2x plus 7x then plus 5 plus 1. and what i like to do is just put a little dot above everything to show that i've used everything just a little technique i use of course you can do whatever you want now once i've written it this way notice how my like terms are next to each other so i can go through here and say well 3 plus 19 is 22 the x squared would come along for the ride right so 3x squared plus 19x squared would be 22x squared and plus 2x plus 7x that would give me 9x right 2 plus 7 is 9. then plus 5 plus 1 we know that it's 6. so this is as simple as we can make this because none of these are like terms with each other okay so we have 22x squared plus 9x plus 6. now another method you can use another common method that you will see and this would work also is to use a vertical addition method just like we would use when we're adding multi-digit whole numbers so the idea here is to line it up by like terms so you could say 3x squared plus 2x plus 5 and this one will line up perfectly for so we'll put plus 19x squared plus 7x and then plus 1. so i'm just going to add in these columns here okay these columns if i go straight down 5 plus 1 is 6 2x plus 7x is 9x and 3x squared plus 19x squared is 22x squared so either way you want to do it you get the same answer i find horizontal addition is a little bit faster because a lot of times i don't end up even writing things out i just kind of eyeball things and i'll say okay well 3x squared plus 19x squared is 22x squared i'll just write that down kind of put a dot over to say i've used it and kind of move on so now let's talk about subtracting polynomials so there is an additional step involved in this process i want you to think back to when we did 5 minus 3 and we wrote this as 5 plus a negative 3. remember you can always change a subtraction operation into addition of the opposite so instead of 5 minus 3 i can say 5 plus negative 3. well that's exactly what we're going to do here we're going to change our subtraction operation to addition and we're going to change the polynomial that's being subtracted away into its opposite how do you change something into its opposite you multiply it by negative 1 or you change the sign so i can just put plus negative 1 here and you're going to see this in a lot of your classes you'll see this plus negative 1 outside of parentheses you're going to wonder why do they keep doing that the reason they do that is to remind you that you need to distribute a negative 1 to each term here and that's going to change the sign of each term for you and then you can just add once you've done that so if we go through and do this the first polynomial here i can just drop the parentheses i don't need to do anything so 7x to the fifth power minus 3x minus 11. now i want to change the sign of each term by distributing my negative 1. so negative 1 times negative 9x to the fifth power would be plus 9x to the fifth power negative one times nine x would be minus nine x negative one times four would be minus four and once you get really good at these what you can do is just change it in here right you can just change it as you're working the problem and it's a little bit quicker right it's the same either way once i've done this i can just drop the parentheses and i can just go through and do my addition remember you can add in any order and you can also relate subtraction to addition of the opposite so if i have minus 3x i could say that's plus negative 3x right so i could rearrange things in any order that i'd like using that technique so what i'm going to do is i'm just going to add on the fly here i'm not going to rearrange stuff i see that my highest exponent is a five so i'm going to start with that so my answer ends up in standard form so if i have 7x to the fifth power and 9x to the fifth power those guys are both going to be like terms you have positive 7x to the fifth power plus positive 9x to the fifth power that's going to give me 16 x to the fifth power so i've used those now i'm going to look for the next highest exponent that's going to be a 1. so you have minus 3x and you have minus 9x so you can basically think of this as plus negative 3x and plus negative 9x so if i have those negative 3x plus negative 9x is going to be negative 12x so we can write this as minus 12x then lastly we have minus 11 and minus 4. again you could write this as plus negative 11 and plus negative 4. so you could say negative 11 plus negative 4 would give me negative 15 so i can write minus 15 here so we end up with x to the fifth power minus 12 x minus 15. the answer is in standard form and there's no like terms here that i can combine anything with so this is as simple as we can make this all right let's kind of turn the heat up a little bit and we'll look at not only a mixed with addition and subtraction we're also going to involve two variables now so we have 14y to the fourth power minus 11x squared then minus we have negative 14y to the fourth power minus 7x squared then plus 10x to the fourth power y to the fourth power then plus we have 9y to the fourth power minus 8x to the fourth power y to the fourth power okay so this guy right here nothing to do there other than just drop the parentheses so i'm going to start by just writing 14y to the fourth power minus 11x squared okay and if you want a lot of people more comfortable just putting plus negative and you can leave it like this or you can put the minus in front doesn't matter okay so now we have this minus that's out here and it's meant to be directly in front of this guy right here we just don't have enough horizontal space on the screen to do that so i'm gonna put plus here and i'm gonna put a negative one here and we're gonna scroll down and i'm going to distribute the negative 1 to each term and that just means we're changing the sign so i'm going to change the sign of negative 14y to the fourth power it's plus 14y to the fourth power change the sign of negative seven x squared it's plus seven x squared change the sign of ten x to the fourth power y to the fourth power it's negative or you could say minus ten x to the fourth power y to the fourth power again you could write minus or you could put plus negative it does not matter okay so now we have plus this guy so i could just drop the parentheses so 9y to the fourth power minus 8x to the fourth power y to the fourth power okay scroll down and get some room going now the highest power here is not so straightforward because you have multiple variables so let's just add in the order that they come in and then we'll show you how to put this in standard form it's something we talked about in the last review lesson but in case you missed that we're going to show it again here so if i start with 14y to the fourth power i've got this guy right here i've got this guy right here and then i've got this guy right here so three of those guys that can be combined so you would do 14 plus 14 which is 28 and then 28 plus 9 which would give me 37 so this would be 37 y to the fourth power okay for the next one i have negative 11x squared i have 7x squared and that's all i have there so negative 11 plus 7 is negative 4. so we can put minus 4x squared and then lastly i'm going to have this guy right here this negative 10x to the fourth power y to the fourth power and negative 8x to the fourth power y to the fourth power so negative 10 minus 8 or negative 10 plus a negative 8 if you wanted to write it like this again doesn't matter you would get negative 18. right so negative 18 x to the fourth power y to the fourth power so this is a valid answer but again we generally want to write things in standard form now when you have multiple variables involved it's generally going to be more acceptable and not right in standard form because people tend to just leave their answer and kind of move on if you have an answer with one variable you should definitely make sure to put it in standard form if you have multiple variables in a lot of cases it's not going to be looked at but just so you know how to do it anyway so the way we want to do this is by degree okay that's the first thing so this term right here has a degree of four right because you have an exponent of four this one has a degree of two from the exponent of two and this guy we sum the exponents so four plus four is eight so we don't have anything that's conflicting here it's straightforward we'd have eight then four then two so we're going to write it in that order so we'd start with negative 18 x to the fourth power y to the fourth power then we would put plus 37 y to the fourth power and then minus 4 x squared and i know you might put this into one of those online calculators and get a different arrangement again this is the proper way to do it it might have a different formula or a different way to calculate it when you work with one variable there's a standard way when you work with multiple variables there's some different things that are going around all right let's take a look at one more problem and then you should have enough review on this to be able to tackle pretty much any problem that deals with adding or subtracting polynomials very very easy concept overall so we have negative 12 x cubed y plus xy to the fourth power then plus we have 10x squared y cubed plus 10x cubed y minus 10x squared y squared then minus negative 3x cubed y minus 5x squared y squared okay so the first guy right here we could just drop the parentheses so i'll put equals negative 12 x cubed y plus x y to the fourth power okay then this guy i have plus this so because of that i can just drop the parentheses so we'll put plus 10 x squared y cubed plus 10 x cubed y minus 10 x squared y squared then i have my minus so i'm going to change that to plus and i'll put my negative 1 here and i'm going to distribute the negative 1 to each term and that's going to remind me to change the sign of each term inside of the parentheses so if i change the sign of this it would be positive or plus 3x cubed y if i change the sign of this i'd have plus 5x squared y squared okay so let's scroll down get a little room going and what we can do here again let's just go in order so i have x cubed y do i have that again i don't have it here not here i have it here not here i have it here and not here so i would have negative 12 plus 10 which is negative 2 then plus 3 which is 1. so this would be 1x cubed y or i could just write x cubed y all right then next we have xy to the fourth power is there any other xy to the fourth power no so let's just put plus xy to the fourth power and then we have x squared y cubed and nothing else with that so plus 10 x squared y cubed plus lastly we have negative 10 x squared y squared and 5x squared y squared so these are like terms negative 10 plus 5 is negative 5. so instead of putting plus let me put minus so we'll put minus 5 x squared y squared now the reason i wrote this example is to give you one where you have a little bit of trouble writing this in standard form if you look at the terms here this one has a degree of four right you have an exponent of three and an implied exponent of one this one has a degree of five an exponent of one an exponent of four this guy has a degree of five right two plus three is five and then lastly this guy has a degree of four so you have two with a degree of four and 2 with a degree of 5 so what do you do well what happens is when the degree is the same we go in alphabetical order so i'm going to look for the x okay because x comes before y i'm going to look for the x with the largest exponent so if i start with a degree of five this guy has an x raised to the power of one this guy has an x squared so this is going to be my winner so that's going to be the leading term so 10 x squared y cubed then this guy so plus x y to the fourth power so i've dealt with these ones that have a degree of five and now i'm going to think about what has a degree of four this one has x cubed and this one has x squared again i'm working in alphabetical order so i'm thinking about the x because x comes before y with the higher exponent so i'm going to go with this one so i'm going to say plus x cubed y and then lastly i'm going to have minus 5 x squared y squared okay so don't let that type of problem trip you up essentially when you have the degree that's the same on more than one term you go in alphabetical order so in this case it was x and y so we looked for x because x comes before y and we looked for the x that had the larger exponent okay so that one's going to win and go in front in this lesson we want to review multiplying polynomials so to multiply two polynomials together we use the associative and distributive properties along with the rules for exponents so the simplest case that we're going to come across is multiplying a single term polynomial which is known as a monomial by another single term polynomial or again monomial so here we have 7x squared times 5x to the fourth power this is a monomial multiplied by a monomial so it doesn't get any simpler than this so you can go through your textbook will probably show you where you can reorder things i can write this as 7 times 5 times x squared times x to the fourth power you know i can group this multiplication together and this multiplication together and it's easy to do when you think about it this way 7 times 5 is 35 x squared multiplied by x to the fourth power x stays the same then we just add exponents 2 plus 4 is 6. so this gives me 35 x to the sixth power now this is a very very slow way to perform this action so although this is technically the way you show it it's not the way you want to do it in practice all you want to do if you see something like this you multiply the parts that you can together so i know that 7 can be multiplied by 5 and give me 35 so i'll start with that so i multiply my coefficient 7 by my coefficient 5. that gives me 35. so that's the first part then i just multiply the variables together i have x squared times x to the fourth power again x stays the same and then we add exponents 2 plus 4 is 6. okay so that's the quicker way to do it multiply the number parts together then multiply the variable parts together and then you'll have your product all right let's take a look at another example so we have negative 4x squared y multiplied by 3xy cubed okay so again let's multiply the number parts together first so you have negative 4 and you have 3. if i multiply negative 4 by 3 i get negative 12. now let's multiply the variable parts together and you can break this up or do it all at once you have two variables you have an x and a y so x squared times x would be x cubed right because if i multiply x squared by x remember this is really x to the first power x stays the same you add exponents you have a 2 and a 1 2 plus 1 is 3. then for the y there if you multiply y to the first power times y cubed you have a y that stays the same and you add the exponent 1 to the exponent 3 and you get a 4. so we end up with y to the fourth power for that part so the answer here is negative 12 x cubed y to the fourth power all right so in some cases you're going to multiply a monomial by a multi-term polynomial we already know how to do this because we've practiced this operation when talking about the distributive property if we see something outside of a set of parentheses we can remove the parentheses by multiplying this guy right here by each term inside the parentheses right we're going to distribute it to each term so i'm going to start out with negative 9x multiplied by negative 5x squared and i'm just going to write this out to make it completely clear what's going on then plus you'd have negative 9x multiplied by 2x then next you'd have negative 9x multiplied by negative 1. okay so what i'm going to do is i'm just going to put parentheses around these just to make it completely clear what's going on and now i'm going to go through a multiply so negative 9x multiplied by negative 5x squared negative 9 times negative 5 is 45. x multiplied by x squared is x cubed again x stays the same we have an exponent there of 1 an exponent there of 2 1 plus 2 is 3. then we have plus we have negative 9x multiplied by 2x this is going to give me negative 18 x squared right negative 9 times 2 is negative 18. so i'll write minus 18 here x times x is x squared all right then we have plus we have negative 9x multiplied by negative 1. so if i multiply something by negative 1 i just change the sign so this would just be plus 9x so we end up with 45x cubed minus 18x squared plus 9x there are no like terms here so i can't simplify anything further and this guy's already in standard form you have x cubed you have x squared and then you have x to the first power so it's written in descending order all right so now if we multiply two polynomials together and neither is a monomial we can multiply each term of the first polynomial by each term of the second polynomial and then combine the like terms so let's take a look at this so i'm going to split this up and say this is going to be 3x multiplied by x squared minus 5x minus 3. so that's the first part it's this guy that's going to be distributed to each term over here then you have minus 1. to make this easy to not mess up the sign i'm going to write this as plus negative 1. so i'm going to put plus and then we're going to have negative 1 outside of the parentheses so outside of x squared minus 5x minus 3. i would advise you if you're not good with signs you're constantly making mistakes don't put a minus anywhere put plus negative right especially if you're doing something like this it'll keep you from making mistakes because a lot of people will put -1 like this and they'll forget about it and then it just creates a problem so let's just keep that as plus negative 1 like this and then we'll be sure to have the correct sign okay so let's go through and do our multiplication so 3x multiplied by x squared is 3x cubed 3x multiplied by negative 5x is minus 15x squared 3x multiplied by negative 3 is minus 9x all right then over here i have negative 1 multiplied by x squared that would be minus x squared i have negative 1 multiplied by negative 5x this is plus 5x and then i have negative 1 multiplied by negative 3 which is plus 3. now one thing you can notice is again if you multiply something by negative 1 you're just changing the sign so instead of going through all this i could have just said okay this is negative x squared plus five x plus three right just change the sign and kind of bring it down okay so now i'm just looking for like terms three x cubed nothing to combine with that negative 15 x squared minus x squared is minus 16x squared negative 9x plus 5x is minus 4x and then we have plus 3. so this is going to be our answer we can't simplify it any further because there's no more like terms and it's in standard form right we have x cubed x squared x to the first power and we have our constant so when we multiply two binomials together this operation happens so often that we have a little acronym to remember how to do it so it's known as foil and i know most of you if you've taken an algebra 1 course or an algebra 2 course or any type of algebra course you've heard of foil it stands for first terms outer terms inner terms and last terms so again this only works if you're multiplying two binomials together it's not something you can use if you're multiplying a binomial trinomial or you know any other scenario so make sure you understand that but it is a good technique to use and let's just look at an example real quick we have a binomial here 2x minus 1 multiplied by another binomial x minus 5. so again binomial two-term polynomial so if i want to use foil i would do the first term so this is the first term and this is the first term okay so it's the first term of each binomial so 2x multiplied by x is 2x squared then i would do my outer terms so what's on the outside so 2x is on the outside here negative 5 is on the outside here so i know this says minus 5 you can again write this as plus negative 5. we could write this as plus negative 1 just to be crystal clear what's going on i need to multiply 2x times negative 5 that's going to give me negative 10x then i'm looking at my inner terms so what's on the inside well i'd have this negative 1 and this x so negative 1 times x is minus x and then lastly i'd have my final terms so that's negative 1 and negative 5. so i say final terms in the acronym it's last terms but it's the same thing so we have negative 1 times negative 5 which would give us plus 5. so we have like terms here in the middle and what ends up happening is we're going to have 2x squared negative 10x minus x is minus 11x and then plus 5. so this is known as foil this is a common thing that you're going to do in algebra and the opposite of this where you factor it you'll have a trinomial and you'll want to go backwards and factor it into two binomials all right let's take a look at another so we have x squared minus 7 multiplied by x plus 3y so again we're going to use foil so we take the first terms so we have x squared times x that's x cubed our outer terms x squared times 3y so that's going to be plus 3 x squared y our inner terms negative 7 multiplied by x is minus 7x and then the last terms or the final terms negative 7 multiplied by 3y is going to be minus 21y so let me kind of move this down so we have enough room to see so we have x cubed plus 3x squared y minus 7x minus 21y all right so let's wrap up the lesson there's not really much to multiplying polynomials you basically just need an understanding of how to use the rules of exponents and how to multiply right if you have that down as long as you understand the distributive property you're basically good to go so for this guy right here we're going to look at a tedious problem something you probably get you have two binomials multiplied together and then it's multiplied by a trinomial so what i would do if i got something like this i would use foil to find the product of this and then you'd multiply that by this okay so i'm going to start out by just finding the product of these two let me highlight them so what's that going to be 3x times 5x would be what i start with because that's my first terms so that would give me 15x squared then my outer 3x times negative 1 is minus 3x my inner negative 7 times 5x is minus 35x my last negative 7 times negative 1 is plus 7. combine like terms here i would get negative 3x minus 35x that's negative 38x okay so you end up with 15x squared minus 38x plus 7. okay so now what we want to do to kind of scooch this up we want to multiply this guy times this guy and how are we going to do that again we just use our distributive property so i can take each term of this polynomial here on the left and multiply it by each term of this polynomial here on the right so let's start by just saying okay we want 15x squared multiplied by 2x squared minus 5x plus 1 then plus we'll have this next term i'm going to write this as negative 38x multiplied by again x squared minus five x plus one and then lastly we have plus seven multiplied by two x squared minus five x plus one so let's get a little room going all right so let's take a look at the first part so 15x squared times 2x squared is going to give me 30x to the fourth power then 15x squared times negative 5x is going to give me negative 75x cubed 15x squared times 1 is just 15x squared then we have negative 38x times 2x squared so this is going to be negative 76x cubed then we'd have negative 38x multiplied by negative 5x this is going to be positive 190x squared then negative 38x multiplied by 1 is just negative 38x or minus 38x however you want to think about that then lastly we have this guy right here so 7 times 2x squared this will be plus 14x squared then 7 times negative 5x this would be minus 35x then 7 times 1 would be 7. okay let's scroll down some room going so let's look for like terms so we have x to the fourth power and nothing to combine with that we have x cubed x cubed and nothing else there we have x squared x squared x squared and then we have x and x okay so let's go ahead and combine like terms so we have 30x to the fourth power nothing to combine with that so we'll start with that then we have negative 75 x cubed minus 76 x cubed so that's going to give me negative or minus 151 cubed then we would have 15x squared plus 190x squared plus 14x squared so 15 plus 190 is 205. if i add 14 to that i get 219 so this is plus x squared then you have negative 38x minus 35x that would be minus 73x then lastly plus 7. so this is our answer 30x to the fourth power minus x cubed plus 219 x squared minus 73x plus 7. it's in standard form and again we can't simplify this any further because we don't have any other like terms in this lesson we want to review special polynomial products so in our last lesson we reviewed multiplying polynomials and essentially when we are multiplying polynomials we're sometimes going to come across problems that occur so frequently it's generally faster to memorize the formula and just apply it to the problem so these frequently occurring polynomial products are generally known as special polynomial products or you might hear some people say just special products so let's start off by talking about the product of conjugates so first and foremost what are conjugates so conjugates will occur whenever we have the same two terms but the sign is different so in this case we have x plus y and then we have x minus y so i have an x and an x a y and a y i have a difference of a sign a plus and a minus you could also have something like 5 x minus 9 multiplied by 5 x plus 9. i have a 5x i have a 5x i have a 9 i have a 9 the sign is minus the sign is plus okay as another example let's say we had something like 2 x squared y plus 7 multiplied by 2 x squared y minus 7. again this part right here is the same in each case this is the same in each case you have a plus and a minus okay so it's the same two terms it's just that you have different signs involved now when we have the product of conjugates it follows a very simple pattern so let me foil this out real quick and then i'll cover this so if i had x plus y multiplied by x minus y so if i started out with the first terms x times x is x squared the outer x times negative y would be minus xy the inner y times x would be plus xy and the last y times negative y would be minus y squared in every case the middle two terms are going to cancel you have minus xy and plus xy that's going to cancel so you're just left with x squared minus y squared so when you see this pattern just think about the first guy squared minus the second guy squared that's all you need to do so if i had something like let's go back to this example of 5x plus 9 multiplied by 5x minus 9. i don't need to foil this out square the first guy what is 5x squared well it's going to look like this because a lot of times when you hear 5x squared you're thinking this right but i'm squaring this whole thing the 5 and the x so put it inside of parentheses and then minus what is 9 squared so 5x squared would be 5 squared which is 25 times x squared then minus 9 squared which would be 81 okay so 25x squared minus 81 and we can foil this out and i can prove it to you just in case you don't believe me so first terms 5x times 5x is 25x squared outer 5x times negative 9 will be minus 45x inner 9 times 5x would be plus 45x you see those would cancel and then last 9 times negative 9 is minus 81. again this is what happens the middle two terms cancel you're left with 25x squared minus 81 which is just this guy squared minus this guy squared okay so let me put equals 25 x squared minus 81. so very simple pattern very easy to do so if you have the product of conjugates it's the first guy squared minus the second guy squared all right let's look at an example so we have x minus seven times x plus seven again i don't need to foil this out first guy squared x squared minus second guy squared second guy is 7 7 squared is 49. very very very easy let's take a look at it all we have 3x cubed plus 5 multiplied by 3x cubed minus 5. so again square the first guy so i'm squaring 3x cubed the whole thing then minus square the second guy i'm squaring five that's just 25. so if i square three i get nine if i square x cubed remember power to power rule x cubed squared is x raised to the power of 3 times 2 which is 6 and then minus 25. so we end up with 9x to the sixth power minus 25. all right for the next formula we're going to look at the square of a binomial and we have two different scenarios we want to cover so we have the quantity x plus y this binomial here this two term polynomial this quantity here is squared okay this is equal to x squared plus 2xy plus y squared then if we have x minus y this quantity squared this is equal to x squared minus 2xy plus y squared so the difference between the two formulas is just the sign in the middle this guy has a minus so you end up with a minus sign in front of the middle term now the reason this is so confusing for students a lot of people make this mistake they say x plus y this quantity squared is equal to what everyone does it x squared plus y squared right this is wrong do not do this okay this is addition you can't distribute an exponent like that what you're thinking of is multiplication if you had x y and this was squared this is x squared y squared you can do that that's legal but you can't distribute it with addition it doesn't work like that you have to expand it so in other words i'd have to take x plus y if this was squared i would say this is x plus y times x plus y okay it's two of these right i have two copies that's what an exponent of two is telling me okay so if i foil this out i would get what x times x is x squared outer would be plus xy inner would be plus xy final term or last terms would be plus y squared so this part right here is what i can combine i end up with two of those right i have x y plus xy that's 2xy very simple formula very hard to remember this is how i remember okay the way i do things is i think about okay i know this guy is going to be squared and this guy's going to be squared that's easy to remember so the first part of this is going to be the first term squared the last part of this is going to be the last guy squared the middle term the way i remember that is i know this exponent 2 is going to be multiplied by each term so 2 times the x times the y so 2 times x y that's it that's how i remember it and if you practice that scenario enough times you can put it on a flash card you could just you know write it in your notes and just go through a bunch of sample problems it's something you'll probably remember for the rest of your life or at least for a good amount of time now when you encounter the scenario with the minus you just have to remember to put a minus in front of the middle term all right let's take a look at an example so nice easy example we have x minus 3 this quantity here is squared so again instead of expanding this and foiling it and going through all that we can use our special products formula i'm going to square the first guy so i'm just going to square x i'm going to square the last guy so i'm going to square 3 but i'm going to put that at the end so 3 squared is 9. now i have a minus sign here so my middle term is going to be minus and this guy is going to be plus so i'm just missing the middle term and again i get that by taking this guy and multiplying it by each term so 2 times x times 3 2 times 3 is 6 6 times x is 6x so end up with x squared minus 6x plus 9. all right let's take a look at another so we have 4x squared plus 5y this is a quantity that's being squared so again i use my technique here square this guy so 4 x squared squared you got to square everything a lot of students make the mistake of just squaring the x part if i square 4 i get 16 if i square x squared i get x to the fourth power so this is 16 x to the fourth power then i'm going to have plus i have my middle term then plus i'm going to square this guy so 5y squared if i square 5 i get 25 if i square y i get y squared for the middle part again just take this guy and multiply it by each term so 2 multiplied by 4x squared times 5y 2 times 4 is 8 8 times 5 is 40 and then you have x squared times y or just x squared y so 16x to the fourth power plus 40x squared y plus 25y squared so you see how easy that is how simple it is to just memorize the formula you see this on a test you can just crank it out really really quickly all right let's talk about the cube of a binomial so this is one that if you memorize it it's going to save you a ton of time okay just for the simple fact that it just takes so long to go through a multiplication where you have three binomials that are being multiplied together you've got to go through and multiply and simplify and just do all this stuff so for this guy we have x plus y this quantity cubed this is equal to x cubed plus 3x squared y plus 3xy squared plus y cubed when we have the x minus y this scenario cubed it's the same thing it's just you have a minus sign in front of the second and final terms okay so this one is a little harder to remember but i have a technique for remembering it basically what i do again i know this guy is going to be cubed and this guy is going to be cubed and those going to make up the first and the final term of this polynomial so this guy cubed gives me this this guy cubed gives me this okay so now i just need to figure out the middle i know that the exponent 3 is going to multiply each of these but i just got to remember that the first time i do it or this term right here i'm going to multiply it by this guy squared and this guy as a first power right then the next time i do what i multiply by this guy is the first power and this guy squared so not that easy to remember but again if you flash card it it's something you can pick up pretty quickly again for this scenario you just have to remember that there's a minus sign in front of the second and the fourth term all right so let's look at an example so we have the quantity 2x plus 3y cubed again to do this i'm going to cube the first term so 2x cubed would be 2 cubed is 8 x cubed is just x cubed then i have a plus i'm going to have a plus some term plus some term plus cube the last guy 3 cubed is 27 y cubed is just y cubed and let me kind of move this answer down because we're going to run out of space let's move that down there okay so for my first middle term i'm taking this guy right here this exponent i'm going to multiply by each term but the first time i multiply it by this term it's going to be squared so i've got to think about 3 multiplied by 2x squared multiplied by 3y okay so first i'd square this 2 squared is 4 x squared just x squared so 4x squared so 3 times 4 is 12 12 times 3 is 36 so this would be 36 x squared y okay now i'm going to do the same thing but now i'm going to square the 3y part and i'm not going to square the 2x so 3 multiplied by 2x times 3y this guy squared and essentially i would have 1. 3 squared is 9 and y squared is just y squared so 9 y squared 3 times 2 is 6 6 times 9 is 54. so i'd have 54 x and this isn't going to fit let me kind of squeeze this down so i'm going to have y squared there okay and i kind of move this down too far all right let me erase this and we can read off our answer as 8x cubed plus 36x squared y plus 54xy squared plus 27y cubed all right let's look at one final problem so we have the quantity 5x minus 2y and this is cubed so remember if i have a minus i'm going to have a minus sign in front of the second and the fourth term so i'm going to go ahead and cube my first term 5x cubed again i'm cubing the 5 that's 125 right 5 times 5 is 25 25 times 5 is 125 x cubed is just x cubed put a minus sign in front of my second term plus sign in front of the third term and a minus sign in front of the last term for the last term it's 2y this guy cubed so 2 cubed is 8 y cubed is just y cubed then for the two middle terms so again i take this guy right here multiplied by each term but the first time i do it i'm going to multiply by this one squared okay so 3 multiplied by 5x squared multiplied by 2y if i square 5x i'm going to get 25x squared so 3 times 25 is 75 75 times 2 is 150 so this is minus 150 x squared y okay and let me erase that so now i'm going to multiply 3 by each term again but now this guy is going to be squared so 3 multiplied by 5x times we would have 2 y squared and 2y squared would be 4y squared so 4 y squared 3 times 5 is 15 15 times 4 is 60 and then times x y squared so our answer here is going to be 125x cubed minus 150x squared y plus 60xy squared minus 8y cubed again you can see if i had to go through and write out 5x minus 2y multiplied by 5x minus 2y multiplied by 5x minus 2y you can you can stop we're done with the video but you can stop at this point and try to go through and multiply that and see how long it takes you versus just using the formula this is a gigantic time saver it's something i definitely recommend that you learn and memorize and use every time you encounter this scenario in this lesson we want to review dividing polynomials so when we start working with polynomials one of the more challenging processes that you work with is dividing polynomials and more specifically it's polynomial long division that gives students a lot of trouble so this is an opportunity if you're not good at this we're going to review the process and you'll be able to work some sample problems and hopefully you'll have this under your belt before we kind of move further into the course so we're going to start with the easiest scenario which is dividing a polynomial by a monomial okay so this guy is really simple you don't need to use any long division here you're just going to set things up as a fraction so if i have 6x squared minus 5x minus 10 and this is divided by 2x this guy right here my dividend okay my division problem will be the numerator for the fraction then this will be over the divisor 2x will be my denominator and what am i going to do here well essentially what i'm going to do is use a little trick from when i work with fractions so i'm going to take each part of the numerator and write it over the denominator so i'm going to go 6x squared over 2x and then i'll go plus negative 5x over 2x and then plus negative 10 over 2x now if you saw this you should know that you can make it into that and because we have inequality there you can go back and forth if i saw 6x squared over 2x plus negative 5x over 2x plus negative 10 over 2x i know that i have a common denominator of 2x so i'm allowed to write 6x squared minus 5x minus 10 over the common denominator of 2x which is what i have here okay so we know that this is legal and now we can just go through and simplify each part separately so what i'm going to look at is what can i do so 6 over 2 is 3 x squared over x would be x to the first power so let me write 3 x then we have plus you have negative 5 x over 2x well i know the x's would cancel there so i would end up with minus 5 over 2. in decimal form you could write that as negative 2.5 but just going to leave it as a fraction then we have plus we have negative 10 over 2x so this would cancel with this and give me a 5. so essentially i would have minus 5 over x so this would be our answer from the division i want you to notice that this is not a polynomial and that's okay right the reason it's not a polynomial is because you have a variable in the denominator and sometimes when you divide a polynomial by another polynomial you get an answer that is not a polynomial like we got here but again that's okay not an issue now if you wanted to check this we know that when we do division something like 6 divided by 3 equals 2 we can go backwards and we can check so i can do 2 times 3 equals 6 right that's how i check my answer do the same thing here if 6x squared minus 5x minus 10 divided by 2x gives me 3x minus 5 halves minus 5 over x well then i should be able to multiply this by this to get this okay let's check and see if that's true so 2x multiplied by the quantity 3x minus 5 over 2 minus 5 over x what does that give me let me kind of erase this so 2x times 3x would be 6x squared then we'd have 2x times negative 5 over 2. so let's think about that we know it would be negative 2x multiplied by 5 over 2 is what the 2s would cancel i'd have 5x there so this would be minus 5x then we'd have 2x multiplied by negative 5 over x i know this is negative then 2x times 5 over x i know that these x's would cancel and basically have 2 times 5 which is 10. so this would be 10. so what did we end up with we got 6x squared minus 5x minus 10 which is exactly what we have right here okay so that tells us our answer is correct all right now let's turn to polynomial long division which is the process that pretty much every student hates it comes from the fact that in elementary school a lot of students hate long division with regular whole numbers and that hatred kind of carries itself through when we start working with polynomial long division i can assure you this is an easy process it's something that you just have to work step by step on so let's start out by just setting this guy up we have 5x cubed plus 24x squared plus 20x minus 5 and this is divided by 5x minus 1. so we're going to use a standard long division process 5 x cubed plus 24 x squared plus 20 x minus 5. so the very first thing is to make sure that both your dividend and your divisor are in standard form so in descending order of the power so 3 2 1 and then constant so we're good to go there so i'm going to put this underneath my kind of long division symbol and then out to the left of this i'm going to put my divisor so my 5 x minus 1. so once i've done this i'm pretty much ready to start and essentially what you want to do is leading term into leading terms so i want to take this guy right here the leading term of the divisor and ask how many times will it go into the leading term of the dividend so in other words i am asking the question what is 5x cubed divided by 5x okay very very simple 5 divided by 5 is 1 x cubed over x is x squared okay so i would take that x squared and i would write it over the slot for x squared so in other words this is kind of the place value for the x cubed this is the place value for the x squared this is the one for the x to the first power and this is for my constant so i'm going to take and write that right here x squared some tutorials some teachers will just put the x squared all the way to the left you can do that if you want to but i was always taught that you want to maintain the place value just like if you were working with whole numbers okay all right so now the next step is to multiply so i'm going to multiply by this term and this term okay so x squared is going to multiply 5x and then also negative 1. so x squared times 5x is 5x cubed x squared times negative 1 is going to be minus x squared the next step that we're going to do is we're going to subtract so what i'm going to do is i'm going to put this inside of parentheses and put a subtraction symbol out in front and what that's telling me to do again whenever you see a subtraction sign in front of parentheses in order to drop the parentheses you've got to change the sign of each term that's why you see a lot of times you'll see plus negative one here right it's to distribute the negative one to each term if you want to drop the parentheses just change the sign so this becomes minus and this becomes plus okay so now we're going to subtract or add or however you want to think about this you're basically going to combine like terms so 24x squared plus x squared is 25x squared and 5x cubed minus 5x cubed is 0. so basically this guy's going to cancel so i'm just left with this and now i want to bring down the next term so bring down the next term so this is plus 20x okay and once i've done that i want to go into the repeat or remainder okay so if i brought something down i want to repeat the process so i start with division again so i'm going to go leading term which is this guy into leading term but i'm not working up here anymore i'm working down here so this is my leading term now so i want to do 25x squared divided by 5x so what is that 25 divided by 5 is 5 x squared over x is x to the first power so my answer is going to go above this 20x so plus 5x and then i multiply again when i multiply i want to multiply by this whole thing 5x multiplied by 5x is 25x squared 5x multiplied by negative 1 is minus 5x now you can do the parentheses if it helps you to remember to change the sign of each term before you combine like terms okay very important because you want to subtract this polynomial away once you get really good at this what i end up doing is just changing the sign and just combining like terms all right so now we know that 25 x squared over 25 x squared that's going to cancel so you don't need to worry about that 20x plus 5x is 25x at this point i'm going to scroll down and get some room going okay so now i want to bring down my next term which is negative 5 and i want to repeat the process because i brought something down so again leading term into leading terms so i'm working here now this is my leading term 5x goes into 25x how many times so what is 25x over 5x we know that the x's would cancel completely 25 over 5 is 5. so i would put plus 5 here and then i would multiply 5 times 5x is 25x and then 5 times negative 1 is minus 5. now you see you have the same thing over itself when i change the sign of each term it's going to become 0. right because if i put parentheses around and put the minus i drop the parentheses by what changing the sign and then if i combine like terms this is 0 and that's 0. this just tells you right here that you don't have a remainder right if i was to multiply 5x minus 1 times x squared plus 5x plus 5 i would get this exactly right there's no remainder now i want to take a moment and kind of go back through the steps with you and if you have a piece of paper and a pen i want you to write down this kind of acronym so you have d m s b r okay d m s b r some of you will remember this from elementary school a lot of people will say dad mom sister brother and then rover okay so this obviously is just a way to remember the dm sbr so dad mom's sister brother rover what does this stand for so dad stands for division okay that's what it represents mom stands for multiply sister stands for subtract brother stands for bring down and rover stands for repeat or remainder okay so when we first start the process we did leading term into leading term that was division then once we got our answer we did multiplication then we subtracted then we brought down the next term then we repeated the process right we did division again then we multiplied then we subtracted then we brought down then we did the repeat again right until we got to the point to where we didn't have anything else to bring down so this rover changes from repeat to remainder which in this case there is no remainder it's just zero okay so that's how you know you're done with your division so you want to write this down and if you follow this kind of step by step it'll help you because sometimes you're like well what's the next step what do i do let's always divide multiply subtract bring down and then repeat or remainder depending on if you brought down something else now let me erase this real quick hopefully you have it down if not you can pause the video i want to check my results and i'm not going to check all of them today but you know how to check them i just want to verify that this guy multiplied by this guy gives me this guy so i would have 5 x minus 1 multiplied by x squared plus 5x plus 5. okay so to do this 5x would multiply by every term here so i'm not going to write this out i'm just going to do it on the fly so 5x times x squared is 5x cubed 5x times 5x is plus 25x squared 5x times 5 is plus 25x now i need to multiply negative 1 by each term here multiplying something by negative 1 just changes the sign so i'm just going to go through and change the sign of each term so i'd have negative x squared minus 5x minus 5. so just combine like terms i'd have 5x cubed i have 25x squared and negative x squared that would give me plus 24x squared i have 25x and minus 5x that would be plus 20x and then i have minus 5. so if i look at this guy right here it completely matches this guy up here 5x cubed plus 24x squared plus 20x minus 5. so again if you're struggling you can always check your answer using multiplication all right let's take a look at one that has a remainder involved just so you can get a little bit of experience with that if you've never seen it so i'm going to set up my division problem the same way i have 7x to the fourth power plus 11x cubed plus 50x squared minus 38x plus 14. this is divided by 7x minus 3. so seven x to the fourth power plus eleven x cubed plus fifty x squared minus thirty eight x plus fourteen let me kind of move this a little bit i'll have to move it down two okay make my long division symbol and then i have seven x minus three out in front and let me kind of scroll down get a little room going all right so again dad mom's sister brother rover so we start with the d step okay divide so i'm going to go leading term into leading term so what is 7x to the fourth power divided by 7x the sevens would cancel x to the fourth power over x is x cubed so this would be x cubed now i wanna multiply x cubed times seven x is seven x to the fourth power x cubed times negative three is minus 3x cubed okay so again if we want to subtract this now instead of putting parentheses around i can just change the sign and combine like terms so we know this is going to cancel 11x cubed plus 3x cubed is 14x cubed and now i want to bring down this term here so plus 50x squared okay so now again after i've brought something down i want to repeat the process okay that's the rover step so now this is my leading term here so leading term into leading term what's that going to be if i had 14 x cubed over 7x 14 divided by 7 is 2 x cubed over x is x squared so this would be plus 2 x squared and i can multiply 2x squared times 7x is 14x cubed 2x squared times negative 3 is minus 6x squared again if i want to subtract i'm just going to change the sign so minus and then plus okay don't forget to do that this is going to cancel 50x squared plus 6x squared is 56x squared bring down the next term so minus 38x let me kind of scroll down get a little room going so again i'm repeating the process now so leading term into leading term so this guy's your leading term now 7x goes into 56x squared how many times so 56x squared divided by 7x so this guy cancels with this guy and gives me 8. this cancels with this and gives me x so just 8x so plus 8x now we want to multiply so if i multiply 8x times 7x i get 56x squared 8x times negative 3 is minus 24x okay so this will be minus this will be plus again again again don't forget to change the signs it's very easy to forget and you'll make a mistake so this is going to cancel negative 38x plus 24x is negative 14x and now we can bring down this guy right here so plus 14. so now we're going to do leading term into leading term and so what is negative 14x divided by 7x and that's going to give me negative 2. right this would cancel with this and give me a negative 2 and x over x cancels and becomes 1. so this is negative 2. so i'll put minus 2 here and let's go ahead and multiply negative 2 times 7x is negative 14x negative 2 times negative 3 is plus 6. and again let's scroll down get some room going so change the sign of this and change the sign of this combine like terms this cancels 14 minus 6 is going to give me 8. so this guy right here is your remainder let me erase everything so when you have a remainder you're going to write it like this you're going to say plus the remainder was 8. so 8 over 7x minus 3. so take your remainder and put it over your divisor okay so just like that all right so to kind of wrap up the lesson let's look at the scenario where we have missing powers so when we divide polynomials and a power is missing we use zero as the coefficient for each missing power so what do we mean when we say missing power well if you have a polynomial that's let's say a fourth degree polynomial meaning the highest exponent is a four then what you expect to have is x to the fourth x cubed x squared x you know and then your constant term well if you're missing one of those powers as an example if i had something like x cubed plus 3x plus 2 i'm missing an x squared right i have x cubed no x squared x to the first power and a constant so what i can do is i can just rewrite this as x cubed plus 0 x squared plus 3 x plus 2. so the 0 isn't adding any value here it's not changing anything 0 times anything is always 0. it's just a placeholder okay it's just something we're going to use to make sure that we can line things up properly and do our division so let's take a look at a problem so we're going to look at 3x cubed minus 2x plus 5 divided by x minus 3. so we set this up the normal way so let's just make a little long division symbol we have 3x cubed i'm missing x squared so i'm going to put plus i'm going to put 0 and then x squared and then i have minus 2x and then plus 5. so from this point on everything is the same right you just need to make sure you put that 0 in as the coefficient for the term that you're basically missing i'm missing x squared so i got 0x squared now over here i'm going to put my x minus 3 and then i'm just going to run through my steps so remember it's dad mom sister brother rover okay dad mom sister brother rover really easy to remember so we start with the dad step or the divide so i want to go leading term into leading term so what is three x cubed over x well x cubed over x is just x squared right so this would be 3 x squared now i want to do my mom step or multiply so multiply 3x squared by x i get 3x cubed multiply 3x squared by negative 3 i get minus 9 x squared okay so now i want to subtract and again you could put parentheses around it or the quicker way is just to realize i'm just changing the sign of everything so change the sign of this and this and then we can just combine like terms so 3x cubed minus 3x cubed is of course zero so that part's going to cancel you think about 0x squared plus 9x squared that's just 9x squared and then i'm at my brother step or my bring down step okay so bring down this negative 2x and then we're at the rover step which means we're going to repeat right in this case because we brought something down so again i'm going to go leading term into leading term so what is 9x squared divided by x so x squared over x is just x right so this would be 9x so plus 9x and now we're at our multiply steps so 9x times x is 9x squared and then 9x times negative 3 is minus 27x and again we're at our subtract stage so we want to put a minus here and a plus here again again again change the signs you've got to remember to do that i can't tell you how many papers i've looked at how many students have helped where they just forget to change the sign and that's what throws their answer off so 9x squared minus 9x squared is 0. that's going to cancel negative 2x plus 27x would be 25x and then we have one more term that we can bring down so let's bring that down and then one more time we're going to go leading term into leading term okay so what is x into 25x or in other words what is 25x divided by x and you know that's just 25. so plus 25 erase this 25 times x is going to be 25x we kind of get a little room going here so this is 25x and then 25 times negative 3 is minus 75 okay so then again we want to subtract so this becomes minus this becomes plus and we can combine our like terms 25 x minus 25 x is zero five plus 75 is 80. okay nothing else to bring down here so what we end up with here now is our remainder right so this 80 here is the remainder so very simple process overall not much different when you end up with a missing term you go through the same steps the same process you just want to make sure you put 0 as the coefficient for the missing term so let me erase this and let me kind of throw this over here so we'll put is equal to this plus our remainder which is 80 over your divisor which is x minus 3. so we get 3x squared plus 9x plus 25 plus 80 over x minus 3. in this lesson we want to review synthetic division so in our last lesson we reviewed the process of polynomial long division now in some cases we can use a shortcut for polynomial long division this shortcut is known as synthetic division so in order to use synthetic division our divisor must be of the form i'm just going to highlight this we have x minus k so some examples would be something like x minus 3 or x minus 7 or you could even do x plus 5 or x plus 20 so on and so forth now the key thing here is to note that in each case we have a coefficient of one okay a coefficient of one on the variable x and the exponent is a one okay so that has to be the case to use synthetic division if you have something like x squared minus 5 you can't use synthetic division because this is a 2. if you have something like 2x minus 1 you can't use synthetic division because this is a 2 okay you've got to have a coefficient of a 1 and you've got to have an exponent of a 1. okay that has to be the case now we want to know the form x minus k so if you have a plus here what you can do is write this as minus a negative okay so i could write that as minus a negative 5. i could write this as minus a negative 20. okay you just want to match this format right here all right so let's jump in and look at a problem what we're going to do is we're going to do the polynomial long division first and then we'll look at the synthetic division and i'll kind of show you the difference between the two and how much faster synthetic division is so if we have 3x cubed minus 31x squared plus 52x plus 32 divided by x minus 8. let's set this up so 3x cubed minus 31x squared plus 52x plus 32. okay so again remember your d m s b r dad mom sister brother rover so it's divide multiply subtract bring down and then repeat or remainder okay that's your steps so we're first going to start with divide so we're going to go leading term into leading term so what is 3x cubed divided by x that's going to be 3 x squared multiply 3x squared times x is 3x cubed 3x squared times negative 8 is minus 24 x squared now in order to subtract a polynomial away i need to change the sign of each term so sometimes you'll see people put parentheses around and put a minus out in front you might see a negative one out in front whatever you need to do to remind yourself that you need to change the sign of each term now you can just combine like terms 3x cubed and negative 3x cubed that's zero see that just cancels negative 31x squared plus 24x squared is going to be negative 7x squared and now we're at the step we've done subtraction we want to bring down okay so that's that b so plus 52x and now because we brought something down our r step is to repeat the process so we're going to go leading term into the leading term here so what is negative 7x squared divided by x that is going to give me negative 7x so negative 7x now we multiply negative 7x times x is negative 7x squared negative 7x times negative 8 is plus 56x again we want to subtract here so i'm just going to change the sign of each so this becomes a plus and this becomes a minus and now we can combine like terms so this is going to cancel negative 7x squared plus 7x squared that's 0. 52x minus 56x is minus 4x and now i just want to bring down this last term and i want to go again leading term into leading term so let me erase this highlighting here so it's clear what is negative 4x divided by x that's going to give me negative 4. so negative 4 is going to go right there let me make that a little better and i multiply now so negative 4 times x is negative 4x negative 4 times negative 8 is plus 32 and of course to subtract here we're going to change the sign of each term so this would become plus this will become minus and of course this would be zero right negative 4x plus 4x is 0 32 minus 32 is 0. so this tells us we don't have a remainder and our answer is just 3x squared minus 7x minus 4. so let me kind of copy this i'll move this to the side and i'm going to erase all this so i'm going to show you that you're going to get the same thing i'm just going to kind of block this off and leave that there and in order to do synthetic division you don't need any of the variable parts so i can get rid of this i can get rid of this i can get rid of this and so what i'm left with is the 3 the negative 31 the 52 and the 32. don't need the plus signs you can get rid of those and you can leave it this spaced out if you want you can crunch it down it doesn't really matter you don't need the x over here when it's in the format of x minus k you just take k okay so because this is x minus eight i just want the eight part i don't need the negative it's already factored in so i just want the eight now if i had something that was x plus five and i rewrote this as x minus a negative 5 well then i take the negative and the 5 part and put it right there okay so that's what i'm looking for if it's x minus k if it's in that format i just want the k part so because this is x minus 8 i just want the 8 part okay so that's very important you don't want to get the wrong sign because you'll make a mistake and you won't get the right answer okay so now what we're going to do is we're going to make a little bar down here and for reference sake i'm going to label stuff you don't need to do this but i'm just going to label it for reference sake i'm going to call this row one i'm going to call this row 2 and i'm going to call this down here row 3 just so i can refer to things that we're on the same page so the very first thing i'm going to do to start this off i'm going to drop this first guy right here down into row 3 and then i'm going to start a sequence that just repeats itself into the end i'm going to multiply this guy right here by this guy so 8 times 3 is 24 i'm going to write the result here in row 2 okay underneath the negative 31. now i'm going to add so negative 31 plus 24 is going to give me negative 7. now the next thing i'm going to do is repeat the process so i'm going to go back to this times this put the answer here 8 times negative 7 is negative 56 now i'm going to add 52 plus negative 56 is going to give me negative 4. then i'm going to repeat the process 8 times negative 4 is going to be negative 32 i'm going to add 32 plus negative 32 is 0. now if you look here you've got a 3 and a 3 a negative 7 and a negative 7 a negative 4 and a negative 4. these are the coefficients for your answer okay so you start with one degree less than your dividend so the dividend has a degree of three right that's the highest exponential power so this guy will have a degree of two so three will be the exponent on x squared negative seven or minus seven will be the exponent on x to the first power negative four will be my constant and zero would be my remainder okay so i end up with three x squared minus seven x minus four exactly what i got here with way less work okay way less work all right let's take a look at another example so we have negative x plus x to the fifth power plus 10x cubed plus seven minus seven x to the fourth power this is divided by x minus two so the very first thing you want to do is make sure that things are in standard form if you see this guy right here it's not in standard form it's all kind of mixed up so i would rewrite my problem to start as x to the fifth power then minus 7x to the fourth power then i would go plus 10x cubed then i'm missing an x squared so if i'm missing something i want to put 0 as the coefficient for that just as a placeholder then minus x then plus 7. then this is divided by x minus 2. and i'm going to enclose these inside of parentheses so that we know what's going on okay let's scroll down and remember i'm just going to take the coefficients so i'm going to take the coefficient here which is going to be a 1. i'm going to take the coefficient here which is a negative 7 the coefficient here which is a 10 the coefficient here which is a 0. this guy is a coefficient of negative 1 and this guy's a coefficient of 7. that's all i need i don't need anything else i need any variables now for this guy right here again if it's in the format of x minus some number x minus k you can remember that so x minus some number just take the number okay if it was in the format of x plus something then you've got to change it into x minus a negative of that something okay but in this case we don't have to do that so we just put the 2 here i put my bar down here again for reference sake i can say row one row two and row three okay just so i can refer to things drop this guy down into row three two times one is two i'm gonna write that here in row 2. go ahead and combine so negative 7 plus 2 is going to be negative 5. again just do the process 2 times negative 5 is negative 10. 10 plus negative 10 is 0. 2 times 0 is 0. 0 plus 0 is zero two times zero is zero negative one plus zero is negative one and then two times negative one is negative two seven plus negative two is going to be five so what do we have as our answer well again the answer will be one degree less than the degree of the polynomial that's your dividend okay so this has a degree of five so this guy's gonna have a degree of four so this 1 is the coefficient of x to the 4th power then we'd have minus 5x cubed you'd have 0x squared and 0x you don't have to write that you can if you want but you can skip it then you'd have minus 1 then 5 would be your remainder so remember how to write this you'd put plus 5 over the quantity x minus 2. okay so this would be your answer right there again super super fast imagine how much longer it would take you if you were doing the whole polynomial long division thing it's just a much quicker way to do things all right let's take a look at one more example and i gave you one where you have a plus in your divisor just so you can practice with that so we have x to the fifth power plus 6x to the fourth power minus 7x minus 36. this is divided by x plus 6. so again very easy to do this take the coefficients so i have a coefficient of 1 i have a coefficient of 6. i'm missing the third power and i'm missing squared so i need to put a 0 there so 0 is the coefficient for x cubed zero is the coefficient for x squared negative 7 is the coefficient for x to the first power and then i have negative 36. make sure to use these placeholders very important then because this is in the format of x plus 6 again i want x minus k so i want x minus a negative 6. this is the same thing as x plus six i haven't done anything illegal so now i can grab this number right here which is negative six okay so again you've got to make sure you get the right sign if this is plus you're going to end up with minus negative if this is minus just take the number okay that's all you need to do so if this was x minus 6 i would just put a 6 here if it's x plus 6 i put a minus 6 there okay very very easy all right let's put our bar here and then again for reference sake i can say row 1 i can say row 2 and row 3. so to start i'm going to drop this one down here into row 3 and i'm going to multiply negative 6 times 1 is negative 6 so we add 6 plus negative 6 is 0. negative six times zero is zero zero plus zero is zero negative six times zero is zero zero plus zero is zero negative six times zero is zero negative seven plus zero is going to be negative seven negative six times negative 7 is 42 if i add negative 36 and 42 i'm going to get positive 6. so for my answer the degree will be one less than the degree of the polynomial that's my dividend okay so this guy has a degree of five so this guy's gonna have a degree of four so let's start off with one x to the fourth power which is x to the fourth power you'd have zero x cubed zero x squared and zero x then you'd have minus seven then you would have plus 6 over x plus 6. so a very very easy process not only something you want to know how to do but something you will need for later on in our course in this lesson we want to review factoring out the greatest common factor which is abbreviated as the gcf so before we kind of jump in and start looking at some sample problems where we're factoring the greatest common factor out from a polynomial we want to first think about what is the greatest common factor for a group of monomial terms so most of you would have learned this in either pre-algebra or kind of your first algebra class that could be algebra 1 or just a general algebra course and essentially it's a very easy procedure just something you might have forgotten to find the gcf for a group of monomial terms you want to first think about the number parts and then you can think about the variable parts the variable part is very very easy the number part takes a little bit of work because you have to factor some things so if i look at 14x squared y 35xy and 56x cubed y to the ninth and we ask what is the gcf i'd start by factoring each number so 14 is 2 times 7 35 is 5 times 7 and 56 is 8 times 7 and 8 is really 2 cubed so i can write this as 2 cubed times 7 or you could write out 2 times 2 times 2 if you wanted to be technical and all i'm looking for here when i'm looking for the greatest common factor go through the prime factorizations and find what's common to everything okay so don't get confused with the least common multiple when you're looking for the gcf it's got to be common to everything so i have a 7 that's common to everything so that's what's going to be the greatest common factor at least for the number parts right if i had 14 35 and 56 and i said what's the greatest common factor it would be seven once we got our number part worked out we think about the variable parts two questions you ask yourself number one is the variable present in each term here i have x x and x forget about the exponent for a minute here i have y y and y so i know i can put an x here and a y here now what will the exponent be on each well it's going to be the smallest exponent that appears on any of the copies so in other words if i look at x i have x squared i have x to the first power and i have x cubed what's the smallest it's right here right it's x to the first power so this guy can stay as just x or you can write an x to the first power for emphasis if you want then in terms of y i have y to the first power i have y to the first power and i have y to the ninth power again either of these is going to be the winner in terms of what's the smallest so you can put a y to the first power there now why does that work well if you think about it the smallest exponent is the least number of occurrences on that variable so if i was to write this out like i do with the numbers x squared is what it's x times x x to the first power is just x and x cubed is x times x times x so what's common to everything this has an x this has an x and this has an x that's common to everything you can't put a second x in there because that's not common everything this one doesn't have it okay it's got to be common to everything so that's why you go with the smallest exponent that appears on any of the copies again two questions ask yourself is the variable present in each of the terms and what is the smallest exponent on the variable if i was to look at a different problem we'll see an example in a second let's just say we change this problem and this guy did not have a y let's just mark it out so this is gone now if that's the case then y would get removed from the gcf because it's not present in each one this guy has a y this has a y but this one doesn't so it's not common to everything so it doesn't go in okay but that's not the situation we have so let's change this back to what it is so our gcf here for these three terms is going to be 7xy all right let's look at another example so we have 8x to the sixth power y cubed z 20x squared z to the fourth and 36 x cubed y to the fifth z squared so what is the gcf or again the greatest common factor so for the numbers i'm going to factor each one so 8 factors into 2 cubed 2 cubed or 2 times 2 times 2 20 factors into what we know it's 4 times 5 4 is 2 times 2 and 36 we know is 9 times 4. so let's write 4 as 2 times 2 and then 9 is 3 times 3. okay what's common everything i've got 1 2 i've got another 2. that's common to everything then i've got a third two here not common anything else i've got a five not common to anything else and i've got two threes here that are not common to anything else now since the two copies of two in other words four is common to everything that's going to be the gcf at least for the numbers right for 8 20 and 36. now in terms of the variable part is the variable present in each term and what's the smallest exponent on that variable you have an x an x and an x you have a y no y so y is not going in you have a z a z and a z so z is going to go in now what's the exponent on x and what's the exponent on z so this guy is a six this guy is a two and this guy is a three so the smallest exponent is a two so this is going to be squared for z i have z to the first power z to the fourth power and z squared so this guy is going to be the smallest so we just leave it as z or you could put z to the first power for emphasis if you want so the greatest common factor here is 4x squared z to the first power or just z so now that we've reviewed how to find the greatest common factor for a group of monomials we want to think about how to factor out the gcf so when i look at something like 6 times the quantity 4 plus 3 this is a very simple example i just want to show you the distributive property 6 multiplied by 4 would give me 24. so let's just write this as 6 times 4 for right now then plus 6 multiplied by 3 would give me 18. now because of the equality here i know that i could go in reverse right i could pull the 6 back out from each term and set it out in front of some parentheses and i'd have what is left here which is 4 plus 3. this property the distributive property allows us to write a product as a sum and it also allows us to write a sum as a product now this might seem like a really easy example because we only have numbers involved but let's suppose we looked at something like this 3x plus 12. how could i write this sum right you have 3x added to 12 as a product so some number multiplied by some quantity well essentially what i want to do is first find the gcf so what's the gcf for 3x and 12 well that's really really easy because 3 doesn't factor and 12 i could write as 2 times 2 times 3 or 4 times 3. so i could just write this as 3 times x like this plus 3 times 4 like this and i can pull the 3 out because that's common so i can write that out in front of some parentheses just pull this guy out and what's left inside i would just have the x plus the four okay very very easy procedure in most cases you're not going to take the time to kind of write things out like this to speed up the process what i would do i would take 3x plus 12 i would find my gcf i know the gcf is 3. so i would just start by writing a 3 outside of a set of parentheses and since i'm multiplying each term here by 3 to get back to this remember multiplication and division are opposites so i can just take each original term and divide it by 3 to get what i'm missing right so in other words if i take 3x and divide it by 3 i get x that should make sense to you because going backwards 3 times x gets me back to 3x if i want this space right here divide 12 by 3 i get 4. again that should make sense because 3 times 4 gives me 12. so this is the quicker way to do this write your gcf outside of the parentheses divide each original term by the gcf by what you're pulling out and that's going to give you each spot all right let's take a look at another one so we have 6x cubed minus 12x squared plus 144x so what is the greatest common factor here so this one's a little bit more challenging so let's think about 6x cubed let's think about forget about the sign there 12x squared and let's think about 144x so i know x is common to everything and the smallest exponent is a 1. so the gcf let's just put an x over here for the number part i have six which is two times three i have 12 which is what it's four times three or two times two times three and i have 144 which most of you know is 12 squared right it's 12 times 12 or 2 times 2 times 3 times 2 times 2 times 3. so what's common to everything it's going to be 6 right 2 times 3 2 times 3 2 times 3. so the gcf here will be 6x so what i want to do now let me erase everything i want to write my gcf which is 6x outside of some parentheses and then i want to divide each term by 6x to get the terms that are missing in here so i've got this one i've got a minus i've got a term here and plus i've got a term here so 6x cubed divided by 6x would be x squared if you have 12x squared divided by 6x 12 divided by 6 is 2 x squared divided by x is x and i've already accounted for the minus sign i already put it there okay so you don't need to worry about that then 144x divided by 6x what is 144 divided by 6 well that's going to give me 24. so we factored out the greatest common factor here and we end up with 6x times the quantity x squared minus 2x plus 24. all right let's look at another one so suppose we had 20xy squared minus 15x squared y minus 100xy so again i'm looking at 20 15 and 100 forget about the signs what's the gcf well to do this quickly 20 i know is 5 times 2 times 2 but just write 5 times 4. 15 is 5 times 3. so i know at this point i can get rid of this 4 and get rid of this 3. really all i'm looking for is a 5 here and i know 100 divisible by 5. so the gcf is going to be 5 right because the 2 wouldn't have been common to everything so you get rid of that and the 3 wouldn't have been common to everything so you can get rid of that so it's irrelevant what the factors of this are other than the fact that it's divisible by 5 or it has 5 as a factor so the gcf is a 5. so the gcf is a 5. now when we think about the variables we have x x and x and we have first power second power and first power so that can stay then we have y y and y we have second power first power and first power so that can stay so my gcf is 5xy so let's write that out in front of a set of parentheses and again to get what's inside you're going to have three terms minus minus so what goes here 20xy squared divided by 5xy 20 divided by 5 is 4 x over x is 1 y squared over y is y 15x squared y over 5xy 15 divided by 5 is 3. x squared over x is x y over y is 1. 100xy divided by five x y one hundred divided by five is twenty x over x is one y over y is one so we end up with five x y multiplied by the quantity four y minus three x minus twenty all right let's go ahead and wrap up our lesson by looking at how to factor out a common binomial factor this is something we're going to use in the next review section where we're talking about factoring by grouping so let's suppose we look at the problem the quantity x plus 1 times the quantity x minus 5 plus the quantity x minus 5 times the quantity x minus 6. so i see that i have a common binomial meaning two-term polynomial so a common binomial factor so i can just pull that out okay just like i've been doing in the other problems i can still just pull this guy out and what happens when i pull it out well inside of parentheses i would have what if i pull out an x minus 5 from here i would just have the x plus 1. so the x plus 1 and you can put that inside of parentheses if you want for now and plus over here if i pull out an x minus 5 i would have the x minus 6 there now these double parentheses here cause a lot of concern for a lot of students so if you want to make this easier on yourself just temporarily put some brackets there we're going to combine like terms inside of here so we can get rid of the brackets in a minute so x plus x is 2x 1 minus 6 is going to be negative 5. so this would be 2x 2x minus 5. and now i could write this with some parentheses here and so you've successfully factored this guy you've pulled out or factored out a common binomial factor now it's easy to prove to yourself that this works if i go through and foil this x times 2x is 2x squared the outer would be minus 5x the inner would be minus 10x so you'd have minus 5x and then minus 10x so you combine like terms there you'd have negative 15x or minus 15x and then the last term is negative five times negative five is plus 25. now if i go through here x times x is x squared the outer would be minus 5x the inner would be plus 1x and the last would be minus 5. so this guy right here in the middle would combine to be minus 4x so i'd have x squared minus 4x minus 5 and plus over here i would have x squared my outer would be minus 6x my inner would be minus 5x so that would be minus 11x if i combine the 2 and then my last would be plus 30. so what does this equal if i'm right it should equal this x squared plus x squared is 2x squared negative 4x minus 11x is minus 15x then negative 5 plus 30 would be plus 25. so i want you to notice that we got the same thing here as we got here so this shows you that we can factor out a common binomial factor we end up with the same result either way the 2x squared minus 15x plus 25. in this lesson we want to review factoring by grouping so most of you know at this point how to factor using the grouping method this is basically something we use with a four term polynomial and what we're going to do is we're going to separate the four term polynomial into two groups of two terms each from each group we're going to pull out the gcf or in some cases the negative gcf and then we're going to look for a common binomial factor okay if you have a common binomial factor you pull that guy out and you will have factored your four term polynomial using grouping so let's start out with an easy example this is one where we don't need to really go through different combinations and what i mean by that is i can take the first two as a group and the last two as a group i don't need to reorder anything so i have 15 x cubed minus five x squared plus six x minus two and if i just take the first two here as a group so 15x cubed minus 5x squared i'm going to put that inside of parentheses so that's one group and plus i have 6x minus 2 so that will be another group put that inside of parentheses and what can i pull out from 15x cubed minus 5x squared well the greatest common factor there is 5x squared so let's pull that out and inside the parentheses i would have what 15x cubed divided by 5x squared is going to give me 3x and then 5x squared divided by 5x squared is 1. so i have the minus there so i'll put minus 1. so then plus what can i pull out from 6x minus 2. so again the gcf of 6x and 2 would be 2. so if i pull that out 6x divided by 2 is 3x and then 2 divided by 2 is 1. so minus 1. so you can see that we have this common binomial factor of 3x minus 1. we could just pull that out okay we can factor this out so we would have 3x minus 1 that's been pulled out in front of a set of parentheses inside the parentheses we would have what's left which is the 5x squared so 5x squared plus the 2 okay plus 2. so we've factored our 4 term polynomial 15x cubed minus 5x squared plus 6x minus 2 as the quantity 3x minus 1 times the quantity 5x squared plus 2. okay so pretty easy all right now let's take a look at a harder example so the first thing you would notice with this we have 18xy minus 6y squared plus 12x minus 9y cubed everything has a common factor to start okay if you look at the example we looked at before let me go back to it if you look at this example again we don't have anything that's common to the four terms other than one or negative one if we look here we do have a common factor of 3 right so 18 is 3 times 6. i know you could break it down further but just think of it as 3 times 6 6 is 3 times 2 12 is 3 times 4 again you could break it down further 9 is 3 times 3. so before i even start i want to pull that out if i don't pull it out then later on i'm going to have additional factoring okay so 3 times inside the parentheses that i have 6xy minus i'd have 2y squared plus i'd have 4x and then minus 3y cubed again i just pulled out a 3 from each term here okay that's all i did now inside the parentheses is where i'm going to be working okay now if i just take these first two terms and these last two terms i'm not going to end up with a common binomial factor so here's where i got to kind of do some trial and error try different combinations now if i look at the gcf of 6xy and 2y squared that's going to be 2y right if i pulled that out what would i have inside i would have a 3x minus y here the gcf is 1 or you could do negative 1 but that's not going to get me anything here right it's not going to match up with the 3x minus y i would have 4x minus 3y cubed or i would have negative 4x plus 3y cubed either way it doesn't work so we're going to need to try a different arrangement here so one thing that i notice is that this guy right here has an x and this guy right here has an x so that's a little hint hint so i'm going to arrange those terms next to each other so don't forget my 3. so i'm going to put my 6xy and my plus 4x that's going to be a group and then you will have negative 2y squared and minus 3y cubed so plus negative 2y squared and minus 3y cubed what i want to do is pull out the gcf from this group here so 3 times you have a 6xy and a 4x i know i can pull out an x and i know i could pull out a 2. so this would be 2x times the quantity 6xy divided by 2x would be 3y and then 4x divided by 2x would be 2. so plus 2 then plus here i know that i need to pull this negative out how do i know because if i look at what my binomial here is it's 3y plus 2 everything's positive so i would want everything in here to be positive so go ahead and pull out a negative so you can write a minus here or you can put plus negative doesn't matter and i'm going to be pulling out what well i'm just going to be pulling out a y squared and then inside i would have 2 right if i pulled out a negative y squared i would just have 2 and then plus if i pulled out a y squared here i would just have 3y now you might say you don't have a common binomial factor you have 3y plus 2 and you have 2 plus 3y these are the same terms they're just in a different order when we add we can add in any order so it's the same thing if you wanted to you could just reorder this stuff you could say this is 3y plus 2. it's the same or you could go back here rearrange these terms put the negative 3y cubed in front and then put minus 2y squared as the last one it doesn't matter it's the same answer either way so put equals so now we're going to factor this guy so this guy is your common binomial factor and if we pulled that out we would have 3 times the quantity this guy comes out 3y plus 2 times inside of parentheses you'd have what's left so you have the minus the y squared and that's our factorization so you have 3 times the quantity 3y plus 2 times the quantity 2x minus y squared now if you didn't pull out the 3 to start okay like we did then what happens is when you get to this stage you're going to have a common factor of 3 okay that you still need to pull out so always look before you start say is there anything i can pull out before i start factoring this by grouping it's going to save you additional factoring in the end all right let's take a look at another one so suppose we had 100xy minus 245 minus 140x plus 175y so before we start is there anything we can pull out this is divisible by five so is this so is this so is this right everything ends in a zero or a five so i know everything's divisible by five this guy right here would be five times twenty this guy right here would be five times forty nine this guy right here would be five times twenty eight and this guy right here would be five times thirty five so there's nothing else i can pull out you notice that this guy this guy and this guy would each have a common factor of 7 but this guy doesn't right 20 is just 4 times 5 so that's not going to work and so there's nothing else we can really pull out other than a 5. so we'd write a 5 out in front then inside the parentheses you'd have 20xy minus you would have 49 minus you would have 28x plus you would have 35 y all right let's erase all this get rid of that and we'll move this up here okay so now let's think a little bit about our grouping if i look at the way it's grouped now in other words these two as a group and these two as a group the first two has a group and the last two is a group will that work well from the first two i can't really pull anything out 49 is 7 times 7 20 is 5 times 2 times 2. nothing i'm really going to be able to do there this has x y this doesn't have any variables at all this guy i know i would have an x and i know i would have a y and i could pull out a 7 but it's just not going to work right so you can eyeball that and see we've got to rearrange things again i'm going to think about the fact that this term right here has an x and this term right here has an x so let's go ahead and rewrite this as 5 times inside of one group i'm going to use brackets here to make this a little easier c so one group could be 20 x y and then minus 28 x and then plus another group could be 35 y minus 49. all right let me close those brackets the reason i went ahead and switch these around if you notice this number part right here this 28 is larger than this number part here 20. so i just followed that same order this number part here 49 is larger than this number part 35 okay so that's where i got that all right so now if we pull out the gcf from this first group what will it be 20 is 4 times 5 or 2 times 2 times 5 if you want to be technical 28 is what it's 4 times 7 or 2 times 2 times 7 so i could pull out a 4 and i could also pull out an x so inside the parentheses there you would have 5 y minus a 7. from here i know i could pull out just a 7. right if you think about 35 and 49 greatest common factor is 7. so i'm going to pull that out and inside i would have 5 y minus a 7. so you can see you do have a common binomial factor so we have this guy right here this 5y minus 7. and let's go ahead and pull that guy out so we'll have our 5 out in front then we'll have our 5y minus 7. then inside the parentheses we'll have 4x plus 7. okay so we factored our polynomial as 5 times the quantity 5y minus 7 times the quantity 4x plus 7. all right let's go ahead and take a look at one more problem pretty easy concept overall just something we need a lot of practice with so if we have 21xy minus 112xz minus 84x squared plus 28yz think about what's common to everything before you even start can i pull anything out variable y is no right you have an x and x and x no x you have a y no y you don't have a z here so nothing's going to be common there so let's start with 21 21 is 7 times 3. so is anything divisible by 3 is anything divisible by 7 112 1 plus 1 is 2 2 plus 2 is 4. that's not divisible by 3. so now let's think about divisibility by 7. 112 divided by 7 is 16. so i could write this as 7 times 16. 84 divided by 7 is going to be 12. so i could write this as 7 times 12 and 28 divided by 7 is 4. so this is 7 times 4. so before i even start i can pull out a 7 and inside i'd have 3xy minus i would have x z minus i would have 12 x squared and then plus i would have 4 y z now is this the correct order meaning can i get a common binomial factor so just eyeballing this you can see that using the first two as a group and the last two as a group will not work and the reason i can tell that so quickly is this guy is x squared over here i'd be pulling out an x and i wouldn't have an x anywhere inside of the parentheses right here i'd be stuck with that x squared right there's no x over here to pull out anything so knowing that i'm going to have to reorder i would look at the fact that you have a z here in this term and you have a z here in this term right if i pulled out the z i would be left with variable y's x and then y if these two terms were next to each other if i pulled out an x i would be left with x and y so the variables work themselves out so let's try that guy okay so you just need to get some practice kind of going through different combinations and you'll become very very good at eyeballing these things and getting the correct order so my 7 goes out in front i'm going to put negative 16 x z plus 4 y z and then i'm going to put negative 12 x squared and then plus 3 x y and again the reason i put them in this order is because the negative 16 is larger than the 4 in terms of absolute value i know negative 16 is a smaller number but in terms of absolute value and then the negative 12 in terms of absolute value is bigger than 3 so that's why i chose that order all right so let's go down a little bit and get some room from the first group here i can pull out a negative for z okay so negative 4 z so this first term would be positive it would be 4 and then just x the second term would be negative right because if i divided this by this negative i would end up with a negative so i'm pulling out a 4 i'm pulling out a z i'm just left with a y all right over here i can pull out a negative 3 x okay pull out a negative 3x and i'm going to be left with inside if i take this guy divided by this guy i'm going to end up with a 4x and of course this guy divided by this guy would give me a negative y okay so close the parentheses there and so we do have a common binomial factor of 4x minus y and if we factor that guy out just like we've been doing we'll have 7 times we'd have 4x minus y and then times what's left you have negative 4z and then minus 3x now to make this look a little bit cleaner it's not necessary this is the correct answer a lot of people will tell you to make your leading coefficient for any polynomial it could be a binomial trinomial whatever it is positive so we could factor out a negative 1 here and to do that i'm just pulling the negative out from here and here and just sending it out here okay so i'm just sending it out and let me kind of make those arrows a little better so let me be more accurate so this guy would come out here and this guy would come out here so we can say this is equal to i put a negative 7 because negative 1 times 7 is just negative 7 times the quantity 4x minus y i pulled the negative out from each so it would be 4z plus 3x inside and so this would be a kind of better answer you would say but if you wrote this on a test it wouldn't be wrong there would be nothing wrong with it but a lot of online calculators a lot of textbooks would make the extra step and write it like this so negative 7 times the quantity 4x minus y times the quantity 4z plus 3x in this lesson we want to review factoring trinomials when the leading coefficient is 1. so anybody that's ever taken an algebra course will have run across the topic where they deal with factoring trinomials into the product of two binomials okay it's very very common so the easier scenario occurs when the leading coefficient is one so in other words if we look at a trinomial which again is a three term polynomial like ax squared plus bx plus c and this is just something generic something you'd see in your textbook if we let a be equal to one okay if we let a be equal to one i'm saying that i have a coefficient of 1 on that leading term so i'd have 1x squared plus bx plus c which really 1 times anything is itself so i could just write it as x squared plus bx plus c this is the easiest case scenario okay and the reason for that is if i'm going to write this as the product of two binomials remember when we multiply binomials together using foil it's first times first right first terms that's the f in foil so how do i get x squared it's just x times x i don't need to figure that out that's already figured out for me i just need to figure out this spot and this spot okay so that eliminates a lot of the work that you're going to come across when you deal with something where the leading coefficient is not a one right if you have something like 5x squared is your leading term or if you had something like negative 27x squared as your leading term you know something like that so let's start out by looking at the product of two binomials using the foil process and then we're going to kind of reverse it and see how we can go back to this so we're going to go from this to a trinomial and then from the trinomial back to this form so doing foil we start with the first term so x times x is x squared so again this is a given when you have 1x squared or just x squared when you go backwards you know you're going to have x and x right so the first terms in each polynomial are a given then the outer you have x times 10 which is just plus 10x the inner you have 1 times x which is plus 1x and the last is 1 times 10 which is just plus 10. so we combine like terms here we have 10x plus 1x so we're going to end up with x squared plus 11x plus 10. okay so if i was given this and i was told to go back to this format what would i do well i know that i would start with x and x that's pretty easy right x times x gives me x squared right how would i get this spot here and this spot here well let's think about it we know that we multiplied the outer terms x times 10 to get 10x and the inner terms 1 times x to get 1x we know those combine to give us 11x so in other words this coefficient here of 10 plus this coefficient here of 1 added together to give me this coefficient here that's on my middle term so 10 plus 1 gave me 11. so 10 plus 1 gave me 11. then notice that the same two integers 1 and 10 multiplied together to give me this last term here so 1 times 10 is equal to 10. so that tells me that what i need to find for these two spots here are two integers that sum to 11 or this middle coefficient here and that have a product of the last term there so two integers again let me repeat two integers whose sum is the coefficient of the middle term okay the coefficient for the variable raised to the first power and whose product is going to be the constant term so give me two integers that fit that criteria if you didn't know this you would think about the last term here which is 10 and you would think about the possibilities now of course you could do negative times negative here but because all the signs are positive i would just think about positive numbers so 10 can be found by 1 times 10 or you could also do 2 times 5 okay 2 times 5 won't work because 2 plus 5 is 7 7 does not equal 11 right we need something that sums to 11. but 1 and 10 does 1 plus 10 gives me 11. 1 times 10 gives me 10. so i know i would need plus 1 here and plus 10 here okay so pretty easy the first time you hear it it might not make sense it might not all register but as we work through a few examples here you'll be like oh this is a very very easy process all right let's take a look at another one so we have x squared plus 3x minus 28 again very easy to factor this i'm just looking for this right here and this right here this guy right here is a given this is going to be x and x i don't need to worry about that this guy right here this last term is a negative 28. so because it's a negative it can only be produced from a positive times a negative so i know i have different signs so to remind myself of that i might as well just say a plus and a minus okay because i'm going to have alternating sides and the order here doesn't matter right i could put the plus here the minus here doesn't matter okay so let's just think about positive 28 we'll work out the signs in a minute and i want to think about how i can multiply two numbers together to get 28. so what can i do i could do 1 and 28 i could do 2 and 14 i could do 4 and 7 and that's basically it so now i want to think about two numbers here that would give me this middle term of plus 3. so that tells me that the bigger number is going to be positive so if i did 28 minus 1 that would be 27 that's not 3 so that's not going to work so you can take this out then if i did 14 minus 2 that would give me 12 so that doesn't work if i did 7 minus 4 that's 3 so that's going to be my big winner so i know that i would need a negative 4 and a positive 7. again the way i know that is if this is plus 3 that's only going to come from a positive 7 minus a 4 or a positive 7 plus a negative 4. so that's why i have a minus 4 and a plus 7. now to check this we can foil x times x is x squared the outer would be minus 4x the inner would be plus 7x and the last would be minus 28. combine like terms here you'd have plus 3x so you'd have x squared plus 3x minus 28 which is again is what we started with right here okay so we know we have the correct factorization all right let's take a look at another one so we have 5x squared plus 10x minus 15. you might see this problem and say whoa we're supposed to be dealing with easy ones what's going on with this 5 in front of the x squared well relax for a minute you might want to look at every term here and see can i pull something out always always always when you're factoring as a rule before you do anything ask yourself the question is there something that i can pull out from each term other than a one or a negative one because that's going to save you from additional factoring later on okay so i notice that everything here is divisible by five if i pull that out inside the parentheses i would have x squared plus 2x minus 3. and now i just need to factor this part okay so i'd have 5 and then inside the parentheses i know this would be an x and this would be an x and then give me two integers whose sum is 2 and whose product is negative 3. 3 is a prime number i can only get 3 from 1 times 3. so it's a question of can i arrange this in such a way that i get a positive 2 as a sum and a negative 3 as a product well yes i can i can do plus 3 and minus 1 right because if i take positive 3 and i subtract away 1 i get 2. if i take positive 3 and i multiply by negative 1 i get negative 3 right and if you want to check this first you would multiply your 2 binomials together so x times x is x squared the outer would be negative x the inner would be 3x sum those together you get 2x and the last would be minus 3. now we need to multiply this by 5 to get back to our original 5 times x squared is 5x squared 5 times 2x is plus 10x and then 5 times negative 3 is minus 15. so this matches this exactly 5x squared plus 10x minus 15 5x squared plus 10x minus 15. all right for the last problem we're going to look at one that kind of trips students up a little bit so we have 6x squared minus 90xy plus 324y squared so the reason this is a little bit more complicated is because it has two variables involved but really if you just kind of follow the simple technique that i'm going to show you here it's not going to be any harder okay so start out by again asking the question can i pull anything out well yeah i can pull out a 6. this is divisible by 6 90 is divisible by 6 324 is divisible by 6. this is 6 times 15 this is 6 times 54 okay i can't pull any variables out because i have an x here and an x here but no x here i don't have a y here but i have a y here and here so nothing that's common to everything so if i pull out the 6 i would have x squared minus 15 x y plus 54 y squared ok so erase this and i'll kind of scooch this up just a little bit bring that up there all right so i'm looking to factor what's inside so i'm going to set up my two binomials so i'm just going to put x here and x here now in terms of the y think about this last term for a minute we end up with y squared here how do you get y squared as a product because remember this has to come from this times this it has to be y times y so go ahead and just write y and y and then forget about it okay you don't need to think about it again if they give you this two variable scenario that's all you need to do take the thing that squared here and again figure out y times y will give me y squared if it was z squared it would be z times z if it was q squared it'd be q times q you know so on and so forth i only now need to think about the negative 15 and the 54 okay so give me two integers that sum to negative 15 and that give me a product of positive 54. i know that has to be a negative and another negative okay two negatives sum to be a negative and two negatives multiplied to give me a positive okay so that's how i know that so thinking about factors of 54 i've got 1 times 54 i've got 2 times 27 none of these would work i've got 3 times 18 and that still wouldn't work but then i come across 6 times 9. so 6 times 9 is 54. now again i want negative 6 and negative 9. so negative 6 times negative 9 is positive 54 and negative 6 plus negative 9 is going to give me negative 15. so that's what i want i want a negative 6 and a negative 9 so i've got my minus 6 here and my minus 9 there and that's it okay you can go through and check this with foil very easy to do and i would just put my 6 out in front x times x is x squared the outer would be minus 9xy the inner would be minus 6xy so if we combine like terms here we have x and x both raised to the first power y and y both raised to the first power so you just want to do negative 9 minus 6 which is negative 15. so minus 15 times x y and then the last we have negative 6y times negative 9y that's plus 54. again y times y is y squared okay so very easy to do and if we want to finish this process 6 times x squared is 6x squared 6 times negative 15xy is minus 90xy and then finally 6 times 54 is 324 and then times y squared so again we end up with exactly what we started with if we factor this guy we end up with 6 times the quantity x minus 6y times the quantity x minus 9 y in this lesson we want to review factoring trinomials using the ac method so in the last review session we talked about the simplest case scenario for factoring a trinomial so that occurs when you have something like ax squared plus bx plus c and a is equal to 1. so if i had something like x squared plus let's say 5x plus 6 we can very easily factor this we know the first position of each binomial is going to be x because x times x the f in foil would give me x squared back to find this and this i only need to figure out two integers whose sum is the coefficient of the middle term so whose sum is five and whose product is the final term or the constant term so give me two integers whose sum is five and whose product is 6. well i would say 2 and 3 right 2 plus 3 is 5 2 times 3 is 6. so this would factor into x plus 2 and times x plus 3. so very very easy very simple stuff in the case where we get something where we can't factor it we consider it to be prime so you might see something like x squared plus 4x plus 7. if you're asked to factor this well when you start the process this is x and this is x give me two integers whose sum is four and whose product is seven well it doesn't exist right so you think about seven as seven times one it's a prime number there's nothing else you can do so 7 plus 1 is 8 that doesn't get me to 4. so when you get something like that you can just stop and say that it's a prime polynomial it cannot be factored using rational numbers now what we're going to talk about today is the more complex scenario the scenario that most students hate and the reason students hate it is because it ends up being a lot more work right it's just very very tedious so if we have a trinomial such as ax squared plus bx plus c now a is not going to be equal to one so the first part of each binomial is no longer a given because this guy right here is not a one okay so in order to use the ac method sometimes people refer to it as the grouping method because you end up doing factoring by grouping we need to understand what a b and c represent in a trinomial so a is always used as the coefficient for the squared variable so that's why you see ax squared b is used as the coefficient for the variable raised to the first power and then c is your constant term all right so let's jump in and look at a problem and i'm just going to walk you through the process it's not difficult it's just something you need to write down and kind of go step by step through so the very first thing you're going to do if we have something like 7x squared plus 51x plus 14 you want to identify again a b and c so a is the coefficient of the squared variable b is the coefficient for the variable raised to the first power and c is going to be your constant term now the very first thing you want to do here is you want to find two integers whose product is a times c so the product of these two integers is a times c and then the sum is going to be b okay so what is a times c it's 7 times 14 which is going to be 98 so let me just write 98 here and the sum is 51. so what you need to do is you need to go through the possibilities you need to go through the factors of 98 so the pairs of numbers that combine through multiplication to give you 98 and see are there two such integers that would give me a sum of 51 and multiply together to give me 98 okay so you can start with 1 and 98 that's obviously not going to work but just for completeness let's do 1 times 98 so 1 and 98 would not work as a sum right everything's positive here so we're looking at two positive numbers 1 plus 98 is 99 you can throw that out the next thing you will come across is 2 and 49 so 2 times 49 would give me 98 and 2 plus 49 would in fact give me 51. so we can stop here we found our true integers so the integers we're looking for are 2 and 49 okay so let's erase all of this all right so some of you might incorrectly assume that these two integers are used like we did in the last method i see students all the time get the two integers and go okay it's x plus 2 times x plus 49 okay that's wrong the two integers are used to expand the middle term okay so we're going to expand the middle term such that we have a four term polynomial and we can use factoring by grouping so in other words what i'm going to do is i'm going to write this as 7 x squared plus 49x then plus i'm going to do 2x then plus 14. now in case you lost me there think about the fact that 49x plus 2x gives you 51x i didn't do anything illegal okay all i did was i took the middle term and i rewrote it using these two integers that we found and so now we have a four term polynomial right we have 7x squared plus 49x plus 2x plus 14. it's the same value it's the same thing we're just doing this so that we can use factoring by grouping all right so now we want to factor by grouping so i'm going to make this first two into a group and the last two into a group so from the first two i can clearly pull out a 7x so inside the parentheses i would have x plus 7. from the second group i can pull out a 2 and inside the parentheses i would have x plus 7. so if i factor out my common binomial factor here of x plus 7 i'm going to end up with an answer of x plus 7 times we'll have 7 x plus 2. so that is the correct factorization of this guy this 7x squared plus 51x plus 14. and of course we're just going to check it we don't need to check every one but i'm going to check this one just so you can see so first terms x times 7x will give me 7x squared the outer x times 2 would be 2x the inner 7 times 7x would be 49x so 2x plus 49x would give you that middle term of 51x and then 7 times 2 would give you 14. okay so we can see this works itself out again our answer is the quantity x plus 7 times the quantity 7x plus 2. all right so hopefully you're not too lost we're going to try another one we're going to do the same process again just identify what's a b and c so this is a this is b this is c right a is the coefficient for the squared variable in this case it's 6. b is the coefficient for the variable raised to the first power here's 41. c is your constant term so i want to find two integers whose product is a times c so product is again a times c in this case that's 6 times 63 which is 378 and we want a sum of 41 right a sum of b or 41. so obviously 1 times 378 would not work if we divided 378 by 2 we get 189 so you'd have 189 and 2 that wouldn't work 378 divided by 3 is 126 that wouldn't work 378 divided by 6 is 63 wouldn't work 378 divided by 7 is 54 that wouldn't work 378 divided by 9 is 42 that also wouldn't work so then we go to 378 divided by 14 and we get an answer of 27. now 27 plus 14 is 41. so we found our two integers right we want 27 and 14. 27 times 14 again would give me 378. 27 plus 14 would give me 41. so let's drag these numbers over here so again what am i going to do with these numbers i'm going to use them to expand this middle term that's all i'm doing very simple process so we have 6x squared plus i'm going to go ahead and write 27x first and then plus 14x and then plus 63. okay so nothing illegal here i just rewrote the middle term i expanded it so we can now use factoring by grouping so let's erase this we don't need that information anymore from the first group i can pull out a 3x so i pull out a 3x inside the parentheses i would have 2x plus 9. and the second group i could pull out a 7. inside the parentheses i would have 2x plus 9 and i have a common binomial factor of 2x plus 9. okay everybody can see that so if we factor this we're going to have a 2x plus 9. this quantity multiplied by 3x plus 7. so this is our factorization we get 2x plus 9 that quantity multiplied by the quantity 3x plus 7. let's take a look at one that has two variables involved so we have 30x squared minus 39xy minus 27y squared and again we're going to approach the problem the same way even though it has two variables it's very very easy the first thing you would notice here is that everything is divisible by 3 right 30 is divisible by 3 39 is divisible by 3 and 27 is divisible by 3. so i can pull the 3 out right before we even start so this would be 10x squared minus 13xy minus 9y squared so just proceed as you normally would inside the parentheses here again this is still going to be considered a this is my b this is my c so give me two integers whose product is 10 times negative 9 or negative 90. so that's the product and then for sum i want negative 13. so that's my sum so running through the factors of just 90 you've got 1 times 90 which is 90. that wouldn't work right you've got 2 times 45 that wouldn't work you've got 3 times 30 that wouldn't work not divisible by 4 but 90 divided by 5 is 18. now you can make 5 and 18 work if you play with the signs if you have a negative 18 and a positive 5 right if you have a negative 18 and a positive 5. negative 18 plus 5 is negative 13 negative 18 times 5 is negative 90. so those are the two integers we're looking for so let's use that to rewrite the middle term so we're going to have 3 times the quantity 10x squared i'm going to put minus 18xy plus 5xy and then minus 9y squared so you can see this y variable didn't really change anything for us you use the same technique the same strategy okay so now i'm just going to factor using grouping so i have 3 and then inside of parentheses the first group i can pull out a 2x that would leave me with 5 x minus 9 y from the second group i can pull out a y that would leave me with 5 x minus 9 y so you have a common binomial factor of 5x minus 9y and if we pull that guy out you would have 3 times you'd have your 5x minus 9y and then you would have your 2x plus your y okay so this guy factors into 3 times the quantity 5x minus 9y times the quantity 2x plus y so again it didn't really affect anything for us to have a second variable involved it's the same process don't get scared if you see two variables very very simple to work with it doesn't change anything the only difference is you have a y here and a y here right because y times y gives you that y squared part okay so let's look at one more and for this one i gave you one that's a bit tricky so we have 8x squared minus 5x plus 35. nothing you can pull out to start so let's just look at what we need we need two integers whose product is 8 times 35 which is 280 and whose sum is negative 5. so this is my product and this is my sum right negative 5. so if we think about 280 let's just go ahead and factor it and i purposely picked big numbers so you get some practice with some things that are you know kind of difficult kind of tedious because a lot of times you're doing these practice problems and they're all small numbers and it takes you seconds to find the answer right i want something that's going to take you a little bit of time so 280 is what it's 28 times 10 10 is 5 times 2 28 is 4 times 7 four is two times two okay so we can have one times two hundred eighty we can have two times one hundred forty we can have four times 70 we can have 5 times 56 we can have 7 times 40 we can have 8 times 35 you can obviously have 10 times 28 you can have 14 times 20 and that's pretty much going to be it now what you're looking for here if the product is 280 positive 280 you would need two integers that both were negative right because you have a negative sum so two negatives would make a positive in terms of being multiplied together and two negatives if they were added together would give you a negative sum so that's what you would need but if you look through this list here nothing would work right so sometimes you get an example where it's prime and it takes more time because you have to go through all the factors write all the pairs and see does anything work well this wouldn't work this wouldn't work this wouldn't work this wouldn't work this wouldn't work wouldn't work wouldn't work wouldn't work so we can go ahead and just say this is a prime polynomial in this lesson we want to review factoring trinomials using reverse foil so for the last two lessons we've been talking about factoring trinomials into the product of two binomials there's a lot of different methods out there to do this this is one of those things that's pretty tedious but over the years it's gotten a lot better with the use of computers it used to be that you would punch this in on your ti 83 or ti-89 or whatever graphing calculator you had and that would factor it for you if you didn't feel like factoring it now you can type this into a variety of online calculators and it calculates it for you in a second okay but you still need to understand the process that you can use to do this because you might be asked to do this on a test for whatever reason and you might not be allowed to use a calculator to do it so if we look at kind of the harder scenario where we have ax squared plus bx plus c and a is not equal to one so in other words this leading coefficient the coefficient for x squared is not one if we look at an example where it's kind of easier where a is one let's say we had x squared minus two x minus three we can factor this so quickly because we know this guy because it's a one here one x squared is only going to be x times x here this is the first or the f and four okay so all i really need to work out is this here and this here that's really easy all i have to do is get two integers that give me a product of negative three and a sum of negative two so for negative three it's either positive one and negative three or negative one and positive three well in this case because i want a negative sum i want a negative 3 and a positive 1. right my outer would be negative 3x my inner would be plus x that would give me a sum of negative 2x and of course 1 times negative 3 is negative 3. so again that's the easier scenario and when we get to the more challenging scenario where a is not equal to 1 something like this guy 7x squared minus 64x plus 9 we learned in the last lesson that we could factor this using something known as the ac method or a lot of people might call it using a grouping method right normally we use factoring by grouping with a four term polynomial so with this method essentially we gather some information and we rewrite the polynomial in such a way that it's a four term polynomial and we can use grouping so really quickly let's just review that real fast and then we'll jump into the reverse foil so to do this ac method we want to find two integers that have a product of a times c and a sum of b so in case you don't know we label this guy right here the coefficient of the squared variable is a we label this guy right here the coefficient of the first degree variable as a b and then we label this guy right here the constant as a c so what is a times c it's 7 times 9 that's 63 so that's what we want for a product and then b is going to be negative 64. so this one's really easy to figure out if you want a sum of negative 64 and a product of positive 63 your two integers would be negative 63 and negative 1. so what i'm going to do i'm going to write those two integers real quick so you have negative 63 and negative 1. you use these two integers to rewrite this middle term so you would have 7x squared minus 63x minus x plus 9. haven't done anything illegal i just rewrote the middle term negative 64x as negative 63x minus x all i did was expand it okay so now that it's a four term polynomial we can use grouping right so from the first group the first two terms i can pull out a 7x and i would have x minus 9. from this group i'm going to pull out a negative 1 and i would have x minus 9. so you can see you have a common binomial factor here of x minus 9 and you go ahead and pull that out you would have x minus 9 and then multiplied by you'd have 7x and then minus 1. so kind of a fast method in a lot of cases depending on how many different factors you need to go through so some of these things when you're factoring you get it right away other things you're there for 10 minutes trying to go through all these different possibilities it's just very very tedious so it's just a luck of the draw right so this method is preferred by most students because with the trial and error method it tends to get pretty tedious depending on what you get so let's erase this let's go through and talk about factoring with reverse foil or again trial and error as some people will say so the first thing i'm going to do is just write my parentheses here and i'm going to think about what's going to go here and here now in this case we're lucky because we have a prime number as the leading coefficient so this is the easiest case when you're doing reverse foil because i know that 7 can only come from 7 times 1 i know that x squared can only come from x times x so i can go ahead and just put a 7x and an x okay if i had a 3x squared i could put a 3x and an x if i had a 5x squared i could put a 5x and an x right so on and so forth but if i had something that was not prime like a 10 well then i could have 10 and 1 or i could have 2 and 5 right so it becomes more problematic because you've got more things to check now i still need to work out this and this so how would i do that well i'm going to look at this last term here and i'm going to say okay what are the factors of 9 because it's got to be that this guy right here multiplied by this guy right here gives me a product of 9. it's got to be plus 9. now this guy is negative which tells me that i need a negative here and a negative there so factors of 9 forget about the signs you've got one and nine and three and three it's the only possibilities so now you would go through and work out the outer and inner right those steps in foil and see which one or which combination more specifically gives you the correct middle term of negative 64x so in other words this would be your outer and this would be your inner so 7x times something plus x times something should be equal to negative 64 x so you can try 1 and 9 and remember we're going to want negatives here so i'm going to put a negative 1 and a negative 9 let's check that so 7x times negative 1 is negative 7 x and then x times negative 9 is going to be negative 9 x if i sum those two together i would get a negative 16x and that is not negative 64x so that doesn't work now i need to flip these around i need to try negative 9 and also negative 1. 7x times negative 9 is going to be negative 63x x times negative 1 is going to be negative x now if we sum these two together we do get negative 64x so this is the correct combination but remember 7x is multiplying negative 9. so 7x is multiplying negative 9. so it's got to be x minus 9 like that and x is multiplying negative 1 so x is going to be multiplying this so this is going to be a negative 1 there okay you can always check your result with foil so we can go back and say okay we know 7x times x is 7x squared the outer here would be negative 63x the inner here would be minus x so that would combine to give me a negative 64x so we're good to go there and then negative 1 times negative 9 is positive 9. all right let's take a look at another one so we have 3x squared plus 29x plus 56. so another case where we have a prime number as the leading coefficient so 3x squared can only come from 3x times x so i'm going to go ahead and start this off by putting a 3x and an x and now i just need to work out this guy and this guy and remember this times this has to give me this so i just think about the factors of 56 and this is positive and this is positive so i know these are each going to be plus okay so for 56 i've got 1 and 56 i've got 2 and 28 i've got 4 and 14 and i've got 7 and 8. okay so this is where it gets pretty tedious because you've got to go through all these combinations and say okay if i put a 56 here and a 1 here does that work well i've got to check the outer and the inner so 3 times 56 is 168 so no way that's too big if you put 56 over here it would still be too big so 1 and 56 will not work you can get rid of that next you've got 2 and 28. so if i put a 2 here and a 28 here 3 times 28 would be too big if i flip that around and put a 28 here and a 2 here even though 28 times x is only 28x the 3x times 2 would be 6x and if you add those two together it's larger than 29x so we can get rid of that guy so now we've got 4 and 14. so if i put a 4 here and a 14 here that's going to be too big right 3 times 14 is 42 too big right you'd have 42x plus 4x which is 46x not going to work if i flipped it around and i put 14 here and 4 here you would have 12x plus 14x right the outer would be 12x the inner would be 14x but that's too small right 12 plus 14 is 26. so that would be 26x you need 29x so that doesn't work so now we're down to our last possibility which is 7 and 8. so if no combination works with 7 and 8 you can go ahead and say this guy is prime so i'll start with a 7 here and an 8 here so the outer would be 24x and the inner would be 7x so that's too big so now let's flip it around and put a 7 here and an 8 here and that's just right right so 3x times 7 would be 21x the inner 8 times x would be plus 8x 21x plus 8x is 29x so that's what we're looking for okay so you can see how tedious that was we had to go through every single possibility until we finally got the one that worked all right let's go ahead and take a look at another one so we have 40x squared plus 92x plus 48 and the first thing you would notice about this problem is that every number here the 40 the 92 and the 48 are even numbers right so we know we could pull something out before we start we know we could at least pull out a 2. now if we went through and factored the numbers we would find that we could factor out a 4 from this whole thing right so the gcf of 40x squared 92x and 48 would be a 4. and if i went ahead and just pulled that out now i'd have a 10x squared plus a 23x plus a 12 inside my parentheses so you always want to pull out stuff that you can right if you have something other than one that's the greatest common factor of your terms pull it out it's going to save you additional factoring in the end so now all i want to do is work on this 10x squared plus 23x plus 12 and i'm just going to write a 4 out here and then i'm going to set up my binomials now i'm going to have two different scenarios okay and let me explain why we have a leading coefficient here of 10 right 10 is the coefficient for my x squared 10 is not a prime number so this is a somewhat more tedious scenario because i can have 10x squared coming from 10x times x where i can have 10x squared coming from 5x times 2x okay either of these is possible right 10x times x is 10x squared 5x times 2x is also 10x squared so we just have to go through and kind of slug it out we think about the fact that all the signs are positive so i know this is plus and this is plus and this is plus and this is plus so i just need to figure out these two final terms and see which one of these is going to work out so if i look at my 12 here i know 12 has factors of 1 and 12. you could do 2 and 6 and then you could do 3 and 4. so let's go ahead and start with this top one let's think about 1 and 12. so if i went ahead and put a 1 here and a 12 here would that give me the correct middle term of 23x well the outer would be 120x and the inner would be 1x so we know that's too large right so that we can just throw that out now this part right here is important so you need to pay close attention can i put a 12 here and a one here no it's not possible so the reason it's not possible is if the original trinomial you're trying to factor does not have a common factor of two in this case then none of the binomials will have one and you might say well we did have a common factor we had a common factor of four but we've already pulled that out that's outside of the parentheses now inside where we have 10x squared plus 23x plus 12 there is no common factor other than one okay so we can't pull anything else out so this guy is not a possibility because between 10x and 12 i could pull out a 2 right there's a common factor of 2 there so this is not a possibility so you can go ahead and scratch that off the list so the next thing you would try is 2 and 6 but for the same reason that i just listed it's not possible if you put a 2 here you got a common factor of 2. if you put a 6 here you've got a common factor of 2. so you can go ahead and cross that off the list so you really just need to try 3 and 4. you can't put a 4 here because you'd have a common factor of 2. so the only possibility left is a 3 here and a 4 here so the outer would be 40x and you know this is going to be too large now because the inner is plus 3x right so that would give me 43x and i'm looking for 23x so this scenario here where 10x and x are your kind of first terms for each binomial not possible so you can go ahead and eliminate that and we're only going to be working with this scenario here now and let's kind of scooch this up and i'll go ahead and put equals now so is it 1 and 12. so i can't put a 12 here again for the same reason i have a common factor of 2. so it would have to be a 1 here and a 12 here does that work the outer would be 5x the inner would be 24x that's going to be too large it's 29x so that doesn't work so i can go ahead and eliminate 1 and 12. i'll just erase it because i don't have anything else to check with it so now i'm going to go with 2 and 6 and that's not possible at all if i put a 2 here i have a common factor of 2 if i put a 6 here i have a common factor of 2 so not possible 3 and 4. i can't put a 4 here so it has to be true that this is a 3 and this is a 4. no other combination is possible so if this does not work we have a prime polynomial so my outer here would be what it would be 15x my inner would be 8x and it just so happens that 15x plus 8x does give me 23x so we have correctly factored this guy as 4 times the quantity 5x plus 4 times the quantity 2x plus 3. and of course you can check this forget about the 4 out here just go ahead and do foil first so 5x times 2x is going to give me the 10x squared the outer again would be 15x the inner would be 8x so 15x plus 8x does give you the 23x in the middle and then 4 times 3 does give you the 12. so you have the 4 outside of parentheses just like you have here if you multiply 4 by each term here you get back to the 40x squared plus 92x plus 48. all right let's take a look at one with two variables involved and if you've watched the last two review sessions you know this is no more difficult so the first thing we should notice here for 45x squared minus 230xy plus 25y squared is we have a common factor of 5 right this ends in a 5 this ends in a 0 and this ends in a 5. so if i pull out a 5 i would have 9 x squared minus 46 x y plus 5 y squared and again now i just want to factor this on the inside so let's put a 5 out in front what are the possibilities i've got a 9x squared so it's going to be 9x times x or it's going to be 3x times 3x right it's got to be one of those two because 9 is either 1 times 9 or 3 times 3 okay that's all there is so when we think about this second variable the fact that it's y squared here in a y here means i just have a y here and a y here okay there's nothing to worry about when you have a variable involved like a y squared or a z squared some second variable that they've thrown in just look at where it is in this position you have a y squared that can only come from y times y okay so just throw it in there and just forget about it now we're just looking at the numbers as if it didn't exist so i also notice that i have a negative middle term and a positive final term so that tells me these two signs are negative so now that i've got all that stuff worked out what are the factors of five well it's just one and five right so if i think about that i can go through and say okay it could be 5 here and 1 here does that work out well the outer would be minus 9 x y the inner would be minus 5 x y does that give me negative 46 x no it gives me negative 14xy so that doesn't work i can switch the order and put 5 here and 1 here now the outer would be negative 45xy so negative 45xy and the inner would be minus xy if we sum these we do get negative 46xy so we don't even need to check this we can get rid of it so we've correctly factored this we get 5 times the quantity 9x you just go ahead and say minus y times the quantity x minus 5y in this lesson we want to review special factoring so a few lessons ago we reviewed special polynomial products or just special products for short in that lesson we looked at a few useful formulas that allowed us to quickly find the product of commonly occurring multiplication scenarios when working with binomials so here we're going to use the same thought process but for factoring instead of multiplication so a lot of the formulas you're going to see in this lesson are just reversing what we've already seen so if we start out with something like the difference of two squares we've already seen the reverse of this so if we have something like x squared minus y squared this is equal to so you take the first thing that's squared so x is being squared so that's going to go in the first position of each binomial and then you take the second thing that's being squared here it's y y is being squared so that goes in the last position of each binomial the second position and then you just have alternate signs now in our lesson on special products we saw this right we called it the product of conjugates or we could say the product of the sum and difference of the same two terms you have an x here and an x here a y here and a y here so it's the same two terms it's just different signs in other words if we looked at something like x plus five this quantity multiplied by the quantity x minus 5 this matches our pattern right we have x and x we have 5 and 5 and then different signs what happens is in the foil process the o and the i are going to cancel right so the outer here is negative 5 x the inner here is 5x so they will cancel and you'll just be left with the first thing squared in this case it's x so x squared minus the last guy which is 5 squared so 5 squared is 25. now if we want to reverse this and we start out with x squared minus 25 how would we apply this formula here the easiest thing to do is to write it in this format so we know that x is squared for emphasis i'm just going to put x inside of parentheses this is not illegal this is not changing anything this is just for emphasis so x is being squared then minus what is 25 in terms of taking the square root of it in other words square root of 25 is 5. so what's being squared there 5 is so i can write this as 5 squared if you break things down into this format to where you can match this completely you can easily recall your formula okay you don't need to do this every time as you get better with this you're not going to have to do that but when you first start it makes it a little bit easier so what's being squared here minus what's being squared here so the first thing that's being squared goes in the first position of each binomial so i'll put an x in the first position of each binomial because that's the first thing that's being squared the last thing that's being squared in this case it's a 5 goes in the last position of each binomial okay very easy to remember and then you just have alternating signs you have a plus and a minus or you can flip that around and you could put a minus and a plus okay very very easy now not all of them are so straightforward like this problem was so let's think about something like 16 x squared minus let's say 811y y squared so what we do here again i just want to think about what's being squared minus what's being squared that will allow me to use my formula well here i have not only a number or coefficient i also have a variable so 16 is a perfect square 16 is 4 times 4. so i'm going to write this as 4x this whole thing is being squared right don't change this into just 4x squared because that would not be correct these are not going to be equal because the 4 is not squared there so you've got to account for that and put this inside of some parentheses so 4x is being squared minus the square root of 81 is 9 and we know y is being squared so 9y and this whole thing is being squared okay so what's being squared minus what's being squared again just follow the formula so this is the first chi squared so 4x is the first chi squared so it goes in the first position of each binomial very very easy 9y is the second thing that's being squared so it goes in the second position or the last position of each binomial and you just alternate the sign so plus and then minus or you could do minus and then plus okay very easy so you get the quantity 4x plus 9y times the quantity 4x minus 9y all right let's take a look at another one so we have 8x to the 10th power minus 50y to the 12th power now if i look at 8 that's not a perfect square right and if i look at 50 that's also not a perfect square that means that there's no rational number that i can multiply by itself to get to 8 there's no rational number that i can multiply by itself to get to 50. but 8 and 50 are each divisible by 2. whenever you're factoring you want to always think about can i pull anything out before i start in this case i can pull out a 2 and i would be left with 4 x to the 10th power minus 25 y to the 12th power now 4 is a perfect square 25 is a perfect square so now we know we can use our difference of two squares formula i'm going to go ahead and write this first part 4x to the 10th power has 2x what does x need to be raised to here remember it's going to be squared so if it's something here and it's going to be squared we use division if it's 10 that we're going to end up with 10 divided by 2 would give me 5. okay so let me just change this to a 5. and then minus we have this 25y to the 12th power square root of 25 is 5 and then of course i would divide 12 by 2 to get 6. so this would be y to the sixth power and this is squared so let me kind of write this a little bit better okay so now we want to use our formula don't forget the 2 out in front so that's just going to stay out in front we're going to set up two binomials here the first guy that's being squared is 2x to the fifth power so that guy is going to go in the first position of each binomial then we're going to have this 5y to the 6th power that's the last guy being squared so that goes in the last position of each binomial and then i just alternate my signs so plus and then minus so this guy is going to factor into 2 times the quantity 2x to the fifth power plus 5y to the sixth power times the quantity 2 x to the fifth power minus 5y to the sixth power all right so now let's think about factoring a perfect square trinomial so whenever you have a binomial squared it turns into a perfect square trinomial so the reason it's a perfect square trinomial remember a perfect square as a number is a number that when you take the square root of it you get a rational number well for a perfect square trinomial when you factor it it becomes a binomial squared okay so this is something like x squared plus 2xy plus y squared which would factor into the quantity x plus y squared and then the other version of this is x squared minus okay you have a minus 2xy plus y squared equals you have the quantity x minus y and this is squared all right so let's look at an example of this and this is relatively straightforward to check essentially the first thing you need to do again if you have something like x plus y is quantity squared this gives you x squared plus 2xy plus y squared so if it's in this format here this part would be squared something squared and this part would be something squared is 5 a perfect square no right 5 is not a perfect square there's no rational number that i can multiply by itself to give me 5. if i want 5 i'd have to do square root of 5 times square root of 5 square root of 5 is an irrational number same thing goes for 125. so at this point it looks like we can't use our formula but we're wrong right the reason we're wrong is that again when you're factoring stuff you always want to look for additional things to pull out so this is a five this is divisible by five it ends in a zero this is divisible by five it ends in a five so i could pull a five out here before i start i'd have x squared plus 10 x plus 25 and now if i look this is something squared this is something squared right i could really write this let me kind of move this down i could really write this as 5 times in parentheses for emphasis i'm just going to put x being squared plus i'm going to put 2 multiplied by this guy right here this 25 i'm going to write as 5 being squared so 2 times 5 times x in other words 2 times what's being squared here which is x times what's being squared here which is 5 and this does in fact give me the 10x so this exactly matches this format here so we can use that to factor so following this pattern it's whatever the first thing is squared plus whatever the last thing is squared so that's inside of parentheses so in this case that's going to be x that's squared here and 5 that squared here and then we just square this binomial okay so this factors into 5 times the quantity x plus 5 squared all right let's take a look at another one so we have 441x squared minus 504xy plus 144y squared now these are numbers that we don't work with very often so for myself if i have 441 usually i'm going to have a calculator so i'm just going to go square to 441 and see if that ends up being a rational number if you do that you end up getting 21. so i could really write this as 21x that amount being squared but i do notice something here i notice that i can pull something out 4 plus 4 is 8 8 plus 1 is 9. 5 plus 4 is 9 and 1 plus 4 is 5 5 plus 4 is 9. if a number's digits sum to 9 is divisible by 9 so that means we can pull a 9 out before we even start 441 divided by 9 would give me 49. so this would be 49x squared minus 504 divided by 9 would be 56. so 56xy and plus 144 divided by 9 would be 16. so you'd have 16y squared close the parenthesis now there's nothing else we can pull out because this is 7 times 7 this is 7 times 8 or 7 times 2 cubed and this is 2 to the 4th power so nothing else that's common to everything all right so again i want to think about is this something squared is this something squared well yeah 49 is 7 times 7. so i could write this as 7x being squared i'm going to put minus forget about this term for a second plus is this something squared yes 16 is 4 squared so 4y this guy squared so is this guy minus 2 times this guy 7x times this guy for y 2 times 7 is 14 14 times 4 is 56 and then you have x times y so you get 56xy so it matches what we're looking for we have the minus here so in our formula we want the 9 out in front of some parentheses so we have a minus inside and we just want the first guy that's being squared which is 7x minus the last guy that's being squared which is 4y and this binomial is squared okay so we factored this guy as 9 times the quantity 7x minus 4y and this is squared all right let's think about the sum of cubes and i want to just make a special note here this if i read the formula is x cubed plus y cubed and it factors into x plus y this quantity times x squared minus xy plus y squared now this is not the same as what we looked at when we saw x plus y inside of parentheses cubed okay a lot of students make the mistake of thinking this is x cubed plus y cubed that is wrong okay wrong you cannot distribute this this is not multiplication okay if you have x y cubed like that with multiplication this is x cubed y cubed okay that's where people get that from but it doesn't work with addition okay when you have this remember we talked about this in special products this becomes x cubed plus 3 x squared y plus 3 x y squared plus y cubed so these two are not going to be the same okay so what we're talking about here is the sum of cubes so in other words i have something cubed plus something else cubed and that's what we're looking at so let's take a look at an example so we have 8 x cubed plus 27 y cubed so is it something cubed plus something else cubed yeah 8 is 2 cubed so i could write this as 2 x that's being cubed plus 27 is 3 cubed so 3 y that's being cubed and then i can follow my formula so let me write the formula down again for you so if we have x cubed plus y cubed this is going to factor into you have x plus y to start and then you also have x squared minus x y plus y squared okay so this one is a little bit more difficult to remember but definitely one that's going to be more beneficial for you all right so if i follow the pattern here the x represents what's being cubed the y represents what's being cubed so what's being cubed here it's 2x so i'm going to start out by putting 2x plus 3y just following the pattern and then we have the guy that's being cubed the first one squared so square 2x i would get 2 squared is 4 x squared is just x squared then minus you have x times y so this guy times this guy 2 times 3 is 6 x times y is x y and then plus you have y or whatever the last thing is that's being cubed being squared so 3y squared 3 squared is 9 y squared is just y squared so this factors into 2x plus 3y this quantity times the quantity 4x squared minus 6xy plus 9y squared all right so we have one more formula to cover we now have the difference of cubes so we have something cubed x cubed minus y cubed something else cubed this turns into x minus y times this other quantity here x squared plus x y plus y squared so the difference here is going to be in the sign right so here i have a minus whereas with the sum of cubes i had a plus and here i have a plus whereas with the sum of cubes i had a minus so the way i remember this is that the first one is the same and the second one is different okay that's how i remember it the last term here is always going to be plus now let's look at an example and let me go ahead and just write the formula down because i know we're going to need it so x cubed minus y cubed is going to factor into again the first guy is the same so i'm going to put x minus y and then the second guy is going to be different so i'm going to have x squared it's going to be different so it's going to be plus you're going to have xy and then you always have a plus here you have y squared okay so not too bad all right so if we have 320x to the sixth power minus 1080y to the 12th power so looking at these two numbers 320 and 1080 neither is a perfect cube but again you always want to think about what can you pull out before you start this guy right here i know it's an even number so i can definitely pull out a 2. it ends in a 0 so i can pull out a 10. this also ends in a 0 so i can pull out a 10. so really i know this would be 32 times 10 okay i know you could do this faster on a calculator but just for the sake of completeness let's just write this out this is 108 times 10. now these are still even numbers 32 is an even number and so is 108. so 108 divided by 4 is 27. so this would be 27 times 4 and this guy right here would be 4 times eight so in other words i could pull a 40 out from each if i pulled out a 40 this guy right here would be what it would be eight which is a perfect cube and then we'd have x to the sixth power and then minus this guy right here if i pulled out a 40 i'd have 27 which is a perfect cube and then y to the 12th power okay so always look for what you can pull out a lot of times when you're doing factoring practice or on a test they've given you something that's not a perfect cube so you say oh i can't use the formula and you kind of move on so let me erase this and put this is equal to so what's being cubed here so 8 is what it's 2 cubed and then what's my exponent going to be again if i'm using the power to power rule something here raised to the 3rd power would give me a 6 here so something times something gives me 6 6 divided by 3 would give me 2. so this is a 2 and then we have minus over here 27 the cube root of that would be 3 right 3 cubed is 27 and then for y to the 12th power again you're just going to divide 12 divided by 3 is 4. so y to the fourth power and this is cubed all right so now we're just going to follow our pattern we kind of move this down all right so we're going to have 40 out in front and then it's x minus y so it's the first thing being cubed minus the last thing being cubed so we're just going to have 2x squared minus 3y to the fourth power and then times you have the first thing that's being cubed now it's squared so 2x squared is squared so i'm going to put this in parentheses so we don't make mistakes so 2 squared is 4 x squared squared is x to the fourth power so this is 4x to the fourth power then we're going to have plus xy so that corresponds to what's being cubed times what's being cubed so 2x squared times 3y to the fourth power 2 times 3 is 6 and then x squared times y to the fourth power okay and lastly we have plus whatever is being cubed last which is 3y to the fourth power is going to be squared again let's do some parentheses so we don't make a mistake 3 squared is 9 y to the 4th power squared is y to the 8th power so y to the 8th power okay so this guy is going to factor into 40 times the quantity 2x squared minus 3y to the fourth power times the quantity 4x to the fourth power plus 6x squared y to the fourth plus 9y to the 8th power so a more difficult or more challenging formula to remember but something that's definitely worth it to put it on a post-it put it on a flash card look at it use it when you're doing your homework and just practice practice practice and it'll become second nature for you in this lesson we want to review factoring by substitution so for most of you watching this this is something you've done before basically the idea here is that normally we work with trinomials and we factor those into the product of two binomials and the trinomials that we're working with have a degree of 2. in the scenarios we're going to look at today this won't be the case right we're going to look at more complicated polynomials or more specifically more complicated trinomials so when you come across these you generally are going to be given problems that allow you to factor them using a little simple substitution technique so to cover this if we have something like x to the 4th power plus 2x squared plus 1 notice that this 4 here the larger exponent is double that of the two okay normally we're working with something like this let's say we had something like x squared plus 5x plus six we all know at this point this can easily be factored into what this first term here would be x this first term right here would be x so i need two integers that sum to five and give me a product of six that's two and three so plus two and plus three so very easy to factor this is the typical scenario we're looking at notice that this two here is double that of the exponent of 1. this 4 here again is double that of the exponent of 2. so we have the same kind of format the same pattern and so when this occurs we can just use a little substitution to make our factoring much easier okay so the way we do the substitution technique is we let another variable a lot of times you see u you might see z you might see q pick whatever you want it doesn't really matter you're going to let that variable be equal to the variable raised to the smaller power so in this case the variable raised to the smaller power we have x squared so we're going to say let's let u be equal to x squared and i'm going to rewrite this in such a way that it's clear where we need to replace things x to the fourth power is x squared squared then plus we'd have two x squared then plus one so in other words i'm going to take and replace everywhere we have an x squared with a u okay because u is equal to x squared so down here where we'd have x squared i'm just going to put a u and that's going to be squared and then plus we have 2x squared so i'm going to put 2 u and then lastly plus 1. so all i want to do is factor this guy and it's pretty easy to do i know the first term for each binomial would be a u right because u times u would give me u squared and then i just need two integers that sum to 2 and give me a product of one well the only possibility there is one and one right one times one is one one plus one is two so that works out so you get u plus one times u plus one or u plus one that quantity squared now you're not done okay because you want this in terms of x because this is the variable you started working with in your original trinomial so all i have to do is go back and resubstitute so since u is equal to x squared i can essentially let me kind of scooch this down i can just write this as x squared right just plug in an x squared for u plus 1 times you'd have x squared plus 1 or of course you could write that as the quantity x squared plus 1 squared let's move this down because it's kind of in the way and let's move this up so we'll say this is equal to this right and you can check this let me kind of scooch these up you can check these if you want x squared times x squared would be x to the fourth power so that would give me that the outer would be plus x squared the inner would be plus x squared so x squared plus x squared is 2x squared so that's where we get that middle term and the last we have 1 times 1 which would give us 1. so checking this with foil we see we got the correct factorization all right let's look at another example so suppose we saw 7x to the fifth power minus 62x cubed minus 9x at this point you have a 5 and you have a 3 and you have a first power this is not the normal pattern but again when you look at things and you're trying to factor always ask yourself the question can i pull something out before i start okay i can obviously pull out an x here and that would give me a 7x to the fourth power minus a 62x squared minus a 9. okay now this matches the format i'm looking for i have the fourth power and i have squared so what i'm going to do now i'm going to put x times inside of parentheses i'll have 7 x squared squared i'm just doing this for emphasis minus 62x squared minus 9 and so i'm going to highlight this and this and i'm just going to say again let's let u let's let u be equal to x squared okay so what i'm going to have is x times the quantity 7 u squared minus 62 u minus 9 and i'm going to factor this so let me let me erase this real quick and i'll just write this off to the side so let u be equal to x squared all right so to factor this inside i'm going to go ahead and use the grouping method or the ac method so give me two integers whose product is 7 times negative 9 or negative 63 so this is the product i'm looking for and whose sum is negative 62. so that's the sum well if i think about this combination here really i'm going to be thinking immediately about negative 63 and positive 1 right so negative 63 and positive 1. so i'm going to write this as 7u squared i'm going to do minus 63u i'm going to do plus u or plus 1u however you want to write that and then minus 9. okay and then i'm just going to factor this using grouping so i'd have x times from the first group here the 7u squared minus 63u i can pull out a 7u that would leave me with u minus 9. from this group here i'm just going to pull out a 1 and that would leave me with u minus 9. and so what's going to end up happening is i have a common binomial factor of u minus 9. so that's going to come out so i would have x times the quantity 7 u plus 1 times the quantity u minus not okay so let me erase all of this we don't need it anymore so now i'm just going to do some back substituting i'm going to put this is equal to i have x times the quantity 7 times u is going to be x squared plus 1 and then times for you i'm going to have x squared and then minus 9. now am i done you might think that you're done but you're not you always want to look for opportunities to factor further in this particular case 7x squared plus 1 i can't do anything with that can't do anything with x but x squared minus 9. what can i do there that's the difference of squares right i have x which is squared minus basically 3 squared right 9 is 3 squared so i can further factor x squared minus 9 as x plus 3 that quantity times x minus 3. so we're going to erase this and we're going to put the quantity x plus 3 times the quantity x minus 3. okay and if you were to go through and do the multiplication here you will end up with this 7x to the fifth power minus 62x cubed minus 9x so very important that when you go through these problems you look for additional factoring because if you would have left this as x squared minus 9 here instead of factoring it further that would be considered wrong right because it's not completely factored all right let's look at another one so we have 9 times the quantity x minus 4 squared plus 30 times the quantity x minus 4 and then plus 25. so this is another opportunity to use our substitution technique notice how we have this guy right here and this guy right here this one is squared and this one is basically raised to the first power so if i let this binomial x minus 4 be equal to a variable like u i can factor using substitution i'll have nine u squared right just plugging in a u where this is plus 30 u plus 25. now i'm going to factor this using the grouping method so this will be equal to i want two integers whose product is 9 times 25 or 225 so the product is 225 and whose sum is 30. so sum is 30. so 225 if i'm breaking that down let me kind of get this out of the way real quick move this over here 225 i know is 1 times 225 that's not going to help me it would be 3 times 75 but that's not going to work it would be 5 times 45 but that's not going to work we could also do 9 times 25 but that won't work and of course we could do 15 times 15 and that would work right 15 times 15 is 225 15 plus 15 is 30. so we found the two integers that we need so let's erase all this and let's go ahead and write this as 9 u squared plus 15 u plus 15 u plus 25 okay so now we can factor this using grouping from the first group i can pull out a 3u and i would have a 3u plus a 5. from the second group i can pull out a 5 and i would have a three u plus a five so i've got this common binomial factor now that i can take out so if i take that out i would have three u plus five times the quantity three u plus 5. basically 3u plus 5 that quantity squared so let's go ahead and erase all this and what i'm going to do is i'm going to replace my u with my x minus 4. so i'm going to say this is equal to you'll have 3 times x minus four make sure to use parentheses then plus five times three times again inside of parentheses x minus four and then plus five okay so let's see what we can do to simplify here so 3 times x is 3x 3 times negative 4 would be negative 12 and then you have plus 5. over here it's the same thing right so i just put 3x minus 12 plus 5. and of course negative 12 plus 5 is negative 7 so this ends up being 3x minus 7 this quantity times 3x minus 7 or you could say the quantity 3x minus 7 squared right either way all right let's take a look at one more of these so we have 8 times the quantity x plus y raised to the power of 2z plus 6 times the quantity x plus y raised to the power of z and then minus 9. so this one's slightly more challenging than the one we just saw what makes it more challenging is you have a variable in your exponent so this kind of trips students up just a little bit so the idea here is still the same we're looking for that pattern right we have this 2z here and we might as well say this is 1z or just z right so this guy the larger exponent is double that of the smaller 2z is double that of just z so again if we see this pattern we can use our little substitution technique now what am i going to let my variable be equal to let's let u be equal to what well to make this crystal clear i'm going to rewrite this i'm going to put this as 8 times the quantity x plus y raised to the power of just c let me kind of scooch this down for a minute and then i'm going to raise this to the power of 2 okay so i haven't done anything illegal power to power rule would put me right back to this right because this is my base it's x plus y that quantity so i would take this exponent here which is z and i would multiply it by this exponent here which is 2 which will get me back to the 2z which i have there okay so i haven't done anything illegal then plus we have 6 times the quantity x plus y this is raised to the power of z and then minus 9. so i'm going to take this guy right here the whole thing the x plus y that quantity raised to the power of z and that's what i'm going to let my variable u be equal to okay so x plus y raised to the power of z okay so i am going to have 8 multiplied by take this whole thing and replace it with a u and that guy's going to be squared then plus i'm going to have 6 replace this with a u and then just minus 9. let me kind of scooch this out of the way so we have some room to work and so now i'm just going to factor this guy 8u squared plus 6u minus 9. i'm going to just use my grouping method because that's easiest for me so i want a product of what 8 times negative 9 is negative 72. so negative 72 is the product and i want a sum of six so six as the sum so two integers that have a product of negative 72 and that have a sum of six so again i like to just work with positives and i work out the sign as i go 1 and 72 wouldn't do anything for me 36 and 2 wouldn't do anything 24 and 3 would not do anything 4 and 18 wouldn't do anything but when i get to 6 and 12 i can make that work right i need a negative product and a positive sum so what i'm going to do is i'm going to let the larger number in terms of absolute value be positive the smaller number in terms of absolute value be negative so i would want a plus 12 and a minus 6. so i'm going to use that to rewrite this guy so i'd have 8 u squared plus 12u minus 6u minus 9 okay so i can erase this i don't need any information anymore and now i'm just going to factor using grouping so over here i know that in the first group i can pull out a 4u so i pull out a 4u that would leave me with a 2u plus a 3. over here i know i can pull out a negative 3 that would leave me with a 2u plus a 3. so we have a common binomial factor here of 2u plus 3. i'm going to go ahead and factor that guy out so this is going to give me the quantity 2 u plus 3 multiplied by the quantity 4 u minus 3. okay so i'm not done let me kind of erase all this i don't need it anymore and let me kind of just slide this up a little bit so all i'm going to do is wherever i see a u i'm going to plug this guy back in so i'm going to say this is equal to let me kind of slide this out of the way so inside of parentheses we're going to have 2 multiplied by in place of u i'm going to put x plus y raised to the power of z so this has to be inside of parentheses because 2 is multiplying the whole thing so x plus y again this is raised to the power of z and then plus 3. and then i have another set of parentheses 4 multiplied by again the same thing i'm plugging in for u so x plus y this guy raised to the power of z and then minus 3. now let me erase all this i don't need any of this anymore this is going to be my answer right this is what this guy factors into so if you start with 8 times the quantity x plus y raised to the power of 2z plus 6 times the quantity x plus y raised to the power of z minus 9 this would factor into or give us an answer of we have the quantity two times the quantity x plus y raised to the power of z plus three multiplied by this quantity four times the quantity x plus y raised to the power of z then minus three you can't make this any simpler you can't really do anything else the reason is is because the exponent here and here is a z right it's not a number where you can kind of do some other things z we don't know what it is it just represents some unknown so this is as simple as we can make this in this lesson we want to review the definition of a rational expression we also want to talk about finding the restricted values for a rational expression so before we jump in and start talking about how to find the restricted values for a rational expression we want to think about the definition of a rational expression and to properly do this let's recall the definition of a rational number so at the beginning of our course we reviewed the definition of a rational number we said that it was any real number that could be formed as the quotient of two integers where the denominator was not zero so something like the number one fourth one is an integer four is an integer so you have the quotient of two integers the denominator's not zero so that's a rational number something like negative seven fifths something like 11 14 something like just the number 10. right 10 could be 10 over 1. right so all of these fit that description the main thing to understand here is that we can't divide by zero so we can't have zero in the denominator okay so that's very important now we think about these as rational numbers but more practically we just call them fractions so in algebra we have an algebraic fraction this is known as a rational expression and what this is is this is the quotient of two polynomials where the denominator is not zero so something like three x minus one over let's say five x squared plus two one polynomial over another so that's a rational expression as another example let's say you had 2x squared plus 5x minus 1 over let's say 3x to the fourth power plus x plus 7 okay so you have a polynomial up here a polynomial down here so one polynomial divided by another again that gives us a rational expression so essentially when you start working with rational expressions the first topic you come across is how to find the restricted values for a rational expression now this topic just basically deals with division by zero we know that we're not allowed to divide by zero if we see something where we have division by zero it's considered undefined so something like 3 over 0 to make that 3 a little better this is undefined okay that would be your answer there in the case and i know some of you will be confused and say no 3 over 0 is 0 that's not right right when you divide 0 by a non-zero number you get 0. so 0 divided by 3 would be 0. okay you can divide 0 by anything you want except 0 you get 0. but if you try to divide by zero it is undefined okay so if i had a rational expression like x plus seven over x minus five what would be my restricted value here well it's anything that makes the denominator zero okay so if i look at my denominator it is x minus five so how do i find the value for x that makes this denominator zero all i need to do is set my denominator equal to 0 and just solve right what am i doing something minus 5 gives me 0 i'm going to solve for that something so that i know what i can't plug in for x it's just that simple so very easy we just add 5 to each side of the equation this will cancel we'll have x is equal to 5 okay and you can easily see if i let x be equal to 5 let's say that we let x be equal to 5 plug in a 5 here and here so you would have 5 plus 7 in the numerator which is 12 over the denominator would be 5 minus 5 which is of course 0 and again this is undefined undefined and this is what we're trying to prevent we don't want to end up with something that's undefined so what we do when we have a rational expression is we find our restricted value or our restricted values depending on how complex your rational expression is and we just put a little note here we say that this is defined for everything except for x when it's equal to 5. so we'll put x cannot be equal to or x does not equal five okay a fancier way to do this is to state the domain okay so the domain in case you haven't seen anything with functions yet we'll talk more about this later in the course the domain is the set of allowable x values so what is allowed to be plugged in here for x well everything in terms of real numbers can be plugged in for x except for 5. so the domain or again the set of allowable values for x would say the set of all real numbers or the set of all elements x such that x does not equal the number 5. okay so anything you want to plug in for x there is fine except for 5 because it makes the denominator 0. all right let's look at another example so we have 9x minus 1 over 3x minus 4. again all we need to do to find the restricted values we take the denominator here and we just set it equal to 0. so we're going to solve 3x minus 4 equals 0. to solve this i add 4 to each side of the equation that cancels i would have 3x is equal to 4 divide both sides of the equation by 3 and i would end up with x is equal to 4 3. okay so again i can write comma x does not equal 4 or i can do it in the fancy way so i can say my domain okay my domain is the set of all elements x and then such that x does not equal four thirds okay all right let's look at another one so we have 2x squared plus 7x minus 4 over 3x squared minus 21x so again i'm only concerned with the denominator here so i'm just going to set this equal to zero and i'm going to solve it now you can get super complicated things in your denominator this guy right here is not complicated at all we can solve it by factoring in case you're not familiar with solving something by factoring the first thing you would do is factor this guy so we can see that we have a common factor of 3x we can pull that out so if we pulled out 3x i would have an x minus 7 here and this is equal to 0. now what i'm going to do to solve this using factoring i'm going to set each factor equal to 0. so i'm going to set 3x equal to 0 and i'm going to set x minus 7 equal to 0 and i'm going to solve each one of those now why does that work we have something known as the zero product property so if you have something like a and it's multiplied by b and it's equal to zero well then one of these statements has to be true a is equal to zero b is equal to zero or both a and b are equal to zero so in other words we know that multiplying by zero gives me zero so it could be true that a is zero and b is not it can be true that b is zero and a is not and it can be true that a and b are both 0. so that's the premise behind the zero product property and essentially all i'm going to do here is just set 3x equal to 0 and solve divide both sides by 3 i get x is equal to 0. i'm going to set x minus 7 equal to 0 and solve add 7 to each side of the equation i get x is equal to 7. so we have two solutions here and thus two values we need to restrict from our domain so if i plugged in a 7 for x or if i plugged in a 0 for x i would get 0 as a denominator and i think we can see that right if you plugged in a 0 here and here you would have 3 times 0 squared minus 21 times 0 0 squared is of course 0 0 times 3 is 0 minus 0 times 21 is 0 0 minus 0 is 0. so the denominator would be 0 if you plugged in a 0 for x okay so we see that also the denominator would be 0 if you plugged in a 7. 7 squared is 49 49 times 3 is 147. so you'd have 147 minus 21 times 7 is also 147 so you get 147 minus 147 which is 0. so we're going to write that x cannot be 0 and x cannot be 7. let me kind of scooch these up i'll scooch this up so i'll put a comma here and you can put a comma there if you want so again all real numbers are allowed except for 0 and 7. so we can write this again in this kind of fancy way the domain is the set of all elements x such that x does not equal 0 or 7. all right let's take a look at one more so very easy concept overall the only thing that can make it more difficult is the tediousness of the denominator right if you get something that's super complex it might not be so easy to solve you can't solve everything with factoring you might have a very very difficult denominator to work with but for this i'm just trying to show you the concept so i gave you things that are pretty easy so we have 2x minus 5 over 2x squared minus 9x minus 5. i'm just going to set again this guy right here equal to 0. so 2x squared minus 9x minus 5 equals 0. so here we're not going to factor out the greatest common factor we're going to factor this into the product of two binomials so i'm going to do this using reverse foil the reason i'm going to choose reverse foil is because this coefficient here this 2 is a prime number so i know this would be 2x and this would be x i just have to work out the correct outer and inner right so i know that i want a product of negative 5. so that can only come from 5 times 1 with the signs being different right so i've got to have either negative 1 times 5 or negative 5 times 1. so knowing that i have a middle term of negative 9x i'm going to want to multiply 2x by negative 5 and i'm going to multiply x by positive 1. all right 2x times negative 5 would give me negative 10x and then 1 times x would give me 1x 1x plus negative 10x would be negative 9x so we factored this one pretty quickly all right so now i'm just going to set each factor this is a factor and this is a factor equal to 0. so i would have 2x plus 1 equal to 0 and i would have x minus 5 equal to 0. so let me kind of slide this out of the way so we have a little room so for this guy let me kind of scooch this down so for this guy i'm going to subtract 1 away to start so this cancels i would have 2x is equal to negative 1 divide both sides by 2 and you would get that x is equal to negative 1 half for this guy just add 5 to each side of the equation very simple you get x equals 5. so let me erase everything kind of scooch these over so we'll say x does not equal negative one half and also x does not equal five okay let me make that a little better so we can again write this using our domain statement so domain is let me make that better the set of all x such that x does not equal two you have negative one half and then five in this lesson we want to review writing rational expressions in lowest terms otherwise known as simplifying rational expressions so before we jump in and start talking about how to simplify a rational expression let's just quickly review the process with fractions so if i wanted to simplify five twentieths what could i do well kind of the easiest way to do it the way that i do it right now i eyeball these things and say okay what's the greatest common factor between 5 and 20. i know 5 is a prime number i know 20 is 5 times 4 or 5 times 2 times 2 so i can just divide the numerator and denominator by 5 right the gcf 5 divided by 5 is 1 20 divided by 5 is 4 so it's 1 4. but the more technical approach to this would be to factor the numerator and denominator completely so 5 doesn't factor i'm just going to write it as 5. 20 would factor into what it's 5 times 4 or 2 times 2 and then i would cancel any common factor that's not 1 between the numerator and denominator so i'd cancel this 5 with this 5 go ahead and write a 1 there and i'd end up with 1 over 2 times 2 which is 4. so my simplified answer here is 1 4. now we're going to use the same thought process when we simplify rational expressions if i have something like 2x squared plus 7x minus 4 over 5x squared plus 20x all i need to do is factor the numerator and denominator completely and then cancel any common factors now before we begin i want to make something absolutely clear if you cancel things here you've got to understand that you can only cancel common factors this is very very common as a source of confusion so if you have something like a multiplied by b over a a and b here in the numerator are factors they're multiplying each other this can cancel with this okay i'm just left with b so if i had something like 2 times 3 over 2 the 2's will cancel i'm left with 3 okay very easy to understand that where students get confused is they have addition involved and they try to cancel if you see something like 2 plus 3 over 2 you cannot cancel this okay please don't do that that's wrong it's a very common mistake you cancel common factors okay so this is wrong so as an example let's say you had something like x plus 5 this quantity multiplied by x minus 1. these are factors x plus 5 is a factor x minus 1 is a factor so this is a factor and this is a factor if this is over x plus 5 this quantity times x minus 3 again this is a factor and this is a factor so i cancel common factors right this and this can be canceled they are factors they can be cancelled but if i had something like let's say x plus 5 plus 1 over let's say x plus 5 plus y or something like that you see students try to cancel this and this this is addition this is addition you can't cancel okay it's got to be cancelling common factors okay that's what we're looking for so please don't get confused between those two now what am i going to do here i'm just going to factor this numerator into the product of two binomials and i'm going to factor the denominator into a monomial times a binomial okay and then i'm going to look to see if i can cancel anything so for the numerator i'm going to use reverse foil here because it'll be a little bit quicker because this guy right here is a prime number 2x squared can only come from 2x times x so this will be 2x times x now i just need to work out this 7x part okay to do that i think about factors of negative 4. we've got 1 and negative 4 or negative 1 and 4 right so 1 and negative 4 or negative 1 and 4. you also have 2 and negative 2 but that won't work why won't it work remember what we talked about in our lesson on reverse foil if i put in a negative 2 here and a positive 2 here or reverse that it doesn't matter you would have a common factor of 2 here there's no common factor of 2 in this guy so this doesn't exist right this can't be possible so we'll erase this and just focus on this scenario now i have 7x i know that that could come from 2x times 4 which would give me 8x and then x times negative 1 would give me negative x 8x minus x would be 7x so i would want a 4 here right because i want 2x times 4 and i would want a minus 1 here because i want x times negative 1 right so again 8x minus x would give me 7x so that's our factorization all right down here it's very very easy we pull out the gcf which is 5 x inside the parentheses we're left with x plus 4. now again i can't stress this enough this is multiplication this is multiplication the quantity 2x minus 1 is multiplied by the quantity x plus 4. these are factors okay these are factors and then these are factors hey these are factors so 5x is multiplied by the quantity x plus 4. we can cancel common factors you have x plus 4 you have x plus 4 that can be cancelled okay so what am i left with after i cancel i'm left with 2x minus 1 over 5x okay so this is the simplified version when you work with rational expressions as we talked about in the last lesson you deal with restricted values you can't divide by zero so any value that you replace the variable with that results in a denominator of zero has to be restricted from the domain if i look at the simplified version here if i plugged in a zero there obviously five times zero will give me zero so i would say x does not equal 0 but that's wrong x can't equal 0 but x also can't equal negative 4. and the reason x can't equal negative 4 is we have to think about this in terms of the original rational expression with the original rational expression if i factor the denominator i have 5x multiplied by the quantity x plus 4. we know 5x being set equal to 0 would give me a result of x equals 0. so we've already restricted that we found that we're done with that this part right here the x plus 4 that cancelled we lost that information so x plus 4 equals 0 if i subtract 4 away from each side i get x equals negative 4. so that's also a restricted value x can also not be negative 4. so don't let that trip you up when you're stating restricted values always go with the original rational expression not the simplified one because the simplified one will lose information okay and you'll end up putting the wrong domain down all right let's take a look at another one so we have x squared minus 4x minus 12 over x squared minus 4. so very easy to factor this guy so for the top i want two integers whose sum is negative 4 and whose product is negative 12. so this is x and this is x for negative 12 i would immediately think about negative 6 and positive 2 right negative 6 and positive 2 and of course that would work out right your outer would be 2x and your inner would be minus 6x negative 6x plus 2x would give me negative 4x so that's good to go for this guy x squared minus 4 it's the difference of two squares so we talked about this this factors into x plus two times you'd have x minus two right whatever squared here minus whatever squared here this is basically x squared minus two squared okay so that's how we get this so i can cancel this with this right x plus 2 in the denominator with x plus 2 in the numerator again if you're thinking about the domain think about it in terms of the original rational expression if i write this as x minus 6 over x minus 2 i've lost the information that this guy presents right because here i know that x can't be negative 2 and i know that x can't be positive 2. with this guy i know that x just can't be 2 right so i'm just going to go ahead and write that x does not equal 2 or negative 2. all right let's take a look at another one and this one's a little bit more tedious so we have 21x squared plus 123x plus 90 over 3x cubed plus 17x squared plus 10x now in the numerator i am going to pull out a 3 to start that would give me 7x squared plus 41x plus 30. and i'm just going to work on this for now i'll come back to the denominator in a minute so to factor the inside here i'm going to go ahead and use reverse foil since 7 is a prime number so 7x and you'd have x over here i'll put my 3 outside so i need to figure out some factors of 30 here so 30 factors into what it's 6 times 5 6 is 3 times 2 okay so i have some possibilities here so to go through the possibilities we have 1 times 30 we have 2 times 15 we have 3 times 10 we have 5 times 6 and that's basically it now 1 and 30 won't work 2 and 15 won't work 3 and 10 won't work but 5 and 6 would work 7 times 5 is 35 and 6 plus 35 is going to give me 41 okay so that's the factorization there let me erase this i don't need that anymore so let me see if i can squeeze this in here hopefully i can 3 times the quantity 7x plus 6 times the quantity x plus 5. all right for the denominator i can pull out an x to start that will give me 3x squared plus 17x plus 10 and then for this part right here again i'm going to use reverse foil because this guy's prime so 3x and x that's the only way you can get 3x squared so i need to figure out how to get a middle term of 17x and the factors of 10 you've got 1 times 10 you've got 2 times 5 and that's it right so if i want 17 i could multiply 3 times 5 and get 15 and i can multiply 2 times 1 and get 2 and 2 plus 15 would give me 17. so we would have x times the quantity 3x plus 2 times the quantity x plus 5. okay all right let's drag this guy down here so we have a little room so we can see that we have a common factor of x plus 5. cancel this with this and i'm left with 3 times the quantity 7x plus 6 over x times the quantity 3x plus 2. now you can leave it in factored form just to show that there's nothing else that will cancel that's very common you could also distribute the 3 to each term here and the x and each term here both answers are valid now in terms of the restricted values you would have three of them right if you look at the original here x can't be zero so x cannot be zero all right i got that from this x here just set x equal to zero three x plus two we would set that equal to zero and solve so this would be three x is equal to negative two divide each side by three you get x is equal to negative 2 3. so x can also not equal negative 2 3. and then lastly you have x plus 5 here which was cancelled again we lose that information in the simplified version so x plus 5 if you set that equal to zero we subtract five away from each side of the equation you get x equals negative five so x also can't be negative five so three restricted values there you have zero that can't be in the domain negative two-thirds and also negative five all right for the last problem we're going to look at one that's a little bit challenging so we have 2x squared minus 9x minus 35 over 2x cubed minus 15x squared plus 6x plus 7. so let's start by just factoring the numerator so i'm going to use reverse foil here this guy is prime so it's going to be pretty easy and i want to get a correct middle term of negative 9x so i know that for 35. forget about the fact that's negative it's going to be 1 times 35 or 5 times 7. so i know 1 and 35 will not work but 5 and 7 would work if i put a minus 7 and a plus 5 because the outer term here 2x times negative 7 would be negative 14x and the inner term here 5 times x would be plus 5x you combine those you would get negative 9x okay so that's what we're looking for there now this is a 4 term polynomial down here this 2x cubed minus 15x squared plus 6x plus 7. pause the video and see if you can factor that using grouping so hopefully you gave that a shot and you'll notice that no matter what you do you cannot factor this using grouping most people would just throw in the towel here and just say okay it can't be simplified but you need to think a little bit deeper here this is a very common problem as you get into more advanced courses you'll be given stuff that doesn't factor using traditional methods so we have to think about something else if i have 6 times 3 and this gives me 18 well i know 3 and 6 are factors of 18 3 and 6 are also divisors of 18 meaning if i divided 18 by 3 i get 6 if i divided 18 by 6 i get 3. so hopefully you can see where i'm going with this if we can simplify here then it must be true that this guy would factor into something that has one of these right so this could be a factor or this could be a factor if neither of them is a factor then we can't simplify so all i actually have to do is do some division right i can do polynomial long division or in this case right here if this turns out to be a factor we can use synthetic division so we're going to start with that because synthetic division is pretty quick so this is in the format of 1x to the first power minus some number so if in that format just grab the number so 7 is going to go out here just like i taught you in the review session we're going to take the coefficients only so my 2 my negative 15 my 6 and my 7 okay let's put a little bar down here let me scroll down get a little room we'll come back up in a second drop down the first number so this goes here 7 times 2 is 14 we add negative 15 plus 14 is negative 1. seven times negative one is negative seven we add six plus negative seven is negative one seven times negative one is negative seven seven plus negative seven is zero this is important here we have no remainder okay we have no remainder so that means that x minus 7 is a factor of 2x cubed minus 15x squared plus 6x plus 7. and we have our answer right here from the division so we know that this would be x minus 7 times you would have what it's one degree less so it would be 2x squared 2x squared and then minus x and then minus 1. okay and if you don't believe me go ahead and multiply these two together and you will see that you get this back okay again all we did just to go through this again because i know some people will get confused this guy right here divided by this guy right here gives me this guy right here so this guy times this guy gives me back to this guy so we just factored it using division okay that's all we did now the question is can i further factor this can i factor 2x squared minus x minus 1. so this would be 2x and this would be x we have negative 1 here so that's pretty simple it's either plus 1 here minus 1 or the other way around the outer here would be negative 2x and the inner would be plus x that's exactly what we want right so let's scooch this down and let's slide this up and we have successfully factored this guy as the quantity 2x plus 1 times the quantity x minus 1 times the quantity x minus 7. now we're going to go ahead and just cancel this with this and we've simplified this guy so it's 2 x plus 5 over you have the quantity 2x plus 1 times the quantity x minus 1. and again we should leave this in factored form just to show that nothing else would cancel if you want to look at the restricted values here again you go with the original one this is in factored form so if i was to set this equal to 0 write the denominator then i could just go through to each factor and solve that equation so two x plus one equals zero subtract one away from each side you get two x is equal to negative one divide each side by two you get x equals negative one half so x cannot be negative one half the other ones are easy you can just solve them by inspection you have x minus one if that was equal to zero x would just be one right so one and then this guy right here which we cancelled x minus seven if that was equal to zero x would just be seven so just to state our final answer here we have this two x plus five in the numerator over the quantity two x plus one times the quantity x minus one and then our restricted values again thinking about the original rational expression x cannot be negative one-half it cannot be one and it cannot be seven in this lesson we want to talk about multiplying and dividing rational expressions so in order to multiply and divide rational expressions we need to think back to our basic math where we talked about multiplying and dividing fractions so to get things started let's just look at a very simple example and then we will take this up a step and we'll think about rational expressions so we have 1 4 times 2 7. how can we solve this problem well basically what we could do is multiply across so 1 times 2 is 2 and then 4 times 7 is 28 so basically numerator times numerator over denominator times denominator but the problem with this is it's quite inefficient right if we just do this without trying to cancel anything we end up with an unsimplified fraction so we have to take the extra step now of simplifying and basically you have a common factor of 2 there that you can cancel out so 2 divided by 2 is 1 28 divided by 2 is 14. so i get 1 over 14 okay as my simplified result now a more efficient way to do this if we think about this we would first make sure that each fraction okay in the multiplication problem is simplified already so that means no common factors other than 1. so if i look at 1 and 4 we're good to go if i look at 2 and 7 we're good to go so nothing can be canceled there then i want to look to cross cancel so in other words i want to look here and i want to look here anything i can cancel no so then i want to look here and i want to look here anything can cancel yes 2 and 4 are both divisible by 2. if i divide 2 by 2 i get 1 if i divide 4 by 2 i get 2. okay so now once you've cancelled everything you can go through and multiply and you're going to get a simplified result so 1 times 1 now would give me 1 and 2 times 7 would give me 14 so now i have 1 14 and it's already simplified so let's use the same strategy now when we're multiplying these rational expressions we're going to think about simplifying each rational expression first and then we're going to look to cross cancel and once we've done both of those steps then we want to move on and do some actual multiplication if it's necessary in a lot of cases so much stuff cancels you don't even need to multiply now before i actually do this problem i want to make sure everybody is on the same page in terms of what can be cancelled and what can't be cancelled this is quite confusing when you get into algebra so if we have something like let's say 3 over 12 we know that we could write this as 3 over 3 times 2 times 2 okay and then i can cancel this with this right i'm basically just saying 3 over 3 is 1 so i can cancel it out right so this ends up being just 1 over 4. now if you saw something like let's say 4 plus 3 over let's say 3 you should know but you might not know that you cannot cancel this 3 with this 3 like this this is a very common error in algebra you can only cancel common factors if this was 4 times 3 like this then you could cancel okay but if it's 4 plus 3 or 4 minus 3 you can't cancel think about this 4 plus 3 is what it's 7 7 over 3 is not the same as if you cancel this and just got four okay so please don't make that mistake now let's take this up a notch and think about this with something like this let's say i had x minus 1 this quantity times let's say x plus 5 this quantity and this was over let's say x minus 1 again that quantity and then times let's say x plus 6 this quantity now a lot of people get confused here because you see there's a minus sign a plus sign a plus sign a minus sign and then when i start canceling stuff they say why can you cancel there these are factors here okay this quantity here of x minus 1 is being multiplied by this quantity x plus 5. so these guys here are what we call factors okay they're being multiplied together and so i can cancel common factors so between the numerator and denominator i have a common factor of x minus 1 so it can be cancelled or removed one way to see this for sure let me just erase this real quick if you thought about this you could plug something in for x you could just pick a number and plug it in let's say you picked three for example well instead of x there i would have a three instead of x there i would have a three instead of x there i would have a three instead of x there i would have a three okay so three minus one would be two so let's say we have two there and then three plus five would be eight so now we have two times eight here we'd have three minus one which is two okay we have two and here we have three plus six which is nine so now it should be crystal clear to you that you can cancel this two with this two right but where did that come from that came from that common factor of x minus 1. so i hope that clears things up for you just pay close attention to what you're trying to cancel it needs to be a common factor okay so with that being said let's look at this first problem we have 3x squared minus 2x minus 8 over 3x squared plus 14x plus 8. this is multiplied by 3x plus 2 over 3x plus 4. so again what i want to do is think about factoring everything simplifying between the rational expressions themselves and then also trying to cross cancel anything i can before i actually do any multiplication so this guy right here and this guy right here those are prime you can't do anything with them but for this one and this one you can factor those okay so i'm going to use reverse foil here if you're more comfortable with the ac method that's fine too so let's go ahead and set up some parentheses here i'm going to do 3x and x because this is a prime number okay so that can only come from 3x and x so all i have to do is work out what's going to go here and here now i have 8 there forget about the fact that it's negative 8 for a moment just think about 8. so 8 is 1 times 8 and it's 2 times 4. so 1 and 8 there's no way to make that work but with 2 and 4 if you think about it let's say you put a 4 here and a 2 here well the outer would be 12x and the inner would be 2x and don't worry about the signs just yet this is not going to work right 12x and 2x no matter how you play with the signs that's not going to work so let's think about the other configuration let's flip this let's put a 4 here and a 2 here well now this would be 6x and this would be 4x and i could make that work right if i made this negative and this positive well this would give me my negative 2x and also if i had a positive and a negative being multiplied together i get my negative okay so what i want here is a negative here and a positive here okay so that would give me my 3x squared minus 2x minus 8. okay so moving on to the next one let's go ahead and put our fraction bar we have this guy here so i'm going to factor it in a similar way i'm going to do 3x and i'm going to do x and again what's going to go here well i have 8 once again so it's only 1 and 8 or 2 and 4. so could i make this work with 1 and 8. well if i put 8 there i would have 24x that's going to be a little bit too big if i put 8 here and 1 here that's going to be too small so 1 and 8 you can throw those out that's not going to work so if i put 4 here i would have 12x and then if i put 2 here i would have 2x so that looks like that would work right i want plus and plus right so the outer would be 12x and the inner would be plus 2x so that would combine to give me the 14x and again 2 times 4 is 8. so we're good to go there so then let's multiply this by three x plus two over three x plus four so once you've factored everything you're pretty much done okay this is the hardest part or the most tedious part if you're good at factoring it's not hard at all so let's come down here and see what we can cancel well i have a common factor of 3x plus 2 here with here okay so if you're confused by this you can always wrap this in parentheses and multiply it by one okay you don't have to do that i'm just doing it to show you this so you can cancel this with this those are again common factors then i'm looking at this 3x plus 4 and this 3x plus 4. so this cancels with this so this right here is a big fat 1 now it's completely cancelled so all i'm left with if you want to solve this problem is just take the x minus 2 over the x plus 4 and that's your simplified answer okay you don't need to do anything else things cancel so you didn't even have to multiply okay let's look at a more tedious problem it's not harder it's just more tedious so we have x squared plus 9xy plus 14y squared over x squared plus 7xy plus 10y squared this is now multiplied by x squared plus 2xy minus 15y squared over x squared minus xy minus 6y squared okay so again the first thing you want to do is factor everything and just try to cancel as much as you can and if you have to you're going to multiply so what i'll do here remember these are meant to look scary but they're actually not so if you have an x squared here and a y squared here just put an x and an x and a y and a y right x times x is x squared y times y is y squared then forget about it i'm going to do the same thing down here so x y x y and if you want to you can do this over here as well so let's go ahead and set this up so let's put these here and i'll do x x x x y y y y and i'm sure those aren't going to be conducive to me fitting everything so let me slide that down just a little bit more and that might be too wide but we'll adjust it as we need to okay so for this one right here it's really easy everything in terms of the leading coefficient is one so i'm just looking for two integers that sum to nine and give me a product of 14 so that's seven and two right so plus seven and plus 2. okay again because you figured out that the y was there you only have to work out this and this that's it so here i've got 7 and i've got 10. so it's got to multiply to 10 and sum to 7 that's 5 and 2. that's really easy so 5 and 2. here it's got to sum to 2 and give me a product of negative 15. so that's going to be positive 5 and negative 3. so positive 5 and this isn't going to fit so let's put x plus 5y and then x minus 3y so positive 5 and negative 3. okay so for this one we have negative 1 you could think of it as a negative 6. so the product is going to be negative 6 the sum is going to be negative 1 so that's going to be negative 3 and positive 2. so negative 3 and positive 2. all right let's see what we can cancel so we have x plus 5y here that can be canceled with x plus 5y here again we're canceling common factors so things wrapped in parentheses that's what we're looking for x plus 2y here can be canceled with x plus 2y here and then x minus 3y here can be canceled with x minus 3y here okay so basically all you're left with is an x plus 7y let me scroll down get some room so i can fit this i'm going to put is equal to i just have x plus 7y that's my numerator for the denominator i'm just left with this x plus 2y so again in a lot of these cases you don't even need to multiply because so much stuff's going to cancel okay so we've done two with multiplication let's switch now to division which is just as easy so basically we remember if we're dividing fractions we keep the first fraction the same we take the reciprocal of the second fraction and then we multiply so this guy is 6 over 11 so the first fraction stays the same change the division to multiplication flip this guy right so the reciprocal this goes up here this comes down here so i would basically have 33 over 4 okay you can already see that this would cancel with this and give me a 3. this would cancel with this so 6 divided by 2 is 3 and 4 divided by 2 is 2. so this is going to give me what it's going to give me 3 times 3 which is 9 over just 2. right because this is 1 it's been canceled so this is 1 times 2 or just 2. so the answer there is just 9 halves very simple process and we're going to do the same thing here so we'll look at the x squared minus 9x plus 14 over 2x squared plus 3x minus 5 divided by x squared minus 2x minus 35 over x squared minus 4x plus 3. so however you want to do this the first thing i'm going to do is just factor this one on the left so i know that this is what this is negative 9 and positive 14. so i need two integers that sum to negative 9 give me a product of positive 14 that's going to be negative 7 and negative 2. so x minus 7 and then x minus 2. okay so i've got that one this one right here is a little bit harder to factor because this guy is not one but it's a prime number so it's pretty quick all right so i'm going to put 2x here and x here now this last term here is negative 5. i'm just going to think about 5 for a second 5 is only 1 and 5. okay it's 5 times 1 or 1 times 5 however you want to think about that okay so the only possibilities are to have and forget about the signs for a moment a 5 here and a 1 here or to reverse that if i multiply 2x times 5 i'm going to get 10x that is too big so it's not possible so we'd have to put a 5 here and a 1 here because this would be 5x okay and then this would be 2x so that would work i just got to get the sign right and basically i want 5x to be the larger one so i want a plus sign here and a minus here right so the outer would be negative 2x and the inner would be positive 5x so this would be your correct factorization now what i'm going to do here is multiply it by i'm going to flip this guy so this is going to be my numerator okay my denominator becomes my numerator so let's factor this guy we'd have an x here and an x here and so what i would do is look for two integers whose sum is negative four and whose product is three so you can go ahead and do negative three and negative 1. so minus 3 and minus 1 and then this is over again this numerator here is now going to be the denominator here so i'm going to factor this guy so give me two integers i'm going to put my x here max here two integers whose sum is negative two and whose product is negative thirty-five well you could do negative seven and positive five so negative seven and positive five okay so let's scroll down now again all you have to do is just flip the second fraction or flip the second rational expression everything else is just the same right you're just multiplying so you're looking to just cancel stuff so i can cancel this guy x minus 7 with this guy x minus 7 those are common factors i can cancel this factor of x minus 1 with this factor of x minus 1 and then can i cancel anything else doesn't look like it so in this case we actually need to multiply now i'm going to tell you two different ways you can do this in every case you are allowed to just say something like x minus 2 times x minus 3. so those two quantities multiplied together and leave it as it is okay some teachers will tell you to multiply the numerator out and leave the denominator factored some want you to have everything factored you just have to check with your teacher to see what's required down here i'm going to do this 2x plus 5 that quantity times x plus 5 okay and just for the sake of completeness let's go through the process and we're going to multiply everything so x times x is x squared the outer would be minus 3x the inner would be minus 2x and the last would be plus 6. so this would be minus 5x here so minus 5x and then plus 6. this is over the first would be 2x squared the outer would be plus 10x the inner would be plus 5x and the last would be plus 25. so if you combine the like terms you'd have plus 15x here so basically you could write this like this in the factored form or like this where you've multiplied it out either way is an acceptable answer all right let's look at one that's very very tedious this is just a lot to do with factoring and basically you have to be good at factoring to be good at this guy so when i look at this problem i see multiplication and then i see division okay so basically you want to turn this into one long multiplication problem you're just going to flip this guy right here okay so basically what i'm going to do i'm going to try to fit this on the screen so x squared minus y squared we know this is the difference of two squares so this is x plus y times x minus y okay so you can factor that then this guy right here this x minus y quantity squared remember i can't say this enough you cannot distribute that in you have to expand this this is the same as having x minus y times x minus y okay now it turns out that if you run through this you're going to get this right so this is the expanded form and this guy is the factored form really you'd write it like this okay so let's see what we need to do with this before we expand it we'll put x minus y i'm just going to put quantity squared so i'm just going to leave it as it is okay then this guy right here up here we cannot factor this this is negative 1 and this is positive one okay so can you give me two integers that sum to negative one and have a product of positive one no you cannot right it's not possible so i'm going to leave this one up here as is so x squared minus x y plus y squared and this is over this guy down here which we already know is going to factor into this right if you expand this you square the first guy so that's how i get x squared then it's minus 2 times the first guy times the second guy so i get minus 2 times x times y then it's plus the last i squared okay so this factors into x minus y quantity squared then i'm going to multiply by i'm going to flip this guy now this guy right here is the sum of cubes so this guy is something you should remember how to factor it's really convenient to remember how to factor it i don't know if i can fit this all on the screen as of right now so let me just write this in unfactored form and then i'm going to go to the next screen down and we'll see if we can get this on here okay and this guy right here this is x minus y to the fourth power so we're going to be able to cancel some stuff right now that's going to give me some space so you'll see here that you have x minus y this quantity squared basically multiplied by x minus y this quantity squared so you could write this out as x minus y times x minus y times x minus y times x minus y okay and you could write this out as the same thing realize your rules of exponents you have the same base here this is your base okay this is your base so really this is x minus y to the fourth power okay so those guys combined would cancel with this completely this would cancel with this and this okay again if you wanted to write this part as x minus y and this says x minus y sorry x minus y times x minus y in each case then you'd see that you have four of those and you could expand that out to exactly this and it would all cancel okay so that is gone now the next part here to go any further let me scroll down a little bit and get some room i'm going to write x plus y times x minus y and then times you have this x squared minus x y plus y squared and then i'll put times you'll have this one over let's go ahead and factor this now so from my special factoring formulas this is going to be x plus y that's the first one then times it's going to be exactly this so x squared minus x y and then plus y squared so now what i'm going to do i'm going to think about this again as a factor so i'm going to cancel this with this and then i'm going to cancel this with this and really all i'm left with now is this x minus y okay so you can see on this type of problem they just want you to really know how to factor in order to go through and just cancel cancel cancel until you end up with something very simple with just x minus y as your answer in this lesson we want to review finding the least common denominator which is abbreviated as the lcd for rational expressions so we should know at this point that again working with rational expressions we follow the same rules that we learned when we work with fractions if we want to add or subtract fractions we know we have to have a common denominator well it's the same thing when we work with rational expressions so in the next review session we're going to talk about adding and subtracting rational expressions so before we kind of jump into that we want to review how to find the lcd right the least common denominator all right so the lcd when we work with fractions is found in what way okay let's just refresh our memory if we have 1 3 4 over 21 and 5 over 36 again the lcd is the lcm or least common multiple of the denominators so 3 21 and 36 okay so you might be really confused at this point because you hear all these different acronyms right you hear gcf you hear gcd you hear lcm you hear lcd so what are all these things and why does everyone get them confused well the gcf and the gcd are the same okay they they mean the exact same thing one is greatest common factor one is greatest common divisor but they really are just referring to the largest number that each number of a group is divisible by meaning i can take each number in that group and divide it by the gcf or the gcd and not get a remainder okay so that's what those are when i think about the lcm it's the least common multiple so the lcm is the smallest multiple of every number in a group okay then the lcd is just going to be the lcm of the denominators okay the lcd is specific to when we're talking about fractions whereas the lcm can refer to any group of numbers that you want to talk about okay so when we look for the lcm of 3 21 and 36 we build it from the prime factorizations of those numbers so 3 does not factor okay 21 is 3 times 7 and 36 is what it's 4 times 9 so it's 2 squared times 9 which is 3 squared so to build the lcm i take each prime factor okay each one and i throw it in the list okay these numbers in this list are going to get multiplied together there's an exception if there's a duplicate prime factor so a prime factor that occurs in more than one prime factorization all i want to do is put the largest number of repeats or the largest number of occurrences if you want to say it that way that will appear in any of the prime factorizations so in other words i have a three here a three here and two copies of three or three squared here so because it's in more than one i go with the largest number of repeats or the largest number of occurrences so when i build my lcm i just put in two okay don't make the mistake of putting in four a lot of people say okay i got a three here or three here and then two threes there so i'm putting in three times three times 3 times 3 no largest number of repeats between any of the prime factorizations is all you need then i have a 2 squared i don't have it anywhere else so i just put in 2 factors of 2. a very simple then i have a seven i don't have that anywhere else so i just throw in a seven very very simple to understand again the only place where it can get confusing is if you have a duplicate factor and again you go with the largest number of repeats in any of the factorizations okay so now i just multiply the numbers on the list and what would that give me so 3 times 3 is 9 9 times 2 is 18 18 times 2 is 36 36 times 7 is 252. so my lcm for the numbers 3 21 and 36 is 252. so this is the smallest multiple for the group of numbers right so the smallest multiple that is a common multiple for 3 21 and 36 now we can say that the lcd because we're working with fractions here right and these are denominators so the least common denominator is going to be equal to 252 okay but if we were just working with the numbers 3 21 and 36 we would not say the lcd we would say the lcm because we're just talking about a group of numbers again because these are fractions we can say the lcd because we are specifically talking about the least common multiple for the denominators okay so i want to make sure that's clear for you all right so let's look at a few examples here we're going to make this a very quick lesson so we have 5x squared minus 1 over x squared minus 1 and we have 2x minus 1 over x squared plus 2x minus 3. don't worry about the numerators at all we're just looking at the denominators and all we want to do with the rational expression is just factor okay just like we did with numbers so what is x squared minus 1 that factors into the difference of squares this is x plus 1 times x minus 1. what is x squared plus 2x minus 3. let's go ahead and factor that guy we know this is x and this is x if this guy is a negative 3 that can only come from 1 times negative 3 or negative 1 times positive 3. so it's going to have to be positive 3 and negative 1 right because the middle term is positive so looking at this guy what's common we have an x minus 1 here and an x minus 1 here so to build the lcd what do i need i'm going to throw in everything right so x plus 1 goes in x minus 1 is here and also here between the two factorizations i've got a maximum of one so when i build my lcd i only need to put in one copy not two i realize it's in both but if it's duplicate you just put in the largest number of occurrences in this case it's only one okay please understand that then i've got my x plus three okay so the lcd here is x plus one times x minus one times x plus three so that's all it is all right let's look at another one so we have two x squared minus five x minus one over x squared plus x minus thirty we have three x squared minus x minus one over x squared minus ten x plus 25. so again all i'm looking to do is just factor the denominators here so if i factored this guy what would i have x here and here two integers that sum to 1 and give me a product of negative 30 well that would give me positive 6 and negative 5 right the hour would be negative 5x the it would be plus 6x so that would give me my positive 1x then this guy over here is a perfect square trinomial how do i know that i've got something squared and something squared right 25 is 5 squared so if i wrote this as x which is what's being squared you got a minus there so minus what's squared here it's 5. so x minus 5 squared right you can check that middle term 2 times x times 5 2 times 5 is 10 10 times x is 10x so you have minus 10x there so we're good to go now i'll probably want to expand this so it's crystal clear for you so now what is the lcd what is the lcd well i've got x plus 6 doesn't occur over here i've got x minus 5 and that does occur over here so i've got one here and i've got two here how many am i putting in please don't answer three the largest number of repeats between this factorization and this factorization occurs here i've got two of them so just put two in when you build your lcd okay so i don't need three i can't stress that enough so the lcd is gonna be x plus six then we're multiplying this by x minus five and then multiplying this by another x minus five or to make this simpler you can just write the x minus 5 part as squared all right let's just look at one more of these i think it's very straightforward and then in the next lesson we're going to review how to add and subtract rational expressions and we're going to use this kind of information or this knowledge right away so we have 12x minus 5 over 3x squared minus 20x minus 32. we have 2x squared minus 7x minus 5 over 3x squared minus 2x minus 8. we have 5x squared minus 1 over x squared minus 4. so what's the lcd what is the lcd so let's start by factoring this guy so i know i would have a 3x here and an x here so i know i have a negative and a negative so that means i have different signs and i'm looking for a product of 32. forget about the fact that it's negative so what can give us a product of 32 we've got 1 and 32 there's absolutely no way to make that work you've got 2 and 16 that's not going to work then you've got 4 and 8 and we can make that work if i put a 4 here and i put an 8 here the outer would be negative 24x and the inner would be plus 4x so that would give me negative 20x that's what we're looking for for this guy right here i can factor this so we would have 3x here and x here so i'm thinking about what i've got a negative eight and so since this guy is negative also i'm gonna have alternate signs so plus and minus or minus and plus if you like it that way it doesn't matter think about the factors of eight you've got one and eight or four and two so would one and eight work no right there's no way to make that work so you would have to go to 4 and 2 and the correct combination would be to put a 4 here and a 2 here so your outer would be negative 6x and your inner would be plus 4x so that guy's factored so now down here we have x squared minus 4 so this factors into x plus 2 times x minus 2 right this is basically x squared minus 2 squared right it's the difference of two squares okay so now that we've got everything factored we want to build our lcd and so we've got a 3x plus 4 so let's put that in there now it's in this guy and it's in this guy so the largest number of repeats between any of the factorizations is one so it just goes in once then i'm going to have my x minus 8 that's not in any other factorization i'm going to have my x minus 2 which occurs here and here but only once in each so i only need to put 1 in and then i have my x plus 2. okay so for the lcd we have 3x plus 4 this is multiplied by you're going to have x minus 8 this is multiplied by x minus 2 this is multiplied by x plus 2. in this lesson we want to review adding and subtracting rational expressions so again before we kind of jump in and look at some examples with rational expressions let's refresh our memory by just working with some fractions right the rules are the same so if we're adding fractions or subtracting fractions and we already have a common denominator we just work with the numerators and we put the result over the common denominator and then we simplify if we can so if i have 1 7 plus 4 7 well then all i need to do is add 1 plus 4 because the denominator is the same so 1 plus 4 is 5. put that result over the common denominator of 7 so we get 5 7. now in this particular case there's nothing to simplify right there's no common factor other than 1 between 5 and 7 so we report our result as just 5 7. now in the case where we don't have a common denominator we need to first get a common denominator now the easiest denominator to work with is the least common denominator in the last review session we talked about how to find the least common denominator if you're a little bit confused about that process please review that video before you move forward but it's a very simple thing essentially if i have 5 9 plus 2 15 to find the lcd i want the least common multiple of the denominators so i want the lcm of 9 and 15 and i just find that by factoring each right so 9 is 3 times 3 and 15 is 3 times 5. and all i want to do is put in each prime factor right from each of the factorizations but i want to make an exception here if i have something that's duplicated between more than one prime factorization i only want to go with the largest number of repeats or the largest number of occurrences in any of the prime factorizations so for the prime factor of 3 you see that you have 2 of them here right when you have the prime factorization of 9 you only have one of them here so i only need to put two in when i build my lcm i don't need three right i just go with the largest number of repeats which is this right here where i have two of them then i have a five here it doesn't occur in the prime factorization of 9 so i just throw that in there so 3 times 3 is 9 9 times 5 is 45. so the lcd or the least common denominator is going to be 45. now what do we do with this information right we don't just get 45 and then turn in the test so the lcd of 45 is used to rewrite each fraction as an equivalent fraction where 45 is its denominator so i'm going to say this is equal to you have 5 9 times what do i need to multiply 9 by to get to 45 well 45 divided by 9 is 5 so i'd multiply this by 5 over 5 right remember you're just multiplying by a complicated form of 1 there then plus you have 2 15. what do i need to multiply 15 by to get to 45 45 divided by 15 is 3 so times 3 over 3. okay so now 5 times 5 is 25 and this is of course over 45 and plus 2 times 3 is 6 and this is of course over 45 and now it's just like the last example we have a common denominator so we just work with the numerator so 25 plus 6 is going to be 31 so you get 31 over the common denominator of 45. now 31 over 45 is already simplified so this is our answer so now let's use the same thought process and do some addition with some rational expressions so we start out with 7x over 5x plus 6 and then plus 2 over x minus 2. so if i look at my denominators to start i have 5x plus 6 nothing i can do to factor and then x minus 2 nothing i can do to factor so if neither one of those is going to factor the lcd is just going to be the product of the 2. so the lcd will be 5x plus 6 multiplied by x minus 2. so to get a denominator here of 5x plus 6 times x minus 2 i just multiply by x minus 2 right so multiply this by x minus 2 but i've got to also do that to the numerator so that it's legal okay so all i'm doing if i wanted to break this up to make it crystal clear i'm just multiplying again by a complicated form of one x minus two over x minus two this can cancel and just be one right it's a complicated form of 1 but it's still like multiplying by the number 1 i'm not changing anything okay so let's redo this so this is times x minus 2 this is times x minus 2 plus then for this one it's the same thing so we have 2 over x minus 2 x minus 2 needs to be multiplied by 5x plus 6 and so i've got to do that to the numerator as well and then i'm good to go so now i just need to do some simplifying let's scroll down so i'm going to use my distributive property to remove the parentheses in the numerator so 7x times x is 7x squared then 7x times negative 2 is minus 14x and plus 2 times 5x is 10x then 2 times 6 is 12. so plus 12. now we have a common denominator here so don't even worry about it we're just going to write it over the common denominator so we're just going to write this over 5x plus 6 and then multiplied by x minus 2. don't even think about the denominator until the end we're just going to work with the numerator so 7x squared nothing to combine with that you have negative 14x plus 10x so that's going to be minus 4x and then you have plus 12 and again this is over the common denominator of you have 5x plus 6 multiplied by you have x minus 2. now i can simplify this if i can factor the numerator and i can cancel one of those factors with something in the denominator right so is 7x squared minus 4x plus 12 something we can even factor well let's think about it so we'd have 7x and we'd have x and i know that i would need what i would need a last term of 12. so if this is positive 12 and this is negative this would have to be negative and this would have to be negative so could it work with 1 and 12 could i do a negative 1 and a negative 12 well the outer and the inner would not work out what about if i had a 12 here and a 1 here would that work no it would not right the outer and the inner would still not work out if 1 and 12 don't work you also have 2 and 6 but you can see obviously that wouldn't work either right 2 and 6 the outer is just too large in terms of absolute value and if you flip it around it's the same and then lastly you have 3 and 4 but that's not going to work either right you can put 3 here and 4 here or you could switch to 2. either way it just doesn't work right they're going to be too large in terms of absolute value to ever work out so this guy right here is a prime polynomial 7x squared minus 4x plus 12 is not going to be able to be factored using rational numbers so this is as simple as we can make our answer we have 7x squared minus 4x plus 12 over the quantity 5x plus 6 times the quantity x minus 2 and we leave our denominator in factored form just to show that we can't cancel anything else i'm just going to leave it that way all right what about 5x over x plus 5 minus 2 over x plus 8. so again we have a scenario where we have x plus 5 that's not going to factor and x plus 8 that's not going to factor so my lcd will just be the product of the two denominators so x plus 5 multiplied by x plus 8 okay very simple so what am i going to do here i'll have 5x multiplied by the x plus 8. and this is over you'll have x plus 5 times x plus 8. and then we have minus you have 2 that's going to be multiplied by x plus 5 that quantity and then this is over you have x plus five times x plus eight okay the main thing here or the main mistake that you might make ensure that you use parentheses around these guys because you've got to distribute everything okay very important to remember that otherwise you're going to get the wrong answer so let me put equals here and equals here and let's kind of continue so 5x times x is 5x squared 5x times 8 is plus 40x now you have minus here so let me put minus and i'm going to put some brackets here because when you subtract something away you've got to subtract away each term okay so let's just go ahead and distribute 2 times x is 2x and then 2 times 5 is 10. so now i can remove these brackets here by changing the sign of each term a lot of times you see people put this plus negative 1 to remind themselves to distribute the negative 1 to each term or just to change the sign so this would be minus 2x and then minus 10. so minus 2x and minus 10 and this is over my common denominator of x plus 5 multiplied by x plus 8. [Music] okay let's get some room going so we have 5x squared nothing to combine with that 40x minus 2x is 38x and then you have minus 10 and this is over you have x plus 5 this quantity times x plus 8 this quantity so this is as simple as we can make this this guy right here i spent the time to show you that the kind of first example the numerator couldn't be factored using rational numbers it's going to be the same thing with this one this is a prime polynomial the 5x squared plus 38x minus 10. so this is as simple as we can make this all right for the last one we have 7x over 5x plus 5 and plus we have 3x over x squared minus 1 and then minus 2 over x minus 1. so what we want to do again is first find the lcd so if i factor the denominator here i get five times the quantity x plus one if i factor the denominator here this is the difference of two squares this would be x plus one times x minus one and then this guy is not going to factor it's just gonna be x minus 1. okay so to build the lcd i have a 5 here no 5 and n in the other one so i'll put a 5 down i have an x plus 1 here i also have an x plus 1 here but between the two factorizations there's only one copy in each so i only need to put one copy in my lcd then i have an x minus one and again between these two factorizations there's only one copy so i only need to put one copy in when i build my lcd so now let's go ahead and write each of these with their denominator and factored form so this is 7x this is plus this is 3x this is minus this is 2. okay so what would i need in this case to get a denominator of 5 times the quantity x plus 1 times x minus 1 well i would need to multiply this by x minus 1 over x minus 1. so let me kind of slide this down and let me get a little room going here so i'm going to put 7x times i need to multiply by x minus 1 and this is over 5 times the quantity x plus 1 times the quantity x minus 1. and plus for this one i'm missing a 5. right i have the x plus 1 part i have the x minus 1 part i just need the 5. so 3x times 5 this is over 5 times the quantity x plus 1 times the quantity x minus 1 and minus over here i'm missing the 5 and the x plus 1. so 2 times 5 times the quantity x plus 1. then this is over again 5 times the quantity x plus 1 times the quantity x minus 1. okay so from this point on it's very simple let's kind of scroll down get some room going so we have 7x times x which is 7x squared then 7x times negative 1 is minus 7x and then plus 3x times 5 is 15x now you have this minus here whenever you're subtracting away something like this again just put a bracket out in front just to remind yourself that you've got to subtract away everything right you're subtracting away each term from this guy so in other words you're going to have 2 times 5 which is 10. i'm going to just write this as 10. so you're going to have 10 times x which is 10x and then 10 times 1 which is 10 so you've got to subtract away the 10x and the 10. so in other words you can put plus negative 1 here and distribute this to each term to remind yourself to remove the brackets and change the sign of each term so this first guy would be negative 10x and this guy would just be negative 10. okay so this is all over again that common denominator of 5 times the quantity x plus 1 times the quantity x minus 1. okay all right so now we have 7x squared and nothing to combine with that we have negative 7x plus 15x which is going to give us 8x and then minus 10x which is negative 2x and then you have minus 10 okay and this is over you have 5 times the quantity x plus 1 times the quantity x minus 1. now this is as simple as you could make this because again this trinomial here is prime right you're not going to be able to factor it using rational numbers in this lesson we want to review complex rational expressions so back when we studied fractions we learned about complex fractions complex fractions have another fraction in their numerator or it could be in their denominator or it could be both in the numerator and denominator so when we come across the topic of algebraic fractions or rational expressions we get the same thing right we just call them complex rational expressions but they follow the same rules that we learned again with complex fractions so to simplify them is very very simple there's basically two different methods you can use one that i would consider to be a little bit slower and one that's a lot faster so if i have something like 6 minus 5 over x and this is over we have 1 plus 5 over x let's first just identify the numerator of this complex rational expression just so we can refer to it so this is our numerator and this guy down here will be our denominator so you can think about this part right here this fraction bar is our main division okay so we've got our numerator our 6 minus 5 over x we've got our main division or our main fraction bar and then we've got our denominator our 1 plus 5 over x so the first method to simplify this kind of the slower way to do this is to simplify the numerator and the denominator separately and then perform the main division so in other words i can say 6 minus 5 over x is equal to what well if i get a common denominator here i would just multiply 6 by x over x so 6 times x over x minus 5 over x would give me what this would be 6x minus 5 over the common denominator of x so that's my simplified numerator so let's erase this and we'll kind of drag this up here now for my simplified denominator we have 1 plus 5 over x so 1 plus 5 over x to get a common denominator i would just multiply 1 by x over x so we know that would just be x over x and then i would basically have x plus 5 over the common denominator of x okay so let's write x plus 5 over the common denominator of x so what do we have here now we have 6x minus 5 over x divided by x plus 5 over x so essentially because this is the main division again i've already labeled that you could really write this as 6x minus 5 over x divided by x plus 5 over x so that makes it crystal clear what you need to do you need to take this guy right here and multiply it by the reciprocal of this guy right here okay so i would say this is equal to let me kind of scroll down just a little bit we'd have 6x minus 5 over x times the reciprocal of this which is x over x plus 5. and what's going to happen is i can cross cancel this x with this x and i'm just left with 6x minus five over x plus five so six x minus five over x plus five so that is the simplified version of this guy right here so that's method one that is kind of the slower way to do things the faster way is to multiply the numerator and denominator of the complex rational expression by the lcd of all fractions involved so this one's pretty simple if i look through i only have this fraction in this fraction and they're the same right so the denominator here is x the denominator here is also x so the lcd will just be x so essentially what i'm going to do let me just rewrite the problem down here so 6 minus 5 over x over 1 plus 5 over x i'm going to multiply this guy by x over x again that's legal because x over x is nothing more than just 1. so let me scroll down get some room going i'm going to use my distributive property here and i'm going to distribute the x to each term so what this would be equal to i would have 6 times x minus i would have 5 over x times x so we know this x would cancel with this x and i'd essentially have 6x minus 5. this is over i'm going to do the same thing over here i'm going to multiply this by this and by this so you would have 1 times x or just x and then plus you would have 5 over x times x we know that we can cancel this x with this x and you would just have a 5 there so look how quick we got the same answer we get 6x minus 5 over x plus 5. again the lcd method is much quicker in almost every scenario so for all the problems we're going to look at after this point we're just going to use that method all right let's go ahead and take a look at another one we have 1 3 minus 3 over x minus 2 and this is over x minus 2 over 36 and then minus we have x plus 6 over x minus 2. so again the idea here with the lcd method is to find the least common denominator of all the fractions involved so you have a 1 3 a 3 over x minus 2 an x minus 2 over 36 and an x plus 6 over x minus 2. so i've got a 3 an x minus 2 a 36 and an x minus 2. so what would be the lcd here again it's the lcm of these individual polynomials so if you looked at 3 x minus 2 36 and x minus 2 again so i know that 36 divided by 3 is 12. so essentially it would just be what it would be 36 times x minus 2. that would be the lcd so my lcd is 36 times the quantity x minus 2. okay very easy to find now all i want to do is just multiply the numerator and denominator by the lcd so 36 times the quantity x minus 2 over 36 times the quantity x minus 2. now this will be a little bit more tedious than the last example so when you do these expect to have a lot of kind of scratch work so again i'm going to distribute this to each term here so the 36 part multiplied by the one third part if i just do this off to the side you can see 36 would cancel with 3 and give me a 12 right so this would basically turn into a 1 this is a 12 so you'd have 12 times the quantity x minus 2. so we would have 12 times the quantity x minus 2 then minus next you'd have this guy times this guy so you can see that this x minus 2 here would cancel with this x minus 2 here okay so what that would give me is what just 36 times 3 which is 108 okay now obviously we can simplify this further but let's just go ahead and distribute down here first and then we'll come back so if this gets distributed to this the 36 here would cancel the 36 here you would have x minus 2 multiplied by x minus 2. so in other words you would have x minus 2 this quantity squared then you have minus if this multiplies by this the x minus 2 would cancel with the x minus 2 there and essentially what i would have is 36 times the quantity x plus 6. okay so now that we've set this up let's kind of scroll down and we're just going to simplify now so in my numerator 12 times x is 12x and then minus you have 12 times 2 12 times 2 is 24 so minus 24 and then you have minus 108 okay so i'm not done up here i can still simplify if i have negative 24 and i subtract away 108 i'm going to get negative 132. so i'm going to write minus 132 and then down here i have x minus 2 this quantity squared so i know that would be x squared minus you would have 2 times x times 2. so basically minus 4x and then plus you'd have 2 squared which is 4. then for this part right here we need to pay close attention to the sign i'm subtracting away this whole thing the easiest way to do this and not make a sign mistake is write this as plus negative 36 and distribute the negative 36 to each term there okay so negative 36 times x would be negative 36x and then negative 36 times 6 would be minus 216. okay so now we want to simplify here x squared i don't have anything i can combine with that negative 4x minus 36x is negative 40x so let's write this as minus 40x and then if we have 4 minus 216 that's going to be negative 212. so this is minus 212 and now we want to see if we can simplify by factoring the numerator and denominator and canceling any common factors so in the numerator i can see that i can factor out a 12 right 132 is 12 times 11. so i can pull out a 12 and i would have x minus 11 inside the parentheses down here can i factor this guy can i factor x squared minus 40x minus 212. well if i can i would set this up as x and x i would need two integers such that the sum is negative 40 and the product is negative 212. so what are the factors of just 212 forget about the signs so 1 and 212 you've got 2 times 106 and then you've also got 4 times 53 now 53 is a prime number so you really can't go any further than that so would any of these work now keep in mind that you have to have opposite signs because this guy is negative and this guy is negative so one of these would have to be plus and one of these would have to be minus well 1 and 212 no way you could make that work 2 and 106 no way you can make that work and then 4 and 53 also no way to make that work so this guy right here is a prime polynomial so it can't be factored using rational numbers so let's go ahead and erase that and we'll just write this over x squared minus 40x minus 212. so this is as simple as we can make this we'll just leave the numerator in factored form that's typically what we would do all right let's take a look at another one so we have one half plus x plus four over four this is over x minus five over two plus x plus four over x minus five so what's the lcd we have a 2 we have a 4 another 2 and an x minus 5. so between 4 and 2 we know it's going to be a 4 right 4 is 2 copies of 2. so in all the prime factorizations you would have just a two just a two and then two copies of two so that would win out and then you have this x minus five so the lcd would be four times the quantity x minus five so i'm going to be multiplying that by both the numerator and the denominator of the rational expression so i'm going to multiply this by this and also this by this so when i start out 4 times the quantity x minus 5 multiplied by one half the four would cancel with the two and this would become a two right so essentially i would have two times the quantity x minus five then plus when this multiplies this my four here would cancel with my 4 here so i would have x minus 5 that quantity multiplied by the quantity x plus 4. all right so down here i'm going to do the same thing i'm going to distribute so i'll put this over 4 times the quantity x minus 5 multiplied by x minus 5 over 2 the 4 would cancel with the 2 and give me a 2. so you would have 2 multiplied by the quantity x minus 5 squared right x minus 5 squared if you wanted to write x minus 5 times x minus 5 that's fine too then for this guy it would multiply in this particular case you would have the x minus 5 here cancel with the x minus 5 here and so you would have plus four times the quantity x plus four okay all right let's get some room going and now we just need to simplify so 2 times x is 2x then minus 2 times 5 is 10 then plus if we have x minus 5 this quantity multiplied by x plus 4 that quantity x times x is x squared the outer would be 4x the inner would be minus 5x so that's minus x then the last negative 5 times 4 is minus 20. so we would simplify this as x squared the 2x minus x would be plus x and negative 10 minus 20 is minus 30. okay let's erase this and we'll just drag this up all right then this is over for this guy right here i have two multiplied by the quantity x minus five that amount squared we follow our formula it's x squared minus 2 times x times 5 so minus 10x and then plus this guy squared which is 25. if i multiply 2 by each term here this would be 2x squared this would be minus 20x and this would be plus 50. okay then plus you have 4 times x which is 4x then plus you have 4 times 4 which is 16. so what can i do here you have 2x squared nothing to combine with that you've got negative 20x and 4x that would give me negative 16x then you've got 50 and 16 that would give me 66. kind of slide this down so it matches up okay so what can we do here to factor in the numerator i need two integers that are going to sum to 1 and give me a product of negative 30 that would be positive 6 and negative 5. in the denominator i know i can pull out a 2 to start so that would give me x squared minus 8x plus 33. now if we want to factor the inside here the x squared minus 8x plus 33 we would need what let me just write that over here we would have x and x the middle term is negative and the final term is positive so you would need a negative and a negative now is there any way to get that with the last term of 33 no there isn't because the factors of 33 would be 1 in 33 or it could be 3 and 11. now you might look at 3 and 11 and say oh if i had a negative 11 and a positive 3 but the signs have to both be negative so that doesn't work right you could only do a 3 here and 11 here both are negative that will combine to give you a negative 14x so that wouldn't work right so this guy's going to be a prime polynomial so this is as simple as we can make this we have our quantity x plus 6 times our quantity x minus 5. this is over 2 times the quantity x squared minus 8x plus 33. let's take a look at one more of these so we have x minus two over y plus one then we have minus we have x plus four over y plus one this is over we have x plus one over x plus four then minus x minus two over x plus four so in this case we have a denominator of y plus one and we have a denominator of y plus one a denominator of x plus four and a denominator of x plus four so the lcd would be x plus four times y plus one okay and again i'm just going to distribute this to each term here so for this guy the y plus one would cancel with the y plus one and i would be left with x plus four multiplied by x minus two and then minus when i distribute it here the y plus one would again cancel with the y plus 1. so now i would have x plus 4 this quantity squared this is over down here when this gets multiplied in each case the x plus 4 is going to cancel right so i would just have the x plus 1 multiplied by the y plus 1 and then minus again this is going to cancel with this so you would have your x minus 2 multiplied by your y plus 1. all right so to simplify in the numerator we'd have x times x which is x squared the outer would be minus 2x the inner would be plus 4x so this is plus 2x the last would be minus 8. then this is minus for this guy right here you've got to be very careful you're subtracting away this whole thing you've got that minus out in front there's no way to really distribute that yet until you kind of do your foil on this so go ahead and put a bracket here and then foil this out or use your special products formula this is x plus 4 squared so we would have x squared then plus 2 times x times 4 so plus 8x and then plus 4 squared is 16. so now i can subtract each one of these away right so you think about that as just distributing the negative to each term a lot of people do plus negative one here and then again they just distribute this to each term just to make it crystal clear what's going on so i'm going to erase all this and i'm just going to change the sign of each term so this is minus this is minus and this is minus so x squared minus x squared is 0. so that's gone 2x minus 8x is going to give me negative 6x so this would be negative 6x over here and then negative 8 minus 16 is minus 24. so negative 6x minus 24. then this is over for this guy right here you have x times y which is x y you have an outer which is plus x you have an inner which is plus y and then you have a last which is plus one then again i'm minusing this so put a minus put some brackets go ahead and do your foil and then again distribute the negative to each term this whole thing is being subtracted away very important you understand that so x times y is x y the outer would be x times one which is just plus x the inner would be negative two times y which is minus two y and the last would be negative two times one which is minus two so what can we do here to simplify i'm gonna distribute this negative to each term so essentially i'm just going to change the sign of each term so this will become negative this is negative this is positive this is positive okay again i can't stress this enough very important you do that otherwise you will get the wrong answer so xy minus xy is zero so those are gone you have x minus x that's 0. so those are gone you have y plus 2y which is 3y so we can erase those and you have 1 plus 2 which is 3. so what i want to do now is try to see what i can cancel between the numerator and denominator i'm just going to factor this guy so in the numerator i can pull out a negative 6 or i could pull out a 6. let's just go ahead and pull out a negative 6 and that's going to leave me with an x plus 4 inside of the parentheses then in my denominator i have 3y plus 3 i can pull out a 3 there and i'd be left with a y plus 1 inside the parentheses so now i can see that i can cancel between this negative 6 and this 3 right i can cancel this with this 6 divided by 3 is 2. so this is basically a negative 2 right there so i can write this as negative two times the quantity x plus four over you could write the quantity y plus one in parentheses or just y plus one it doesn't really matter so we see we have a simplified answer here nothing else we can cancel again you've got negative 2 multiplied by the quantity x plus 4 over y plus 1. in this lesson we want to review negative exponents and the quotient rule for exponents so not that long ago in one of our previous review sessions we reviewed some rules of exponents right we talked about the power rules for exponents and we also talked about the product rule for exponents and we learned about an exponent of zero right we learned that if we took a non-zero number and we raised it to the power of 0 we always end up with the number 1. so today we're going to start off by looking at negative exponents and then we'll move into the quotient rule for exponents and you'll see that we're going to need these rules as we start to review radical expressions okay so with negative exponents we're going to look at this using a pattern and it's the same pattern we use to show how we get a non-zero number raised to the power of zero to be equal to one so i like to work with the number two it's nice and small if you're kind of teaching exponents so if i started out with something like two to the first power we know this is 2. so let's just start here then 2 squared is what well i could take 2 and i could multiply it by 2 right because i have 2 factors of 2 now and that would be 4. so to increase the exponent by 1 to go from 1 to 2 i take this number here and i multiply by 2. let's do that again if i now go to 2 cubed what do i do i can take this guy right here this 4 and i can multiply by 2 i can multiply by 2 and i'll get to 8. 4 times 2 is 8 so 2 cubed is 8. okay very very easy to understand that now if we reverse the process let me kind of erase what i've put here if i want to decrease my exponent by 1 starting at the 3 if i want to go down to 2 i would take this 8 the result from 2 cubed and i would basically divide it by 2. so 8 divided by 2 gives me 4. then if i wanted to go from 2 squared to 2 to the first power again i'd start at 4 and i would just divide by 2. right 4 divided by 2 would give me 2. so following this same pattern when we get to 2 to the power of 0 right 1 minus 1 is 0. all i'm going to do is again just divide by 2. so 2 divided by 2 gives me 1. okay this is what we've already learned any nonzero number raised to the power of 0 is 1. and the reason it can't be zero is zero to the power of zero would end up being zero over zero which is undefined okay so as we continue to move down now we would have two to the power of negative one right if i decrease zero by one i get negative one so what am i gonna do i'm just gonna divide one by two so this is gonna give me one divided by two you could basically write that as one-half okay one-half now as i keep going what's going to happen is you're going to see a pattern emerge so 2 to the power of negative 2 would be what it would be one-half divided by 2. okay so divided by 2. so what is one-half divided by 2. using fractions here we know we would multiply by the reciprocal of this so dividing by 2 is the same as multiplying by half so this is basically 1 over 2 squared right 1 over 2 squared or 1 4. but again i want to write this as 1 over 2 squared because you're going to see something interesting in a minute let me erase this as we go one more time to complete this pattern i'm going to divide by 2 so 2 to the power of negative 3 is what it's 1 over this is 1 over 2 squared times another half so this exponent here just increases by 1. so this is 1 over 2 cubed so 1 over 2 cubed or 1 8. now you'll notice that the exponent down here matches the exponent here in terms of absolute value the exponent here matches the exponent here in terms of absolute value again here and here so that's going to lead us to the following rule so for any real number x as long as x is not equal to 0 if we raise it to the power of negative a it's equal to 1 over x raised to the power of a so in layman's terms basically you take the reciprocal of the base and you make the exponent positive so in other words if i wanted 2 to the power of negative four i would take the reciprocal of two which is one half and i would make the exponent part which is negative four here positive okay so you get one over two to the fourth power or just 1 16. very very easy rule to implement so let's take a look at a few easy examples we have 6 raised to the power of negative 2. so all i want to do is take the reciprocal of the base 6 so that's 1 over 6 and i want to make my exponent positive right now it's negative 2 it's just going to be a 2. you could leave this as 1 over 6 squared or you could say this is 1 over 36 okay both answers are correct now one thing i want to show you before we kind of get deeper into the lesson a common technique to do this is to write this over 1 and then as you drag an exponential expression across a fraction bar what happens is you leave the base the same and you change the sign of the exponent so this would be 6 squared down here and this would just be a 1 up here right and to make that crystal clear you could start off by just multiplying this by 1 so the 1 is still there in the end okay so that comes in really handy when you're working with something like let's say i gave you 1 over 2 to the power of negative 3. well instead of going through kind of a lot of mess there you can just drag this up here and say this is 2 change the sign of the exponent to positive so this is 2 cubed okay if you wanted to do this kind of the long way what you would do is you would say this is 1 over 1 divided by you would have 2 raised to the power of negative 3 is 1 over 2 cubed and so now what you would do is you're dividing 1 over 1 by 1 over 2 cubed right so it would be 1 times the reciprocal of this which is 2 cubed which equals 2 cubed okay so a very very lengthy process when you have this scenario where this guy's in the denominator so it's just easier to drag it across the fraction bar keep the base the same change the sign of the exponent all right so let's look at another one so we have 3 to the power of negative 4 again all i want to do is take the reciprocal of the base so 1 over three change the sine of negative four to positive four and i have one over three to the fourth power or you could say one over eighty-one suppose we have something like x to the power of negative z how could we write this without a negative exponent take the reciprocal of the base so 1 over x and this would be raised to the power of positive z okay so this is 1 over x to the power of z all right let's go ahead and take a look at another one so we have 3x to the power of negative 5 y to the power of negative 2. this is over we have z to the power of negative 4 q to the 8th power so if i wanted to simplify this or write it without any negative exponents what i could do is i could leave 3 unchanged in the numerator i have this x to the power of negative 5. how can i write that without a negative exponent well i could just drag this across the fraction bar keep x the same so again keep the base part the same change the sign of the exponent so right now it's a negative five we would make this a positive five we're going to do the same thing for y raised to the power of negative two again just drag it across the fraction bar this becomes y raised to the power of what change the sign from negative to positive so you get y squared down there now in terms of my denominator here i have z raised to the power of negative 4. so if i drag this into the numerator same thing applies right if you cross the fraction bar you just keep the base the same so z stays the same change the sign of the exponent so from negative four it will become positive four and then my q to the eighth power here i don't need to do anything with that other than just leave it in the denominator it does not currently have a negative exponent so i don't need to do anything with it so i end up with 3z to the fourth power over we have x to the fifth power y squared q to the eighth power all right for the next one we have inside of parentheses 2 times x raised to the power of negative 2 times z raised to the power of negative 1 times q so this whole thing inside of parentheses is raised to the power of negative 3 then this is all over 3n to the fourth power so to simplify here i'm going to use my power to power rule so each part here is going to be raised to the power of negative 3. so i would have 2 raised to the power of negative 3 i would have x to the power of negative 2 raised to the power of negative 3. so what do i do there i keep my base x the same i multiply exponents negative 2 times negative 3 is positive 6. z to the power of negative one is raised to the power of negative three keep z the same multiply negative one times negative three that's three and then q is raised to the power of negative three okay and this is over 3 n to the fourth power okay so what do we have here that has a negative exponent so we have this guy and we have this guy so let's just drag those across the fraction bar so if i take 2 to the power of negative three and i place it in the denominator two stays the same and the exponent changes to positive three x to the sixth power i leave that in the numerator z cubed i leave that in the numerator this guy q to the power of negative three drag it across so q comes down here the exponent changes from negative to positive so i get q cubed and then i have 3 n to the fourth power so let's clean this up we know 2 cubed is 8 and 8 times 3 is 24. so i'm going to write this as x to the sixth power z cubed over i'm going to have 24 and then times q cubed and then times n to the fourth power okay that would be my answer all right let's take a look at another one so we have x squared y to the power of negative one z to the fourth inside of parentheses and this is raised to the power of negative seven so again i'm going to use my power to power rule so x squared is raised to the power of negative seven x stays the same two times negative seven is negative fourteen y to the power of negative one is raised to the power of negative seven y stays the same negative one times negative seven is seven and then z to the fourth power is raised to the power of negative 7 z stays the same 4 times negative 7 is negative 28. so if i want to deal with this guy right here with a negative exponent and this guy right here y to the 7th power just stays in the numerator x to the power of negative 14. i'm just going to drag the x down here into the denominator negative 14 becomes positive 14. i'm going to drag this z down here into the denominator negative 28 becomes positive 28 so my answer here is y to the seventh power over x to the 14th power z to the 28th power all right so now that we have a great understanding of how to work with negative exponents let's move on and talk about the quotient rule for exponents this is very very simple basically when we divide with like bases we keep the base the same and we subtract the exponent in the denominator away from the exponent in the numerator okay very important there that you follow that direction when we looked at the product rule for exponents it didn't matter how you added because addition is it's irrelevant right the commutative property tells me if i add 2 plus 3 or 3 plus 2 i get the same result subtraction is not commutative so you've got to make sure you do this in the right order so if i have something like x to the power of a over x to the power of b this is equal to x raised to the power of a minus b so why is that the case well let's look at an example so let's look at this 2 to the 7th power over 2 cubed so following our rule the base 2 would stay the same and i would subtract 3 the exponent in the denominator away from 7 the exponent in the numerator so this would be 2 raised to the power of 7 minus 3. we know that 7 minus 3 is 4 so this would be 2 to the 4th power or 16 if you wanted to write that now let's think about why this is the case what is 2 to the seventh power it's seven factors of two two times two times two times two times two times two times two one two three four five six seven factors of two okay what is 2 cubed 2 times 2 times 2 okay well if i write these over each other like this it is very clear that i can basically cancel three of these guys right because i can cancel this with this this with this and this with this so i've removed three factors of two what did i do here i started with seven and i removed three and i ended up with four same thing here okay so that's all you're really doing and it works either way if i wrote if i wrote let's say 2 to the third power or 2 cubed over 2 to the fifth power the exponent of the denominator is larger here so i have more factors of 2 in the denominator it's still going to work so 2 times 2 times 2 times 2 times 2 this is 2 times 2 times 2. so following this this cancels this cancels this cancels i'm basically going to have 1 over 2 squared or 1 4 right if i follow my formula what do i have base stays the same subtract the exponent in the denominator which is 5 away from the exponent in the numerator which is 3. so three minus five is negative two so this equals two to the power of negative two we all know at this point that this is what reciprocal of the base which is one over two make the exponent positive so it's one over two squared it's exactly what we have here okay so it works the same way whether you have a larger number of factors in the denominator or a larger number of factors in the numerator you just follow the pattern all right let's go ahead and take a look at an example so we have 5 squared times x squared times y to the fourth power and this is over we have 5 cubed times x to the ninth power times y cubed so if we wanted to simplify this we would look for like bases we have 5 squared over 5 cubed so 5 would stay the same and then again i would subtract the exponent in the denominator so the 3 away from the exponent in the numerator which is a 2. so i would have 5 raised to the power of 2 minus 3. and we know that's negative 1 but let's replace that in a minute let's just keep going for now so then we have x squared over x to the ninth power so x stays the same and then we're going to raise this to the power of what it's this guy the 9 that's being subtracted away from this guy the 2. so we have 2 minus 9. we know that's negative 7 but again we'll do that in a minute then we have y to the fourth power over y cubed so y stays the same and then we want to do 4 minus 3. right because this is the guy in the denominator this is the guy in the numerator you want the guy in the denominator which is 3 to be subtracted away from the guy in the numerator which is 4. so this is 4 minus 3. so this ends up being 1. again 2 minus 3 is negative 1 so 5 to the power of negative 1. you have x raised to the power of 2 minus 9 so x raised to the power of what that's negative 7 and then y raised to the power of 4 minus 3 which is just y so now if i think about how to simplify this if i want to get rid of 5 raised to the power of negative 1 that's what it's just 1 5th but again a simpler way to do that is just to set up a little fraction bar y stays in the numerator because it does not have a negative exponent this guy right here if i drag it into the denominator the five part stays the same the exponent changes from negative one to positive one and a positive one as an exponent is implied right so you don't need to write it then i have x to the power of negative 7. again if i drag this into the denominator i'm going to have my x parts stay the same my negative 7 will change to positive 7. so we basically have y over 5 x to the seventh power all right let's take a look at one more so we have 7 that's multiplied by x to the power of negative 2 times y cubed times z to the power of negative 4. this is over we have inside of parentheses x multiplied by y to the fourth power multiplied by z squared this is all raised to the power of negative 2. so the first thing i want to do is kind of simplify down here so let me just rewrite this i have 7x to the power of negative 2 y cubed z to the power of negative 4. down here i've got to use my power to power rule so x would be raised to the power of negative 2. y would be raised to the power of 1. keep that the same 4 times negative 2 is negative 8. z would stay the same multiply 2 times negative 2 that's negative 4. so now i'm just looking at i have a 7 up here nothing's going to change with that i have x to the power of negative 2 over x to the power of negative 2. what is that that's just 1 right if i have x to the power of negative two over x to the power of negative two same thing over itself is going to be one okay unless it's zero and we're assuming that x is not zero here so i can show you this also by saying keep x the same take this guy right here negative 2 and subtract it away from negative 2. what is negative 2 minus a negative 2 this is negative 2 plus 2 which is 0. x to the power of 0 is 1. okay so that guy's cancelled so you can cancel this same thing for the z you have z to the power of negative 4 z to the power of negative 4 cancel it so all i have left is a y raised to the power of 3 over a y raised to the power of negative 8. so you're going to subtract this guy away from this guy okay so if i subtract negative 8 away from 3. so 3 minus a negative 8 be careful there you want to minus the negative 8. this is the same thing as 3 plus 8 which is 11. okay so you get 7y to the 11th power now another way you could have done this you could have had 7 up here you could drag the y up so you could have y cubed if you drag the y down here up into the numerator it's y the exponent changes from negative 8 to 8. then you can just use your product rule right for exponents so you have 7y raised to the power of 3 plus 8 which is 11 okay so a lot of different ways to kind of do the same thing whatever you're more comfortable with if you like dragging things across the fraction bar cool just keep the base the same change the sign of the exponent okay if it goes across from numerator to denominator or denominator to the numerator you also have the method of just taking the reciprocal of the base and making the exponent positive so either way you want to do it in this lesson we want to review rational exponents so in this section that we're in currently we're exclusively going to be reviewing radical expressions so it's very important to understand how to convert between the radical notation and kind of the exponential notation that's involved when you work with radicals so we're going to kind of start out with this first rule and then we're going to work through some examples we have that the nth root of x is equal to x raised to the power of 1 over n and basically what we're doing here is we're taking this index here and we are placing it in the denominator of this fractional exponent so you have a 1 over n where again n is the index of your radical so if we look at something pretty easy like a square root if we have the square root of x we can write this as x raised to the power of one half remember when we work with square roots there's an implied index of two we don't write it but it's there right so this guy's a two that means this guy's going to be a two right so if i wanted to prove this mathematically that the square root of x is equal to x raised to the power of one-half how can i do that well we know that the square root of x multiplied by the square root of x is equal to x right and that's true for any non-negative real number x so if i had something like the square root of 3 multiplied by the square root of 3 this would give me 3 right if i wanted to write this in exponential form instead of writing the square root of x like this i would just write x to the power of one-half and i would multiply by x to the power of one-half and i would say this is equal to what well i would use my product rule for exponents the base is the same i have x i have x so i'm going to put an x here and then i'm going to add my exponents so i have a one-half plus another one-half what's that going to give me i know that the denominator there is a 2 in each case so i have a common denominator already so i just need to add the numerators 1 plus 1 is 2 if i had 2 over 2 it would just be 1. so this is x to the first power or just x so you can see that whether you write this in kind of radical notation by putting the square root symbol or writing this with exponential notation where you have a one-half as the exponent it's the same exact thing mathematically so as an example here we showed the square root of three times the square root of three was three it would also be true that three to the one half power times three to the one half power is equal to three again same thing here base is the same so three would stay the same you would add the exponent one half to the exponent one half you would get one three to the first power is just three so let's practice a few easy examples if i had something like the square root of two how could i write this using an exponent well again i could just take this guy right here this 2 that's going to be the base for my exponential expression the index here which is an implied 2 would be the denominator of my fractional exponent so i would have a one over a two or just a one half but it's easy to remember when you have the square root of something it's just raised to the one half power if you have the cube root of something right the index is three it's raised to the one third power if you have the fourth root it's raised to the one fourth power you know so on and so forth so square root of 5x again 5x the whole thing here the 5 and the x is going to be raised to the power of one half you know from working with exponents that you could also write this as 5 to the power of one-half times x to the power of one-half so all of these mean the same thing we have 5x as a radicand we're taking the square root of 5x all right what about something like this we have the square root of 3x plus 5. so we have to be careful here this whole thing would be inside of parentheses and we'd raise it to the one-half power now can i distribute like i just did in the last example no you cannot cannot cannot distribute exponents over a sum i've probably said this in like five videos already very important that you don't do something like this please don't say this is 3x to the power of one-half plus 5 to the power of one-half that is absolutely wrong this is wrong please don't do this okay it's a very common mistake and i'll just keep mentioning it till it gets through your head okay you can't do that you can only do that with multiplication if we had something like let's say the square root of 3x multiplied by 5y something like that and of course this could be simplified and we could just put the square root of 15xy but in this case i could write this as what inside of parentheses 15xy raised to the power of one-half because this is multiplication here 15 is multiplied by x which is multiplied by y you can use your rules for exponents so the one half can actually be distributed there so it can be applied to each of these so 15 would be raised to the power of one half x would be raised to the power of one half and y would be raised to the power of one half so that's okay that's legal because it's multiplication but when you have a sum involved like this you can't do it okay so make sure you understand that so writing this with a fractional exponent or using rational exponents if you want to say that you get 3x plus 5 inside of parentheses the whole thing's raised to the power of one-half all right so again if you're working with higher level roots it's the same thing okay so if i have the cube root of x this is x raised to the 1 3 power this index here is always going to match this denominator here okay so if i had the cube root of 2 i could just write this as 2 to the power of 1 3. very very simple what if i had something like this i have the cube root of 12 x squared y squared z notice how it's all multiplication here okay so i could write this as 12 x squared y squared z inside of parentheses and raise this guy to the one-third power and again because it's all multiplication here there's no sum involved i can put 12 to the power of one-third i can put x squared to the power of one-third so power to power rule that would be x to the two thirds power and we'll talk about what that means in a minute then we would have y squared raised to the one third power so power to power rule two times one third is two thirds and then you'd have z raised to the power of one third okay all right let's evaluate a few of these so we have 64 raised to the one-third power so to do this you could either realize you're taking the cube root of something or you can convert it into radical notation a lot of people will convert to radical notation to evaluate so we'll just go ahead and convert this to the cube root of 64 like this i know off the top of my head that the cube root of 64 is 4 but if you didn't know that again you could just factor 64. most of you would just use a calculator which is perfectly acceptable in today's age but if you wanted to do this without a calculator you would say okay 64. i know that's 4 times 16 4 is 2 times 2 16 is what it's 4 times 4 and 4 is 2 times 2. so you can see that if you wanted a number that was multiplied by itself three times that gave you 64 you would be looking for four right you have four times 4 times 4 so again the cube root of this would be 4. okay pretty easy what about 625 raised to the power of 1 4 well again if this part right here is your denominator that's the index of the radical okay so 625 raised to the power of one-fourth is the fourth root of 625. again another common problem so i know off the top of my head this is five many years of working with this but again if you didn't punch it up on a calculator at 625 raised to the power of one-fourth or you could do the fourth root of 625 same answer you get five or again go ahead and just factor this guy this we can say ends in a 5 so it would be 5 times 125 125 is 25 times 5 25 is 5 times 5 okay so what number multiplied by itself four times gives me the radicand of 625 well five right i've got one two three four of these guys so i can just answer this and say this is going to be five okay pretty simple so if we have the nth root of x and it's raised to the power of m in exponential notation we can write this as x raised to the power of m over n okay all you're basically doing is just using your power to power rule if this guy right here is x raised to the power of one over n okay so we've already defined that if this is now raised to the power of m again just using the power of the power rule this is x raised to the power of m times one over n which is just m over n okay very easy now because of the commutative property of multiplication you're going to see that it doesn't actually matter if you have the nth root of x to the nth power or if you have the nth root of x and this thing is raised to the power of m why doesn't it matter well in this case right here what do i have i end up with x to the power of m and this is being raised to the power of 1 over n well if i use my power of the power rule i get x raised to the power of m over n same thing is up here same thing as we're going to get here right we already know this scenario because we've already shown that this is equal to x raised to the power of m over n but again it's the same thing because of the commutative property of multiplication you can multiply in any order it's not going to change typically this is how you'd want to do it though because you want to work with a smaller value so let me show you that real quick so let's say you had 4 raised to the 3 halves power and you're asked to evaluate this you want to know what it is so you can either do 4 cubed and then raise that to the power of one half or you can also do four raised to the power of one half and then you can cube that if you use your power to power rule here either way you end up with four raised to the power of three halves right one half times three this way or three times one half that way you end up with three halves right again commutative property multiplication so doing it this way 4 cubed is 64 if i take the square root of 64 or again raise this to the one half power i get 8. but notice i had to work with a really big number there i mean 64 isn't a really big number but it's a bigger number than i would work with here in this scenario the square root of 4 is 2. if i cube 2 i also get 8. it's the same result but this guy allowed me to work with a smaller number and therefore in a lot of cases it's less tedious and it's just easier so for myself this method is preferred okay this is my preferred method but again in math there's a lot of different ways to do things so whichever way you want to do it it doesn't matter because you'll get the same result either way all right let's go ahead and take a look at negative 27 raised to the power of five thirds so essentially i'm going to write this as negative 27 this whole thing i'm going to include parentheses here because i have a negative involved this is going to be raised to the 1 3 power and then i'm going to raise this i'll just use brackets here to the power of 5. so to write this more simply or in a more recognizable form i'm going to write the cube root of negative 27 and then i'll raise this to the fifth power so most of you know the cube root of negative 27 would be negative three all right so negative three again that's because negative three times negative three is nine nine times negative three is negative 27 then we'd raise this to the fifth power and that will give me what negative 243. negative three times negative three is nine nine times negative three is negative 27 negative 27 times negative three is 81 and then 81 times negative three is negative 243. all right what about something like 16 to the power of negative 3 4. again what i'm going to do i'm going to take 16 and i'm going to start by raising it to the 1 4 power and then i'm going to raise this guy to the power of negative 3 okay so we know how to work with negative exponents we'll get to that in a second first and foremost let's just do this as the fourth root right because this guy's a four of sixteen and then again this guy's raised to the power of negative three so the fourth root of 16 is 2 right 2 times 2 is 4 4 times 2 is 8 8 times 2 is 16. so this is 2 raised to the power of negative 3 which again is take the reciprocal of the base so this becomes 1 over 2 make this guy positive so this is 3. so this ends up being 1 over 8 because 2 cubed is 8. now there's another way you could have done that let's erase everything we know that when we work with a negative exponent we take the reciprocal of the base and we make the exponent positive so i could start out because i have a negative exponent here writing this as 1 over 16 raised to the power of 3 4. i can do that to start so this is 1 over the 4th root of 16 which is going to be cubed so this is 1 over 2 cubed so this is 1 over 8. that's another way to do it but again in all of mathematics you're going to have many different ways to do things you pick one that works for you you pick one that you like and that you can remember and you do it that way all right so let's look at a few examples here and these are just going to be examples that help you kind of bring your knowledge of exponents together it's always good to get some practice with exponents so that kind of later on in the course when we look at some really challenging materials you don't get tripped up on stuff that's trivial like exponents so if we have something like 2 raised to the power of negative 2 times x raised to the power of negative 3 4 times y raised to the power of negative 5 8 and this is over x raised to the power of 3 4 times y raised to the power of negative 1 half so this whole thing is raised to the power of negative 3. so i'm going to start by just using my power to power rule and everything notice that there's no addition involved this is all multiplication so i'm going to have 2 raised to the power of negative 2 raised to the power of negative 3. 2 stays the same multiply exponents negative 2 times negative 3 is positive 6. then i have x which stays the same negative 3 4 times negative 3 would be positive 9 4. y would stay the same negative 5 8 times negative 3 would be positive 15 8. this is over same thing applies in the denominator we have x there that stays the same 3 4 times negative 3 would be negative 9 4. and then you have a y and then negative 1 half times negative 3 would be positive 3 halves so now that we've done that let's go through and use the quotient rule for exponents which we talked about in the last lesson and see what we can simplify so we know that two to the sixth power is going to be 64 right 2 times 2 is 4 4 times 2 is 8 8 times 2 is 16. 16 times 2 is 32 and the 32 times 2 is 64. you can leave that as 2 to the 6th power or just put 64. doesn't matter then you have x and x so same base let's just write it you want the exponent in the denominator here which is negative 9 4 to be subtracted away from the exponent in the numerator which is 9 4. so 9 4 minus a negative nine fourths okay we know if we subtract away a negative we're adding a positive so you might as well just put plus nine fourths so nine plus nine would give me 18 so i would have 18 fourths 18 4. so we can simplify this 18 and 4 both have a common factor of 2 this would be 9 halves okay so 9 halves then next i have my y so i have y then i'm going to write here again this guy right here is going to be subtracted away from this guy right here so you would have 15 8 minus three halves you got to get a common denominator going so i'm going to multiply this by four over four so this would be a 12 8 right so i'm going to write this as 12 8. 15 minus 12 is 3 so this would be 3 8. okay now let's take a look at one more here so we have 4x raised to the power of 5 3 multiplied by the quantity x raised to the power of negative two thirds minus eight x raised to the power of negative five thirds so a lot of you might get confused with this problem i have students that have tutored in the past and they look at this problem they go i can't distribute that because you said you can't distribute when there's a sum or a or subtraction well no that's when you're talking about exponents right i can still distribute this guy this is multiplication here i'm multiplying this by this difference or if you wanted to write this as plus negative you could say by this sum right so that's okay it's perfectly legal it's just when you have something like let's say x squared plus y squared and you have an exponent like 1 3 you can't distribute the exponent okay that's what you can't do but i can still distribute this so what would i have well i'm going to just work with this as 4 nothing to combine with that x to the 5 3 power multiplied by x to the negative 2 3 power x stays the same you would add the exponents so what is five thirds plus negative two-thirds five plus negative two or five minus two is three three over three is what you have three over three is just one so this is just 4x then we have plus i'm going to just treat this as negative 8x to the power of negative 5 thirds so 4 times negative 8 is negative 32. so i can just go ahead and write minus if i want so minus 32 then i know i'll have an x and what will my exponent be you have 5 thirds plus negative 5 thirds so 5 thirds plus negative 5 thirds or basically 5 3 minus 5 thirds that's 0. so what is x to the power of 0 it's 1. so i can just say this is 4x minus 32 as my answer some of you will like to factor that and say this is 4 times the quantity x minus 8 that's fine as well so either of these answers would be fine in this lesson we want to review radical expressions so in the last lesson we reviewed rational exponents which are also known as fractional exponents now i want you to recall these allow us to convert between radical notation and exponential notation so what we want to do is just quickly recap that process because we're going to need that operation for our lesson today so we have something like the nth root of a we should recall that we can write this in exponential form by basically taking the radicand a and writing that as the base of our exponential expression and so a will be raised to the power of 1 over the index in this case the index is n so 1 over n so if i had something like let's say the square root of 4 this has an implied index of a 2 so essentially in exponential notation i just take this 4 and i say it's raised to the power of 1 over again the index is 2 so 1 over 2 or 1 half so we know that taking the square root of a number is like raising it to the power of one half similarly if we take the cube root of a number it's like raising it to the power of one-third so something like let's say the cube root of 27 this would be equal to the 27 here is the base of the exponent and it's raised to the power of 1 over 3. now we have another rule which we're going to rely on heavily throughout our course if you have the nth root of a and let's say a is raised to a power now so a is raised to the power of m this is equal to this part's the same we're just going to have a raised to the power of m over n so everything's the same except for the fact that you have an m in the numerator now over your n your index in the denominator now you might also see this written in a different form you could also say this is the nth root let me make that n a little better so the nth root of a and this whole thing is raised to the power of m and essentially because of the power to power rule you could really write this as a raised to the power of m over n as well but essentially for this lesson we're going to deal with this scenario so we don't really need to worry about this we just want to reference it maybe later on if it comes up so we can use this rule to kind of simplify here i want you to realize that 4 is nothing more than 2 squared right so we know the square root of 4 is 2 because 2 squared is 4. but to simplify this using exponents we can say the square root of 2 squared like this and we'll go ahead and say this is equal to we have 2 which will be raised to the power of 2 over 2 which is basically 2 raised to the first power so i want you to realize that in most cases you're going to be able to cancel an exponent here with an index here we'll see later on in the lesson that when you have a negative involved there and you have an even index you need to kind of think about what's going on first but in general you can just cancel between here and here if your number here is non-negative okay so we'll talk about the exception later on in the lesson so as this one goes we have the cube root of 27 we know 27 is 3 cubed so i could write this as the cube root of 3 cubed and again if i wanted to i could cancel this with this but i'm going to show that process with exponents so i can say this is 3 raised to the power of 3 over three so this exponent over this index and this is equal to three over three is one so three to the first power or just three right and here i could just really write this as two all right so the purpose of this lesson is to give a basic overview of radical expressions most of you or if not all of you understand how to take the square root the cube root the fourth root of some number but it's important to kind of go back and make sure you understand all the vocabulary and all the different scenarios that we're going to encounter throughout the course so we'll start with something very basic we're going to show the parts of a radical so if you have the nth root of a the n right here is known as the index and for a square root the index is left blank so if you have a square root it's implied to be a 2 right so something like again the square root of 4 you don't show a visible index but it's known or understood to be a 2. but if i have a cube root a 4th root a 5th root some higher level root i need to make it clear that i have that level and so i want to put a visible number there so if i have the cube root of 8 i want to make sure to put a 3 there to identify that this is a cube root okay so very important that you do that then this guy right here this number or variable or expression that's underneath the radical symbol is known as a radicand and then the symbol itself is just called a radical symbol radical symbol okay then when we talk about the whole thing together it's just known as a radical so when i say a radical i'm talking about all parts of this together so what does a radical allow us to do well we know about taking a number to a power and essentially when we have a radical expression it's reversing the operation of raising to a power so just like addition and subtraction or inverses multiplication division or inverses you have radicals that are inverses with kind of raising to a power so it basically asks the question what number raised to the index is equal to our radicand so in this case we have the square root of 49. so i'm going to put a visible index of 2 here just for our demonstration purposes but again you don't need that and essentially i'm asking what number so let me write this out what number raised to the power of the index and the index in this case is 2. so what number raised to the power of 2 gives us the radicand which in this case is 49 so that's the question we're asking each time if i had the cube root of 8 as another example i would just say what number raised to the power of three because my index is a three gives us the radicand is eight okay if i had the fourth root of let's say 625 so it would be what number raised to the power of 4 gives us the radicand of 625 so on and so forth so it's a very simple process to understand now how do i go about figuring this out well if i don't have a calculator i need to factor 49 happens to be a perfect square so it's built from the product of a rational number times itself we'll talk more about the definition of a perfect square in a minute but essentially when you factor 49 you only get 7 times 7 or you also get negative 7 times negative 7. so to answer this question what number raised to the power of 2 gives us 49 it's 7 or also negative 7. when we work with radicals and we have an even root so a root that's a square root a fourth root a sixth root so on and so forth we're gonna get two different roots you're gonna get a positive or principal root and then also a negative root so this guy right here the square root of 49 written like this would just be 7 because this is the principal or positive square root of 49. then to notate the negative square root of 49 i just put a negative out in front so i would say the negative square root of 49 and this equals negative 7. so for this guy it's asking what positive number squared gives me 49 that's 7 right because 7 squared is 49. this guy is asking what negative number squared gives us 49 in this case that's negative 7 because negative 7 inside of parentheses squared is positive 49. and then we have a shorthand to include both of these in one so we put plus or minus like this the square root of 49. this just means to give me the principal square root of 49 which is 7 and the negative square root of 49 which is negative 7 so i'm going to write this as plus or minus 7. okay so very very compact notation there and something we're going to use throughout the course so what if i got something like the square root of 144 and i wanted to figure out what that was without using a calculator what could i do so most of you know that 144 is also a perfect square it's 12 times 12. but if you didn't know that you could just factor the number so 144 you just use your divisibility rules so this guy ends with 44. 44 is divisible by 4 right 44 divided by 4 is 11. so that tells me the number 144 is divisible by 4. so what would it be it would be 4 times 36 4 is 2 times 2 2 is a prime number so let's just circle those and then 36 is 9 times 4 4 again is 2 times 2 2 is a prime number so let's circle those and then we know that 9 is 3 times 3 and let me make that better and we'll circle these threes here so we can see that we have 2 times 2 times 2 times 2 where you could really say you have 2 squared times 2 squared again times 3 squared and i did this to where each number there had an exponent of 2 because we're looking for something that's squared so now we can use our rules of exponents to write this as 2 times 2 times 3 being squared 2 times 2 is 4 4 times 3 is 12 so this is 12 squared so that's one way to find the square of a number if you don't have a calculator it's unrealistic at this point in time to not ever have a calculator because we all have these phones in our pockets that can basically do anything but in case you wanted to know how to do it that's what you would do so we would say the principal squared of 144 is 12. if i was asked for the negative square root of 144 that would just be negative 12. what about something like the negative fourth root of 625 well again i'm looking for a negative number that one raised to the fourth power gives me the radicand of 625 so i know that an even number of negative factors gives me a positive so that part would be taken care of i know this would just be a negative so 625 if i factored that it ends in five so it would be 125 times five 125 is 25 times 5 and 25 is 5 times 5. so we can see this is 5 to the fourth power to get us to 625 but also negative 5 to the fourth power would be 625 as well because you would have negative 5 times negative 5 times negative 5 times negative 5. so an even number of negative factors would give me a positive product okay so the negative fourth root of 625 is negative 5. if you got asked for the principal fourth root of 625 that would be positive 5. all right so when we work with odd roots like a cube root of fifth root a seventh root so on and so forth we don't need to worry about having a plus or minus there when you have an odd number of negative factors you get a negative result so there's not going to be two roots here so we have something like the cube root of negative 64 we would just think about what number cubed what number raised to the third power gives us negative 64. well i know it's negative right because negative times negative times negative would give me a negative so i just need to work out 64 and think about the factors there i know it's divisible by 4 it's 4 times 16 and i know 16 is 4 times 4. now i can go through and say this is 2 times 2 and this is 2 times 2 and this is 2 times 2 but for the purposes of what we're doing we've already found that 4 to the 3rd power or 4 cubed is 64. so the cube root of negative 64 is just negative 4. what about something like the fifth root of negative 243 well for this guy again i'm looking for what number raised to the fifth power is going to give me negative 243. so think about the fact that this will be negative and to factor 243 you've got 81 times 3 you've got 9 times 9 and this is 3 times 3 right and this is 3 times 3. so you've got 1 2 3 4 five factors of three 243 is three to the fifth power right so the fifth root of negative 243 is negative three so when we work with radicals we're basically going to have three different scenarios so we should recall that our real number system breaks up numbers into either rational numbers or irrational numbers so when you work with radicals you're either going to have a value that represents a rational number it could represent an irrational number or it could represent something that is not a real number so we'll get to the last part of that in a moment for right now let's talk about how you can tell if you have a rational number or an irrational number so you've probably heard the term perfect square perfect cube perfect fourth so on and so forth so what is a perfect square a perfect square is a number whose square root is a rational number so if we went through kind of the first few whole numbers let's say we had one squared this is 1 2 squared this is 4 3 squared this is 9. let's just skip a few and go to 10 10 squared is 100. let's skip a few more and go to 20 20 squared is 400. so i would recommend you memorize from 1 all the way to 20 squaring those numbers what's the result it's something you're going to need to go back and forth with but essentially each of these results here are perfect squares right so if you think about the square root of 1 it's 1 right it's a rational number the square root of 4 is 2 a rational number the square root of 9 is 3 a rational number the square root of 100 is 10 a rational number and the square root of 400 is 20 irrational numbers so these are perfect squares because when i take the square root i get a rational number so if i have something like the square root of let's say 400 this represents the rational number 20 okay you could also have a scenario where you have a fraction that's a perfect square right so it doesn't have to be limited to a whole number or an integer you could have something like the number one fourth this is a perfect square because the square root of it is one half okay so the square root of one-fourth is a rational number one-half so one-fourth is a perfect square now you could expand this definition to having a perfect cube or a perfect fourth whatever it is for a perfect cube you think about cubing a number 1 cubed is 1 so 1's a perfect cube 2 cubed is 8 so 8 is a perfect cube 3 cubed is 27 so 27 is a perfect cube so on and so forth so if the number when i take the cube root of it gives me a rational number it's a perfect cube if a number when i take the fourth root of it gives me a rational number it's a perfect fourth right so on and so forth now if i'm taking a square root of a number and that number is not a perfect square i have a radical that represents an irrational number so for example 5 is not a perfect square there is no rational number that i can square to get 5. so if i take the square root of 5 this represents an irrational number this is irrational and you can punch that up on a calculator and you're going to get a nasty decimal and it's a decimal that does not terminate and it does not repeat the same pattern forever so at some point because you're limited in space your calculator is either going to truncate or it's going to round right it's going to do one of the two and it's going to give you an approximation of the square root of 5 but it won't actually be equal and the way you can prove that to yourself is you can take the number they give you and you can square it and you won't exactly get 5. you'll just get something that's close then the same thing goes if i take the cube root of something that's not a perfect cube so let's say i took the cube root of 12. 12 is not a perfect cube there's no rational number that i can cube to get 12 okay so that means if i take the cube root of that it's going to be irrational okay so the last scenario you're going to deal with is a radical that represents something that is not real so we know that we have imaginary numbers in the complex number system that we can use to deal with this and we'll talk about that in a few lessons for right now we just need to understand this concept because it will come up a lot throughout our course so if we take the square root of a negative number or an even root of a negative number we get a problem an even number of negative factors always gives me a positive so if i try to take the square root of negative 25 i'm basically asking the question what number squared or what number multiplied by itself gives me a product of negative 25. well there's no number in the real number system that does that right because if you square a number and it's negative you get a positive if you square zero you get zero and if you square a positive you keep the positive so this is never going to happen in the real number system so this is not real similarly the fourth root of negative 625 this is not real now i don't want you to get confused by what we looked at earlier you can have a negative outside of the radical so if i have the negative fourth root of 625 which we looked at earlier this is just what negative number raised to the fourth power gives me 625 we know from earlier this is negative 5. these two are not the same notice how in this guy i have the negative inside it's part of the radicand here my radicand is positive so i'm good to go my negative is on the outside so make sure you understand the difference between the two because this is a common source of confusion now you might get in a scenario where you see an odd root of a negative number this is allowed right we talked about these earlier in the lesson we looked at some examples something like the cube root of negative 8 this is just negative 2. so this is a real number because negative 2 cubed is negative 2 times negative 2 which is 4 times negative 2 again which gives you negative 8. you can raise a number to an odd power and get a negative it's okay then something like the fifth root of negative 3125 this is just negative 5 because negative 5 raised to the fifth power would give me negative 3125. all right so let's talk a little bit about what happens when you work with variables it's very important to understand that we can't take the square root or the fourth root or again an even root of a negative number so when you work with variables you don't know what the variable represents so you have to be cautious so if i have the nth root of x i have to say that okay i can't plug in a negative value if n is even right so if n is even then i have to assume that x is greater than or equal to zero otherwise i'm allowing for the possibility of a non-real solution okay so you have to be very careful and you might not have something that's just x you might have an expression there which we'll see in a moment so if n is odd [Music] then x can be any real number x is any real number okay it doesn't matter because again an odd number of negative factors is a negative so that's okay so if i have the square root of negative 81 this is again not real not real the fifth root of negative 32 is just negative two then something like the fourth root of negative two thousand four hundred one this is a four this is a negative here can't take the even root of a negative number this is not real okay not real and again in a few lessons we'll learn how to deal with taking the square root of a negative but for right now we're just going to say it's not real let's talk about one more thing here and this is very important also so if i have the nth root of a to the n power it's equal to the absolute value of a when n is even and if i have the nth root of a to the n power it's equal to a when n is odd so this might seem like a peculiar rule but i'm going to show you why it works in just a moment so if i had something like the square root of x squared we've been shown already that we can basically cancel what the index here with the exponent so most of you will cancel this with this and just say this is equal to x well that's only true if x represents a non-negative real number if that's the case then you're fine suppose we had something like let's say the square root of 7 squared well if this is a 2 here this is perfectly valid you can cancel this with this and it's just set right 7 square would give me 49 square root of 49 would give me 7. so that's okay but if i had the square root of negative 7 squared now i've got a problem if you try to cancel this with this and say this is negative 7 that is wrong okay this is wrong this is where you need your absolute value so we would say we would cancel this back and say this is equal to the absolute value of x so in other words in this case it's the absolute value of seven which is seven here it's the absolute value of negative seven which is positive seven okay why negative 7 squared is positive 49. so if i have the square root of negative 7 squared this becomes the square root of positive 49 which is 7 okay not negative 7. so you're basically doing the squaring operation first before you take the square root so it takes your negative into a positive and then it's basically going to give you the absolute value of what was being squared okay so that's how we have this rule it's very important to understand when you're working with variables what restrictions are there if your textbook tells you hey i want you to assume that all the variables are non-negative real numbers then go ahead and cancel away right in this case if we have that assumption you can cancel this with this and say this is x if you don't have that assumption then you better make sure you put the absolute value of x because on a test the teacher will be looking for that and they'll mark it wrong so something like the cube root of negative 11 cubed well here i've got an odd index so i can just cancel right it's not an issue because if i cube a negative i get a negative right so if i take the cube root of that i would still have a negative so here i can just cancel this with this and this is just negative 11. all right so this is no problem no issue this is okay as long as you have an odd index with this guy i have an even index so the fourth root of negative six to the fourth power again i can't just go through here and cancel what i got to do is think about the fact that this would be equal to the absolute value of negative 6 which would just be 6 okay and the reason for that is if i take negative 6 to the 4th power i get 1296 positive 1 296. if i take the fourth root of that i go back to six right so i just want the absolute value of this guy that's being raised to that power that's going to end up giving me the correct answer right so please don't just cancel when you have an even index let's look at something a little bit more complex but following kind of the same thought process so the first thing i want to show you is that throughout the course you will often see something like the square root of x minus 5. so for this guy right here we need to figure out where it's defined right if we don't want to take the square root of a negative then we've got to take our radicand here which is an expression and we've got to say okay this has got to be greater than or equal to 0. i would just solve this inequality so i would add 5 to each side and i would have x is greater than or equal to 5. so we would kind of put this restriction and say x is greater than or equal to 5 here because if it's less than 5 let's say it was 4 so you would have the square root of 4 minus 5 which is negative 1 and the square root of negative 1 is not real okay not a real number all right so down here we have something that's kind of similar to what we're looking at a minute ago if we have the square root of the quantity x plus a squared so some expression being squared here we basically say that expression that's being squared we're going to take the absolute value of that for the answer so let me show you here if we had something like the square root of x squared minus 10x plus 25. a lot of you know this is a perfect square trinomial it can be factored into a binomial squared so this is the square root of x minus 5 that quantity squared and essentially what i'm saying here is that you can't just cancel this with this it's not equal to x minus 5. it's only going to be equal to that if x is greater than or equal to 5 okay the case where the result under here is not negative and let me show you why it's equal to the absolute value of x minus 5. if x represented a value that gave me a positive here inside the parentheses let's say x was 8 as an example so i could plug in an 8 there 8 minus 5 is 3 absolute value of 3 is 3. so i know i would get 3 as an answer so plugging in an 8 there 8 minus 5 is 3 3 squared is 9 the square root of 9 is 3. so i'm good to go if i plugged in something that made it negative though think about this if i plugged in let's say a 3. 3 minus 5 is negative 2. so i'd plug in a 3 here negative 2 the absolute value of that is positive 2. so this should be my answer so plug in a 3 there 3 minus 5 is negative 2. negative 2 squared is 4 okay positive 4 the square root of positive 4 is 2. if i didn't have that absolute value there think about the fact that if i just went through and cancelled i would just be left with x minus five if i plugged in my three there three minus five is negative two so i would have the incorrect answer of negative two okay so that's why you've got to have those absolute value bars there it's to include scenarios where this guy ends up being negative you square the negative and you get a positive then when you take the square root you've got to involve the absolute value operation okay so very important to understand that all right so the square root of x squared minus 4x plus 4 again we have another perfect square trinomial i would factor this and say this is the square root of x minus 2 squared and i could say this is equal to the absolute value of x minus 2. okay again if we wanted to just define this for x being greater than or equal to 2 then we can cancel so if i put a restriction here and say that x is greater than or equal to 2 well then i can just go through and cancel i can say the square root of x minus 2 squared i can cancel this with this and i can say this is x minus 2. that's perfectly legal because no matter what x is greater than or equal to 2 so if i put a 2 there i get 0 square root of 0 is 0. so i'm ok if i put a number larger than 2 there i'm good to go i can't put anything less than 2 because that would give me a negative here and that would result in me having to use absolute value bars so as long as you put this restriction in here or the restriction is given to you then you're good to go okay you don't have to go through and use absolute value bars but if i take this away then i've got to make sure that this is the absolute value of x minus 2 to account for the fact that this expression could be negative we're squaring the negative we get a positive we take the square root again you need the absolute value operation for that scenario in this lesson we want to review simplifying radicals so one of the first steps to simplifying a radical is understanding the product rule for radicals this rule allows us to multiply two radicals together with the same index so if i had the nth root of a times the nth root of b the index is the same we'd have the nth root of a times b so again notice that this part this part and this part are the same right it's got to be the same index then we just multiply the radicands together so under the radical symbol you have this radicand multiplied by this radicand okay that's what we have there so a couple of examples if i had something like the square root of 2 times the square root of 3 this would be what got a square and a square root so it's the square root of 2 times 3 which is 6. okay if you had a higher level root it's the same thing if you had the cube root of let's say seven times the cube root of let's say five this is the cube root right index stays the same of seven times five which is 35. so very very simple process very easy to understand what we're going to need to do is kind of reverse this to simplify so in other words if i see something like the square root of 40 i want to write this in such a way that i can simplify so what do i mean by that well let's take 40 for example and let's factor it so 40 factors in 8 times 5 and 5 is obviously prime 8 factors into 4 times 2 and we know 2 is prime 4 factors into 2 times 2 and again we know 2 is prime so what am i looking for if i have a square root i'm looking for factors of 40 that are a perfect square in other words i'm looking for factors of 40 where their square root is a rational number so 4 is a perfect square because it's 2 times 2 right so i can write this as the square root of 4 multiplied by 2 times 5 is 10 so the square root of 10 okay haven't done anything illegal notice that if you had been presented with the problem the square root of 4 times the square root of 10 by the rule we just learned we could write this as the square root of 40. now this wouldn't be considered simplified but we could still do it now let's erase this i want you to think about what you can do here the square root of 4 we just said was 2. so what i want to do to make this more simple is just replace that with a 2. so i'm going to put 2 times the square root of 10. so the square root of 40 and 2 times the square root of 10 represent the same value it's just that this guy right here 2 times the square root of 10 is considered simplified all right let's take a look at another example so suppose you had the square root of 54 and you wanted to simplify again just go ahead and factor 54. so we know this is 9 times 6 and you can pretty much stop there you know 9 is 3 times 3 right so 9 is a perfect square 6 is going to be 3 times 2 but nothing you can really do with that so i could write this as the square root of 9 multiplied by the square root of 6. let me erase this get this out of the way and we know the square root of 9 is 3 so we're just going to write that so this is 3 times the square root of 6. again square root of 54 and 3 times the square root of 6 represent the same number it's just that this guy right here 3 times the square root of 6 is considered simplified all right we can also do this process with higher roots suppose you have the cube root of 128. so the first thing i'd want to do again is just factor the number and 128 a lot of you do know that that is 2 to the seventh power but if you didn't you could start by saying okay it's 64 times 2 64 is 16 times 4 4 is 2 times 2 16 is 4 times 4 and 4 is 2 times 2 and so you have 1 2 3 4 5 6 7 factors of 2. now if i want the cube root of something i am looking for factors of this number right here of this radicand that are going to be a perfect cube meaning the cube root of the number is a rational number so if i look here i can do 2 times 2 2 times 2 and 2 times 2 right so i would have 1 2 3 of those guys in other words i would have four to the third power which is 64 times two so i could write this as the cube root of 64 which we know is four times the cube root of two okay all i did was break this down just because it's a cube root doesn't mean the process is any harder or more difficult it's the same thing instead of looking for a perfect square like you would with a square root you're looking for a perfect cube because you have a cube root okay so all we're going to do is just replace this guy right here with a4 okay cube root of 64 is 4 then times the cube root of 2 okay so again 4 times the cube root of 2 is the simplified version of the cube root of 128. all right so for the next one we have one with a variable involved so we have the square root of 50x cubed so the square root of 50 is very easy to simplify we pretty much mastered that at this point we should know that 50 is 25 times 2 and 25 is a perfect square so i could write this as the square root of 25 times the square root of 2. pretty simple we know that this guy right here the square root of 25 is just 5. all right but we need to deal with this part right here so off to the side if i have the square root of x cubed what does that simplify into well there's a lot of different ways to think about this first and foremost can you simplify the rule is if this exponent here is larger or equal to this index here yes you can simplify so if i have x cubed here and my index here is a 2 right i'm taking the square root of that i want to think about the next exponent going down that's going to be divisible by 2 my index y well if i rewrote this as the square root of x squared times the square root of x this guy right here would simplify to x okay how do i know that well again if i wrote this using exponential notation this is x squared raised to the power of one-half so this is what it's x raised to the power of two over two which is one okay that's the simplest way to kind of think about it another way to think about it is to visibly write an index here and just say this guy right here the 2 the exponent there is going to be divided by this guy right here the index 2 divided by 2 is 1 so you would just have one of those guys so to simplify this i would break this up and say this is the square root of x squared times the square root of just x okay and again we already know that this part right here can just be written as x so we're just going to say that this square root of 25 is 5. square root of x squared is x and then times what's left over the square root of 2 times the square root of x is written as the square root of 2x so we end up with 5x times the square root of 2x all right let's take a look at another one with a variable in it so we have the fourth root of 80x to the ninth power all right so i'm looking for a number here that is a factor of 80 and that's a perfect fourth okay so if i was to factor 80 what would i get well it just so happens i would get 16 times 5 and 16 is 2 to the fourth power we don't need to go through the whole factorization so i'm going to write this as the fourth root of 16. times the fourth root of five okay this part right here would simplify to two right two to the fourth power is 16 so the fourth root of 16 is 2 right 16 is a perfect fourth then for x to the ninth power again we gotta think about this so x to the ninth power if i have the fourth root of that first question can i simplify the nine is larger than the four so the answer is yes now if i want to do this think about the next exponent going down that's divisible by 4. if i think about 9 9 is not divisible by 4 but 1 down 8 would be so i'm going to rewrite this as the fourth root of x to the eighth power times the fourth root of x right x to the eighth times x is x to the ninth so this is completely legal now i can either use exponents or again i can just divide this by this 8 divided by 4 is 2 so this would end up being x squared okay so put times the fourth root of x to the eighth power times the fourth root of just x and again this is going to simplify to x squared and again if you want to see that using exponential notation that's one of the ones that really helps students if they're struggling if you have the fourth root of x to the eighth power this x to the eighth power would be here it's raised to the 1 4 power just think about the fact that this guy right here is always going to divide this guy right here so it's the 8 here the exponent divided by the index here that's why i do that so quickly when i use my power to power rule this becomes x raised to the power of 8 4 which is x squared all right so now let's go ahead and write this we know the 4th root of 16 is 2 so we'll put 2 we know the fourth root of x to the eighth power is x squared so we'll put that and then times the fourth root of what do we have left we have a 5 and then times an x so times the fourth root of 5x so our answer here is 2x squared times the fourth root of 5x all right so for the next part let's talk about the quotient rule for radicals so if we had the nth root of a over b this is equal to the nth root of a over the nth root of b so this would help us if we had something like let's say the square root of 4 over 9 right now that looks kind of hard to solve so you can break this up and say it's the square root of 4 over the square root of 9. so you're attacking the numerator and denominator kind of separately there the square root of 4 is 2 the square root of 9 is 3. so we get 2 3 as our answer okay pretty simple so suppose you saw something like the square root of 7 64. so all i'm going to do is just break this up and say this is the square root of 7 over the square root of 64. the square root of 64 is 8. so i'll write the square root of 7 over 8. i can't do anything else with the square root of 7. 7 is a prime number nothing else i could do so this is as simple as you can make this square root of 7 over 8. all right let's look at another one with a variable involved so we have the fourth root of 625x to the fifth power over y to the 12th power okay so i'm just going to break this up so i would have the fourth root of 625 x to the fifth power over the fourth root of y to the twelfth power so the denominator is very easy right again if this guy is greater than or equal to this guy then think about if they're divisible by each other so 12 is divisible by 4. i don't need to do anything else 12 divided by 4 is 3 so i can just write this as y cubed i'm done okay i don't have to do anything else up here i've got the fourth root of 625 times the fourth root of x to the fifth power if i wanted to break it up so the fourth root of 625 times the fourth root of x to the fifth power now i know i can simplify here because the 5 is larger than the 4 we'll get to that in a second the 4th root of 625 is 5 right because 5 to the 4th power is 625 so let's go ahead and just simplify this we know that's a 5. so we would have a 5 multiplied by the fourth root of x to the fifth power over y cubed now how do we simplify this again if this guy which is a 5 is greater than or equal to this index which is a 4 we know we can simplify again think about is 5 divisible by 4 no the next number down that would be would be a 4. so i'm just going to break this up and say this is 5 times the fourth root of x to the fourth power times the fourth root of just x over y cubed okay so this guy right here again if i divide this by this i get one so we would have five times just x right it'd be x to the first power then times the fourth root of x and then over y cubed all right now let's talk about how we can multiply radicals with a different index okay this is something that's pretty tedious but you need to know how to do it in case it comes up so we had something like the square root of 7 multiplied by the cube root of 3 what do you do you can't just multiply the radicands together and you know throw some index in there randomly what you have to do is find the least common multiple between the indexes so if this is a 2 and this is a 3 the least common multiple is 2 times 3 or 6. so i'm going to write this in exponential notation to make it really easy so this is 7 to the one-half power times three to the one-third power and all i'm going to do is write this with a common denominator okay the common denominator is going to be a 6 because that's going to be my common index so multiply this by 3 over 3 multiply this by two over two you're going to get seven raised to the power of three over six times three raised to the power of two over six and you can write this in exponent form like it is now or you can go back to radical form whatever you want to do so we can essentially say this is 7 cubed and we're going to take the 6th root of that times we're going to say this is 3 squared and we're going to take the sixth root of that okay and you can even combine them under one radical doesn't matter however you want to do it for most of these you're going to end up using a calculator anyway you just need to know how to do it so what is 7 cubed that's 343 what is three squared that's nine so you would have the sixth root of 343 times nine which is the sixth root of three thousand eighty seven three thousand eighty all let's talk about one last concept before we wrap up the lesson so we have something known as the power rule for radicals and you can see this in radical notation or exponential notation so if you have a raised to the power of one over n and then you raise this to the power of one over m a stays the same you're using your power to power rule basically you multiply the exponents you get a raised to the power of one over m times n in radical notation if you have the nth root of the nth root of a basically this radicand here of a is going to stay right so when you write your radical here the radicand is still a you get your index as the product of these two indexes so m times n so suppose we had the fourth root of the cube root of five so based on what we just saw we know we would have a radicand of five and we know we would just multiply these indexes together so four times three would be twelve this is the 12th root of 5. but again if you're struggling you could start out by writing this inside guy right here as 5 to the power of 1 3 and then this guy is being raised to the power of 1 4 right so this guy 5 stays the same power to power rule 1 3 times 1 4 is 1 12 5 to the 1 12 power is the same thing as the 12th root of 5. all right for the last problem we're going to look at the seventh root of the square root of 14x and again all i'm going to do is i'm going to take this guy right here this 14x i'll write that as the radicand and then for my index this guy is a 2 this guy is a 7. 7 times 2 is 14. i'll have the 14th root of 14x again if you're struggling you can write this as 14x this whole guy is going to be raised to the one-half power and then that's going to be raised to the one-seventh power so this would be equal to 14x would stay the same and this would be raised to the power of one half times 1 7 which is 1 14. this whole thing would be in parentheses so you could really write this as 14 to the 1 14 power times x to the power of 1 14 which is essentially what we have here right if it's inside of parentheses like this it looks like that but this is the same thing they're equivalent because i could write this as the 14th root of 14 times the 14th root let me make that better again the 14th root of x but you wouldn't want to do that so we would just leave our answer in this format here and say that this is the 14th root of 14x in this lesson we want to review operations with radicals so before we jump in and start working on some problems let's just take a minute to review the concept of like radicals so before we do this let's think briefly about like terms so we know if we have something like 3x and 7x these are like terms right if i add 3x plus 7x all i need to do is add 3 and 7 that gives me 10 and the common variable of just x comes along for the y right these can be added because they're like terms if i had something like negative 5 x squared y squared and let's say 3 x squared y squared again these are like terms you have the same variable parts exactly you have x squared and x squared y squared and y squared so if i added these guys let's say i did negative 5 x squared y squared plus 3 x squared y squared again all i do is negative 5 plus 3 just those numbers those coefficients that gives me negative 2 and then times the variable parts that are the same so that kind of common variable part so x squared y squared okay that just comes along for the ride if we got let me erase this something where we didn't have like terms let's say it was 3x to the fourth power minus 2 x squared well i can't just subtract here i can factor out an x square but that really doesn't do anything for me so there's nothing i can really do to make that simpler right i can't subtract there because they don't have like terms i have an x to the fourth power and an x squared so i just kind of leave that alone the same concept is going to apply when we work with radicals we've got to have the exact same radicand okay that's the number under the radical symbol and we've got to have the same index so if i had something like 4 times the square root of let's say 11 and let's say i had 5 times the square root of 11 these are like radicals okay this guy right here is a square root it's got an implied index of 2. this guy is a square root it's got an implied index of 2 and the radicand is 11 and the radicand is 11. so same index same radicand those are like radicals okay as another example suppose we had something like 5xy multiplied by the cube root of 2x and let's say we had negative 11xy multiplied by the cube root of 2x so these guys are like radicals you have the cube root and the cube root right so the index is the same a 3 and a 3. and then the radicand is the same it's a 2x and a 2x okay so that's what you're looking for when you have like radicals to give you an example something that's not like radicals suppose you had i don't know let's say the cube root of x squared y and then let's say you have the cube root of 2 x squared these are not like radicals because although the index is a 3 in each case the radicand is not exactly the same you have 2x squared y and then just 2x squared okay so again the index has to be the same and the radicand has to be the same for you to have what's considered like radicals so when we do have like radicals we can add or subtract by basically using the distributive property that's kind of the slow way if you want to kind of factor things out the fast way is just to work with the numbers that are multiplying the radical so just like we saw when we combined like terms if you had 3x plus 5x what would you do you would do 3 plus 5 that's 8 and then times the common variable x well it's the same thing here if i have 2 times the square root of 7 plus 5 times the square root of 7 i'm just going to do 2 plus 5 that's 7 times this common radical which is the square root of 7. and i can show this officially using the distributive property it's just a little bit slower i could factor out the square root of 7 from each of these okay so if i factor that out i'd have the square root of 7 out in front of some parentheses and then inside i would have 2 plus 5 right i'd have a 2 left here and a 5 left there and if i did this operation 2 plus 5 is 7 so i would get 7 times the square root of 7. but again you don't want to go through this each time you add or subtract radicals you just want to work with the numbers that are multiplying the radicals so again 2 plus 5 is 7 and then times this common radical of the square root of 7. okay very very simple all right suppose we had 4 times the cube root of 11 minus 13 times the cube root of 11. again you have like radicals you have a cube root in each case so the index is a 3 you have an 11 and an 11 as the radicand in each case so those are like radicals so all i need to do is work with the numbers that are multiplying the radical so i have a 4 minus a 13. 4 minus 13 is negative 9 so i'm just going to put negative 9 times the cube root of 11. okay it's just that simple all right let's take a look at another one so suppose we had 3 times the 4th root of 11xy minus 7 times the 4th root of 11xy so just because we have variables involved doesn't change anything i have the same index i have a 4 and a 4 and i have the same radicand i have 11xy and 11xy so everything has to match in this case it does so i'm going to do 3 minus 7 which is negative 4 and then times i'm going to have the fourth root of 11xy all right let's take a look at some harder examples now so suppose you see negative 3 times the square root of 2 plus 2 times the square root of 162 plus 2 times the square root of 128. now at first it looks like you can't simplify anything because you don't see like radicals right you see a square root of squared and square root but you have a 2 162 and 128 as your radicands so what you need to do here is simplify everything first and then determine if you have like radicals this guy right here this 2 times the square root of 162 you know 162 is 81 times 2 right 81 is a perfect square so i know this can be simplified this is 2 times the square root of 81 times the square root of 2. the square root of 81 is 9. so i could say this is 2 times 9 or 18 times the square root of 2. so let's put equals you have negative 3 times square root of 2 plus this guy is going to be 18 times square root of 2. now what about this 2 times square root of 128 so 2 times square root of 128 this guy right here would factor into 64 times 2. so i could write this as 2 times the square root of 64 times the square root of 2. we know that 64 is a perfect square it's 8 times 8. so i could say this is 2 times 8 or 16 times the square root of 2. so i'll put plus 16 times the square root of 2 and now that we have everything in its simplest form we can see that we do have like radicals we have a square root a square root and square root we have a radicand of 2 a radicand of 2 and a radicand of 2. so all i'm going to do is just combine the numbers that are multiplying that common radical so negative 3 plus 18 is going to be 15 15 plus 16 is going to give me 31. so this is 31 multiplied by the square root of 2. all right let's look at one that's a little bit more challenging so we have x squared y cubed times the cube root of 128 x to the fourth power y to the fifth power minus you have the cube root of negative 54 x to the 10th power y to the 14th power so this one's going to be tedious i'll warn you in advance it just takes a while to simplify so i'm just going to look at each one separately and then we'll kind of come back up so if i had x squared y cubed times the cube root of 128 x to the fourth power y to the fifth power let's think about each part separately so let me kind of scroll down and get some room so 128 i know that's 2 to the seventh power so let me write the cube root of 2 to the 7th power now if this exponent here is a 7 and this index is a 3 then i want to think about what number going down from 7 is going to be divisible by 3 it will be 2 to the 6th power so let me go ahead and break this up into the cube root of 2 to the sixth power times the cube root of 2. this guy right here all i got to do is divide this by this 6 divided by 3 is 2 so this would be 2 squared or 4. so i'm going to go ahead and write this i'm going to keep this part the same x squared y cubed this guy is going to simplify to a 4 so i'm going to put that on the front and then times the cube root of 2. now i'm not done because i still need to work on those variables there so if i had the cube root of x to the fourth power what would this be again i think about this exponent and this index what number going down from 4 would be divisible by 3 3 would be so i could say this is the cube root of x cubed times the cube root of x this guy would simplify to what 3 divided by 3 is 1 so that would be x okay so if i multiply that by x squared this would now become x cubed and out here i would have times the cube root of x which i can just write that underneath here okay so now let's deal with the cube root of y to the fifth power again what number going down from 5 would be divisible by 3 3 would be so i could write this as the cube root of y cubed times the cube root of y squared this guy right here would simplify to y so y times y cubed would be y to the fourth power and this guy right here there's nothing really i can do with it so i'm just going to write a y squared underneath here and i'm basically done simplifying this guy so let me erase all of this and i'm ready to drag this up so i'm just going to paste that in here so this part right here is going to simplify to this so now let me put minus and let me deal with this down here so i have the cube root of negative 54 x to the 10th power y to the 14th power so let's think about that guy now so thinking about negative 54 if i have the cube root of that let's go ahead and separate this into the cube root of negative 1 times the cube root of what the thing about 54 it's 27 times 2. 27 is a perfect cube 27 is 3 times 3 times 3 so then times the cube root of 2. so this can be simplified the cube root of negative 1 is negative 1. this can be simplified the cube root of 27 is 3 and this can't be simplified so i'm going to say this is equal to you would have negative 3 times the cube root of 2. now again i've got to deal with these variables so i've got this x to the 10th power so the cube root of x to the 10th power again to deal with this i want to think about what exponent going down is divisible by 3 9 would be right so i could write this as the cube root of x to the ninth power times the cube root of x so 9 divided by 3 is 3 so this guy would simplify into x cubed so let me slide this negative 3 down and i'll put my x cubed next to that let me kind of move this over a little bit and what i'm left with is the cube root of x so i'll just write an x under there so we can erase this for now and now lastly i have the cube root of y to the 14th power again what number here going down from 14 is going to be divisible by 3 it's going to be 12. so i can write the cube root of y to the 12th power times the cube root of y squared okay so this guy right here is going to simplify to y to the fourth power again 12 divided by three is four so that would be y to the fourth power this guy right here can't do anything with it so i would write y squared underneath here because i have the cube root of y squared okay so we've simplified the other guy now let's erase this and let's go ahead and drag this up okay so we'll put that there now notice we have minus a negative so we're going to go ahead and just write plus here okay minus the negative is plus a positive so now i do have like radicals i have the cube root and the cube root i have 2xy squared and 2xy squared another thing is you've got to have like terms right because these guys right here have variables involved so they got to be like terms for you to add anyway but you do have that as well right you have x cubed x cubed y to the fourth and y to the fourth so all i need to do is add 4 plus 3 that's 7 and then times x cubed y to the fourth and then times this common radical which is the cube root of 2 x y squared and that's my answer let's take a look at another one that's pretty tedious so we have x multiplied by the fourth root of 256 x to the fifth power y to the sixth plus y multiplied by the fourth root of 625 x to the ninth y squared so let's try to again just separate things and simplify and then come back up so x multiplied by the fourth root of 256 x to the fifth power y to the sixth power so let me scroll down get a little room going if i think about 256 it is 2 to the 8th power so it's a perfect 4th right the fourth root of 2 to the 8th power 8 divided by 4 is 2 so this is 2 squared right so this guy would be 2 squared or 4 times x times the fourth root of x to the fifth power y to the sixth power then if i think about the fourth root of x to the fifth power again this is pretty easy at this point what number going down from 5 is going to be divisible by 4 4 would be so i write this as the fourth root of x to the fourth power times the fourth root of x this guy is just x right 4 divided by 4 is 1. so if i had x multiplied by x it would be x squared so you can get rid of this guy and you're just left with the fourth root of just x so go ahead and erase this 5 here and that'll be done there so what about the fourth root of y to the sixth power again if you think about this number and this number going down from six a four will be divisible by four so i can say this is the fourth root of y the fourth power times the fourth root of y squared okay so this guy is going to just be y because 4 divided by 4 is 1. so i'll put a y out here so that's gone and then i'm thinking about the fourth root of y squared so i'll change this to a 2 here okay so let's erase all this and this and let's go ahead and drag this back up to the top all right for this guy right here we have y times the fourth root of 625 x to the ninth power y squared so let's put plus we already know 625 is a perfect fourth the fourth root of 625 is 5. so i can go ahead and write this as 5 y and then i'll put a little bit of space here so let me erase this and now i'm just thinking about what is the fourth root of x to the ninth power what is the fourth root of y squared so what is the fourth root of x to the ninth power and what is the fourth root of y squared well this guy can't really be simplified other than the fact that this part right here and this part right here they share a common factor of two but we'll deal with that in the end so for right now i'm just going to put the fourth root of y squared like that and i won't have to worry about this for now i just want to simplify this so i know that this 9 here is not divisible by 4 but an 8 would be so i can split this up and say this is the fourth root of x to the eighth power times the fourth root of x so this is what's going to go underneath here and notice you do have like radicals now so i can get rid of that the fourth root of x to the eighth power is going to be x squared right 8 divided by 4 is 2. so this becomes x squared here and to write this in the same order i'll put x squared first and then y the order doesn't matter but i don't want that to trip you up so look at what you have you have 4x squared y and 5x squared y so those are like terms and then you have the fourth root of xy squared and the fourth root of xy squared those are like radicals so all we need to do is just add 4 plus 5 is 9 and then times x squared y and then this is times the fourth root of x y squared now between this guy right here and this guy right here you have a common factor of 2. so you can further break this down and say that you have 9 x squared y times the fourth root of x times if i think about y squared and we have the fourth root of that well in exponent form this is y squared raised to the one fourth power so this is y raised to the power of two over four or y raised to the one-half power so then you can put times the square root okay the square root of y okay and if you wanted to make that a little bit clearer we can put a multiplication sign there because i know it's kind of jambled up so let's write this as 9x squared y i'll put times the square root of y and i'll put times the fourth root of x all right let's take a look at some easier problems now so a lot of times when you work with radicals you do involve multiplication and this guy right here is just going to be using my distributive property so the square root of 5 multiplied by the quantity the square root of 7 plus 2. so just distribute the square root of 5 to each term the square root of 5 times the square root of 7 is just the square root of 35 nothing to simplify there the square root of 5 times 2 is just 2 times the square root of 5. okay pretty easy all right for this guy we're going to use foil we have basically two binomials here we have 4 plus the square root of 3 multiplied by 2 minus the square root of 3. so first terms 4 times 2 is 8 outer terms 4 times negative square root of 3 is going to be negative 4 times square root of 3 inner terms square root of 3 times 2 is 2 times square root of 3. then for my last terms i have the square root of 3 times the negative square root of 3 so that's just going to be minus 3. so what can i combine 8 minus 3 is 5 and then negative 4 times square root of 3 plus 2 times square root of 3 is negative 2 times square root of 3. so you get 5 minus 2 times square root of 3 as your answer let's do one more here and you'll probably notice that we can use our special products formula pattern here we have the square root of 13 minus 11 times the square root of 13 plus 11 so this is basically the product of conjugates right you have the same term here and here you have the same term here and here and you have opposite signs right a plus and then a minus so all i need to do i know the two middle terms would drop out i just need to square the first guy so if i squared the square root of 13 this would give me 13 okay so i would have 13 and then minus if i square 11 i get 121. so what is 13 minus 121 that's going to be negative 108 okay negative 108. in this lesson we want to review rationalizing the denominator so in order to present a simplified radical as an answer you have four criteria that you need to meet so first and foremost we say that the radicand that's the number under the radical symbol has no factor raised to a power greater than or equal to the index now what does this mean in layman's terms we already talked about this a few lessons ago essentially if you had something like the square root of 20 the index here is a 2 on a square root so you don't want any factors of this radicand this number 20 to be a perfect square so with 20 we know this would factor into 4 times 5 and 4 is a perfect square so we could write this as the square root of 4 times the square root of 5. this guy would represent the rational number two so we would just write this as two times the square root of five and that would be considered simplified okay if you had a cube root then you don't want any factor of the radicand to be a perfect cube if you had a fourth root you don't want any factor of the radicand to be a perfect fourth you know so on and so forth all right the next item on the list the radicand has no fractions we learned about the quotient rule for radicals that's how we can deal with that if you had something like let's say the square root of 12 over 25 we could break this up into the square root of 12 over the square root of 25 we know the square root of 25 is 5. so let's just write that down there in the denominator for the square root of 12 we can simplify because it's 4 times 3 right that's 12. so we have the square root of 4 times the square root of 3. so this guy right here again square root of 4 is just 2 so i can end up writing this as 2 times square root of 3 over 5. okay then this guy right here is the subject of today's lesson we have no denominator contains a radical so if i had something like let's say 4 over the square root of 2 we don't want a radical in the denominator that's considered not simplified so the process we use to clear a denominator of any radicals is known as rationalizing the denominator okay we'll talk about that just in a second now lastly we see that exponents in the radicand and the index of the radical have no common factor so what this means is if you had something like let's say 5 to the sixth power and let's say you have the ninth root of that well between the nine and the six there you have a common factor of three so you can quickly say okay if i divide nine by three i get three and if i divide 6 by 3 i get 2. so this would be the cube root of 5 squared or basically the cube root of 25. now if that doesn't make sense for you doing it that way you can always convert this to exponential form you can say i have 5 to the sixth power raised to the 1 9th power and of course you can use your power to power rule so this is 5 raised to the power of 6 times 1 9 which is 6 9 and from there you can just simplify the fraction six and nine again each have a common factor of three divide six by three you get two divide nine by three you get three so this ends up being five raised to the two thirds power which again is just the cube root of 5 squared which is the cube root of 25 okay it's just simpler to divide here than to go through all this with the exponential form but if it's something you struggle with some of you can't mentally comprehend yet then go ahead and take the time convert it to exponential form and then you'll be good to go let's talk a little bit about rationalizing the denominator now it's a pretty easy concept overall so if i had something like 5 over the square root of 2 realize that you have a radical in the denominator and again if you want to present a simplified radical this is not allowed so right now this square root of 2 is an irrational number okay it's an irrational number so in decimal form it does not terminate it does not repeat so to simplify this all i really need to do is multiply the numerator and denominator by the square root of 2. and let's think about why that works for a second the first thing is the square root of 2 over the square root of 2 represents the number 1 right it's just a complicated form of 1. i can always multiply by 1 and i leave my number unchanged so i'm not doing anything illegal in the numerator 5 times square root of 2 is just 5 times square root of 2 nothing fancy there in the denominator what's going to happen is square root of 2 times square root of 2 is the square root of 2 squared or the square root of 2 times 2 which is just going to be 2 right it's the square root of 4 which is 2. so this is going to be equal to 5 times square root of 2 over 2. so now this 2 is a rational number so i've changed my denominator from an irrational number to a rational number henceforth the term rationalizing the denominator so basically to do this especially if you're just working with a square root just make sure your square root is simplified first and then once that's done just multiply the numerator and denominator by that square root so let's look at another example so if i had something like 6 over the square root of 3x this guy right here is already simplified so i'm just going to multiply this by the square root of 3x over the square root of 3x and what would i get 6 times the square root of 3x is just 6 times the square root of 3x down here the square root of 3x times the square root of 3x is just 3x now i'm not done because i can simplify further between the numerator and denominator i have a common factor of 3. so this divided by this will give me what that's gone this would be a 2 so we end up with 2 times the square root of 3x over x and that's my final answer i can't do anything else but again notice how i don't have an irrational number in the denominator i just have an x all right for the next one we're going to look at one with a higher level root so we have 7 over the cube root of 2 and essentially what i want to end up with is something in the denominator that's the cube root of something that's a perfect cube so if i think about 2 2 times 2 would be 4 4 times 2 would be 8. so i want to end up with the cube root of 8 because 8 is a perfect cube so i'm going to multiply this by the cube root of 4 because 2 times 4 would give me 8. and i got to do the same thing to the numerator to make it legal so i would have 7 times the cube root of 4 over again in the denominator the cube root of 2 times the cube root of 4 would be the cube root of 8 which is just 2 right so i'd have 7 times the cube root of 4 over 2. all right for the next one we'll look at 15xy over the cube root of 25x squared y so if i think about 25 it's 5 squared i know x squared is just x squared and i know y is y to the first power i need a perfect cube under here so what i'm going to need is 5 cubed i'm going to need x cubed and i'm going to need y cubed so what am i missing to get to 5 cubed i just need a 5. so i'm going to have the cube root of 5 and then what am i missing to get to x cubed i need an x what am i missing to get to y cubed i need a y squared okay so everything under there would basically be a perfect cube after it's multiplied so if i go through here and i'll deal with the numerator in a second if i go through here and multiply 25 times 5 is 125. x squared times x is x cubed and y times y squared is y cubed so if i take the cube root of this again this is a perfect cube i'm going to end up with 5 x y as my denominator okay so once i've figured that out the numerator part is pretty easy i'm going to have 15xy multiplied by the cube root of 5xy squared so i'm just going to write this here we have 15 xy times the cube root of 5xy squared and again one more time down here so 15xy times the cube root of 5 x y squared and we can do some simplifying here we see between numerator and denominator we have a common factor of 5. so this becomes 3. we have a common factor of x so that's gone and then we have a common factor of y so this guy will cancel with this guy and i'm basically just left with a 3 multiplied by the cube root of 5 x y squared i'll make that a little better all right let's take a look at another one so we have 2x minus 5y over the fourth root of 125xy cubed z squared again if i have a fourth root here i'm looking to make this into a perfect fourth down here so for 125 i know that's five cubed for x i know that's x to the first power we have y cubed and we have z squared so what do i need to do to make this into a perfect fourth five cubed i'm missing a five x to the first power i'm missing an x cubed y cubed i'm missing a y and z squared i'm missing a z squared so i need to multiply the numerator and denominator by the fourth root of 5 x cubed y z squared okay let me erase this so this is equal to i've got to be careful here because when i multiply this by this this is two terms here okay so i've got to make sure that each term is multiplied by this so i'm going to use some parentheses i'm going to put my fourth root of 5 x cubed y z squared out in front and inside of parentheses i'll have 2x minus 5y again with two terms there this has to be multiplied by each term then this is over if i multiply these guys i'll have the fourth root of 125 times 5 is 625 x times x cubed is x to the fourth power y cubed times y is going to be y to the fourth power and z squared times z squared is z to the fourth power so notice how we have a perfect fourth down here so let's go through and keep simplifying and i'll scroll down get a little room going so i'll have my fourth root of 5 x cubed y z squared multiplied by the quantity 2x minus 5y then this is over this guy right here the fourth root of 625 is 5. then the fourth root of x to the fourth power is x the fourth root of y to the fourth power is y and the fourth root of z to the fourth power is z now i can go through and i can distribute this to each term if you want but it's not really necessary because this is as simple as we can really make it but for the sake of completeness we will clear these parentheses so we'll say we have 2x times the fourth root of 5 x cubed y z squared then minus you're going to have 5 y times the fourth root of 5 x cubed y z squared then this is all going to be over you're going to have that 5 x y z all right let's talk a little bit about rationalizing a binomial denominator so for this guy right here it's a pretty easy concept you're going to use your formula for multiplying conjugates to clear a binomial denominator that has a radical in it so what do i mean by that i want you to recall that if you have x plus y multiplied by x minus y remember these guys are called conjugates it's the same first terms it's the same last terms you have different signs so a plus and a minus when you multiply these two together the middle two terms drop out so you end up with just the first term squared minus the last term squared so what's going to happen is i'm going to multiply this by we'll have the square root of x minus 3 over the square root of x minus 3 okay so what i did was in my denominator i now have the product of conjugates it's the same two terms so square root of x and square root of x 3 and 3 but different signs so what's going to happen is the middle two terms are going to drop out they'll cancel and you'll be left with the first term squared square root of x squared is x minus the last term squared 3 squared is 9. okay so if you want to go through that the long way square root of x squared of x is x the outer would be minus 3 times square root of x the inner would be plus 3 times square root of x so you can see those would cancel and the last would be minus nine okay so this is gone and you're just left with x minus nine right which is exactly what you have there then in the numerator we basically have one times this quantity which is just that quantity so you get the square root of x minus three up there okay nothing you can really do here to simplify further so we just have the square root of x minus 3 over x minus 9. all right let's take a look at another one so we have the square root of x over 3 minus the square root of x again i'm just going to multiply the numerator and denominator by the conjugate of the denominator so i would have 3 plus the square root of x over 3 plus the square root of x again it's the same two terms in the same order just change the signs okay that's all you've got to do so in the denominator i would just square the first guy 3 squared is 9 then minus square the last guy square root of x squared is just x and then in the numerator we would basically have 3 times square root of x and then plus square root of x times square root of x would be x nothing else i can really do to simplify here so 3 times square root of x plus x over 9 minus x is your answer all right for the last one we'll look at one that's a little bit challenging so we have 3x over we have 2 plus the square root of x plus y so the idea is to treat this is the first guy and this is the second guy so if i want to multiply by conjugates here i'm just going to say this is 2 minus the square root of x plus y and then down here i'm going to put 2 minus the square root of x plus y okay so when you go through and multiply it's the same principle you'd square the first guy which is 2. so that's 2 squared which is 4 and then minus you'd square this guy right here so if i squared the square root of x plus y i would just get x plus y but you've got to be careful because you've got to subtract the whole thing away okay so this would be minus x and then minus y in the numerator you would have 3x times 2 which is 6x and then minus you would have 3x times the square root of x plus y now nothing you can do to simplify this further you just get 6x minus 3x times the square root of x plus y over 4 minus x minus y in this lesson we want to review the addition property of equality so for the majority of you taking the course you fully understand how to solve a linear equation one variable but again it might have been some time since you've kind of gone through the steps and you might not have fully ever understood it so it's just good to get a review in before we look at anything that is more challenging so a linear equation one variable is just defined as something like ax plus b equals c where a is said to not be equal to zero so you have this guy right here this a which is the coefficient of x you have b and c which are constants and all of those can be any real number that you want so a b and c can be anything you want with the exception of a which cannot be zero the reason a can't be zero is because it's multiplying the variable and if zero was multiplying x well then you'd have no variable involved so that would be an issue for us now a linear equation in one variable is sometimes called a first degree equation because the largest exponent on the variable is a one right and some of you remember that quadratic equations where the largest exponent on a variable is a two are sometimes called a second degree equation okay so it's the same concept there so a few examples of a linear equation one variable something like 2x minus 5 equals 11 where a here would be 2 b would be negative 5 and c would be 11. or something simple just like x equals 19. right here a would be 1 b would be 0 and c would be 19. right so something like that or you could have some fractions involved you could have something like 1 4 x plus 1 9 equals 23 17. okay here a would be one-fourth b would be one-ninth and c would be 23 17. so when we work with equations we need to understand the basic definition so we're saying that two algebraic expressions are set equal in value so this algebraic expression here of 2x minus 3 is set equal to this algebraic expression here of 7. so this guy and this guy are each algebraic expressions but the whole thing because of the equality is known as an equation right so the 2x minus 3 is equal to 7. now when we solve an equation and we'll talk about this more in a moment we're basically finding the value or the values that when we replace the variable or the variables gives us a true statement when we say gives us a true statement it means that the left side of the equation has the same value as the right side of the equation so they are the same so in this example here i've given you the solution so x is equal to 5. that means that i can plug a 5 in here for x and that will give me the same value on the left as i have on the right so let's try that out so we have 2 multiplied by plug in a 5 for x then minus 3 equals 7 and we just simplify each side 7 on the right i can't do anything with that on the left 2 times 5 is 10 so you'd have 10 and then minus 3. 10 minus 3 is 7. let me kind of write that a little better again 10 minus 3 is 7 so we would say 7 is equal to 7 and that's true 7 is the same or is equal to 7. so this is a true statement now if you plug something in for the variable and it's not the correct solution you will not get a true statement so let's say that i said that x was 4 as an example so let's erase this and this and this we'll put a 4 here 3 times 4 is 8 8 minus 3 is 5. 5 does not equal 7. so this would be false okay this would be false so when you work with a linear equation one variable you can always check to make sure you got the right answer you just plug your proposed solution in for your variable and then you make sure you have the same value on the left as you have on the right so let me put this back to the way that it was so that it's correct again so again we said this was a 5 and so this would be 10 and this would be 7 and this would be true so in most cases we are not going to be just simply given our solution we have to go through a series of steps to obtain our solution again when we solve an equation we are finding the value or the values that make the equation true your ultimate goal when you're solving an equation is to get the variable you're looking for in this case it's x equal to some number that's when it's solved okay x equals some number that is the value i can plug back in for x again that makes the equation true so if you want to write something down you would say that your goal is to isolate isolate the variable okay so isolate the variable on one side of the equation so there's a few properties we can talk about real quick that'll get us all the way there so the first thing is that adding a number and its opposite will always give you zero the second thing is that adding zero to a number leaves it unchanged and the third thing is that we can add or subtract the same value to or from both sides of an equation and not change the solution so that last property is known as the addition property of equality and that's the focus of our lesson today so if i have something like again x minus 7 equals 13 and i want to isolate the variable i want to undo what's being done to x if it's subtraction i want to think about addition if it's addition i want to think about subtraction if it was multiplication like we'll see in the next lesson i want to think about division if it was division i want to think about multiplication here i'm subtracting away 7 so to undo that i just want to add 7 because negative 7 plus 7 gives me 0. again the additive inverse property and then if i had x plus 0 on this side that's just equal to x right 0 is the additive identity so let's go through this guy again i'm subtracting away 7 so i would start by adding 7 to kind of counteract that i'm subtracting away 7 i add 7 but then to make it legal i've got to add 7 on the right side as well so i did this to this side and this to this side you think about that as maintaining the balance if i add 7 to the left i've got to add 7 to the right and then i'm good to go so now it's pretty simple if i have minus 7 plus 7 we can think about that as negative 7 plus 7 that's just 0. so i have x plus 0 is equal to 20 which is just x equals 20 right so that's our solution there nice and easy and you might see people or you might see your textbook write your solution like this this is known as solution set notation it's not something we're going to use too much throughout the course but i just want you to know that it does exist in case you see it you're not wondering why you know it's notated differently or wondering if you made a mistake it's just different notation so we can always check this again by plugging in a 20 for x and making sure we have the same value on the left as on the right so i would have a 20 minus a 7 equals 13. 20 minus 7 is 13 so 13 equals 13. so yes this is true this is true we have the right answer and again for this one it's so easy you could just eyeball it what number minus 7 gives us 13. well 20 does right 20 minus 7 is 13 so we know we have the correct answer here all right let's take a look at a number so we have x plus 11 equals negative 34. again our goal is to isolate isolate the variable okay so how do i do that i have plus 11 here if i'm adding 11 to x how can i undo that just subtract 11 away right if i have 11 minus 11 or 11 plus negative 11 i get 0 right so it's gone right i'd have x plus 0 which is just x so i'd have x plus 11 and then plus negative 11. or if you want to you could just put minus 11. it doesn't match the same thing and then is equal to you have negative 34 because i did it over here i've got to do it over here okay you've got to maintain that balance again that's the addition property of equality you can add the same number to both sides of the equation so over here i can just cross this out and put a zero right it's just basically canceled you're left with x and this is equal to negative 34 plus negative 11 is going to be negative 45. again we can check this we think about plugging in a negative 45 for x there so if i had negative 45 plus 11 does that equal negative 34 yes it does negative 45 plus 11 is negative 34. so you get negative 34 equals negative 34. so this is true we can just check that off and say true all right let's look at one that's a little bit more challenging so we have 2x minus 7 equals x minus 3. so you can still just solve this with addition or subtraction and again our goal is to isolate the variable always think about that we're isolating the variable so how do i isolate the variable here we have 2x minus 7 equals x minus 3. well the first thing is i've got to get all the variable terms to one side and then i want all the numbers on the other right that's going to give me the possibility of having x equals some number right now i have an x involved in this guy here and an x involved here so i've got to move this over here and the way i can do that is i can use the addition property of equality i can add negative x to both sides of the equation so i can say this is 2x minus x minus 7 is equal to x minus x minus 3. again all i did was i subtracted x away from each side or again you could write this as plus negative x it doesn't matter whatever is more convenient for you x plus negative x is 0. so this is gone i now have 2x minus x minus 7 is equal to negative 3. now i can simplify because 2x minus x is just x so i have x minus 7 is equal to negative 3. and now all i need to do is think about the fact that negative 7 is being added to x or you can think about that as 7 being subtracted away from x so i want to add 7 to each side of the equation so i'm going to do this in vertical format because it's a little bit simpler so i'm going to put plus 7 here and then plus 7 here it's just a more compact way to do it this is going to cancel and i'm going to end up with x is equal to over here negative 3 plus 7 is going to be positive 4. so if we wanted to check this let me erase everything so i would plug in a 4 for this x here and this x here so it's got to be for each occurrence of the variable okay so i would have 2 multiplied by 4 minus 7 is equal to 4 minus 3. 2 times 4 is 8. so you'd have 8 minus 7 over here 4 minus 3 is 1. 8 minus 7 is 1. so you get 1 equals 1. this is true okay so we know our solution here x equals 4 is correct in this lesson we want to review the multiplication property of equality so in our last lesson we reviewed the addition property of equality now this property allows us to add or subtract the same value to or from both sides of an equation so generically if we have something like let's just say a equals b we could say with the addition property of equality i can add c to each side of the equation and maintain that equality right it's still true so this allowed us to solve basic equations with just addition and subtraction right one step equations very very simple something like x minus 11 is equal to let's say 17. what i want to do when i solve an equation is just isolate the variable on one side so if i want to isolate x here on the left side i've got to undo what's being done to it here i'm subtracting away 11 so the opposite of subtracting away 11 is adding 11. so i'm going to use my addition property of equality to just add 11 to each side of the equation so i would have x minus 11 plus 11 equals 17 plus 11. so again notice how i did it to the left and also the right that's what made it legal over here negative 11 plus 11 is 0. so i can just say this is x over here so i've isolated the variable and then here 17 plus 11 is going to give me 28. so x here would just be 28 right 28 minus 11 does give me 17 so this is in fact the correct solution now we're going to see more complicated linear equations obviously and we need more tools than just the addition property of equality so in this lesson we're going to review something known as the multiplication property of equality and this property basically tells us that we can multiply or divide both sides of an equation by the same non-zero number and not change the solution okay so the reason i say non-zero number is if you multiply both sides by zero you get zero on each side right it's done for if you divide by zero it's undefined so you have to put zero in as a restriction there okay so you've got to maintain that so essentially what we're saying here is that if a is equal to b then a multiplied by c is equal to b multiplied by c so you maintain the equality even though you've multiplied both sides by that same value c we would specify here that c does not equal zero okay so how can i use that to solve something like 3x equals 12. well let's erase this real quick and again i want to write down my goal here which is always always always to isolate the variable okay we want to isolate the variable very important to remember that stay focused on that goal well if i want to isolate the variable x what's being done to x right now x is being multiplied by 30. so how do i undo what's being done to the variable well if i'm multiplying by 3 i just need to divide by 3. and again the multiplication property of equality lets me divide both sides of the equation by 3 so that's legal to do as long as i do it by both sides so 3x divided by 3 is equal to 12 divided by 3. now 3 over 3 is 1 right any non-zero number divided by itself is 1 we already know that multiplying by 1 leaves a number unchanged so essentially this becomes 1 times x or just x right 1 is the multiplicative identity so i can just write x here i don't need 1x this is equal to 12 divided by 3 which is just 4 so i get x equals 4 as my solution right so i can plug a 4 back in for x here and verify that i got a true statement so 3 times 4 is equal to 12. we know that is correct right 12 equals 12 so we can go ahead and say that this guy is true okay check that off and say yes x equals 4 is the correct solution all right let's take a look at another one so we have x divided by 5 or you can say x over 5. it's in fractional form this is equal to negative 20. so again what's being done to x we want to undo that so that we can isolate isolate the variable if you just write this down each time it's very very simple to go through the steps and figure it out if i want x by itself and it's being divided by 5 just undo it right what can i do to undo division multiplication so i would multiply both sides of the equation by 5 so x over 5 times 5 is equal to negative 20 times 5. and so this 5 would cancel with this 5. 5 over 5 is just 1 1 times x is just x this equals negative 20 times 5 is negative 100 okay so we're good to go there now you can check this plug in a negative 100 for x so negative 100 divide that by 5 this should be equal to negative 20 and it is right you get negative 20 equals negative 20. so yes this is a true statement so we can check that and say yes this is true all right let's take a look at another one so for this one we're going to do some simplifying first we have 3x plus 6 plus 4x equals negative 15. so in any equation you can always simplify things before you get going and in this case i know 3x plus 4x is 7x so this is 7x then plus 6 which is equal to negative 15. now i can't solve this in one simple step but i can solve it in 2. i'm going to use both of my properties here i'm going to use my addition property of equality and i'm going to use my multiplication property of equality now a lot of you would say okay 7 is multiplying x let's go ahead and undo that well you can solve it that way but it's going to be much more complicated the first thing you want to do is get rid of this plus 6 part here okay you always want to isolate the variable term first and then you can isolate the variable itself so if i want to get rid of this 6 here since it's being added to 7x i just subtract it away so 7x plus 6 minus 6 is equal to negative 15 minus 6. so over here if i look at this 6 minus 6 is 0 so i have 7x is equal to negative 15 minus 6 is going to be negative 21. so now i recognize this format from the last two problems i have a single operation that i can do to get x isolated 7 is multiplied by x that's why you say 7x and this is equal to negative 21. well if i'm multiplying x by 7 all i need to do is divide by 7 to undo that so this would cancel with this 7 over 7 is 1 1 times x is just x and then negative 21 divided by 7 is negative 3 okay so x equals negative 3 here so we can plug in and check so let me erase all this since we have more than one instance of x here we've got to make sure we plug a negative 3 into each copy okay so we have 3 times plug in a negative 3 for x and then plus 6 plus 4 times plug in a negative 3 for x and then we have is equal to negative 15. 3 times negative 3 is negative 9 plus 6 plus 4 times negative 3 is negative 12. so you have a plus negative 12 or you just put minus 12. this should equal negative 15. negative 9 plus 6 is negative 3 negative 3 minus 12 is negative 15. so you get negative 15 equals negative 15 which is true right this is true so you can go ahead and check this guy off and say x equals negative 3 is the correct solution all right let's take a look at one more of these so we have negative 22 plus 56x minus 35 is equal to negative 169. so again i can simplify on the left here negative 22 minus 35 is negative 57. so i would have 56x minus 57 is equal to negative 169. so before i do anything here i want to realize that i want to isolate the variable term before i isolate the variable itself it just makes things easier so what i'm going to do since i'm subtracting away 57 from the 56x i'm just going to add 57 to both sides of the equation okay that's all i'm going to do that's because subtracting weight 57 and adding 57 those operations would cancel and become 0. so on the left i would have 56x right that's going to cancel and be 0. and this is equal to negative 169 plus 57 is negative 112. so now i can isolate x by just dividing both sides by 56 right since 56x means 56 is multiplying x to undo that operation i just divide by 56 so divide the left side and also the right side so x would be equal to you have negative 112 divided by 56 which is negative 2 okay so that's our solution and again we can check this by plugging in a negative 2 for x there let me erase all this and let's drag this up here and one note i want to give you if you simplify the problem that's fine we always want to check it in terms of the original problem in case you made a mistake when you were simplifying that will help you catch the mistake right if you plug it back into something that you incorrectly simplified you might verify something that wasn't correct okay so we want to plug this in right here in the original equation so we get negative 22 plus 56 times negative 2 just plug in a negative 2 for x then minus 35 equals negative 169 and so 56 times negative 2 would be done first that's going to give me negative 112. so you would have negative 22 minus 112 minus 35 equals negative 169. negative 22 minus 112 is negative 134 and then negative 134 minus 35 is going to be negative 169. so you get negative 169 is equal to negative 169. so this is true right this is true and we can say that this guy is the correct solution right x equals negative 2. in this lesson we want to review solving linear equations in one variable so over the course of the last two lessons we reviewed the addition property of equality along with the multiplication property of equality and we looked at how to solve some kind of single step equations and towards the end of the last lesson we looked at some things that were a little bit more complex but in this lesson we want to put everything together and show a four-step method that will enable us to solve any linear equation in one variable so any equation of the form ax plus b equals c where again a does not equal zero so the method starts with step one we want to simplify each side separately so this is just where we combine any like terms we clear any parentheses we might have basically we get things as simple as we can the second step is to move all variable terms to one side and then move all constants to the other okay so the way you're going to do this is you're going to use your addition property of equality then the third step is to isolate the variable you're going to do that using the multiplication property of equality and the last step is to check the result there's absolutely no excuse if you have time to get a wrong answer when you're solving one of these equations it is so easy to check the result you can make sure right then when you're taking your test that you got the correct answer before you turn it in okay so it's very important to check all right for the first example we have 3x plus 12 is equal to negative 8 times the quantity negative 4x minus 6 plus 4 multiplied by the quantity negative 7x minus 4. so the first thing i want to do again is simplify each side separately on the left 3x plus 12 i can't do anything with that that's as simple as i can make it on the right i can clear my parentheses negative 8 times negative 4x is going to be 32x negative 8 times negative 6 is going to be plus 48. then 4 times negative 7x would be minus 28x and then 4 times negative 4 is minus 16. so let's scroll down and get some room going now what i want to do is combine like terms over here because i can still simplify so on the left i still have 3x plus 12. on the right 32x minus 28x is going to give me 4x then 48 minus 16 is going to give me 32. so now each side is as simple as i can make it so my second step is to move all variable terms to one side and all constants to the other so a variable term is a term with a variable so 4x is a variable term and so is 3x and then constants are just numbers so we would say 12 is a constant and so is 32. so i want to move 12 to the right side by just subtracting 12 away from each side of the equation and so let's just go ahead and say we have 3x plus 12 minus 12 would be zero but let's just go ahead and show it so plus 12 minus 12 equals 4x plus 32 minus 12. so again this guy would cancel and just be zero so the left side now is just 3x and this is equal to 4x plus we have 32 minus 12 which is 20. so i want to move this guy over here now to the left so i would just subtract 4x away from each side of the equation so let's just go ahead and say minus 4x and minus 4x let's just go ahead and do this vertically because it's a little bit easier this is going to cancel right 4x minus 4x is just 0. on the left side 3x minus 4x is negative x or if you're more comfortable you can write this as negative 1x that's fine too this is equal to 20. now my third step now that i have a variable term equal to a number is just to isolate the variable so what's being done to the variable in this case you have a negative x or again a negative 1 times x so you can multiply both sides of the equation by negative 1 or divide both sides of the equation by negative 1. in this case it really doesn't matter let's just go ahead and divide by negative 1 on each side of the equation and so negative 1 over negative 1 is 1. so you have x is equal to 20 over negative 1 is negative 20. so this is our solution x equals negative 20. so let me erase everything and we're going to check so again we said our solution was x equals negative 20. and i'm just going to plug in a negative 20 everywhere i see an x okay we're going to evaluate each side and make sure that we get the same value so on the left you'd have 3 times negative 20 plus 12 is equal to then on the right you'd have negative 8 times the quantity you'd have negative 4 multiplied by negative 20 then minus 6. so that quantity there so then plus you'd have 4 times the quantity you have negative 7 times again negative 20 and then minus 4 okay so that quantity there and let's just go through and simplify so 3 times negative 20 is negative 60 and the negative 60 plus 12 is negative 48 so let's write negative 48 over here on this side negative 4 times negative 20 is 80 80 minus 6 is 74. so you would have negative 8 multiplied by 74 then plus over here you'd have negative 7 times negative 20 which is 140 and then 140 minus 4 and 136 so plus 4 times 136. so this is negative 48 is equal to negative 8 times 74 is negative 592 then plus 4 times 136 is 544. so if you sum negative 592 and 544 you're going to get negative 48. so you would get negative 48 is equal to negative 48. so the left and the right side are the same value so we go ahead and say this solution is correct x equals negative 20. all right let's go ahead and take a look at another one so we have 11 times the quantity 2x plus 3 is equal to negative 3x plus 9 times the quantity 1 plus 3x all right so again we want to simplify each side separately to start so on the left i'm going to multiply 11 by 2x that's 22x and plus 11 times 3 is 33. this equals you have negative 3x then plus 9 times 1 is 9 and plus 9 times 3x is 27x so on the left i can't really do anything else i just have 22x plus 33. on the right i can combine like terms i have a negative 3x and a 27x that would be 24x and then plus not so now that i've simplified each side as much as i can i want to move all the variable terms to one side and move the constants to the other what i'm going to do is i'm going to move this guy over here and i'm going to move this guy over here okay so the way i'm going to do that is i'm going to subtract 24x away from each side of the equation [Music] so this guy right here is going to cancel 24x minus 24x is 0. so that's gone then over here i'm going to subtract 33 away from each side of the equation and again 33 minus 33 is 0 so that's gone so on the left i have 22x minus 24x which is negative 2x let me kind of straighten that out just a little bit and this equals 9 minus 33 is negative 24 negative 24. so now i have a variable term equals some number okay so essentially at this point i just want to isolate the variable itself so i want to isolate x and to do that since negative 2 is multiplying x i'm going to divide both sides of the equation by negative 2. so let me scroll down just a bit and so what's going to happen is this would cancel with this negative 2 over negative 2 is 1. so you'd have 1 times x which is just x this is equal to negative 24 over negative 2 is positive 12. so our solution here is x equals 12. again let's check this so i'm going to erase everything so again we said x is equal to 12. so i'm going to plug in a 12 here and here and also here so we'll have 11 times inside of parentheses we have this quantity 2 times 12. plus 3 okay so that quantity and then this should be equal to you have negative 3 times you have 12 plus 9 times this quantity so inside the parentheses we have 1 plus 3 multiplied by 12 okay so that's the quantity there and let's go ahead and simplify on the left side first to start so 2 times 12 is 24 24 plus 3 is 27. so you would have 11 times 27. so this is equal to over here negative 3 times 12 is negative 36 and then plus 3 times 12 is 36 36 plus 1 is 37 so you would have 9 times 37 all right so let's see what we got here so 11 times 27 is 297 this should be equal to 9 times 37 is 333 plus negative 36. so negative 36 plus 333 is 297 so you get 297 is equal to 297 so that tells us that our solution here x equals 12 is correct all right so pretty straightforward let's just go ahead and look at one more of these guys and then we'll move on kind of to the next concept so we have 8 multiplied by the quantity 7x minus 9 is equal to negative 4 times the quantity negative 3x plus 10 and then plus 12x so again i want to simplify each side separately so 8 times 7x is 56x and then minus 8 times 9 is 72 this equals negative 4 times negative 3x is 12x and then negative 4 times 10 would be minus 40 and then we have plus 12 x so on the left side i can't simplify any further i'm just going to copy what i have so 56x minus 72 on the right side 12x plus 12x is 24x and then minus 40. so now each side is as simple as i can make it now i want to move all the variable terms to one side all the numbers to the other so this guy i'm going to move over here and this guy i'm going to move over here okay all i'm going to do i'm going to do in just one step so i'm going to say minus 24x on each side of the equation so this is gone and i'm going to add 72 to each side of the equation so this is gone okay so on the left side i have 56x minus 24x which is 32x on the right side i have negative 40 plus 72 which is 32. so now i have a variable term 32x equals some number 32. so it's easy to find my solution now i just want to isolate the variable x and if 32 is multiplying x i just want to divide both sides of the equation by 32 so i can undo that so this is going to cancel with this and i'm left with x is equal to 32 over 32 is just 1. so x equals 1 is my solution here so let's erase everything and check so again x equals 1 is our solution so i'm going to plug a 1 in here and here and here so i would have 8 times let's just do this on the fly because 1 is easy to work with 7 times 1 is 7 7 minus 9 is negative 2. so you'd have 8 multiplied by negative 2. and let's just go ahead and write that as negative 16 and we can kind of move on so over here if i plug in a 1 inside the parentheses negative 3 times 1 is negative 3 negative 3 plus 10 is 7. so you'd have negative 4 times 7 and that's negative 28. then you'd have plus 12 times 1 is 12. so what is negative 28 plus 12 that is negative 16. so again we have a true statement because we have the same value on each side of the equation and so x equals 1 is the correct solution in this lesson we want to review solving equations with decimals or fractions so in the last lesson we reviewed a four-step method to solve any linear equation in one variable in this lesson we're going to talk about how to clear an equation of fractions or decimals just to make it cleaner and easier to work with this process of clearing an equation of either fractions or decimals is completely optional you can work with fractions in an equation you can work with decimals in an equation you get the same answer either way so this is just something that if you want to do what you can if you don't you don't have to so let's start out with an example with some fractions involved we have negative two thirds x plus three halves and this is equal to negative five has x minus 13 six so the basic idea here if you want to clear an equation of fractions you want to multiply both sides of the equation by the lcd the least common denominator of all the denominators involved remember this is legal because i can always multiply both sides of the equation by the same non-zero number now if i look at my denominators here i have a 3 a 2 a 2 and a 6. so what would the lcd be the lcd here would be 6. so all i need to do is multiply both sides of the equation by 6. so let's go ahead and set that up so outside of some parentheses i put my six and inside i'll put my negative two-thirds x plus three-halves this is equal to six is going to multiply by the right side as well so i'll set up my parentheses so negative 5 halves x minus 13 6 and i'll put the 6 outside here so i'm just going to use my distributive property here and multiply 6 by each term inside of the parentheses on each side of the equation so to start out 6 is going to get multiplied by negative 2 3 x so 6 times negative 2 3 x we know that the 6 would cancel with the 3 and give me a 2. 2 times negative 2 is negative 4. so i can go ahead and erase this and put negative 4x here so next we would see that 6 is going to multiply by 3 halves so i'm going to put plus 6 times 3 halves and what we'll have here is that the 6 will cancel with the 2 and give me a 3. 3 times 3 is 9. i'm going to go ahead and write this as plus 9 here so this is equal to i'm basically doing the same thing over here multiplying the 6 by each term so 6 times negative 5 has x and we can see that the 6 would cancel with the 2 and give me a 3. 3 times negative 5 is negative 15. so you'd have negative 15x there so negative 15x and then lastly you have 6 that's multiplied by negative 13 6. so i'll go ahead and put minus you have 13 6 times 6 and i'll cancel this 6 with this 6 and i'm just left with 13 there so now i have an equation that is fraction free and i can go through my normal procedure again if this is something that you don't want to do you don't have to realize that i have to spend the time to clear the fractions and i could just work with them if i want to so at this point i just want to move all the variable terms to one side all the numbers or constants to the other so i'm going to go ahead and add 15x to both sides of the equation so what's going to happen is this is going to cancel then additionally i'm going to subtract 9 away from each side of the equation and this is going to cancel let me scroll down get some room going negative 4x plus 15x is 11x so that's the left side of the equation on the right side negative 13 minus 9 is negative 22. so what i'm going to do is now finish this up by isolating my variable x since 11 is multiplying x i'm just going to divide both sides of the equation by 11 and that's going to give me a final answer of x equals negative 2. now you can check this if you want i firmly advise that you check everything if you have time in the interest of time for this video i'm not going to check all of them i'm going to check just this first one okay so all the additional problems we're going to look at today i'll leave it up to you to check so to check we want to plug a negative 2 in for each occurrence of x in the original equation so let me erase everything we're going to go back up all right so what we want to do remember our answer is x equals negative 2. we want to plug negative 2 in for each occurrence of x so there and also there so i would have negative two thirds multiplied by negative two then plus three halves this is equal to negative five halves multiplied by again negative two and then minus 13 over 6. okay so up here we have negative 2 times negative 2 that's 4. so i would have 4 thirds plus over here 3 halves and this is equal to over here negative 5 times negative 2 is 10 so 10 halves and then minus 13 over 6. okay scroll down get a little room going so the idea here is to simplify each side and make sure you have the same value on each side okay so i need to get a common denominator going so between three and two my lcd would be six right so four thirds times two over two plus three halves times three over three i would basically have 8 plus 9 over the common denominator of 6 8 plus 9 is 17. so let's go ahead and write this side on the left as 17 over 6. and on the right i would get a common denominator by just multiplying this by 3 over 3. 10 times 3 is 30 so you would have 30 minus 13 over the common denominator of 6 30 minus 13 is 17 so you would also have 17 over 6 on this side as well so same value on the left as we have on the right that tells us that our solution x equals negative 2 is correct okay so we can check that one off so for the next problem we're going to look at something a little bit more tedious we have some parentheses involved we have 1 3 multiplied by the quantity negative three halves x plus one this is equal to ninety-five twelfths plus five-thirds x so the idea here is that if you're trying to clear an equation of fractions and you have parentheses involved go ahead and clear the parentheses using your distributive property first and then clear the fractions so 1 3 times negative 3 has x would give me negative 3 6 x then plus 1 3 times 1 is 1 3 this equals 95 12 and then plus five thirds x now between my denominators here i have a six i have a three i have a 12 and i have a three so the lcd is just going to be 12. my lcd is 12. so all i got to do is multiply both sides of this guy by 12 and i will clear all the fractions so 12 times negative 3 6x would be what i know that 12 would cancel with 6 and give me 2. 2 times negative 3 is negative 6. so this would be negative 6x then plus 12 times 1 3 12 times 1 3 is basically 12 divided by 3 which is 4. so this is 4 and this equals you would have 12 times 95 12. we know the 12s would cancel there you're left with just 95. okay and then plus you have 12 times 5 thirds x so 12 times 5 thirds x 12 cancels with 3 and gives me 4 4 times 5 is 20. so this would be 20x okay so now that we've cleared our fractions let's just proceed into our normal procedure for solving a linear equation let's go ahead and subtract 4 away from each side of the equation and also let's go ahead and subtract 20x away from each side of the equation so this is going to cancel and this is going to cancel negative 6x minus 20x is negative 26x this equals 95 minus 4 is 91. so to solve this equation for x i'm just going to divide both sides of the equation by negative 26 since negative 26 is what is multiplying x now we will not get an integer or whole number here we're going to end up with a fraction as an answer so we're going to get x is equal to if i think about 91 and negative 26 i know if i'm dividing here i'm going to get a negative right because positive over negative is negative now 91 is basically 7 times 13. so i could write this as 7 times 13. i could write negative 26 as negative 1 times 13 times 2. so we can all see that this 13 would cancel with this 13. you'd be left with 7 over negative 2 or you just write negative 7 halves however you want to do that essentially you could also write this in decimal form as negative 3.5 but however you want to notate it the answer is x equals negative 7 halves now i'm not going to check this one in the interest of time but if you wanted to you could pause the video go back up to the original equation plug in a negative 7 halves for each occurrence of x verify that the left and the right side are the same value all right so let's talk a little bit about how to clear an equation of decimals this is a very very simple process and again you don't have to do this this is not one that's going to really save you a lot of time especially if you have a calculator okay so remember that when you're multiplying by 10 or a power of 10 such as 10 squared 10 cubed 10 to the fourth 10 to the fifth you know so on and so forth this moves the decimal point one place to the right for each zero and the power of 10. and remember also that if you have this notation so 10 squared the 2 there that exponent on 10 tells me how many zeros i have 10 squared is 100 it has two zeros ten cubed is a thousand it has three zeros ten to the fourth power is ten thousand it has four zeros right so on and so forth so if this guy right here is telling me that i have 10 squared two zeros i would just move my decimal point two places to the right if i multiplied by it so if i had something like let's say 3.68 and i multiplied by 10 squared all i've got to do is take this decimal point here and move it two places to the right this would give me an answer of 368 okay so very easy to multiply by powers of 10 whether they're given as a whole number or they're given an exponential form it's very easy to do all right so let's take a look at an example here so we have negative 3.1 x minus 1.1 x and this is equal to negative 2.2 x minus 8. so the general idea here is to look at your equation look at all the decimal numbers that you have involved so what is the largest number of decimal places that you have on any given decimal number so in this guy right here this negative 3.1 x i have one decimal place again that's places after the decimal point here i have one decimal place here i have one decimal place and if i think about this as negative eight or i could just think about it as subtracting weight however you want to think about that number as a whole number or as an integer it's not a decimal number right you could put point zero if you want but basically this zero doesn't add any value so this is really not a decimal number so we don't even look at it i'm going to erase this and just kind of look at these three numbers here so the largest number of decimal places is just one so if i wanted to clear this equation of decimals i could just multiply both sides of the equation by 10 and i would move all the decimal points one place to the right i would have no more decimal numbers so let's go ahead and do that i would have 10 multiplied by negative 3.1 x minus 1.1 x and then is equal to over here i would have negative 2.2 x minus 8 again multiplied by 10. so 10 times negative 3.1 would just be negative 31. i just move this one place to the right then my x comes along for the ride then minus 10 times 1.1 would be 11 right move this one place to the right then times x so this left side just becomes negative 31x minus 11x on the right 10 times negative 2.2 would be negative 22. again just move this one place to the right and you've got times x and then minus 10 times 8 is just 80. right you can think about that as putting a decimal point there just moving one place to the right so it's just 80. so now i have an equation that is decimal free right so all basically whole numbers or integers involved so on the left negative 31x minus 11x is going to be negative 42x on the right i just have negative 22x minus 80. nothing i can really do to simplify there to solve this i'm just going to add 22x to each side of the equation and what's going to happen is this is going to cancel negative 42x plus 22x is going to give me negative 20x this will be equal to negative 80. if i've got negative 20 that's multiplying x i can isolate x by dividing both sides by negative 20. right the coefficient of x or what is multiplying x so now i'm going to cancel this with this and say this is x is equal to negative 80 over negative 20 is 4. okay so that's going to be my solution again if you want to stop and check go back up and plug in a 4 for each occurrence of x and you will see that you get the same value on the left as you do on the right all right let's take a look at one more of these pretty easy concept overall so we've got negative 0.9 x minus 59.76 and this is equal to negative 10.8 multiplied by the quantity x plus 8 and then minus 4.9 x so again if you're clearing fractions or clearing decimals and you have some parentheses involved clear those first and then go to your procedure so i'm going to rewrite this as negative 0.9 x minus 59.76 and this is equal to negative 10.8 times x is negative 10.8 x and then negative 10.8 multiplied by 8 is negative 86.4 so minus 86.4 and then minus 4.9 x so what i want to do is just multiply everything by the power of 10 that would clear all of the decimal numbers okay so we want to make those into integers or whole numbers so this has one decimal place this has two this has one this has one and this has one so the largest number of decimal places occurs here with two so to clear that decimal i need to multiply by a hundred right because 100 has two zeros i would move two places to the right one for each zero so i'm going to multiply both sides of the equation by 10 squared or again 100 as a whole number so 10 squared multiplied by negative 0.9 x minus 59.76 and this is equal to over here i'm going to do the same thing so negative 10.8 x minus 86.4 minus 4.9 x again multiplied by 10 squared or 100. so all i'm going to do for my multiplication here it's very very easy just moving each decimal point two places to the right so this guy is going to go one two places to the right so if you had 0.9 forget about the negative a second and i move this two places to the right i'm gonna end up with the number zero nine zero the zero in front of the nine doesn't add any value it's just ninety so what i'm going to end up with here because it's negative i'm gonna have a negative 90 and then x then minus this guy goes two places to the right i get 5976 then over here if i multiply 10 squared by each term my negative 10.8 x i would have negative one zero this goes one and then a second place to the right so negative one thousand eighty x then minus this would go two places to the right so eight thousand 8640 and then minus this goes two places to the right so 490 x so now let's go ahead and just combine some like terms on the left i can't really do anything so i'm just going to write negative 90x minus five thousand nine hundred and seventy six on the right negative one thousand eighty x minus four hundred ninety x is going to give me negative one thousand five hundred seventy x and then minus eight thousand six hundred forty all right let's scroll down get some room going so what i want to do now i'm going to add 1570x to each side of the equation and what's going to happen is this is going to cancel and i'm going to add 5976 to each side of the equation so this is going to cancel so negative 90x plus 1570x is 1480x this equals negative 8640 plus five thousand nine hundred seventy six is negative two thousand six hundred sixty four all right so the last thing i wanna do is isolate my variable x so i'm gonna divide each side of the equation by one thousand four hundred eighty this is not going to be a whole number or an integer as an answer essentially i know that it's going to be negative right x will be equal to something that's negative because i have a negative over a positive if i factored each of these i would find that we have a common factor of 296. so if i divide 2664 by 296 i would get 9. so this would be negative 9 over if i divided 1 480 by 296 i would get 5. so you get x equals negative 9 fifths or as a decimal number you could say negative 1.8 right if you did that division so this would be my answer either one of those you could write x equals negative nine fifths or x equals negative 1.8 again if you want to check you go back up you plug either of those in for x in the original equation and you verify that the left and the right side are equal in this lesson we're going to review identifying types of equations so in this short lesson we're going to talk about the three types of equations that we're going to come across in most cases we're going to work with something known as conditional equations so these equations are true under certain conditions but false under others so in the case of a linear equation of one variable we have something like negative 3 times the quantity x plus 3 is equal to negative 13 minus x so let's just solve this for x we would simplify each side first so negative 3 multiplied by x would be negative 3x the negative 3 multiplied by 3 would be minus 9. this would be equal to we have negative 13 minus x let's move all the variable terms to one side and all the numbers to the other so i'm going to add x to each side of the equation and i'm going to add 9 to each side of the equation so what's going to happen is this is going to cancel and this is going to cancel negative 3x plus x is negative 2x this is equal to negative 13 plus 9 is negative 4. so negative 2x is equal to negative 4. i can solve that by just dividing both sides of the equation by negative 2 and i get that x is equal to 2. okay so that's my solution so for this equation it's conditional right it's true if x equals 2 but it's false under any other circumstance so let me prove this if i plug in a 2 here and here i'm just plugging in for x i will get a true statement so let's erase this and we would have negative 3 multiplied by the quantity 2 plus 3 is 5. so times 5 this is equal to negative 13 minus 2. so negative 3 times 5 is negative 15 and this should be equal to negative 13 minus 2 which is negative 15. so negative 15 equals negative 15 so yes this is true so right now this is a true statement so we're good to go but that's when x equals 2. so what if i tried a different value for x we would see that we get a false statement right so let me erase this and let's suppose i chose the value of negative 2 for x so negative 3 times you'd have negative 2 plus 3 which is going to give me 1. this equals negative 13 minus a negative 2 is plus 2. so negative 3 times 1 is negative 3 negative 13 plus 2 is negative 11. this is false negative 3 does not equal negative 11. so you can see that our equation is true when we replace the variable x with the correct value of 2 but false under any other circumstance so again this is the definition of a conditional equation so the next type of equation that you're going to come across is known as a contradiction so these are equations that don't have a solution and we're going to have a lot of these throughout the course so something like 3 times the quantity 3x minus 1 is equal to 9x plus 5. so 3 times 3x is 9x and then minus 3 times 1 is 3 this equals 9x plus 5. so what i'm looking at here is the fact that 9x is over here on the left and 9x is over here on the right if you have the same value on each side of an equation you can just get rid of it why do i say that well look at what i can do i can subtract 9x away from each side and what's going to happen is it cancels here and it cancels here it's just gone what i'm left with is nonsense negative 3 equals 5. that is false so if you go through this process of solving an equation and you're left with nonsense something like a false statement like negative 3 equals 5 you have a contradiction so in this particular case you would say that there's no solution you could put no solution your book might show the symbol for the null or empty set so when you think about sets if your set contains no elements right if you had no solution you wouldn't have any elements in the solution set the way you notate that is with this symbol here so this means no solution if you see it in your book okay no solution you might also see empty set braces this is probably less popular but you might see that as well what you don't want to do is combine the two you don't want to use set braces with this symbol inside that is incorrect okay and for that we'll talk about that when we get to set theory all right let's take a look at something else so we also have an equation that's known as an identity so negative 2 plus 2x is equal to negative 2 times the quantity 1 minus x all right so what do we do here so we have negative 2 plus 2x is equal to negative 2 times 1 is negative 2 and then negative 2 times negative x is plus 2x so you'll notice that you have the exact same algebraic expression on each side of the equation two ways you can think about this the first way is no matter what i plug in for x it doesn't matter what value it is i'm doing the same thing to it on each side so i'm always going to get the same result it's always going to be true for all real numbers no matter what i do the other way you could think about this is i could go through the process of solving it and i could say okay well i can subtract 2x away from each side so this is going to be gone right that's gone i can add 2 to each side so now this is going to be gone so what am i left with i'm left with 0 is equal to 0 which is true so if in the process of solving your equation you end up with something that's just true always well then you have an identity okay you have an identity okay and an identity is true for all real numbers so you can say all real numbers okay that's going to be your solution in this lesson we want to review solving proportion equations so in this lesson we're going to talk a little bit about how to solve proportion equations it's a pretty simple concept again it's something that most of you already know how to do we're going to start off by just talking about the equality test for fractions so essentially you can tell if two fractions are equal by basically checking the cross products so if i have something like 1 4 and 3 12 i can tell if they're equal by multiplying the denominator of one fraction by the numerator of the other so 4 times 3 is 12 and then doing the same thing over here so the denominator of one fraction by the numerator or the other 12 times 1 is 12. if you get the same result in each case then you have a proportion right you have two fractions or two ratios or it could be two rates that are equal in value so that's a proportion so these are equal i could say 1 4 is equal to 3 12. you can also see that that's equal because if i multiply 1 by 3 i get 3. if i multiply 4 by 3 i get 12. so essentially all i really did here was take 1 4 multiply it by 3 over 3 which is the complicated form of 1 and i got 3 12. okay so of course 1 4 and 3 12 are equal all right let's look at a number so we have 9 14 and we have 20 27 so again we can check this guy by just cross multiplying so i can multiply 14 times 20. it's pretty easy to do you multiply 14 times 2 you get 28 you put a 0 at the end you get 280 and i'd multiply 27 times 9 which would be 243. so these two values these two cross products are not equal so these two fractions are unequal so we can say 9 14 is not equal to 20 20 sevenths so let's talk a little bit about solving proportion equations we already reviewed previously how to solve equations with fractions so one way you could solve this if you wanted to you could multiply both sides of the equation by the lcd of all the denominators but a quicker way when you just have one fraction equal to another is just to cross multiply right to set everything up so in other words if i have x over 38 is equal to 2 over 19 all i have to do is cross multiply right because if we have two fractions that are equal the cross product should be equal so that means that 38 times 2 which is 76 should be equal to 19 times x or 19x so now all i got to do is solve this in one step divide both sides by 19 and i get x is equal to 4. you could check that pretty easily just plug in a 4 for x there you would have 4 over 38 is equal to 2 over 19. of course this is true if i take 2 when i multiply by 2 i get 4. if i take 19 and multiply by 2 i get 38. so essentially it's like i did this i took 2 over 19 and i multiplied it by 2 over 2 and i got 4 over 38 right so x is going to be 4 in this case what about something like x over 60 is equal to negative 7 over 15. again we can just set this up by cross multiplying so 15 times x is 15x this is equal to negative 7 times 60 negative 7 times 6 is negative 42 and then just put a 0 at the end so it's negative 420 and now we just want to divide both sides by 15 so that i can isolate my variable x and so what am i going to get negative 420 divided by 15 would be negative 28 so x is equal to negative 28 okay and of course you can check that you could plug in a negative 28 for x there so you would have negative 28 over 60 is equal to negative 7 over 15. again you can kind of eyeball this and see if i multiply negative 7 times 4 i get negative 28 if i multiply 15 by 4 i get 60 okay so we can see that that is correct right negative 28 over 60 would reduce to negative 7 over 15. so those two are equal all right let's take a look at another one so we have 9 over x minus 7 is equal to 6 over x so i'm going to cross multiply here just like i normally would but i've got to be careful 6 is multiplying x minus 7. it's multiplying that quantity so i would have 6 times the quantity x minus 7. the 6 has to multiply the x and it's got to multiply the negative 7. a common mistake is just to put 6x minus 7 like that okay that's wrong the 6 has to multiply both of those terms so 6 times the quantity x minus 7 is equal to x times 9 or 9 x so i'm just going to kind of clean this up on the left 6 times x is 6 x and then minus 6 times 7 is 42 this equals 9x let me subtract 9x away from each side of the equation and let me add 42 to both sides of the equation so what we're going to have here is that this cancels and this cancels and let me just kind of bring this up here to the top we're going to say 6x minus 9x is negative 3x and this is equal to we're going to say 42. if we divide both sides of the equation by negative 3 we know we would have x is equal to 42 over negative 3 we know that's negative and 42 over 3 is 14. so this would be negative 14 as our answer now you check this if you want let me erase everything we'll plug in and make sure the left and the right sides are equal let me just kind of scooch this up and i'm going to plug in a negative 14 here and here so what you'd have is 9 over you have negative 14 minus 7 negative 14 minus 7 would be negative 21. now before i go any further i know i can simplify this each is divisible by 3. 9 divided by 3 is going to be 3 negative 21 divided by 3 is going to be negative 7. so let's just write this as negative 3 7. so that's what i'm looking for here if i plug in a negative 14 i'd have 6 over negative 14. so each is divisible by 2. if i divide 6 by 2 i get 3. if i divide 14 by 2 i get 7 and of course it's negative there so let me just throw that out in front you see that you get negative 3 7 is equal to negative 3 7 okay so we get the same value on the left as we get on the right so we can say x equals negative 14 is our correct solution all right let's take a look at another so we have 5 times the quantity 3x plus 1 over 7 is equal to 9x minus 3 over 3. so again i'm just going to set this up by cross multiplying so 7 multiplies the quantity 9x minus 3. so 7 times the quantity again in parenthesis 9x minus 3 is equal to 3 multiplies in this case it's easier just to multiply the 3 by the 5 first so 3 times 5 is 15 and then times this whole quantity 3x plus 1. so now let me simplify each side 7 times 9x is 63x and then minus 7 times 3 is 21. this equals 15 times 3x is 45x and then 15 times 1 is 15. so let me subtract 45x away from each side of the equation and what's going to happen is this will cancel let me add 21 to both sides of the equation and so this guy right here is going to cancel so 63x minus 45x is going to give me 18x and this equals 15 plus 21 is going to be 36. so let me erase all this i don't need any of this anymore all i'm going to do is divide both sides of the equation by 18 now so that i can isolate x so let me drag this up here i'm going to divide both sides by 18. so again that i can isolate x 18 over 18 is one so i just have x is equal to 36 over 18 is 2. okay so that's my solution now again i like to check things i think you should too plug in a 2 for x here and here and make sure you get a true statement so if i had 5 times the quantity three times two is six six plus one is seven so this is five times seven over seven seven over seven is one so this is basically five on the left so nine times two is eighteen eighteen minus three is fifteen so you'd have fifteen over three which is 5 as well so you get 5 equals 5 so x equals 2 is the correct solution here all right let's take a look at one more of these and then we'll just kind of look at some word problems so we have negative 12x plus 5 over 2 is equal to 7 times the quantity x minus 9 minus 3 over negative 11. so again we set it up in the usual way we're going to cross multiply we've got to be careful here 2 is going to multiply by this numerator you've got to multiply 2 by each part so you can kind of set this up using brackets so 2 times inside of brackets i'll just put my numerator so 7 times the quantity x minus 9 and then minus 3. this equals you're going to set this part up so negative 11 times you've got your quantity here your negative 12x plus your 5. so you can go through here and you can distribute this to each part or you can clean up what's inside and multiply by 2 whatever you want to do 2 is going to multiply by 7 times the quantity x minus 9 that's just 14 times the quantity x minus 9. then minus 2 times 3 is just 6 and this equals negative 11 times negative 12x is 132x the negative 11 times 5 is minus 55. now on the left hand side 14 times x is 14x and then minus 14 times 9 is 126 okay then minus another 6. so if you had negative 126 minus 6 that would be negative 132. so that's my left side 14x minus 132 and this equals on the right side you've got 132x minus 55. okay so let's go ahead and solve this guy i'm going to add 132 to each side of the equation so we know that this is going to cancel then i'm going to subtract 132x away from each side of the equation so this is going to cancel so 14x minus 132x is negative 118x and this equals you've got negative 55 plus 132 which is 77. so let's go ahead and finish this guy up let's divide each part by negative 118 and this will cancel with this you'll have x is equal to negative 77 over 118. now is this something that i can simplify 77 is 7 times 11 7 is prime 11's prime 118 is not divisible by 11 and it's also not divisible by 7 so this is as simple as you can make it so x is going to be negative 77 over 118. now this is something you can check just like we have throughout this lesson but because it's so tedious and in the interest of time i'm not going to check it but if you want to you can go back you can pause the video plug in a negative 77 over 118 for each occurrence of x and you'll see that the left and the right side are equal all right so now let's turn to some word problems and these are typical examples that you would see if you were in a section where you're talking about solving proportions so we have that jacob drove his car 300 miles using 20 gallons of gas how many gallons of gas would it take for jacob to travel 1200 miles so this is a very easy problem to set up and solve essentially what we want to do is take kind of the first information they give us and set up a unit rate right so it says jacob drove his car 300 miles using 20 gallons of gas so we have 300 miles [Music] per 20 gallons 20 gallons of gas and you can write of gas if you want i'm just going to leave 20 gallons in the denominator and so what i would want to do or the easiest thing to do here is just to get a unit rate so 300 the number in the numerator divided by 20 the number in the denominator would be 15. so i could basically say that it's 15 miles per 1 gallon of gas okay so that's a unit rate now with the second part we're asked the question how many gallons of gas would it take for jacob to travel 1200 miles so again because the miles per gallon here we're going to assume it stays the same i can set up a proportion equation so i can say this is equal to the miles are going to be in the numerator so i'm going to put 1200 miles in my numerator over here and this is over the unknown here is the gallons of gas that would be used so since gallons or gallon in this case is in the denominator i'm going to just use x as a variable to represent the unknown number and then my units would be gallons okay so gallons so once you've got this set up it's very easy you're just going to cross multiply if your units are the same in the numerator which in this case it's miles and miles and your units are the same in the denominator so you have gallon and gallons same units one singular ones plural you're good to just cross multiply with the number parts or in this case it's going to be a vary so i would take 1 and multiply it by 1200 so that's 1200 and then set this equal to 15 times x or just 15x so all i need to do now is just solve the equation for x i can really quickly do that by just dividing both sides of the equation by 15. so what is 1200 divided by 15 well that's going to give me 80 right so i would get that x is equal to 80 and x here represents the number of gallons it's going to take for him to go 1200 miles okay so let's erase everything and we can just answer this how many gallons of gas would it take for jacob to travel 1200 miles we'll say it would take 80 gallons of gasoline all right let's take a look at one more of these so the distance between smithland and creek falls is 2400 miles on a given wall map this distance is scaled down to 4.8 feet on the same wall map how many feet would be between port charles and river ridge two cities which are 4 000 miles apart so again another easy one essentially it tells us that we have this given wall map and basically the scale is proportional right so it takes a certain distance in the real world and it scales it down so between smith and creek falls it's 2 400 miles so we're going to say that 2400 miles over on this wall map it's scaled down to 4.8 feet so 4.8 feet we're just going to set up our proportion again on the same wall map we have this city port charles and river ridge they're 4 000 miles apart but we don't know how this is going to be expressed on our map so to figure this out we're just going to put 4 000 miles here in the numerator because that's their distance in real life just like we had over here and this will be over x feet okay this is going to be the kind of distance on the map how many feet will it be so to figure this out we're just going to cross multiply we're going to take 4.8 multiplied by 4000 which would give us 19 200. we set this equal to x times 2400 so 2400x and to solve this guy for x i'm just going to divide both sides of the equation by 2400. if i divide 19 200 by 2400 i'm gonna get eight right so x here is going to be equal to eight so i can erase this got all my information now instead of saying x feet this would really be eight feet so i can go back and just answer this we would say they would be eight feet apart on the wall map in this lesson we want to review solving literal equations so so far in our course we've only looked at equations with one unique variable so something like 2x minus 5 is equal to 11. we know if we solve this we get x equals some number right so if i quickly solve this i could add 5 to each side of the equation we know that this is going to cancel we would have 2x is equal to 11 plus 5 is 16. to isolate x here we just need to divide both sides by 2 and we would find that x is equal to 16 divided by 2 is 8. so you see you have x equals some number there okay when you're working with just one unique variable in most cases you get a variable equals some number but when we talk about literal equations we're talking about equations with more than one unique variable so a lot of times we will work with more than one unique variable especially as we get further along into algebra so if we see something like let's say 2x plus let's say 5y and let's go ahead and say this is equal to 20. now i can't get this in terms of a variable equals just some number because i have two variables involved but i can still solve for one of the variables i can choose to solve for x i can also choose to solve for y so if i wanted to just solve for x what would i do well if i want to isolate this guy the first thing i always want to do is isolate the variable term that contains the variable that i'm trying to isolate so i would start by just isolating the 2x so to do that i would subtract 5y away from each side of the equation so i would have just 2x on the left i know that this would cancel over here 5y minus 5y that's 0. so this equals now i'll have my minus 5y and then plus my 20. now how can i get x by itself well i just have the 2 that's multiplying x so no change here i just divide both sides of the equation by 2. but now over here i'm dividing this whole expression here i have my negative 5 y plus 20 the whole thing is divided by 2. and let me scroll down get a little room going so over here 2 over 2 is 1 so i just have x and this is equal to you have your negative 5 y plus 20 over 2. now there's different ways to write that you could split that up as negative 5 y over 2 plus 20 over 2 so i could also say x is equal to negative 5 y over 2 plus 20 over 2 and of course 20 over 2 is 10. so you could also write it like this either answer is acceptable okay but i'm just showing you that you can write things in terms of one variable so i could say x equals something i can also solve this equation for y so if i want to solve for y again the first thing i want to do is isolate the variable term with y involved so because 2x is being added to this 5y i just subtract 2x away from each side of the equation very very easy this is going to cancel we would have 5y is equal to you have your negative 2x and then plus your 20 and of course to get y by itself i'm just going to divide both sides of the equation by 5. and so you'd have y is equal to you have your negative 2x plus 20 over 5. and again you could write that differently if you want you can split that up and say this is y is equal to negative 2x over 5 plus 20 over 5 and 20 over 5 is just 4. so either way you want to write it both these answers would be acceptable we're just showing you that you can solve for one of the variables right so now this equation is in terms of y y equals sum expression before we looked at x equals some expression so when you get more than one unique variable involved you're not going to have a variable equals some number you're going to have a variable equals some expression so normally when we talk about solving literal equations we're talking about taking a formula and putting it in terms of one of the variables so this is very useful when we look at application problems right so word problems so we're just going to go through a few typical problems that you would see and basically we just need to understand how to solve something for a specified variable so the first formula we're going to look at is for the perimeter of a rectangle so if you have a rectangle you have a length here and a length here so on the top and the bottom and then on the sides you have your width right so length is labeled with l width is labeled with w okay so the perimeter or the distance around a rectangle is given as two times the length plus two times the width okay so what we want to do here is just solve for w and then also solve for l right now you'll notice that it's solved for p or put in terms of p because p is isolated on one side of the equation right you have p equals some expression in this case that expression is 2l plus 2w all right so to solve for l let me just rewrite p equals 2l plus 2 w and if i want to solve for l again i just want to start by isolating the 2l right always start by isolating the variable term which contains the variable you want to isolate so all i need to do is subtract 2w away from each side of the equation so i'm going to have that p minus 2w is equal to 2l and if i want to isolate l there what's being done to it it's being multiplied by 2 so just divide both sides of the equation by 2. and i'm going to flip this around so 2's cancel there you get l is equal to p minus 2w over 2. okay so this is in terms of l now so if i wanted to put this in terms of w we'd use the same thought process so i want to begin by isolating the term with the w involved so to do that i would just subtract 2l away from each side of the equation so i would have that p minus 2l is equal to 2w and again to isolate w i just divide both sides of the equation by 2. okay very very easy 2 over 2 is 1 so you get w is equal to so w is equal to you get p minus 2l over 2. and all i'm doing here is i'm just switching the variable to the left that's perfectly legal right if i wrote this here as w on the right and this is equal to p minus two l over two i can switch this around and say this is w is equal to p minus two l over two okay it's the same thing i just like my variable on the left i don't like it on the right but you can write it either way they're both acceptable all right let's take a look at another formula from geometry so we have the area of a trapezoid here so a which stands for area is equal to h which stands for height that's this guy right here times the quantity you have an upper case b which is down here that's the lower base then plus a lowercase b which is up here that's your upper base so then we divide this by 2. so what if i wanted to solve this guy for uppercase b what could i do well remember you want to isolate the variable term with that involved first right now it's wrapped inside of a set of parentheses so one thing i could do is just remove the parentheses with my distributive property to kind of get things started so we would have a is equal to you have h multiplied by again uppercase b plus lowercase b and this is over 2. so i have a is equal to h times uppercase b is h times uppercase b then plus h times lowercase b is h times lowercase b and this is over 2. now if i want to isolate again this term here to start what i'm going to do is just clear my denominator i can easily do that by multiplying both sides of the equation by 2. so this would cancel let me kind of go up here i would have 2a is equal to you would have h times uppercase b plus h times lowercase b okay so what do i do from here well i want to isolate again this term right here because it's got my uppercase b involved so all i'm going to do is just subtract h times lowercase b away from each side of the equation and so let me kind of scroll down and get some room going i'm going to have 2a minus h times lowercase b is equal to h times uppercase b so to get uppercase b by itself i'm just going to divide both sides of the equation by h and this would cancel so let me write my answer over here so it's out of the way i have uppercase b is equal to you have 2a minus h times lowercase b over h okay so that's just solving this for uppercase b if you wanted to solve it for lowercase b or you wanted to solve it for h you would go through the same thought process so let's go ahead and solve this for h let's erase everything so again we'll rewrite this formula down here area or a is equal to you have h times the quantity uppercase b plus lowercase b and this is over 2. how can i get h by itself well before i go through and use my distributive property let's just clear the denominator let's multiply this side by 2 and this side by 2. so this cancels so now i have 2a is equal to i have h again multiplied by the quantity big b or uppercase b plus small b or lowercase b now if i want h by itself what can i do h is being multiplied by something here erase this and imagine that you had 4h and you wanted h by itself you would just divide both sides by 4 right what's multiplying h let's go back now this quantity here is multiplying h so all i've got to do is just divide both sides of the equation by that quantity big b plus little b so big b plus little b and i've got my answer right this is going to cancel and so what i'm left with here i'm going to put h is equal to you have 2a over bigger b or capital b plus lower b or smaller b okay so now it's in terms of h all right let's go ahead and wrap up this problem by solving the equation for the lowercase b now and again the idea is always the same if i'm solving for the lowercase b i just want lowercase b on one side of the equation and i want an expression on the other okay so very very easy so we have that a is equal to we've got h multiplied by the quantity you've got your uppercase b plus your lowercase b and this is over two and let's get some room going here now there's many ways to start this problem you can start by using your distributive property to remove the parentheses because ultimately you need to get to that b and it's inside of parentheses right now so that's one thing you can do another thing you can do is you can clear the denominator first so we can do that and let's just start by that so we'll go ahead and cancel this with this and we'll do 2a is equal to we've got h times the quantity you've got your uppercase b plus your lowercase b now you can use your distributive property to remove parentheses that's fine that's one way to do it or another way and probably a faster way you can just divide both sides of the equation by h right because if you divide both sides of the equation by h then you can go ahead and remove those parentheses and you'll have access to this lowercase b here so i'll have 2a over h on the left this is equal to now i just have the capital letter b plus the lowercase b on the right okay so to solve for the lowercase b now is very very easy i just have this capital letter b that's being added to it so i'm just going to subtract the capital letter b away from each side of the equation and so i'm going to end up with 2a over h minus the capital letter b and this is equal to the lowercase b okay and if you like the variable on the left like i do you can rewrite this you can say lowercase b is equal to 2a over h and then minus uppercase b now you'll notice that we don't have a common denominator here if we want a common denominator we can get one we can just multiply this by h over h if we were to use the distributive property we would have gotten one so this is just a matter of personal preference it's the same answer either way so i can write this as lowercase b is equal to 2a minus uppercase b times h all over the common denominator of h or again it's the same thing if you say lowercase b is equal to 2a over h minus uppercase b all right let's look at some other problems so suppose i wanted to solve this for just x so z times the quantity ax minus bd plus nq equals nx minus a how can i solve this for x i have an x here and i have an x here well if you get this scenario you've got to get all the terms with an x involved to one side and everything else to the other so let's go ahead and distribute this z to each term inside the parentheses so you would have z ax minus z bd then plus nq and this is equal to nx and then minus a so i want this term and this term because they each have an x involved on one side of the equation i want everything else on the other so what i'm going to do is i'm going to subtract nx away from each side of the equation and i'm going to add zbd to each side of the equation and then i'm going to subtract nq away from each side of the equation okay so a lot of stuff on the screen here let's go through this so this is gone i've cancelled it and this is gone i've canceled it so on the left i'm going to have zax minus nx then this will be equal to on the right this is going to cancel so i'll just have negative a plus z bd minus n cubed let me make that q a little bit better okay so how can i solve this for x a lot of you if you see this will be perplexed what can i do here well think about this if you had something like 6x minus 4x we know we can combine like terms there but why can we do that well because of the distributive property i can pull the x out and i can say this is 6 minus 4 inside so this becomes x times 2 or 2x if i use the same thought process here i can pull my x out because it's common to each right so if i pull that out inside the parenthesis i would have za minus n now i can't combine those but that's okay you'll see in a minute i can just divide by that because that's what's multiplying x and this is equal to you have your negative a plus your z bd and then minus your n q now again this is multiplying x so it's just like if i had 4x is equal to 20. if 4 is multiplying x i can divide both sides of the equation by 4 so that x is equal to 5. same thought process here all i want to do is divide both sides of the equation by what's multiplying x so it's the quantity z a minus n and i'm going to divide by that over here as well so z a minus n and it looks messy but this is what we need to do when we have a lot of variables involved so i've solved this for x now because this is going to cancel you have x is equal to you have your negative a plus your z bd then minus your n q over you have your z a and then minus n all right let's go ahead and take a look at another one so we have 5x minus the quantity 2q plus d is equal to a times the quantity x plus 1. so we want to solve this for x and again the idea here is if i want to solve it for x i've got to first get all the variable terms with x involved on one side and everything else on the other so because i have an x here wrapped inside of some parentheses let's go ahead and use our distributive property get everything simplified so we have 5x i have this negative out in front of these parentheses so again you can think about this as plus negative 1. the negative 1 would multiply by each term so you'd have minus 2q and then minus d and this equals the a is going to multiply by each term so ax and then plus a okay so i want all my again terms with an x involved on one side so let's go ahead and subtract ax away from each side of the equation and let's add 2q to each side of the equation and let's add d to each side of the equation okay so this would cancel here this would cancel here and this would cancel here so on the left hand side i've got 5x minus ax and this is equal to on the right hand side i've got a plus 2q plus d all right so now what do i have here on the left how can i isolate x well again i think about the process i used to combine like terms if i have 5 x minus let's say 6x well i can factor out the x there and inside the parentheses i would have 5 minus 6 we know 5 minus 6 is negative 1 so this is negative 1x well here i can basically again do the same thing i can factor out the x because it's common to each so i can have x times the quantity 5 minus a this is equal to a plus 2 q plus d since this quantity 5 minus a is multiplying x to get x by itself i just divide both sides by that quantity and i'll have x by itself so i will have x is equal to we'll have a plus 2q plus d over your 5 minus a in this lesson we want to review solving linear inequalities in one variable so in the traditional sense when we think about an inequality we're thinking about a statement that two algebraic expressions are not equal in value so something like 6 is less than 12 is a way to notate that 6 is a smaller number than 12 or something like negative 4 is greater than negative 11 is a way to notate that negative 4 is larger or you could say greater than negative 11. so when we work with inequalities we have strict inequalities and we have non-strict inequalities so strict inequalities you have strictly less than and strictly greater than with non-strict inequalities which we'll see in a moment you have less than or equal to and then you have greater than or equal to so you're allowing for the possibility of equality so when we have an inequality with a variable involved we need to understand interval notation and we need to also understand how to graph an interval on the number line so if we say something like x is greater than 5 how could we notate this on the number line well first let's think about x equals 5. if x equals 5 then x can only be one value can only be 5. so on the number line i could just put a dot there right a filled in circle however you want to think about that but if x is greater than 5 i don't want to go through and put a dot at every possibility right because it's anything larger than 5. so it could be 6 it could be 7 it could be 8 it could be 9 it could be 2 million so i want to basically shade all of these and i would just shade the number line and i would shade that arrow to say this continues forever but i've got to have a point known as a boundary okay a boundary is what separates the solution region from the non-solution region so over here this is my solution region okay anything on the right side of 5 satisfies this inequality so if i took 6 and plugged it in 6 is greater than 5 that's true okay that's true if i took 1 million and plugged it in 1 million is greater than 5 that's true then anything on the left of 5 is in the non-solution region right because it wouldn't work if i took a 3 for example and i plugged it in is 3 greater than 5 no this is false okay this is false so 5 itself is our boundary it's where it separates the solution region from the non-solution region so at that point you either want to use a parenthesis or a bracket you use a parenthesis when it's not included and you use a bracket when it is included okay so in this case it's not included because we have a strict inequality so what we want to do we want to put a parenthesis facing in the direction of the solution region so this is how you would graphically notate that x is greater than 5. 5 is not included that's our boundary but anything larger than 5 is included and that's why it's shaded now some of you will see a different notation at 5 you may see an open circle to show that 5 is not included so you might see something like this and then everything is shaded to the right and that's okay too no problem there it's just a different way to notate it in this course we're basically going to stick to this so we're going to put our parenthesis here and then shade everything to the right but feel free to do it either way now you also need to think about interval notation and it basically follows the same concept as we just looked at with graphing so you're going to think about your smallest value you're going to think about your largest value and you're going to separate them with a comma now in this particular case you don't really have a smallest value right because if you think about it five is your boundary but anything larger than five works so 5.00001 works or five point you know you can extend a million zeros and then put a one that works so for the smallest value we just put 5 because that's as close as we can really get and then we're going to put a parenthesis to the left of that to say that 5 is not included right so this just means not included not included but we know automatically that anything larger than 5 would be included okay then we put a comma and then our largest value again because there is none we use infinity this is just a concept so we put our infinity symbol to say this is going to continue forever in the right direction you always use a parenthesis with infinity okay so again not included so 5 comma infinity with a parenthesis next to each so this is graphically and this is with interval notation now let's look at a less than so x is less than negative 3. so again x can be any value that's less than or to the left of negative 3. so negative 3 is my boundary my solution region is to the left of negative 3 so i want to shade everything to the left of negative 3. so i'm going to shade this arrow and all of this to negative 3 and at my boundary i'm just going to put a parenthesis facing to the left okay to show that negative 3 is not included but again because i've shaded to the left everything that's less than negative 3 would be in interval notation it's the same concept we're going to put a negative infinity here to represent our smallest value because there is none i always use a parenthesis with either negative infinity or positive infinity then comma i'm going to put negative 3 for my largest value and again because that's not actually my largest value because it's not included i'm going to put a parenthesis there so you have negative infinity you have negative 3 you have a parenthesis next to each let's quickly look at non-strict inequalities so you have less than or equal to and greater than or equal to so these allow for the possibility of equality so x now is greater than or equal to five so it's basically the same thing but we allow for 5 to be part of the solution so i use a bracket when that occurs so all i need to do here is just put a bracket at 5 and shade everything to the right and the bracket just tells me that 5 is included now in interval notation i can put a bracket a 5 comma infinity and again the bracket just says that 5 is included okay it's part of the solution now and it's worth noting here that we also have another notation that we can use for this some textbooks when you have a non-strict inequality we'll put a filled in circle at the boundary so to include five instead of using a bracket they'll put a filled in circle okay so the filled in circle means it's included and then we would shade everything to the right like this but again in this course we're going to stick to using a bracket so i'm going to put a bracket at 5 and just shade everything to the right you can use any method you want they're both the same just different ways to show the same concept then for x is less than or equal to negative 3 again i'm going to use a bracket because negative 3 is part of my solution so i'm going to put a bracket at negative 3 facing to the left and just shade everything to the left here okay in interval notation i've still got my negative infinity and then comma i'll have a negative 3 with a bracket next to it so negative 3 is included so that's why i have my bracket with positive or negative infinity i always use a parenthesis all right so now let's move into what a linear inequality in one variable is going to look like so we have stuff like ax plus b is less than c where a does not equal zero so essentially here this is generic we have a b and c which are any real number with the exception that a cannot be zero and we could basically replace this less than with a greater than or less than or equal to or greater than or equal to so something like 3x plus 5 is greater than 11 or something like negative 7x plus 2 is less than or equal to 16. so essentially exactly what we saw when we solved linear equations in one variable with the exception of we're flipping the equality sign into an inequality sign all right so suppose we came across an inequality such as x minus 11 is less than negative 14. so before we solve this let's think about an equation so x minus 11 is equal to negative 14. how could we solve this we know about the addition property of equality that tells us we can add or subtract the same number 2 or in the case of subtraction from both sides of an equation and not change the solution so to isolate x here all i need to do is add 11 to both sides of the equation right because i'm subtracting away 11 here if i add 11 i'm going to basically cancel this with this and i'm left with just x so here i would have x is equal to negative 14 plus 11 is negative 3. so we can use the same process when we solve inequalities so let's erase this and let's write our problem down here so we have x minus 11 is less than negative 14. so when it's used with inequalities it's called the addition property of inequality and essentially this property allows us to add or subtract the same value to or in the case of subtraction from both sides of an inequality without changing the solution set okay so i'm using the same technique here i want to isolate x and since i'm subtracting 11 away from x i just add 11 to both sides of the inequality and again this cancels and i get x is less than negative three okay very very easy so we would want to notate our solution again using interval notation and then graphically so to notate this in interval notation i know that i would put a parenthesis and a negative infinity and then comma i would have a negative 3 there and again because negative 3 is not included i'd put a parenthesis so graphically i would find negative 3 so that's right here i'd put my parenthesis in and then i would just shade everything to the left okay so all of this going to the left including the arrow all right let's take a look at another one so we have x plus 9 is greater than or equal to 15. so again i want to get x by itself i always want to isolate x on one side of the inequality so to do that i can just subtract 9 away from each side of the inequality and this guy right here is going to cancel i'll have x is greater than or equal to 15 minus 9 is 6. so again i've got a greater than or equal to there so i want to use a bracket so let's find 6 on our number line we want to put a bracket to show that it's included and then we're going to shade everything to the right that's our solution region okay in interval notation i would put a bracket and then put a 6 this would be my smallest number and then comma you would have infinity that would represent the largest number right there is none this continues forever so now let's talk a little bit about the multiplication property of inequality so you know that we have a multiplication property of equality that we use with equations so if i have something like 5x is equal to 15 we know we can multiply or divide both sides of an equation by the same non-zero number and not change the solution so since 5 is multiplying x here i can divide both sides of the equation by 5 and i get that x is equal to 3 right very easy you have the same property with inequalities but you have an exception here that you really need to pay attention to with the multiplication property of inequality you can essentially multiply both sides of an inequality by the same positive number okay positive number and not change the solution okay you can also divide both sides of an inequality by the same positive number not change the solution what about negative numbers well we'll talk about that in a minute so let's just look at this inequality real quick so 5x is greater than 15 we know we want to isolate x 5 is multiplying x so we can go ahead and divide both sides of the inequality by five and we would get that x is greater than three in interval notation we would put a parenthesis a three and then we would have an infinity symbol and a parenthesis okay graphically i could put a parenthesis at 3 to show that it's not included 3 is the boundary and we're going to shade everything to the right let's take a look at another one so now we have 7x is less than or equal to 49. so 7x is less than or equal to 49. again if i want to isolate x if i want to isolate x i can just divide both sides of the inequality by 7 and i'll get x is less than or equal to 7. okay very very easy in interval notation you would have a negative infinity and you would have a seven here but again because this is less than or equal to next to the seven i put a bracket okay because it's included so graphically i would look at seven i would put a bracket and then i would shade everything to the left here this is my solution region so all of this all right so let's talk about the case where you're multiplying or dividing an inequality by a negative number so for this i want to just show you real fast what happens when you multiply an inequality by a negative number all right so suppose we started out with something like 3 is less than 7. we know at this point that this is a true statement 3 is less than 7. if i multiply both sides of this guy by positive 2 so let's say i multiply by 2 over here and i multiply by 2 over here so 2 times 3 is 6. 7 times 2 is 14. 6 is less than 14 that's still true okay so nothing really changed there still have a true statement but let's say i multiplied both sides by negative 2. what would happen well i'd have negative 6 over here and i would have negative 14 over here now this is false now this is false so what happened here well the smaller number when it got multiplied by a negative became a smaller negative and therefore a larger number the bigger number when multiplied by a negative became a bigger negative and therefore a smaller number okay a bigger negative represents a smaller number because it's further to the left on the number line so when you multiply or divide an inequality by a negative you've got to flip the direction of the inequality symbol so to make this true when we multiplied here we had to flip this guy so instead of a less than it had to become a greater than and then it becomes a true statement okay so we want to keep this in mind when we solve something like this so we have negative 9x is greater than or equal to 18. so negative 9x is greater than or equal to 18. okay so i know that x needs to be isolated by dividing both sides of the inequality by negative 9. what i've got to do is flip the direction of the inequality symbol right now i have a greater than or equal to i want a less than or equal to so negative nine over negative nine is one you just have x so x is less than or equal to eighteen over negative nine is negative two okay if you forget to flip the direction of the inequality symbol you will absolutely get the wrong answer okay so you've got to make sure if you're multiplying or dividing both sides of the inequality by a negative value you flip the direction of the inequality symbol very important so graphically if x is less than or equal to negative 2 negative 2 is the boundary so i'm going to put a bracket at negative 2 and i'm going to shade everything to the left in interval notation i'd put negative infinity comma negative 2 i'm putting a bracket next to negative 2 because it's included and a parenthesis next to negative infinity all right let's take a look at a number so we have negative 1 3 x is less than 2. again negative 1 3 x is less than 2. how can i isolate x well i need to multiply both sides by the reciprocal of negative 1 3 basically i need to multiply both sides of the inequality by negative 3. so times negative 3 and times negative 3. now i am multiplying both sides by a negative so i've got to flip the direction of the inequality symbol right now it's a less than it's going to become a greater than i know this is going to cancel and i'll just have x and over here 2 times negative 3 is negative 6. so we get that x is greater than negative 6. in interval notation i would put a negative 6 next to a parenthesis again because it's not included then comma you have the infinity symbol because this continues forever in the right direction and then a parenthesis next to that so graphically at negative 6 we'd put a parenthesis and then we would just shade everything going to the right okay again including this arrow because the solution region continues forever in the right direction all right so now let's talk about how to solve any linear inequality in one variable you're going to follow again the same steps you used for an equation with just a minor difference the fact that if you multiply or divide by a negative you need to flip the direction of the inequality symbol so the first thing is to simplify each side separately so combine like terms clear any parentheses do all the simplification you need to do the next thing is to isolate the variable terms on one side and the constants on the other so again you can use the addition property of inequality to do that then you want to isolate the variable so you're going to use your multiplication property of inequality to do that and then lastly you're going to check but as i'm going to show you when you check an inequality it's much more tedious okay there's a lot more involved than when you checked an equation so for this one you kind of put an asterisk next to it and say if time permits all right let's take a look at an example so we have 2x plus 9 times the quantity x minus 5 is less than 4 times the quantity 4x minus 15. so we have 2x plus 9 times the quantity x minus 5 is less than 4 times the quantity 4x minus 15. so what i want to do here is first simplify each side so let's go ahead and do that so over here i have my 2x and then plus 9 times x is 9x and then 9 times negative 5 is negative 45. this is less than 4 times 4x is 16x and then 4 times negative 15 is minus 60. all right 2x plus 9x is 11x and then minus 45 and this is less than 16x minus 60. so now i've simplified each side as much as i can and it's time to move all the variable terms to one side all the numbers to the other so i'm going to subtract 16x away from each side of the inequality so that's going to cancel i'm going to add 45 to each side of the inequality that's going to cancel 11x minus 16x is negative 5x and this is less than negative 60 plus 45 is negative 15. okay so what i want to do now is divide both sides of the inequality by negative 5. but again because i'm dividing by a negative i've got to flip the direction of the inequality symbol right now it's a less than it's going to become a greater than so this becomes x is greater than negative 15 over negative 5 is going to be positive 3. so x is greater than 3. so let me erase all this and so x is greater than 3. okay so in interval notation we know this would be a parenthesis a 3 and then an infinity here and a parenthesis graphically i would put a parenthesis at 3 and then again i would shade everything to the right now i want to teach you how to check this guy okay i'm not going to check anything after this one because it's just so tedious but essentially what you want to do is replace your inequality symbol in both the original inequality and your solution with an equals so i would erase this and put equals and i would check x equals 3. this is my boundary again we talked about the concept of a boundary it comes up all the time when you work with inequalities it separates the solution region from the non-solution region so you can see at 3 here if i just circle it everything to the right of 3 is in the solution region and satisfies the inequality everything to the left of 3 is in the non-solution region 3 is included for a non-strict inequality but not included for a strict inequality so in other words if i change this to a less than or equal to this would become a greater than or equal to and this guy would be included i'd have a bracket there but because we had a strict inequality there it's not included and so it's part of the non-solution region okay but it's still considered the boundary because it separates the two we check our boundary so essentially i plug a three in for each occurrence of x so i'd plug a 3 in here and i'd plug a 3 in here and i'd plug a 3 in here and i'm looking to see if i get a true statement so 2 times 3 is 6 plus 3 minus 5 is negative 2 9 times negative 2 is going to be negative 18. so i can put 6 minus 18. this should be equal to 4 times 3 is 12 12 minus 15 is negative 3 you would have 4 times negative 3 so 4 times negative 3 is negative 12. 6 minus 18 is negative 12 so you're good to go here okay so the boundary is good so now what i want to do is i want to grab a number that's in the solution region and i want to check that so this is a less than so if i check something in the solution region it should result in a true statement so let's just grab 5 so we'll plug a 5 in here and here and here so 2 times 5 is 10. plus 5 minus 5 is 0 0 times 9 is 0. so you just have 10 over here so this is less than 4 times 5 is 20 20 minus 15 is 5 and then 5 times 4 is 20. so is 10 less than 20 yes this is true so you're good to go there then you want to check something on the other side of the boundary so something in the non-solution region let's just choose 0 because it's easy to work with so we'll plug a 0 in there and there and also here okay all right so 2 times 0 is 0 so that's gone 0 minus 5 is negative 5 9 times negative 5 is negative 45. so this is less than 4 times 0 is 0 0 minus 15 is negative 15 4 times negative 15 is negative 60. so is negative 45 less than negative 60 no it isn't right negative 45 is a smaller negative and therefore a larger number so this is false okay and that's good because we pick the number in the non-solution region it should not work so we verify that our solution here which is x is greater than 3 is correct all right let's take a look at another one so we have negative 5 times the quantity 5x minus 11 minus 9 is greater than or equal to 7 times the quantity 9x minus 6. so again i'm going to start by simplifying each side so negative 5 times 5x is negative 25x then negative 5 times negative 11 is plus 55 then minus 9 and this is greater than or equal to 7 times 9x is 63x and then minus 7 times 6 is 42. so on the left side 55 minus 9 is 46 so i can say this is negative 25 x plus 46. this is greater than or equal to we have 63x minus 42. scroll down get some room going now i just want to get all the variable terms to one side all the numbers to the other so let's subtract 46 away from each side and let's go ahead and subtract 63x from each side so this is going to cancel and this is going to cancel negative 25x minus 63x is negative 88x this is greater than or equal to negative 42 minus 46 is negative 88 okay so to get x by itself we divide both sides of the inequality by negative 88 and this guy's going to flip because i divided by a negative so you would get x is less than or equal to 1 okay x is less than or equal to 1. so let's erase all this so x is less than or equal to 1. if x is less than or equal to 1 then at 1 i'd put a bracket and then i would shade everything to the left and then in interval notation i would have a negative infinity and then comma i would have a one and then just make sure you put a bracket there because one is included now let's take a look at another one so we have negative 3 times the quantity 3x plus 1 and then is greater than negative 2x plus 3 times the quantity x minus 2. so again i'm going to simplify each side to start negative 3 times 3x is negative 9x and then negative 3 times 1 is minus 3. this is greater than you have negative 2x and then plus you have 3 times x which is 3x and then minus 3 times 2 is 6. so on the right side here negative 2x plus 3x is just x so we'll just put x minus 6 like that now i'm going to get all the variable terms to one side and all the numbers to the other so i'm going to subtract x away from each side of the inequality i'm going to add 3 to each side of the inequality and this guy is going to cancel and this guy's going to cancel so negative 9x minus x is negative 10x and this is greater than negative 6 plus 3 is going to be negative 3. so to finish this up let me kind of erase this to finish this up i'm just going to divide both sides of the inequality by negative 10 and again i've got to flip the direction of the inequality symbol so negative 10 over negative 10 is obviously 1. so you have x is less than you have negative 3 over negative 10 which is just 3 10. so in interval notation i'll have my negative infinity and then i'll have 3 tenths and then again i've got to use a parenthesis next to 3 10 because it's not included so graphically we have this in a scale that represents the integers 3 divided by 10 is 0.3 so let's just say it's about right here i'm just gonna write a little marking there and say this is three-tenths and in other words i would just go ahead and put a parenthesis there and just shade everything to the left okay if you're working with a pre-drawn graph like that just find out a area that makes sense put a little marking and notate it in this case we're just notating it to say we have three tenths and we just draw a little arrow going there to say hey this is three tenths put your parenthesis in this case because it's not included and then just shade everything to the left all right for the last one we'll look at one with some fractions involved so we have four thirds times the quantity x plus five halves is greater than or equal to you have negative two times the quantity negative four x over three plus five thirds so if you want to clear the fractions let's go ahead and clear the parentheses first so four thirds times x is just four-thirds x then plus four-thirds times five-halves would be twenty over six this is greater than or equal to negative two times negative four thirds x would be positive eight thirds x then negative two times five thirds would be minus 10 thirds okay so to clear the fractions i'd multiply both sides of the inequality by the lcd okay and what's that going to be we have a 3 a 6 a 3 and a 3. so i'll put a 6 here and here and let me kind of scooch this down so 6 times 4 3 x think about 6 divided by 3 is 2 2 times 4 is 8 so this would be 8 x then plus 6 times 20 over 6 the sixes would cancel you just have 20. then you would have a greater than or equal to 6 times 8 thirds x 6 would cancel with 3 and give me 2 2 times 8 is 16. so this is 16x and then minus 6 times 10 thirds 6 would cancel with 3 and give me 2 2 times 10 is 20. so let's move all the numbers to one side and all the variable terms to the other so i'm going to subtract 20 away from each side of the inequality and i'm going to subtract 16x away from each side of the inequality so this is going to cancel and this is going to cancel 8x minus 16x is going to give me negative 8x this is greater than or equal to negative 20 minus 20 is negative 40. so very easy at this point we divide both sides of the inequality by negative 8 and again we're dividing by a negative so we've got to flip the direction of the inequality symbol this will be a less than or equal to now so you get x is less than or equal to negative 40 over negative 8 is 5. so let's erase everything so again x is less than or equal to 5. so an interval notation we know that would be we have a negative infinity here on the left and then we have a 5 on the right with a bracket next to it okay because 5 is included find 5 on the number line and we're going to put a bracket there and we're going to shade everything to the left our solution region in this lesson we want to review solving a three part inequality so in the last lesson we reviewed how to solve any linear inequality in one variable in this lesson we just want to talk a little bit about three-part inequalities and kind of set the stage for our next lesson where we're really going to start talking about compound inequalities so if we have something like negative 11 is less than x minus 5 which is less than 4 how could we solve this for x let's rewrite this first we have x minus 5 that's in the middle we can say that this is greater than negative 11 and we can use a special keyword here we could say and it's less than 4. so we don't have to write it like this we can kind of split this guy up using that keyword and and i'll show you how to do that in a minute the first way i want to show you how to solve this is to basically do the same thing to each part so you have a part right here on the left you have a part right here in the middle and you have a part right here on the right so that's why we say a three part inequality my goal is to isolate the variable x in the middle i want x to be between two numbers and so what i want to do is add five to each part because i'm subtracting 5 away here in the middle so if i add 5 in the middle and then i add 5 on the right and then i add 5 on the left what's going to happen is this negative 5 and this plus 5 here those are going to cancel right negative 5 plus 5 is 0. so we get x in the middle and we could say this is greater than negative 11 plus 5 is going to give me negative 6 and we could say and it's less than 4 plus 5 is 9. so realize what this is saying this is saying that x is between negative 6 and 9. in other words x has to be greater than negative 6 and also less than 9. so let's kind of look at this on the number line so we can fully understand this concept let's kind of drag this up here and let's take a look at this guy so if i wanted a number that was between negative 6 and 9 and it's not inclusive because these are strict inequalities i would find negative 6 on my number line which is right here i would put a parenthesis facing to the right then i would find 9 which is right here i would put a parenthesis facing to the left and then i would shade everything in between so the area that i shaded there that represents all the solutions that are possible for x if i grab any number in that range any number that's in the shaded area it will work let's grab one let's say we chose 3 as an example so we would have negative 6 is less than 3 which is less than 9. is that a true statement is 3 greater than negative 6 yes is 3 less than 9 yes so this is true both parts have to be true if i grab something outside of the shaded area let's say i grabbed negative 7 as an example what's going to happen is it'll be true in one scenario but false in the other and so it's false for the three part inequality so let's say we did negative 6 is less than negative 7 which is less than 9. okay is that true is negative 7 greater than negative 6 nope that's false right that's false so you can mark the whole thing as false even though you could say that negative 7 is less than not that part is true but the whole thing's got to be true it's got to be larger than negative 6 and also less than 9. so the numbers that fit that criteria are in this shaded area here okay where negative 6 and 9 are kind of the end points they're not included those are the boundaries and everything in between works as a solution now the other way you can think about this is you can split this guy up using the keyword and that i mentioned okay so you could take this guy x minus 5 and you could say this is greater than negative 11. okay so you solve that as one problem you use your keyword and then you're going to take the other side so you're going to say x minus 5 and you're going to set that as less than 4. so you solve these individually and then you put them back together with that keyword and so what i would do is just add 5 to each side of the inequality here so negative 11 plus 5 is negative 6. this is less than this cancels you just have x so then again you have the keyword and here you add 5 to each side of the inequality so you get x is less than 9. so we have that x is greater than negative 6 and x is less than 9. so in other words both of these have to be true it's the same thing it's just a little bit easier to understand when you put x in the middle and you say x is greater than negative 6 and less than 9. right because this way it's crystal clear that x is a number between negative 6 and 9 with negative 6 and 9 not being included because again these are strict inequalities if you want to besides graphing this guy you can of course write it in interval notation and so i can say that my solution here i'd put a negative 6 next to a parenthesis because the negative 6 is not included comma i'd put a 9 and again a parenthesis next to 9 because it's not included so this is kind of two different ways to notate this and then you have a way to graph it right so that's kind of the solution for this three part inequality in the next section we're going to talk more about compound inequalities with and and you're going to see that in some scenarios you can't start out writing them as a three-part inequality they're going to have to be broken up okay because you're going to have more than just a variable expression in the middle you're going to have variable expressions all over the place and so it's not going to allow you to basically write it in this format this is the easier scenario so let's go ahead and take a look at another one all right for the next one we have negative 63 is less than or equal to 10x plus 7 which is less than or equal to 77 or you could use your and keyword you could say 10x plus 7 is greater than or equal to negative 63 and it's less than or equal to 77 okay so you could break it up if you wanted to in this specific situation because you just have this variable expression in the middle it's just easier to work it by kind of doing the same thing to each part if you split it up you got to put it back together so it's just extra work so we have negative 63 is less than or equal to again 10 x plus 7 which is less than or equal to 77 okay so let's solve this guy very very easy i want to isolate x in the middle so since 7 is being added to this term 10x i want to subtract 7 away from each part okay so from each part so that's going to cancel so in the middle i'll just have 10x and then on the left hand side negative 63 minus 7 is negative 70. on my right hand side 77 minus 7 is going to be positive 70. now i want to isolate x right here since i'm multiplying x by 10 what i'm going to do is i'm going to divide each part by 10 right so i can isolate x so in the middle we'll have x and so it's going to be greater than or equal to negative 70 divided by 10 is negative 7 and it's less than or equal to 70 divided by 10 is 7. so we see that x is greater than or equal to negative 7 and it's less than or equal to 7 okay so let's erase all this i'll kind of drag this up to the number line so in interval notation you have non-strict inequalities here so you're going to put a bracket next to negative 7 then comma you'll have a positive 7 and another bracket graphically i'm going to find negative 7 i'm going to put a bracket i'm going to find positive 7 i'm going to put a bracket and i'm just going to shade everything in between there so the solution here for x is any value that's between negative 7 and 7 with negative 7 and 7 included right because we have again non-strict inequalities all right for the next one let's take a look at an example with some fractions involved so we have negative 16 fifths is less than or equal to negative seven halves x plus 19 fifths which is less than you have 159 fifths so the first thing i'm going to do is clear the fractions again if you want to work with the fractions it doesn't really matter for me i like to not work with fractions because it just makes the inequality a lot cleaner so the lcd here you have 5 you have 2 you have five and you have five it's basically five times two or ten so if i multiplied each guy here by ten what would i have well ten multiplied by this guy right here the five and the ten would cancel give me a two two times negative 16 is negative 32. so you'd have negative 32 is less than or equal to if i multiply this guy by 10 10 divided by 2 is 5. 5 times negative 7 is negative 35 and then times x then over here if i multiply this by 10 10 would cancel with 5 and give me 2 2 times 19 is 38 so plus 38 and then this is less than if i multiply this guy by 10 10 divided by 5 is again 2 2 times 159 is 318 okay 318. all right so from this point it's pretty easy i just want to isolate x in the middle so all i need to do is start out by just subtracting 38 away from each part and negative 32 minus 38 is going to be negative 70. and this is less than or equal to you have negative 35x this canceled and this is less than 318 minus 38 is going to be 280. so now all i want to do is isolate x in the middle so i'm going to divide each part by negative 35 but i've got to be very careful here because again when i work with inequalities if i multiply or divide by a negative i've got to flip the direction of the inequality symbol so that means that this guy right here has to get flipped and this guy has to get flipped so this one is a less than or equal to will become a greater than or equal to this is a less than it will become a greater than so x will be in the middle negative 70 over negative 35 is going to be positive 2 and 280 over negative 35 is negative 8. okay now although this is technically correct because your symbols are each pointing towards the smaller number of negative 8 you generally want to order this the same way you would see things on a number line negative eight is less than x which is less than or equal to two okay typically you're just working with less than or less than or equal to symbols when you work with a three-part inequality because this is not technically in the order of the number line although it is valid because the signs are both pointing to the smaller number okay but this is generally what we do now let me erase everything all right so to put this in interval notation negative 8 is not included so i'm going to put a parenthesis and then a negative 8 and then 2 is included so i'm going to put a 2 and then a bracket now graphically i'm going to follow the same format i'm going to put a parenthesis at negative 8 and i'm going to put a bracket at 2 and i'm just going to shade everything in between okay everything in between so x is greater than negative 8 and also it's less than or equal to 2. so any number in the shaded region including 2 because again we use a non-strict inequality here but not including negative 8 because we used a strict inequality there all right let's take a look at one more so we have negative 1 is less than or equal to 2x minus 5 over 6 which is less than or equal to 2. all right so the idea here is to start out by just multiplying each part by 6 so i can get rid of that denominator so let's just copy our problem here we have negative 1 is less than or equal to 2x minus 5. again this is over 6 which is less than or equal to 2. all right so if i want to clear that denominator of 6 i can multiply each part by 6. and what's going to happen is this guy over here is going to cancel right 6 over 6 is 1 so that's gone on the left 6 times negative 1 is negative 6 so this is less than or equal to you'll have your 2x minus 5 and this is less than or equal to 2 times 6 is just 12. so let's get some room going all i want to do now is just add 5 to each part very very simple so that this part goes away negative 6 plus 5 is negative 1 so this is less than or equal to now you just have 2x in the middle and this is less than or equal to 12 plus 5 is 17. okay so the last thing here is to isolate x and to do that i'm multiplying x by 2 so i just divide each part by 2. so what i'm going to end up with is that x is going to be greater than or equal to negative one-half and it's less than or equal to 17 halves okay so let's erase everything so in interval notation i'm going to basically use a bracket with each of these numbers i'd have a bracket and then a negative one half and then comma i'd have a 17 halves and then a bracket now when we go to do our graphical solution here our number line is arranged in a scale with integers involved that's okay we know that negative one-half would be pretty much about right here so i'm just going to put a marking there in between zero and negative one basically where i see the halfway mark and i'll just put a little arrow there saying this is negative one half and i'm going to put a bracket there okay facing to the right all right so 17 halves that's basically going to be 8.5 so i'm going to go halfway between and nine i'm going to put a little mark there draw my arrow and say this is 17 halves again i'm just going to put a bracket there and now i can shade everything in between so we can say that x is greater than or equal to negative one half and it's also less than or equal to 17 halves so x can be any value between negative one-half and 17 halves with negative one-half and 17 halves included in this lesson we want to review solving compound linear inequalities with and or or so in our last lesson we talked about how to solve a three-part inequality and essentially we looked at problems like let's say negative four is less than you have something like three x plus five which is less than twenty okay now we can leave it like this or we learned that we could split this up using the special keyword and okay so we can also write the same problem as negative 4 is less than 3x plus 5 and then we would say and you would have 3x plus 5 is less than 20. now you wouldn't go through the trouble of splitting this up because this is kind of more complex or more tedious not more complex more tedious to solve it in this manner to solve a three-part inequality all you need to do is do the same thing to each part your goal is to isolate x in the middle okay notice how you just have a variable expression the 3x plus 5 in the middle you don't have a variable on the left and you don't have a variable on the right okay so all i need to do is just subtract 5 away from each part let me kind of slide this down for a moment and let me slide this up so negative 4 minus 5 is going to be negative 9. so this is less than 3x and this is less than 20 minus 5 is 15. okay so now to isolate x i just need to divide each part by 3. very very simple and what i'm going to have is let me kind of erase this we don't need it 3 divided by 3 is 1 so you just have x in the middle this is greater than negative 9 divided by 3 is negative 3 and it's going to be less than 15 divided by 3 is 5. so essentially what we're saying here is that x is a number that is between negative three and five with negative three and five not included okay so it's very easy to look at it in this format here and understand the context of what we're talking about if i looked at a number line again if x is greater than negative 3 and less than 5 well i know that graphically i could just put a parenthesis at negative 3 and a parenthesis at 5 and i could shade everything in between right those numbers that i shaded there represent solutions for x if i take any number that's shaded and i plug it in there i'm going to get a true statement right so if i plug in a 2 we could say that 2 is greater than negative 3 and it's also less than 5 right both have to be true that's the key when you're working with something with and okay both have to be true if i plugged in something outside of what i shaded let's say i just took 6. both are not going to be true here 6 is greater than negative 3 that part's true but 6 is not less than 5 6 is a larger number so that part's false so the whole thing's false okay so when you work with an inequality and you have and involved you've got to think about satisfying both requirements you can think about it as looking at a job listing that says okay you've got to be 21 and have a valid driver's license okay you've got to have both otherwise no job for you right if you're 18 and have a valid driver's license that sucks you don't qualify if you're 22 but you don't have a valid driver's license that sucks you don't qualify i want to jump into some more complex problems with and these are problems that we typically cannot write as a three-part inequality so you see that you have more variables involved with a three-part inequality you can break that up and use and but you just have one variable expression in the middle you have a number on the left side and a number on the right side with these guys you have 7 minus 17x is less than negative 13x minus 5 and then you have and you have negative 11x minus 14 is less than or equal to negative 12x minus 5. so you can't write that as a three-part inequality so you're going to have to go through the method of solving each one individually and then looking at the intersection of the solution sets in other words the areas that satisfy both inequalities okay that's what you're looking for all right so let's start with the first one we have 7 minus 17x is less than negative 13x minus 5. all right let's get some room going we'll come back up so all i want to do is just subtract 7 away from each side of the inequality and so i know that this is going to cancel and negative 5 minus 7 is negative 12. i want to add 13x to both sides of the inequality that's going to cancel and let me put my less than symbol there negative 17x plus 13x is going to give me negative 4x so to solve this i just want to divide each side by negative 4 and remember if i work with an inequality what do i have to do if i divide by a negative i've got to flip the direction of the inequality symbol so this cancels with this i've got x is now greater than okay my less than became a greater than negative 12 over negative 4 is 3. so let's erase this and go back up so the solution for this part is that x is greater than 3. okay now let's solve this part what i want to do is i want to add 14 to each side of my inequality that will cancel i want to add 12x to each side of my inequality that will cancel negative 11x plus 12x is x so x is less than or equal to negative 5 plus 14 is going to give me 9. so we have this x is less than or equal to 9. so let's think about this using my keyword and essentially what i have is that x is greater than 3 and x is less than or equal to 9. so again it goes back to my job listing scenario you've got to be 21 and you've got to have a driver's license well here x has to be greater than 3 and it's also got to be less than or equal to 9. let's copy this real quick and i want to show you this on a number line real quick so i've got two of these guys and i'm just gonna basically do one on the top and one on the bottom and we'll see the intersection or the overlap between the two graphs so if x is greater than three i'd put a parenthesis at three to shade everything to the right then for x is less than or equal to 9 i'm going to put a bracket at 9 and i'm going to shade everything to the left all right so essentially what i'm doing is i'm looking for the overlap so you see this guy right here if we kind of just put a little dash going down here 3 is not going to be included right so essentially i'm just going to copy this marking and put that here and everything to the left of 3 is not included because in this solution it's only things that are larger than three so it's got to work for both and since nothing to the left of three and including three doesn't work in this one it can't work in this one so let me erase this shading here it's anything that's less than nine so it can go all the way down to the point of three but it can't be three itself so let's just shade from here to here okay so this is the overlap of the two solutions okay so that's what we're looking at and essentially i can write this with x in the middle i can say that x is going to be larger or greater than 3 and less than or equal to 9. and it's also very convenient to write this in interval notation i can go ahead and say i have a 3 and i have a parenthesis next to that because it's not included and then comma i'm going to have my 9 and then a bracket next to 9 because 9 is included but again it's got to satisfy both so again if i took a number like 6 and i plugged it in there 6 is greater than 3 and it's also less than or equal to 9. so that's good if i took something outside of that if i took 0 0 is not greater than 3 that doesn't work 0 is less than or equal to 9 but it doesn't satisfy the first one and if it doesn't satisfy both it doesn't work again just think about the example with the have to be 21 and have to have a valid driver's license you got to have both or you're not getting a job okay so that's the idea all right let's take a look at another one with and so we have 11x minus 2 is less than 18 minus 9x and we have 16x plus 7 is greater than 1 plus 15x all right so all i want to do is just solve these individually and then again just put the solutions together so i have 11x minus 2 is less than 18 minus 9x you scroll down get a little bit of room going and we'll come back up so i'm going to add 2 to each side of the inequality so that's going to cancel 18 plus 2 is 20. i'm going to add 9x to each side of the inequality and this is going to cancel so i'll have the less than symbol there 11x plus 9x is 20x to get x by itself very easy i just divide both sides of the inequality by 20 and i get that x is less than 1. okay 20 over 20 is obviously 1. so let's erase everything so we'll say here that x is less than 1. all right let's solve this guy now so if i have 16x plus 7 is greater than 1 plus 15x again i'm just going to go ahead and subtract 7 away from each side of the inequality that cancels i'm going to subtract 15x away from each side of the inequality and this is going to cancel so on the left 16x minus 15x is going to be x this is going to be greater than 1 minus 7 is negative six so we have x is greater than negative six again we have this keyword and so we're looking for the intersection of those two solution sets so x needs to be less than one and okay and x needs to be greater than negative six so you can think about this again by putting x in the middle and saying it's greater than negative six and less than one okay so it's between negative 6 and 1 not including either so let's take this to the number line all right so in interval notation essentially all i would need to do is put a parenthesis a negative 6 a 1 and a parenthesis right because negative 6 and 1 are not included so you've got a parenthesis next to each one i'm just basically going to match this format on my number line x is between negative 6 and 1 and negative 6 and 1 are not included so a parenthesis at negative 6 a parenthesis at one i'm just going to shade everything in between all right let's take a look at one more of these with and then we're going to move on to the key word or so we have 5x plus 15 is less than or equal to 8x minus 12 and then we have and we have 15x minus 3 is less than 14x minus 4. so we'll start out with the guy on the top we have 5x plus 15 is less than or equal to we have 8x minus 12. so i'm going to subtract 15 away from each side here so that's going to cancel you're going to have 5x is less than or equal to you'll have 8x and then negative 12 minus 15 is negative 27. i'm going to subtract 8x away from each side of the inequality so that'll cancel and i'll have 5x minus 8x which is negative 3x and basically this is less than or equal to negative 27. so if i divide each side here by negative 3 again what do i got to do i've got to flip the direction of the inequality symbol so this becomes a greater than or equal to so i would have x is greater than or equal to negative 27 divided by negative 3 is positive 9. so we've got x is greater than or equal to 9. so we have x is greater than or equal to 9. okay for this guy right here i would add 3 to each side of the inequality and then i would subtract 14x from each side of the inequality so that's going to cancel and that's going to cancel 15x minus 14x is x and this is less than negative 4 plus 3 is negative 1. so let's erase this we're going to say that x is less than negative 1 here so again when we put our kind of and statement in between these two we get that x is greater than or equal to nine and x is less than negative one okay x is less than negative one now before we kind of think about this let's look at a number line real quick and see where we get a problem here so let's paste this guy right here let's first graph x is greater than or equal to 9. so at 9 i would put a bracket and i would shade everything to the right let's graph x is less than negative one so at negative one i would put a parenthesis facing to the left and i would shade everything to the left here okay all of this area now do you see an overlap between the two solution sets no you don't you can't have a number that is less than negative one and greater than or equal to nine it's not possible okay so when this happens and you're using the keyboard and you don't have a solution so for the compound inequality you just say no solution right individually for each inequality there's a solution but for the compound inequality again for the overlap of the two solution sets it's not possible so you just put no solution a lot of teachers will like you to put the symbol for the null or empty set it's that right there so you can put that as well but you're basically just saying that there's no solution all right so now let's think about the key word or when we're talking about compound inequality so let's go back to our example with the job with and we said we had to be 21 and have a valid driver's license okay you had to have both but let's say we replaced it and said you had to be 21 or have a valid driver's license what does that mean it means if i have a valid driver's license it doesn't matter how old i am i can apply for the job and i can get it it also means if i don't have a driver's license and i'm 21 i can apply for the job and i can get it if i don't have either right if i'm not 21 and i don't have a driver's license well then i don't get a job right but if i have at least one of them then i'm okay if i have both of them okay so that's the situation with or right it can be either of them or it can be both of them okay so if we look at 9x plus 20 is less than 16 plus 11x or negative x plus 13 is greater than 5x plus 19. okay so we're going to solve each one individually again so let's write this one down here we have 9x plus 20 is less than 16 plus 11x let me get a little room going come back up so what we want to do is subtract 20 away from each side and let's just go ahead and subtract 11x away from each side and we can cancel this and we can cancel this and what do we have here 9x minus 11x is negative 2x this is less than 16 minus 20 is negative 4. so to get x by itself i just divide both sides by negative 2 and i've got to flip the direction of the inequality symbol because i divided by a negative right now it's a less than it's going to become a greater than we would have x is greater than negative 4 over negative 2 is 2. so we've got x is greater than 2. so let's erase all this so x is greater than 2. okay for this one right here we can subtract 5x away from each side and we can subtract 13 away from each side so this cancels and this cancels you would have negative x minus 5x that's negative 6x this is greater than 19 minus 13 is going to be 6. we would divide each side by negative 6 and again i'm dividing by a negative so i've got to flip the direction of the inequality symbol it's a greater than right now so it's going to become a less than so x is less than 6 over negative 6 is negative 1. so we would have or x is less than negative 1. so let's erase this and again all i'm going to do here i'm just going to drag this down to the number line so if x is greater than 2 i could put a parenthesis at 2 and i can just shade everything to the right for x to be less than negative 1 i could put a parenthesis at negative 1 and i could shade everything to the left okay think about the fact that if you add and here you would not have a solution there's no overlap but we don't have and we have or remember if i'm going for my job and there's an or i can be 21 or have a valid driver's license i can get it either way so here x can be greater than two or it could also be less than negative one either of these works okay either of these works the only numbers that don't work are from negative one to two with negative one and two included okay so in interval notation to kind of show this concept we need to use a union symbol some of you have seen this you've talked about sets before we'll talk about sets more in depth later on in the course but for right now you have this u looking symbol and it stands for the union okay it's not an actual u but it looks like a u so this is the union so i'm gonna place this between the two intervals to combine them as a union okay so what i'm going to do is i'm going to say that x can be from negative infinity up to and not including negative 1. so that's kind of the first solution here that's this one right here that's the one coming from the left coming from negative infinity up to but not including negative 1. so i'm always going to put that on the left because it's on the left in the number line right and then i want to use my union symbol so i want to use my union symbol again looks like a u it's not a u but it looks like one and then i'm going to look at the solution on the right in terms of the number line so it's going to start with any number greater than 2. so i'm going to put a 2 next to a parenthesis and then comma it goes out to positive infinity okay so it's the again the union the union of the two solution sets you've got this guy right here again coming from negative infinity up to but not including negative one it's the union with this guy right here anything larger than 2 again going out to positive infinity all right let's take a look at one more of these very very easy concept overall so we have 15 plus 13x is greater than or equal to you have 5 plus 18x then or you have 1 plus 5x is greater than 4x minus 6. again i'm just going to write this one down here so 15 plus 13x is greater than or equal to you have 5 plus 18x so what i want to do i want to subtract 15 away from each side so that's going to cancel and i want to go ahead and subtract 18x away from each side and make that 8 a little bit better so that's going to cancel all right let me get a little room going here so 13x minus 18x is going to be negative 5x this is greater than or equal to 5 minus 15 or 5 plus negative 15 is negative 10. and so i'm going to divide each side by negative 5 so that i can isolate my variable x negative 5 over negative 5 is obviously 1. you would have x is remember we divided by a negative so this gets flipped x is less than or equal to 2 right negative 10 over negative 5 is positive 2. so x is less than or equal to 2. so again x is less than or equal to 2. okay for this guy right here i'm going to subtract 1 away from each side that cancels i'm going to subtract 4x away from each side that cancels 5x minus 4x is x this is greater than negative 6 minus 1 is negative 7. so you just get that x is greater than negative seven all right so let's think about this again using our number line a lot of times it's very easy to visualize the solution so we have x is less than or equal to two or we have x is greater than negative seven so let's go ahead and copy this all right so let's think about x is greater than negative seven so greater than negative seven i'm just gonna put a parenthesis at negative seven and i'm gonna shade everything going to the right including my arrow and for x is less than or equal to 2 i would go to 2 and i would put a bracket there and i would go ahead and shade everything going to the left let me use a different color here so i would shade everything going to the left so what you'll see here is that your solution set would be all real numbers so any real number would work in at least one of them so the ones that would work in both fall in the category of between negative 7 and 2 with negative 7 not being included because it's a strict inequality and 2 being included because it's a non-strict inequality so any number in that range would satisfy both okay but outside of that range we would fall in the category of one of them so it would work in one and not work in the other but again because we have the or it only has to work in one of them so i can take any real number that i would like and i would get a correct answer if i plugged it into the compound inequality again because of the or okay so i can just say that my solution here is all real numbers okay now one way you can notate this we can just erase kind of these markings here you can just shade the entire number line graphically so i can just shade this whole thing and just say that they all work okay everything from negative infinity to positive infinity in interval notation that's exactly what you would put you would put from negative infinity to positive infinity so all real numbers in this lesson we want to review solving absolute value equations so before we jump in and start looking at problems where we're solving absolute value equations we want to make sure that we have a good understanding of the concept of absolute value so this is something we reviewed at the beginning of the course so if i said i want the absolute value of the number 7 what am i asking for well the absolute value of a number is the distance from the number to 0 on the number line so if i want the absolute value of 7 this is equal to 7 because if i started at 7 on the number line and i counted how many units i'd have to go to get to zero we all know it would be seven right start here you go one two three four five six and then lastly seven units to get to zero similarly if i wanted the absolute value of negative 7 it's also 7 because if i start at negative 7 and i want to get to 0 i got to go 7 units to the right so 1 2 3 4 5 6 and then finally 7. so these two numbers 7 and negative 7 are known as opposites okay they have the same absolute value in each case it's 7 but they lie on opposite sides of the number line so 1 is going to be negative and one's going to be positive right so you have something like negative 2 and positive 2 or negative 2 3 and positive 2 3 or negative 2 million and positive 2 million right so on and so forth when we take the concept of absolute value and we start working with variables you really have to think a little bit so if i had something like the absolute value of x equals 5 what could i plug in for x there and get a true statement so the absolute value of x is equal to 5. so in other words i am asking for a value that when plugged in here i would take the absolute value of it meaning i want to find the distance between that number and 0 on the number line and it should be equal to 5. well what numbers have an absolute value of 5 well it's only going to be 5 and negative 5 right so 5 and negative 5 okay so that's what i can plug in for x if i plug in a 5 there and i evaluate i get 5 equals 5. that's true but again that's not the only solution because negative 5 the opposite of 5 also has an absolute value of 5. so if i plugged in a negative 5 for x you would get 5 equals 5 as well which is also a true statement so what ends up happening is we're going to say that our solution for this equation is going to be what it's going to be that x could be equal to 5. then we have that keyword or that we talked about in the last lesson with compound inequalities so x equals 5 or x equals negative 5. again the or just tells me that either works if i plug in a 5 here take the absolute value i get 5 5 equals 5 that's true if i plug in a negative 5 absolute value of negative 5 is also 5 5 equals 5 that's also true so either one works that's why we use the or keyword now in most cases we're not going to get something as simple as the absolute value of x equals some number okay we're going to have some absolute value of an expression equal to some number or later on we'll see things get more complicated than that but essentially if you have an equation in this format you have the absolute value operation isolated on one side with an algebraic expression inside the absolute value bars and it's equal to some non-negative number like k then we can go ahead and set up a compound equation with or we could say that this guy right here the ax plus b could be equal to k or ax plus b could be equal to the negative of k all right now it's hard to understand this when you look at it in this generic formula but i want you just to jot this down in your notes anyway and let's look at the first example we're going to kind of relate back to this so we have the absolute value of x minus 3 is equal to 4. so you see that you have the absolute value operation isolated on one side you have some non-negative number on the other okay so we're good to go there so essentially what my formula told me to do was to take this expression here so x minus 3 and set it equal to this number so equal to k remember in this case k is just 4. so equal to 4 then or you take your expression again so x minus 3 could be equal to negative 4. okay remember you said negative k now it's just negative 4. now why does this thought process work well think about it if i find a value for x that when i plug it in here and i subtract away 3 it's going to give me 4 basically i'm going to be evaluating the absolute value of 4 and that is 4. so i'm good to go there alternatively if i find a value for x then when i subtract away 3 i get negative 4 well i'm going to be evaluating the absolute value of negative 4 which is also positive 4. so that's what i'm looking for i'm looking to get a result inside of the absolute value bars here of either 4 or negative 4. it doesn't matter which one it is because in each case i will get a value of 4. so let's go ahead and solve this really quickly on this one i would just add 3 to each side of the equation i'm going to get that x is equal to 7. for this one i'm going to add 3 to each side of the equation and i'm going to get that x is equal to negative 1. so let's check these it's very easy to do so i'm going to plug in a 7 for x here so i would have the absolute value of 7 minus 3 is equal to 4. so 7 minus 3 is 4. so i would have the absolute value of 4 equals 4 and of course that's true 4 does equal four so that's a big check okay so that works now let's check the other scenario we have x equals negative one so if i plug in a negative one for x right there so negative one minus three we're taking the absolute value of that this should be equal to 4. negative 1 minus 3 is negative 4. so you'd have the absolute value of negative 4 is equal to 4 and we know that's true right the absolute value of negative 4 is 4 so you get again 4 equals 4 so this one's a big check 2. so you can see how that ends up working right you take this expression right here this guy and you set it equal to this number and then the negative of that number you solve those two equations you kind of put it together with that keyword or and then you're going to have your solution so there's a few different ways to notate the solution you can kind of remove these check marks you don't need them anymore you can just put an or there between them so x equals 7 or x equals negative 1 that's fine you can use the solution set notation and say you have negative 1 and 7 as the elements or basically solutions that work for your equation and then lastly you could also do this you could say x equals 7 comma negative 1 right lots of different ways to notate this but just make sure that you're indicating that each one is a solution to your equation all right let's take a look at another one so we have the absolute value of 5x minus 4 is equal to 6. so again we're following the same thought process we're going to take this expression inside the absolute value bars so 5x minus 4 we're going to set it equal to 6 and we're going to have or we're going to have 5x minus 4 is equal to the negative of 6. okay so what i want to do i want to solve each one of these so i want to add 4 to each side of the equation here that's going to cancel i'll have 5x is equal to 10 divide both sides by 5 and i'll get that x is equal to 2. over here i just want to add 4 to each side that's going to cancel i'll have 5x is equal to you'll have negative 2 divide both sides by five and you'll have that x is equal to negative two-fifths all right so let's erase these so i'm gonna go ahead and say that x is equal to two or x is equal to negative two-fifths like that again you can write it that way you can use the solution set notation a lot of books do that or you can also just say that x equals two comma negative two-fifths right each is a solution now again if you want to check this you could plug a 2 in for x and you could also plug a negative two-fifths in for x and you would see that the left and the right side are equal so let's go ahead and check it so the absolute value of 5 times 2 and then minus 4 is equal to 6. so 5 times 2 is 10 10 minus 4 is 6. the absolute value of 6 equals 6. so 6 equals 6 so yeah that's true you can check that one off then the other one is negative two-fifths so negative two-fifths so five times negative two-fifths the fives would cancel you'd have negative two minus four which is negative six so you basically have the absolute value of negative six which we know is six so six equals six here so this one checks out as well so again x equals 2 or x equals negative 2 fifths let's look at one that's a little bit more complicated it's just more complicated because you don't have the absolute value operation isolated on one side so we have 5 multiplied by the absolute value of negative 7x plus 3 plus 3 equals 58. so your goal if you get something like this is to isolate the absolute value operation so how can we do that it's very very easy just think about what's being done to that absolute value operation you have multiplication right you're multiplying by five and you have addition three is being added to it so let's go ahead and subtract three away from each side to start we'll get rid of that we'll have 5 times the absolute value of negative 7x plus 3 and this equals 58 minus 3 is just 55. now i can just divide both sides by 5. so i divide both sides by 5 and this would cancel with this and so i have the absolute value of negative 7x plus 3 is equal to 55 over 5 is 11. so now i have it in the format that i've been looking at and again once i have it like this it's very very easy i take my expression so negative 7x plus 3 i set it equal to 11 and then i say or i take my expression so negative 7x plus 3 i set it equal to negative 11. okay just solve each one very very simple so i'm going to subtract 3 away from each side of the equation i'm going to get negative 7x is equal to 11 minus 3 is 8. then i'm going to divide both sides by negative 7 and i'm going to get that x is equal to negative eight sevenths over here let me subtract three away from each side that cancels you'll have negative seven x is equal to negative fourteen divide each side by negative seven and you get that x is equal to positive 2. so let me erase this so again x equals 2. so you can leave this as your solution again you could write it in solution set notation you could put the comma there's a zillion different things you can do whatever you're more comfortable with so let's go ahead and check this real quick and then all the rest of them i'll leave that up to you to check them on your own so i would have 5 multiplied by the absolute value of i'd have negative 7 times negative 8 7 so we know that this would cancel with this the negatives would cancel so i'm basically just going to have an 8 there so i'd have 8 plus 3 which is just 11. so then plus 3 equals 58. so the absolute value of 11 is 11. so you have 5 times 11 there that's 55 and then 55 plus 3 is 58 so this one's good to go over here you have the 2 that's going to be plugged in for x so 5 times the absolute value of negative seven times two is negative fourteen negative fourteen plus three is negative eleven but again because i'm taking the absolute value there just becomes positive eleven so five times eleven then plus three equals fifty-eight we already know that's true 5 times 11 is 55 55 plus 3 is 58 so we're good to go in each case again our solution is x equals negative 8 7 or x equals 2. all right let's take a look at another one so suppose we had 4 multiplied by the absolute value of negative 7 minus 10x then we had plus 5 and this was equal to 17. again my goal here is to isolate the absolute value operation on one side and then to have a number on the other okay so what i want to do i want to just subtract 5 away from each side of the equation to start so we know this would cancel so we'll say we have 4 times the absolute value of we have negative 7 minus 10x and then this is equal to 17 minus 5 is going to give me 12. scroll down get some room going all right so to isolate the absolute value operation i have 4 that's multiplying it so all i need to do is divide both sides of the equation by 4 so we know that again this would cancel i'll have the absolute value of negative 7 minus 10x this will be equal to 12 divided by 4 is 3. so now we have this in the format of our absolute value operation equals some number so the absolute value operation here is equal to some number so i can set up my compound equation with or so i just take the expression inside in this case it's negative 7 minus 10x i'm going to set it equal to 3 then or i'm going to take my expression inside so negative 7 minus 10x and i'm going to set it equal to the negative of 3. so negative 3. and then i'm just going to solve these two equations here and i'll have my solution so on the left i'm just going to add 7 to each side of the equation i'll have that negative 10x is equal to 3 plus 7 is just 10 and now we can divide both sides of the equation by negative 10 and of course this guy is going to cancel i'll have x is equal to 10 over negative 10 is just negative 1. so that's one solution let's get our other solution over here on the right so again i'm going to add 7 to each side of the equation this is going to cancel i'll have negative 10x is equal to negative 3 plus 7 is going to be 4. so now let me scroll down get a little bit more room here if i divide both sides of the equation by negative 10 then this guy right here is going to cancel with this guy and i'll have x is equal to 4 over negative 10 we know that's negative they're each divisible by 2. 4 divided by 2 is 2 and 10 divided by 2 is of course 5 so we get x equals negative 2 5. so i can write this as x equals negative 1 or my key word or x equals negative 2 fifths again you can do the solution set notation or you can just list them with a comma whatever you want to do all right so let's take a look at some special case scenarios that come up when we're solving absolute value equations so the first one we're going to talk about is the situation where you have no solution so something like the absolute value of x is equal to negative five this guy would not have a solution so no solution why is that the case well we should know that the absolute value of some real number x is greater than or equal to zero because if i plug in a 0 here i'd get 0. if i plugged in a negative here i'd get a positive and if i plugged in a positive i would get a positive right this stems from the definition of absolute value when you take the absolute value of the number you want the distance from zero to the number on the number line so if it's zero it's just going to be zero if it's some negative again you'd get a positive if it was a positive it would just be a positive okay so you're never going to get a negative result so this guy would always be no solution when we go back to our generic formula here i said earlier that the absolute value of some expression ax plus b if this is equal to k k has to be non-negative so if you have isolated this guy right here the absolute value operation on one side and on the other side you have a negative value you can stop and say there's no solution what you don't want to do is say there's no solution before you've isolated the absolute value operation on one side that can lead to problems okay so you need to make sure you isolate first all right let's take a look at an example so we have 9 times the absolute value of 5x minus 1 then plus 10 is equal to negative 44. so the idea here is that we want to isolate the absolute value operation on one side of the equation and then we want a number on the other but if that number on the other side happens to be negative we're not going to have a solution so don't just look at the negative over here on the right hand side and assume that this guy is going to be a no solution to start you've got to isolate the absolute value operation before you can make that determination so i'm going to start by just subtracting 10 away from each side of the equation and we'll cancel this we'll have 9 times the absolute value of 5x minus 1. and this is going to be equal to negative 44 minus 10 is negative 54. so at this point you pretty much know it's going to be no solution but let's continue let's go ahead and divide both sides by nine and of course this will cancel we'll have the absolute value of five x minus one is equal to negative 54 over positive nine is negative six so we know at this point there's no solution right there's no solution we have isolated the absolute value operation on one side and we have a negative number on the other so we can just say that there's no solution like this this is what i always do i just write no solution a lot of textbooks again will use the symbol for the null or empty set which looks like this that's fine too if you want to write that it's basically anything you want to do so again i just want to emphasize here why there's no solution there's no value i can plug in for x there multiply by 5 and then subtract away 1 and then take the absolute value of it and get negative 6. it doesn't matter what this value is once i take the absolute value it's either going to be 0 or positive okay so it can never ever ever be negative 6 so that's why there's no solution now let me show you one thing real quick just want to make sure we're crystal clear on this before we move on if for example i change this problem and i'm not going to go all the way through but if i change this problem to a negative out in front you will have a solution okay let's work through that real fast i'm not going to go all the way through but i'll just show you if i subtract 10 away from each side here that would cancel i'll have negative 9 times the absolute value of 5x minus 1 is equal to negative 44 minus 10 is negative 54. now if i divide both sides by negative 9 what's going to happen let's scroll down get some room well we see that this guy right here cancels with this guy so i've isolated the absolute value of 5x minus 1 but now negative 54 over negative 9 is positive 6. so i would have a solution here i would have to set 5x minus 1 equal to 6 and then 5x minus 1 equal to negative 6 so i would have that compound equation with or and i would get two solutions there i'm not going to go through that process but i'm just making sure that you understand that just because i start with a negative on the right hand side doesn't mean that i have no solution you've got to get to the point where you've isolated the absolute value operation on one side and a number on the other if the number is zero then you're just going to have one solution which we'll talk about in a minute if it's positive you're going to have two solutions and if it's negative you're going to have no solution all right let's think about another scenario we're going to come across so suppose the absolute value of x was equal to just 0. well in this case you're not going to have to set up a compound equation with or you're just going to have one equation the absolute value of x equals 0 only if x is equal to zero okay that's the only possible solution there because the only number that has an absolute value of zero is just zero zero doesn't have an opposite if this was the absolute value of x is equal to two again x could be 2 or it could be negative 2 because each have an absolute value of 2. but when you get 0 there you only have 1 unique number that has an absolute value of 0 and that's just 0. so let's look at this equation here so we have 4 times the absolute value of 2x minus 14 then plus 5 is equal to 5. so again our goal is always to isolate the absolute value operation and when i subtract 5 away from each side you're going to immediately see you have a problem you get 4 times the absolute value of 2x minus 14 and this is equal to 0. so if i divide both sides by 4 we know 0 divided by 4 is just 0. so i'm going to end up with the absolute value of 2x minus 14 is equal to 0. so this is a special case scenario okay you do have a solution here but you only have one okay you're not going to get two so you need to pay close attention to what to do here so all we need to do is say that 2x minus 14 is just equal to zero we're just going to solve that equation so i'm going to add 14 to each side and of course this is going to cancel we'll have 2x is equal to 14. then we're going to divide both sides by 2 and i'm going to get that x is just equal to 7. so just one solution here x equals 7. so lastly let's think about what happens when we have two absolute value operations set equal to each other so something like the absolute value of ax plus b is equal to the absolute value of cx plus d so the way we think about this is two numbers or expressions can only have the same absolute value if they're equal to each other or if they're opposites of each other so we would say ax plus b is equal to cx plus c so in the first scenario just remove the absolute value bars and set the two expressions that were inside of the absolute value bars just equal to each other then or next you're going to remove the absolute value bars and you're going to put parentheses around one of the expressions and basically negate it right so you're going to put a negative sign in front of it or make it into its opposites you can even put this as a negative 1 out in front just to make it crystal clear what you're doing and it would be just as valid if i said or and i put a negative out in front of ax plus b and said this is equal to cx plus d just don't make them both negative only make one of them negative because you need them to be opposites so let's look at our example we have the absolute value of 4x minus 5 is equal to the absolute value of 9x plus 7. so again i'm going to drop the absolute value bars so i'm just going to say this is 4x minus 5 is equal to 9x plus 7. so i'm going to solve that then you have or you have or i'm going to drop the absolute value bars so i'm going to have 4x minus 5 is equal to you have 9x plus that so i'm going to choose one of these expressions it doesn't matter which one you want wrap it in parentheses and put a negative out in front okay you could do this one or again doesn't matter you can do this one you will get the absolute same answer just for the sake of what we did earlier i'm going to go ahead and just wrap this around here and just be done with okay so for the first one i'm just going to solve this i'm going to write this down here so we have 4x again 4x minus 5 is equal to 9x plus 7. let me scroll down and i'll come back up so to solve this is very very easy we just add 5 to each side of the equation that cancels 4x is equal to 9x plus 12. we subtract 9x away from each side of the equation and this is going to cancel so what i'm going to have is 4x minus 9x is negative 5x this is equal to 12. we can go ahead and just divide each side of the equation by negative 5 and we're going to find that x is equal to negative 12 fifths so let's erase all of this we're going to keep that x equals negative 12 fifths so i don't need this equation anymore i'm just going to put x equals negative 12 fifths and then i'm going to have my or here i'm going to do it vertically and then i'll have x equals something here and we'll get that from this okay so here i'm going to distribute my negative to each term again if you have trouble with that put a negative 1 there and then distribute okay that's the easiest way to remember so 4x minus 5 is equal to negative 1 times 9x is negative 9x and then negative 1 times 7 is minus 7. now what i want to do is add 5 to each side of the equation and add 9x to each side of the equation so this is going to cancel and this is going to cancel 4x plus 9x is 13x this is equal to negative 7 plus 5 which is negative 2. if we divide each side by 13 we're going to get that x is equal to negative 2 13. so x equals negative 2 13. in this lesson we want to review solving absolute value inequalities so in the last lesson we talked about solving absolute value equations in this lesson we're just going to go one step further and talk about solving absolute value inequalities now before we kind of jump into some problems let's just take a moment here and think about absolute value as a concept so if i asked you for the absolute value of something like the number 5 what am i asking you for remember the absolute value of a number is the distance from 0 to that number on the number line so if i want to know the absolute value of 5 i'm asking how far away is 5 from 0 on the number line we know with any positive number it's just the number right so 5 would be 5 units away from 0 and we can show that it's 1 2 3 4 5 right and it doesn't matter that i started at 0 and went to 5 i could start at 5 and go to 0. we're not measuring anything to do with direction we're measuring the distance okay that's what we're focused on so the absolute value of 5 would just be 5. the absolute value of again any positive number is just the number so if i wanted the absolute value of let's say 8 it's just 8. if i wanted the absolute value of 1 million it's just 1 million right so on and so forth now when we work with absolute value we've got to be aware of some of the restrictions or things that happen when we take the absolute value of a number if i take the absolute value of a negative i end up with a positive right so the absolute value of negative 5 would be what let's go to negative 5 on the number line and i'm also 5 units away from 0. so i would go 1 2 3 4 5 units to get to 0. so this answer here is also 5 right so i took the absolute value of something that's negative and i got a positive because i'm measuring a distance with absolute value i can never get a negative result okay you can think about distance as it relates to you driving in a car let's say you want to go to the grocery store you can either get in your car and decide not to go right you just play some tunes in the car and then you get lazy and decide hey i'm not going today or you can go some positive amount right but you can never go a negative distance the only thing you can do besides getting a positive result is you can get 0 as a result okay if i took the absolute value of 0 i would get 0. and that's because 0 is 0 units away from itself on the number line so the main thing to understand here is that the absolute value operation is always going to give you a result that is non-negative so zero or positive okay very important to understand that because it will come up later on in the lesson now we also want to think about one other thing here i want you to notice that 5 and negative 5 have the same absolute value right if i did 8 and negative 8 those would also have the same absolute value okay so numbers like these 5 and negative 5 or 8 negative 8 or 2 million and negative 2 million these are known as opposites okay they're going to have the same absolute value okay because they're the same distance away from zero but they lie on opposite sides of the number line one's on the negative side one's on the positive side so we call them opposites all right so now that we understand the concept of absolute value let's think a little bit about how we can solve an absolute value inequality we're just going to start with some very simple examples here and then we'll get into some more challenging material so we're going to start with the absolute value of x is greater than 5. so using our understanding of absolute value what am i asking for what can i plug in for x here that would satisfy the inequality well remember if i'm taking the absolute value of something i want to know the distance from that to 0 on the number line the absolute value of x is greater than 5 is saying that x can be any value whose distance from zero is greater than five so to find this i could start at zero and i could go five units to the right and i could shade from five on but five wouldn't be included because it's got to be strictly greater than 5. so i'll put a parenthesis at 5 or you can use an open circle if you want if that's what you're doing in your class and i want to shade everything to the right here because that would be part of my solution right so if i plugged in a 6 there absolute value of 6 is 6 6 would be greater than 5. that's true okay so that would work if i plugged in an 8 the absolute value of 8 is 8 8 is greater than 5 that's true so any number larger than 5 would work that's pretty obvious everybody gets that but also i've got to look in the other direction of 0 because again when we think about absolute value it's a distance so i can also travel to the left i can go 1 2 3 4 5 units here so anything less than negative 5 would work as well negative 5 does not work so i'm going to put a parenthesis there and again you can use an open circle if that's what you're doing and then i would shade everything to the left here and then we could try some numbers out in this range let's say we picked negative seven so we plug in a negative seven for x the absolute value of negative seven is seven seven is greater than five that's true so what we have here for our solution is a compound inequality with or okay so we can basically say that our solution is that x is greater than 5 or remember our keyword or x is less than negative 5. so we could also notate this in interval notation right it would be the union of the two solution sets so i would have from negative infinity out to but not including negative 5 and then the union with you have anything larger than 5. so a parenthesis next to 5 again out to positive infinity okay notice how i used a parenthesis with each with negative 5 and 5 because they're not included all right so let's look at a general rule that we can follow when we're solving these guys and this first rule i'm going to give you is just the simplest case scenario we'll look at a different rule in a minute that will cover more challenging problems so we saw that the absolute value of x is greater than 5 led to x is greater than 5 or x is less than negative 5. so in general the absolute value of x is greater than k so this is a greater than case where we say that k is greater than 0 okay this has to be the case to use this rule then what's going to happen is we set up a compound inequality with or so x can be greater than k or you could say the negative of x because the absolute value will make that positive is greater than k now what happens here when you see this in your textbook what they do is they divide both sides by negative 1 so that it's solved for x and so you're going to get that x is less than remember if you divide by a negative you've got to flip the direction of the inequality symbol so a greater than becomes a less than and then k over negative 1 is just negative k so what we see is that x can be greater than k or x can be less than negative k okay so this is what you're going to see and again when we worked with this first example here where we said that the absolute value of x is greater than 5 we got just this x was greater than 5 or x was less than negative 5 okay so that's where that comes from so if i gave you a problem like let's say the absolute value of x is greater than 14 well then you can very quickly say that x is greater than 14 or x is less than negative 14. okay the first scenario you just remove the absolute value bars so in this case it would just be x is greater than 14. the second scenario you just think about it like this flip the direction of the inequality symbol so my greater than becomes less than and then change the sign of the number out here so from 14 i get negative 14. so if you could just remember that you flip the sign and then make the number negative you're good to go okay that's all you need to remember so let's look at our rule when it's more complicated so when we get more challenging problems we're not going to always have the absolute value operation isolated to start so we've got to get it in this format of the absolute value operation on one side and a number on the other okay notice how i've said that k is greater than zero again if it's zero or less than that we'll talk about those special cases at the end of the lesson but i want to isolate this to where the absolute value of some expression ax plus b is greater than k so this is going to set up to this expression ax plus b is greater than k or ax plus b is less than the negative of k all right so same process for the first one i just remove the absolute value bars i copy what's there it's very simple for the second one i remove the absolute value bar so the expression doesn't change i flip the direction of the inequality symbol so now it's a less than and then i make that number negative so instead of just k it becomes negative k all right so let's look at an example all right so for this one we have the absolute value of 2x plus 3 is greater than 17. so the first case is i just remove the absolute value bars so i'll have 2x plus 3 is greater than 17. then or my second case i remove the absolute value bars so 2x plus 3. i flip the direction of the inequality symbol so it's a less than and then i make this number negative so it's going to be a negative 17. okay and if you have trouble remembering that you can always just make the expression over here you could always say it's the negative of this guy so the negative of 2x plus 3 is greater than 17 when you divide both sides by negative 1 you're going to get it into that format so this would flip and this becomes negative okay so that's another way you can kind of remember it all right so let's quickly solve these two inequalities for this one i would just subtract 3 away from each side i would get 2x is greater than 17 minus 3 is 14. divide both sides of the inequality by 2 i get x is greater than 7. okay over here if i subtract 3 away from each side of the inequality that cancels i get 2x is less than negative 17 minus 3 is negative 20. divide both sides by 2 and we're going to get that x is less than negative 10. okay so we got that x is greater than 7 or x is less than negative 10. so let's erase this okay so let me put my or here so you can write it like this as your solution you can of course do interval notation so from negative infinity up to but not including negative 10. so it's the union with you've got not including 7 and then out to positive infinity okay so negative 10 is not included you've got a parenthesis there 7 is not included you've got a parenthesis there now graphically on our number line we can put a parenthesis at 7 and we can shade everything to the right okay all that and then in terms of x being less than negative 10 we don't have a notch for negative 10 but we can just make one real quick so we'll say this is negative 10 and let me make that a little bit better and then i'm just going to put a parenthesis there and i'm going to shade everything to the left all right let's take a look at another so suppose we have 10 times the absolute value of negative 4x minus 5 plus 2 is greater than 72. so again the idea here is to isolate the absolute value operation on one side have a number on the other so what i'm going to do is i'm just going to subtract 2 away from each side of the inequality to start that cancels you've got 10 times the absolute value of negative 4x minus 5. and this is going to be greater than 72 minus 2 is 7. so now since 10 is multiplying this absolute value operation i can just divide both sides of the inequality by 10. so that cancels now i've isolated the absolute value operation i've got the absolute value of negative 4x minus 5 is going to be greater than 70 over 10 is 7. okay so let me erase this all right so we know our scenario here with the greater than we want to remove the absolute value bars so negative 4x minus 5 is greater than 7 then or we want to remove the absolute value bars so negative 4x minus 5. we want to flip the direction of the inequality symbol so this is less than we want to make this negative so this is negative 7. okay so let's solve our compound inequality let's add 5 to each side of the inequality that cancels you get that negative 4x is greater than 12. we divide both sides by negative 4. remember this has the flip we divided by a negative so this is a less than now so you've got x is less than 12 over negative 4 is negative 3. over here if i add 5 to each side of the inequality that cancels you're going to get negative 4x is less than negative seven plus five is going to be negative two we divide both sides by negative four again if i divide by a negative i've got to flip that guy so this becomes a greater than so we have x is greater than negative two over negative four is positive one-half all right let's erase everything so we can write our solution as x is less than negative three or x is greater than one-half again in interval notation you're coming from negative infinity up to but not including negative 3. so the union with you don't include one-half out to positive infinity okay so again negative three and one-half are not included you have a parenthesis next to each so now graphically at negative 3 i'd put a parenthesis facing to the left and i would just want to shade everything going to the left here so all of that and then at one half which would be about right here let's say i'm going to say that's one half i'm going to go ahead and put a parenthesis there so let's put a parenthesis there and again let's just shade everything to the right okay so that's my solution region all right now let's talk about the other scenario which is when we have a less than so if i have the absolute value of x is less than 5 i'm going to follow the same procedure but i'm going to use an and compound inequality so in other words i'm going to remove the absolute value bars so i'm going to say x is less than 5. then i'm going to say and then the next one i'm going to remove the absolute value bar so i'm going to have x i'm going to flip the direction of the inequality symbol so this is greater than i'm going to make the number negative so this is a negative 5. so it's the exact same process but we're using an and now okay because we're going to have a between so x will be between negative 5 and 5. you could really write this as a three part inequality and say that negative 5 is less than x which is less than 5 or you can read that as x is greater than negative 5 and less than 5. now conceptually why does this make sense i'm asking for all the numbers whose absolute value or whose distance from 0 is less than 5. again if i started at 0 and i went 5 units to the right i couldn't be 5 because it's got to be less than 5 so i'd put a parenthesis there facing to the left if i started at 0 and went the other way again i go 5 units it couldn't be negative 5 but it could be anything greater there so it would have to be between negative 5 and 5 on the number line those are the only values that would satisfy this if i plugged in a negative 2 the absolute value of negative 2 is 2 2 is less than 5. that works if i took something outside of the range let's say i took 6. the absolute value of 6 is 6 6 is not less than 5. so it's got to be in that range of numbers to work and satisfy that inequality if we see something more complex we can just write down this rule so i isolate my absolute value operation on one side it's the same thing we saw with the greater than but just now i'm going to have an and so it's the exact same procedure so the absolute value of ax plus b is less than k where k is greater than zero we say we have ax plus b is greater than negative k and ax plus b is less than positive k so again the first scenario just remove the absolute value bars so that's this here the second scenario remove the absolute value bars flip the direction of the inequality symbol so you get a greater than and then make this number out here negative okay the only difference is you're using an and statement so really if you want to make this easier on yourself just write it as a three part inequality so the expression goes in the middle and it's greater than the negative of this and it's less than the positive of this all right let's take a look at an example so suppose we have the absolute value of 8x plus 4 and this is less than or equal to 52. well this guy's already isolated for me so i don't have a whole lot of work here all i've got to do is follow my rule so i'm going to drop my absolute value bars i'll have 8x plus 4 is less than or equal to 52. then and okay i'm going to drop my absolute value bars so 8x plus 4 flip the sign so this becomes a greater than or equal to make 52 negative so it's negative 52. now i would advise when you solve these to put it as a three-part inequality because it's just easier to work with so i'm going to put that negative 52 is less than or equal to 8x plus 4 which is less than or equal to the number 52. so if i want to solve for x in the middle i would subtract 4 away from each part let me scroll down get some room going so on the right hand side 52 minus 4 is 48 and then this cancels so i've just got 8x in the middle on the left hand side negative 52 minus 4 is negative 56. all right so to get x isolated in the middle i'm going to divide each part by eight let me make that a little better so negative 56 divided by 8 is negative 7. this is less than or equal to 8 over 8 is 1 so we just have x here in the middle this is less than or equal to 48 over 8 is 6. so i have that x is basically between negative 7 and 6 and negative 7 and 6 are both included here because we have non-strict inequalities so let me erase all this and again we can write this in interval notation we can use a bracket here at negative 7 and then a bracket at 6 to show these are both included on the number line i can use a similar process so at negative 7 i put a bracket at 6 i'm going to put a bracket and let me make that a little better and again i'm just going to shade everything between here so from negative 7 to 6 all those numbers work including negative 7 and 6. all right let's look at another one so we have 6 times the absolute value of negative 3x plus 9 then plus 4 is less than or equal to 94. so again i've got to isolate my absolute value operation so i can use my little rule so i'm going to subtract 4 away from each side of the inequality that cancels i've got 6 times the absolute value of negative 3x plus 9 this is less than or equal to 94 minus 4 is 90. so now i'm going to divide both sides by 6 because that's what's multiplying this absolute value operation so let me get some room here this cancels we'll have the absolute value of negative 3x plus 9 is less than or equal to 90 over 6 is 15. okay let me get a little bit more room so now we're ready to set up our compound inequality with and so again first scenario just remove the absolute value bars so negative 3x plus 9 is less than or equal to 15. then and okay you've got negative 3x plus 9. again remove the absolute value bars flip the sign so it's a greater than or equal to make this negative so this is negative 15. and again it's easier to just write this as a three part inequality just makes it faster to solve so negative 15 is less than or equal to negative 3x plus 9 which is less than or equal to 15. all right let's solve this guy so we're going to subtract 9 away from each part here negative 15 minus 9 is going to be negative 24. this is less than or equal to you've got your negative 3x here and this cancels and this is less than or equal to 15 minus 9 is 6. all right so to get x by itself let's divide each part by negative 3 and remember even though it's a three-part inequality we've still got to flip these signs very important you do that okay otherwise you will get the wrong answer so x is going to be in the middle by itself i'm going to flip the signs to greater than or equal to negative 24 over negative 3 is positive 8 and 6 over negative 3 is negative 2. so i'm just going to reorder this and say this is x is greater than or equal to negative 2 and it's less than or equal to 8. so let's erase everything all right so let's go ahead and write this in interval notation again our endpoints are included here so i'll put a bracket and then a negative 2 and then i'll put 8 and then a bracket okay so this is interval notation again negative 2 and 8 are both included so we use our brackets then on my number line i could put a bracket at negative 2 and a bracket at 8 i could just shade everything in between so that's my solution region again from negative 2 to 8 with negative 2 and 8 included all right now let's talk about the special case scenarios that come up basically you have to think about things if you get a right side of 0 or a negative you have to think about what's going on okay if i get something like the absolute value of x is greater than or equal to zero this is true for all real numbers okay so if you get the absolute value isolated on one side and it's greater than or equal to the number zero it's true for every real number because the absolute value operation is either zero or some positive amount so if the result from your operation is zero you're okay right it works if it's some negative it still works right because you're going to get a positive if it's positive it still works so every real number would be greater than or equal to 0 when it's run through the absolute value of operation so this is all real numbers okay for an answer for the absolute value of x is greater than or equal to negative 5 again absolute value is always non-negative so it's always going to be 0 or higher and it's always going to be greater than the number negative 5 right it will never be equal to it but if it's a negative value it'll always be greater than that so this is also true for all real numbers okay so if you see a negative over here and you've got a greater than all real numbers is your solution all right what if you've got a less than and a negative over here well here you can't get a solution right the absolute value of x is less than negative five that's not possible okay it can never be less than a negative it can't be less than zero okay it can be zero or positive that's it so for this guy there's no solution okay now this one's a tricky one okay what if you saw the absolute value of x is less than or equal to zero well we know that it can't be less than zero but it can be equal to zero so this one will have one solution you basically need to set x equal to zero and that's your solution if x equals zero here that works right the absolute value of zero is less than or equal to 0 right 0 is less than or equal to 0 that is true okay but if i use any other number it does not work okay so what i want to do with this type of scenario if i've isolated the absolute value operation and it's less than or equal to 0 i just set that expression inside the absolute value bars equal to 0 and i solve and that's going to be my solution all right let's look at a few examples so we have negative 7 times the absolute value of 10x plus 4 plus 1 is less than or equal to 43. so again we want to start by isolating the absolute value operation so i'm going to subtract 1 away from each side of the inequality that's going to cancel i've got negative 7 times the absolute value of 10x plus 4 again is less than or equal to 43 minus 1 is 42. all right let's get some room going so now i'm going to divide both sides of the inequality by negative 7. this cancels i'm going to have the absolute value of 10x plus 4. now i divided by a negative so this has got to get flipped so it's going to become a greater than or equal to 42 over negative 7 is negative 6. now you can stop here the absolute value operation is isolated and it's greater than a negative value what did we say about that the absolute value operation is always going to be 0 or some positive amount so it will always be greater than negative six so this is true for all real numbers so you can stop and say all real numbers in interval notation you can put from negative infinity to positive infinity and if you wanted to show this on a number line you could basically just shade the entire number line okay this would be your graphical solution just shade the entire thing all right let's take a look at another so suppose we had negative 4 times the absolute value of 6x minus 8 then minus 6 is greater than 98. so again i want to isolate this part right here first very important you do that so let's add 6 to both sides of the inequality that cancels you've got negative 4 times the absolute value of 6x minus 8 and this is greater than 98 plus 6 is 104. let's divide both sides by negative 4. so we know this cancels i've got the absolute value of 6x minus 8. this is going to get flipped so it'll become a less than then 104 over negative 4 is negative 26. now can the absolute value of something ever be less than negative 26 no it cannot again it's got to be 0 or some positive amount so it doesn't matter what i plug in here for x whatever this result here is inside the absolute value bars once i take the absolute value it's going to be 0 or positive so it can never ever ever be less than negative 26 so there's no solution here no solution okay and again you can put that symbol for the null or empty set all right let's take a look at one more so we have 4 times the absolute value of 8x minus 2 is less than or equal to 0. so again i can divide both sides by 4 so that i can isolate the absolute value operation so this cancels i get that the absolute value of 8x minus 2 is less than or equal to 0 over 4 is 0. now the only way again that this works is if we set this expression here 8x minus 2 equal to 0 because it can never be less than zero that's not ever going to work right because once we take the absolute value it's going to be zero or something positive so we can get it to equal zero but we can never get it to be less than zero so there's only one solution here so 8x minus 2 is equal to zero add two to each side of the equation you'll have eight x is equal to two we're going to divide both sides by eight we're going to get that x is equal to two over eight is one over 4 okay or 1 4. so the only way to make this true is if x is 1 4. this is not a range of values if we looked at this on the number line you would basically just go to 1 4 let's say this is one fourth here and you would basically just put a dot there and say my solution is just that number in this lesson we want to review translating phrases into algebraic expressions and equations so for the majority of you you've already taken an elementary algebra course and so you have at least some experience with solving a word problem where you need to set up and solve a linear equation in one variable so these problems are usually referred to as applications of linear equations and this is going to be the topic of our current section of the course so for most students word problems are extremely challenging and the reason for this is in addition to solving an equation they now have to read through a problem interpret the problem correctly and create an equation based on the situation given now one of the roadblocks to success with word problems is being able to understand the phrases that you come across in your word problem you have to be able to correctly translate those phrases into the required algebraic expressions and equations that are needed to get the correct solution all right so we're going to look at some of these phrases and just go into how we can translate them into again expressions and equations we're going to start out today by looking at some common addition phrases and so first we have for a phrase more than so more than is telling us we need to add so as an example we have 11 more than a number now anytime i see a number or the number i can replace it with a variable like x or y or z whatever you want to use i'm just going to use x one strategy is just to kind of line out where you see a number and just put a variable i'm just going to put x there and if i read it now it says 11 more than x well that just means if i had x i want 11 more than that so that can be represented with x plus 11 right this is 11 more than x then you have sum of so the sum of a number and 5. so again when i see a number it's going to line that out and put x let me kind of change my ink here so the sum of x and 5 that's just x plus 5. then next we have total of so the total of 9 and a number so let me line this out and let me put x here so now we have the total of 9 and x so that just tells me i have x plus 9 or 9 plus x so next we have increased by and then we have for our example a number increased by 17. so where i see a number i'm going to cross that out just put x so if x is increased by 17 it's just x plus 17. then we have plus a number plus 2. so again let's line this out a number let's put x here and it says x plus 2 that's just x plus 2. very straightforward then lastly we have added 2 and then we see 15 is added to a number let me line this out where it says a number let me put x here so 15 is added to x so we have x plus 15. okay so it's very easy when you're working with addition because addition is commutative the order that you add in does not affect the result so in this case where we have x plus 11 i could reorder that and say 11 plus x it doesn't change anything same thing here x plus 5 could be 5 plus x x plus 9 could be 9 plus x you know so on and so forth so with addition and also multiplication they're commutative the order does not matter with subtraction and division the order does matter so we've got to pay closer attention to the way things are worded when we work with subtraction and division so let's look at an example of some addition phrases and we have this sample problem here so if a number is increased by 7 the result is 19. find the number so again if i see something like a number or if i see the number i'm just going to replace it with a variable so i'm going to use x but again you can use whatever you want so i'm just going to put that in each place and down here i can even specify that i'm going to let my variable x be equal to the unknown number let me make that e a little bit better okay so what am i saying here if x let me write that down is increased by we know this is a key phrase that means plus so x plus we have our number seven it says the result here is 19. so when you see is in this context here it means equals okay so we're going to put equals here and then we have 19. so once again if x is increased by we have our x here increase by again means plus then we have 7 7 is here the result is that means equals and then we have our 19 so our 19 is here so we want to find x and all we need to do is just solve this guy so i'm going to subtract 7 away from each side of the equation and that's going to give me that x is equal to 12. and you can read back through the problem and determine if you got the right answer right if 12 is increased by 7 the result is 19. well yeah that's true 12 plus 7 or 12 increased by 7 is 19. so if we want to find the number the number is just 12. okay let's take a look at another one so if a number is added to 12 the result is negative 11. we want to find the number so again if i see a number or i see the number i'm just going to replace that with a variable and i'm going to use x again i can label this down here and just say that we're going to let x be equal to the unknown number okay just for clarity and again if x so i'm going to read it like that so if x is added to 12. so is added to just means plus so x plus and then 12 the result is okay in this context this means equals then we have negative 11 okay so if x is added to 12 so x plus 12 the result is or the result equals negative 11. so that's how we got our equation then we want to find x or find the number so again i just want to solve this guy let me subtract 12 away from each side of the equation i would have that x is equal to negative 23 right and again reading back through the problem if negative 23 is added to 12 the result is negative 11 that is true right so our number here is negative 23 okay this is what we're looking for all right let's look at some subtraction phrases again you've got to pay close attention with subtraction because not that available equal to 5 plus 10. with addition i can change the order with subtraction i cannot okay so one of the phrases is minus okay we all know minus mean subtraction something like 12 minus a number so i'm starting with 12 and i'm subtracting away a number so again every time i see a number i can line that out and just put x so 12 minus x is 12 minus x again if you put x minus 12 there it's not correct right this would be wrong this would be wrong okay so you've got to make sure you go in the correct order 12 is what i'm starting with and then minus my variable x or my unknown number then we have less than so 14 less than a number okay let me line this out and put x here so again very important to interpret this correctly we want 14 less than x so that means i'm starting with x and i want 14 less than that so i'm going to subtract away 14. so this translates into x minus 14 not 14 minus x okay next we have less so a number less eight okay again a number i'm going to cross that and put x so x less eight that means i'm starting with x and i'm taking away eight from that okay another one is subtracted from so again a number subtracted from 19. let me line this out and put x so we have x subtracted from 19. so pay attention there it's subtracted from 19 so i'm starting with 19 and i'm subtracting away x okay so this is 19 minus x then we have decreased by so a number decreased by four so let's line out a number and put x so if x is decreased by 4 it's just x minus 4. then we have difference of so the difference of 5 and a number so you've got to go in order here remember when we talk about the difference that's a result of subtraction so the difference of 5 5 is going to go first and a number let me line this out and put x the difference of 5 and x would be 5 minus x if it said the difference of x and 5 that would be x minus 5. so you've got to go in order then we have takeaway so a number is taken away from 15. so again for a number i'm just going to put x so x is taken away from okay it's taken away from 15. so 15 starts out and then we take away our x okay so again with subtraction the order matters so you've got to correctly interpret these otherwise you will not get the right answer so let's look at some examples all right so we have that 4 less than a number is 29 find the number so again wherever i see a number or the number i'm just going to replace that with a variable again i'm going to use x and again for the sake of completeness i'm just going to say we're going to let x be equal to our unknown number okay so 4 less than x right think about that that means i have x and i want 4 less than that so x minus 4. then we have is in this context that means equals and then 29. so to find the number i would just add 4 to both sides of the equation and we get that x is equal to 33 right so if i had 4 less than 33 that would be 29 so the number is 33. all right next we have the difference of a number and 7 is 13. find the number so again where i see a number and the number i'm just going to replace that with a variable so we'll use x again and again for completeness i'm just going to say that we're going to let x be equal to our unknown number okay so what can we say here the difference of we have x and 7. so the difference of x and 7 x comes first so we'd have x and then we have minus we've got our seven and then we've got our is okay this means equals and then we have 13. so essentially we have x minus 7 equals 13. to solve this for x we add 7 to both sides of the equation we get x is equal to 20. so the difference of 20 and 7 is going to be 13 that's correct right 20 minus 7 is 13 so we can say our unknown number here is 20. all right let's take a look at some multiplication phrases so we have times we all know that times means multiplication so 19 times a number again i can mark out a number and put x 19 times x is just 19x or you put 19 times x like this whatever you want to do then the product of we know the product is a result of multiplying two or more numbers together so the product of 11 and a number so again we can mark out a number and put x so the product of 11 and x is again just 11 times x then multiplied by 4 multiplied by a number mark out a number and put x 4 multiplied by x is 4 times x and then twice we know that's like doubling something or multiplying by 2. so twice a number so if we put twice x that's just 2 times x tripled a number is tripled we know if we multiply by 3 we're tripling something so we can mark out a number and put x x is tripled that's 3 times x and then of here we have one half of a number so you can mark out a number and put x generally you're going to use of when you're working with fractions you want a fractional amount of something so i want 3 4 of something 5 6 of something or in this case one half of something so one half of x is one half times x now again with multiplication it's commutative right the order that i multiply in does not affect my result so let's look at some examples so when a number is multiplied by 19 the result is negative 266. find the number so again when a number and here we have the number i'm going to replace that with x so let's put an x here and let's put an x here and again for completeness i'm just going to say let x be equal to my unknown again my unknown number okay so when x is multiplied by 19 so that's just 19x the result is again this guy right here is just going to be equals negative 266. so this tells me i have 19x that's equal to negative 266. i can solve this very easily i can divide both sides by 19 and i'm going to get that x is equal to negative 14. so again find the number it's going to be negative 14 because when negative 14 is multiplied by 19 the result is negative 266 that's true so negative 14 is my unknown number here right now we're saying that one half of a number is 28. find the nump so again a number and the number i'm just going to replace that with a variable like x so one half of x would just be one half x and then we're saying is okay this is going to be equals and then we have 28. so find the number all i've got to do is solve this equation but let me write this right here and say that we're going to again let x be equal to our unknown number again this is just for the sake of completeness and to multiply both sides by 2 here would give me a solution for my equation right i can multiply this side by 2 and this side by 2. this guy here would cancel i'd have x is equal to 28 times 2 is going to be 56 so that's my unknown number right one half of 56 is 28 that's correct all right so we also have some division phrases we want to remember that division is not commutative so the order that we divide in is going to change the result right so something like divided by and then we have a number divided by 3 so again i'm just going to line out a number and put x if that's divided by 3 we would have x over 3 right x divided by 3. so essentially x goes in the numerator it's your dividend and 3 goes in the denominator it's your divisor right so x divided by 3 would be x over 3. it would be wrong if you put 3 over x right if i put 3 over x that would be incorrect so again the order matters here then we have quotient of so the quotient of a number and 7 it's going to line up this part that says a number and put x so the quotient of x and 7 would be x over 7 okay then we have ratio of the ratio of a number and 20 again just going to line out a number and put x so the ratio of x and 20 would just be x over 20. all right so let's look at a few problems with division so we have the quotient of a number and 13 is 18 find the number so again wherever i see a number or the number i'm just going to replace this with our variable x and again for completeness i'm just going to say down here we want to let x be equal to our unknown number so the quotient of x and 13. that's just x over 13. then we have is right in this context that means equals then we have 18. so to find the number let's just solve the equation we'll multiply both sides by 13 and we'll get what this cancels we'll get x is equal to 234 right 18 times 13 is 234. so when we say find the number the number is going to be 234 because the quotient of 234 and 13 is 18. right now we have a number divided by 8 is 17. find the number so again for odd number or for the number i'm just going to put my variable x so x divided by 8. so x divided by 8 is again this means equals so we have equals then we have 17 again find the number or find x so let me write down here that we're going to let x be equal to our unknown number again just for completeness and so to solve the equation here i'm just going to multiply both sides by 8. 17 times 8 is 136. so these guys cancel you get x equals 136. so that's going to be my unknown number again if 136 is divided by 8 the result is 17. okay so 136 again is the number for all the problems we looked at we used is for equals but there's other things you can see so it's basically any phrase that implies sameness okay so some examples would be again we use is so five times a number is seven so again if you line this out and you put x here 5 times x is okay that is means equal so 5 times x is or equals 7. then here we just have the word equals so 4 added to a number equals 17. so that's pretty straightforward line out your number and put x 4 added to x which is x plus 4 equals 17. then we have is the same as so a number less five is the same as twice the number so let's line this part out here and put x let's line out this part here and put x so x less 5 that's x minus 5 is the same as okay that's basically us saying equals right it's implying sameness so is the same as as equals and then we have twice x so 2 times x let's go ahead and take a look at two examples that kind of tie everything together and in the next lesson we'll start looking at some common application problems again that deal with linear equations in one variable so when 2 is multiplied by a number and then 8 times the sum of 11 multiplied by the number and 1 is subtracted away the result is the same as negative 6 times the sum of 10 and 10 times the number find the number so this is a lot more complex than the little simple ones we looked at earlier we're just kind of putting things together so let's go through and again i'm just going to let x be equal to my unknown number here so i'm going to highlight everywhere i see a number and just put x there so everywhere i see our number of the number so let's read back through when 2 is multiplied by x and then 8 times the sum of 11 multiplied by x and 1 is subtracted away the result is the same as so we know this part right here where we say is the same as that's going to be equals negative 6 times the sum of 10 and 10 times again for the number i'm going to put x and then find the number i'm going to put x here all right so let's read back through this now so we have when 2 is multiplied by our number x so i'm going to put 2x then we have and then so and then means afterwards we have eight times the sum of so whenever you have the sum of in this type of problem you're going to multiply 8 by a set of parentheses okay because you're going to have a sum so i'm going to put 8 out here times a set of parentheses i don't know what's going to go inside yet but i'm just going to set that up so 8 times the sum of you've got 11 multiplied by our x so we've got 11x and then and 1. so and 1. and now it says is subtracted away so this was subtracted away from our 2x so let's just read through this part again and make sure we got it so when 2 is multiplied by x we have our 2x and then 8 times the sum of 11 multiplied by x and 1 is subtracted away so 8 times the sum of we have 11x and 1. so 8 times the sum of 11x and 1 that's what we have here and again that's subtracted away so that's why we have minus 8 times the quantity 11x plus 1. then we have the result is the same as so we know this is equals then we have negative 6 times the sum of so again i'm going to put negative 6 here if i have times the sum of something i'm going to use parentheses so then we have 10 and 10 times x so 10 plus 10x and this is basically going to be our equation right so again to go through this part again we have negative 6 times the sum of so negative 6 times the sum of 10 and 10 times x so negative 6 times the quantity 10 plus 10x is what that translates to okay let's go ahead and solve this equation so i'm going to have my 2x here and then i'm going to distribute the negative 8 to each term so you would have minus 88x and then minus 8 this equals i'm going to distribute the negative 6 to each term so negative 60 plus negative 60x on the left side i can clean up because 2x minus 88x would be negative 86x then you have minus 8 is equal to negative 60 plus negative 60x so you can just put minus 60x whatever you want to do doesn't matter let me kind of get some room going here so let's go ahead and add 60x to each side of the equation let's add 8 to each side of the equation so we know that this is going to cancel here and this is going to cancel here negative 86x plus 60x is negative 26x and this equals negative 60 plus 8 is going to be negative 52. now if i divide both sides of the equation by negative 26 i'm going to get that x is equal to positive 2 okay so 2 is my unknown number so let's go back up let's write this as our unknown number and let's check this so when 2 is multiplied by 2 so 2 times 2 is 4 and then 8 times the sum of 11 multiplied by 2 and 1. so we're going to have 8 multiplied by the sum of 11 times 2 which is 22 and 1. so 22 plus 1 is 23 23 times 8 is 184 so we'd have 184 here now it says that this is subtracted away so 4 minus 184 is going to be negative negative 180 and it says the result is the same as so we're going to put equals here you've got negative 6 times the sum of you've got your 10 and 10 times x and again x is 2 so you'd have negative 6 times the sum of you'd have 10 plus 10 times 2 is 20. so 10 plus 20 is 30 so you'd have negative 6 times 30 which is negative 180 so the same value is on the left as we have on the right so 2 is our correct solution here right that's going to be our unknown number so let's take a look at one more of these so we have when a number is divided by four and then added to three the result is the same as seven less than the sum of the number and one then find the number so again wherever i see a number so where we see a number here and then the number here and then the number here i'm just going to put x in each place and then again for completeness i'm just going to say we're going to let x be equal to our unknown our unknown number okay so reading back through now when x is divided by four so that's x over four and then add it to three so then just plus three the result is the same as so that means equals right so i can just kind of highlight this where it says is the same as and i can just put equals there so this would be equals here it says 7 less than 7 less than the sum of x and 1. so we want the sum of x and 1 which is x plus 1 and then we want 7 less than that so just subtract 7 away from that and this will be our equation okay so let's copy this and we'll just come down here to a fresh page so we're not in the way of everything and what i'm going to do is just kind of clean up first so i'm going to add 1 and negative 7. 1 plus negative 7 or 1 minus 7 is going to be negative 6 so i would have x over 4 plus 3 equals x minus 6 and i can subtract 3 away from each side of the equation kind of make that a little better and let me go ahead and cancel this so we have x over 4 is equal to x minus if we have negative 6 minus another 3 or negative 6 plus negative 3 you're going to get negative 9. so this is x minus nine and then now i can clear this denominator here by multiplying both sides of the equation by four so these would cancel i just have x on the left on the right you have four times the quantity x minus 9 so that's 4x minus 36 okay so to solve this guy for x what i'm going to do is i'm going to subtract 4x away from each side of the equation so that's going to cancel x minus 4x is going to be negative 3x and this equals negative 36. so now we can divide both sides of the equation by negative 3 and we're going to get that x is equal to 12. so let's go back up so we're going to say that our unknown number is 12. so 12 here okay so let's check when 12 is divided by 4 that would give me 3 right 12 divided by 4 is 3 and then add it to 3 so 3 plus 3 would be 6. we say the result is the same as or equal to 7 less than the sum of you'd have x as representing the number so 12 and 1. so the sum of 12 and 1 is 13 and if i had 7 less than that i would have 6. so again same value on the left as you have on the right so you can say that 12 is in fact the correct solution that's your unknown number in this lesson we want to review applications of linear equations and talk about consecutive integer problems all right so once again when we have an application problem that involves setting up and solving a linear equation in one variable we have a little six step procedure we can use so for solving a word problem the first thing we want to do is read the problem and determine the question or get questions to be answered then we want to assign a variable to represent the unknown and then in these type of problems we can always express any other unknowns in terms of this original variable then we're going to write an equation this will be based on the situation given we're going to solve the equation then we're going to write the answer in terms of the question or again the questions being asked then lastly we're going to check the answer using the words of the problem we always want to make sure that our answer is reasonable all right so let's just jump in and look at a problem these problems are very easy to understand and solve so you really shouldn't have too much of an issue with them so two pages that face each other in a book have a sum of 663 what are the page numbers so to get an idea of what's going on here if you think about a book and you think about two pages that face each other you could say this page is one and this page is two right they're going to differ by one or you could say that this page is let's say 500 and this page is 501 you know something like that so we know that the smaller page could be represented with a variable like x so let's let x be equal to the smaller page okay let me make that better okay so then what can we say about the larger page well it's just one more right in the instance where we had one and two one was the smaller two is the larger it's just a difference of one okay so what i can say is that we could say then x plus one is the larger okay the larger page okay so what we're told is that the sum of the pages is 663. so it's very easy to solve this i just take my smaller page which is x and i add it to my larger page which is x plus one and this equals 663 okay so all i got to do now is just solve my equation very very easy so if i just remove the parentheses here because i don't really need them x plus x is 2x and then plus 1 equals 663. let's go ahead and subtract one away from each side of the equation we know this will cancel and you'll have 2x is equal to 662. let's now divide both sides of the equation by 2 so that we can isolate the variable x so we know that this will cancel with this and i'm going to end up with x is equal to 662 divided by 2 is 331 okay so again when you work with a word problem you can't just say x equals 331 and turn in the test right you got to make sense of it so let's erase all this i'll show you on our little picture what's going on all right so if we come back up here again we said x was equal to 331 this is the smaller page so in our book we could say this page here is 331 and then the next page the page that faces that one is going to be 332 okay and you could check that is the smaller page 331 well if the larger page is 332 if you sum those two amounts together you do get 663 okay so everything is consistent with what we're told in the problem so let's go back up again what are the page numbers so to answer that question we could say the page numbers are again 331 and 332. [Music] okay nice and simple all right let's take a look at another one so the sum of three consecutive even integers is 30. find the integers so in this particular case you've got to understand what it means by saying consecutive even integers so with that we're talking about again even numbers are numbers that are divisible by 2 right so this is 0 2 4 6 8 10 12 things like that so if i want consecutive even integers it could be 2 4 6 8 10 something like that it could be starting at 12 so 12 14 16 18. whatever number you start at you're just increasing by two each time to get to the next even integer okay so if i want three consecutive even integers i can let the smallest just be x so let x be equal to the smallest even integer okay i make these e's a little better so what can we say about the middle one all right we have a smallest we have a middle one and we have a largest well again i'm just increasing by two if i started at two and i went to four i'm increasing by two right then if i go to six i'm increasing by 2 again so we can say then x plus 2 is our middle even integer okay let me make that better and then what we could say then if i add another two so x plus two plus another two would be x plus four so x plus four would be the largest even integer okay let me make these better so we're told the sum is 30 right it's right here so all we really have to do is take x which is the smallest even integer add it to x plus 2 which is the middle even integer and add that to x plus four which is the largest even integer and set that to thirty solve that equation and we'll be good to go so we have x plus x plus two plus x plus four you can wrap these in parentheses to kind of separate them if you want but it really serves no purpose so i'm not going to do that and this is equal to 30. so let's solve this guy real quick and we'll have our answer so x plus x plus x is 3x and then 2 plus 4 6 so plus 6 this equals 30. let's go ahead and subtract 6 away from each side of the equation and this is going to cancel we're going to have 3x is equal to 30 minus 6 is 24. and then i want x by itself so let's divide both sides of the equation by 3 and we'll get what this cancels with this you've got x is equal to 8. okay so let's erase everything again the smallest even integer was represented with x so we know that's 8. so then the next even integer would be what we add 2 to that so we get 10 and then add 2 to that to get to the largest and that would be 12. okay so the integers would be 8 10 and 12 is that consistent with the problem 8 plus 10 is 18 and then 18 plus 12 is 30. so we're basically good to go right they're all even numbers and they're in a consecutive sequence so i can just say find the integers the integers are the integers are we'll say 8 10 and 12. well let's take a look at another one and this one kind of trips students up so you just want to pay close attention here so the sum of five consecutive multiples of seven is 245 find the multiples so the idea here if you don't remember what a multiple is from pre-algebra just note that there's multiple definitions of this but essentially if i take the number 7 and i multiply it by an integer so by 1 or 2 or it could be negative 7 or whatever i want to multiply it by as long as it's an integer the result there is a multiple of seven so generally in pre-algebra or a lower level algebra course you kind of restrict this to multiplying the number which in this case is seven by whole numbers which are one or larger right so i would say multiples of seven you'd have seven times one is seven then seven times two is fourteen then seven times three is twenty-one right so on and so forth so if i start with seven then i increase by seven i get to fourteen increase by 7 to get to 21 you know so on and so forth so i go 28 35 42 49 right this would continue forever now in this particular case all we need to understand is that each time i go to the next multiple in the sequence i just add 7 right so to get from 7 to 14 i add 7. to get from 14 to 21 i add 7. so if the sum of five consecutive multiples of 7 is 245 the first multiple could just be x so let's say let x be you say this is the smallest or the first multiple [Music] you'd say of seven so then each time i increase to the next multiple i'm just going to add 7 right just to follow this pattern so we could say then x plus 7 is what it's the next number in the sequence so we'd say the next largest [Music] okay and then following this pattern you're just going to keep adding 7. so we could say then x plus 7 plus 7 or x plus 14 would be what it would be your next largest so you just keep going until you have five of these guys so we would say then let me make that e better you'd have x plus 14 plus 7 which is x plus 21 is again the next largest so you have one two three four you need one more so then we could say x plus 28 is now going to be your largest let me write that out so this would be your largest and i'll be specific here and say multiple of seven okay so now what do we want to do we're told the sum of these guys is 245 so essentially what i got to do is just take x plus x plus 7 plus x plus 14 plus x plus 21 plus x plus 28 and just set it equal to 245 solve that for x and i'll have my answer okay pretty easy overall so again i have x plus the next guy in the sequence is x plus 7 plus the next guy in the sequence is x plus 14 plus the next guy is x plus 21 and then lastly the largest is x plus 28. you've got one two three four five multiples that you're representing this is equal to 245. okay so on the left side x plus x plus x plus x plus x is 5x so it's 5x and then plus you've got 7 plus 14 plus 21 plus 28 which is 70. okay so then this equals 245. if i want to solve this for x i'm going to subtract 70 away from both sides of the equation so this guy is going to cancel you'll have 5x is equal to 245 minus 70 is going to give me 175. and now what we can do is divide both sides of the equation by 5 so we can isolate x and we'll get that x is equal to 175 divided by 5 is 35 okay so let's make sense of this so x is 35. let's go back up so we said x was the smallest multiple of seven so the next largest would be plus seven so we would get 35 add seven you would get 42 again keep adding seven so you get to 49 you would get to 56 and then finally you would get to 63. so these are your multiples of 7. so let's erase this all right so let's say the multiples of 7 are again you have 35 35 42 49 56 and 63. now if you sum these you get 245 right we know that they're multiples of seven that's very easy to see but if we sum these do we get 245 well 35 plus 42 is 77 then if i add 49 to 126 then if i had 56 i get 182 and then finally i've had 63 i do get 245 so when i sum these numbers i do get 245 so my solution here is correct all right let's take a look at one more problem so if six is subtracted from the largest of three consecutive odd integers and then multiplied by two the result is 23 less than the sum of the smallest and twice the middle integer what are the integers so when you get a problem like this you just have to stop and go step by step okay if you try to go too fast you might miss a parch this is one of the ones that's just meant to kind of confuse you so the first thing is it says consecutive odd integers think about the fact that if you have consecutive even integers it's 2 4 6 8 10 right so on and so forth odd integers are integers that are not divisible by 2 so if they're consecutive it's going to differ by 2. so something like 1 three five seven nine i'm just increasing by two once i start with an odd integer so to kind of model this let's let x let's let x be the smallest odd integer and then we can say that x plus 2 is going to be the middle one so this is the middle odd integer i'm going to run out of room so let's take this to another page and so we can also say then x plus 2 plus 2 which would be x plus 4 is the largest okay the largest odd integer okay so now that we've modeled the odd integers that sequence let's go back up and let's think about what it's saying if six is subtracted from the largest of three consecutive odd integers okay so six is subtracted away from the largest so the largest we said was x plus four so six subtracted away we'd have x plus four minus 6 which would translate into x minus 2. okay so let's just write that so let's put this as x minus 2 and it says again if 6 is subtracted away from the largest of three consecutive odd integers which we've already done and then okay so and then so it's done after the fact multiplied by two so we're going to multiply this guy by two this is a quantity x minus two is a quantity so i've got to multiply the quantity x minus two by two then it says the result is so that always means equals so again the result is you have 23 less than the sum of the smallest and twice the middle integer okay so this is a lot to unpack so we know that we're subtracting 23 away from something because it says the result is 23 less than so let's go back here i'm just going to put minus 23 out here because we're going to do something and then we're going to subtract 23 away now it says the sum of the smallest which we said was x and twice the middle integer the smallest which we said was x and twice the middle integer we want that sum the middle integer is x plus 2. so again i've got to use some parentheses here so let me kind of drag this over so 2 times this quantity x plus 2 okay so let's read through this again and make sure we understand where this came from so again if six is subtracted from the largest of three consecutive odd integers and then multiplied by two so essentially we started with the x plus 4 is the largest odd integer we subtracted away 6 that's how we got x minus 2 and then we multiplied by 2 to get the left side then it tells us the result is so that's where we got our equal sign from if you say the result is you're basically saying is equal to so on the right side we have that it's 23 less than the sum of the smallest which are represented with x and twice the middle integer now the middle integer is x plus two so twice the middle integer is two times the quantity x plus two so that's where you got this from the sum of the smallest integer which is x and twice the middle integer which is two times x plus two so we sum those two together and then we need to subtract 23 away from that because it says the result is 23 less than okay so now that we have this set up again these are kind of a pain to get to this point but once you have it it's pretty easy you're solving an equation so clean up on the left 2 times x is 2x then minus 2 times 2 is 4 this equals you have x plus 2 times x is 2x and then plus 2 times 2 is 4 and then minus 23. okay so let's think about this on the left side nothing i can do on the right side i can take x and add to 2x that would give me 3x and then 4 minus 23 is negative 19. again over here i'm just going to write 2x minus 4. nothing's going to change i'm going to subtract 3x away from each side of the equation and i'm going to add 4 to each side of the equation so what's going to happen is this guy right here is going to cancel and this guy is going to cancel 2x minus 3x is negative x and this equals negative 19 plus 4 is going to be negative 15. now to get x by itself i can multiply both sides by negative 1 or divide both sides by negative 1 makes no difference you're going to get that x is equal to 15. okay so let's erase all this so remember x was the smallest odd integer so if we go back up what are the integers we'll say the integers are let me make the e better so we have 15 the next one would be 17 right if it's consecutive odd integers and then the last one would be 19. so let me put and 19 here and 19. okay does this make mathematical sense normally with these problems you just get a straight sum so it's very easy to check with this one you have to do a little bit of work so if 6 is subtracted from the largest of three consecutive odd integers so the largest is 19 if i take 6 away i'd have 13. then it tells us and then multiplied by 2. so 13 multiplied by 2 is 26. so this is 26 and it says the result is so this is equal to so we have 23 less than the sum of the smallest so the smallest is what it's 15 plus twice the middle integer the middle integer is 17. so if i take 2 and multiply by 17 i get 34 and again it's 23 less than this so i'm going to subtract 23 away from this sum so 15 plus 34 is 49 so you would have 49 minus 23 which is 26 right you end up with the same value on the left as you do on the right so we've successfully completed checking this guy right we can say our answer is correct the integers are going to be 15 17 and 19. in this lesson we want to review applications of linear equations and look at some examples of age problems so again whenever we're working with an application problem that involves a linear equation in one variable we can use our little six step procedure so for solving a word problem we have that we want to read the problem and again determine the question or questions to be answered then we want to assign a variable to represent the unknown with these problems we're going to express other unknowns in terms of this variable then we're going to write an equation this is based on the situation given to us in our problem and then we're going to solve the equation then we're going to write the answer in terms of the question or again questions asked and then lastly we're going to check the answer using the words of the problem we always want to make sure the answer is reasonable so for age word problems this is another extremely common type of word problem in algebra this type of problem is generally going to give us the sum of the ages of family members friends or some group of associated people it's going to ask us to find their individual ages now it doesn't always set up this way but that's the most common scenario all right so let's take a look at the first example so we have that albert jamie and jessica are siblings and have a combined age of 50 years so jessica is three times as old as albert but half as old as jamie how old is each sibling so this is our main question here right how old is each sibling so we need to find out how old is jamie how old is albert and how old is jessica okay so we are only given the fact that they have a combined age of 50 years and we've given some comparisons here it tells us jessica is three times as old as albert but half as old as jamie so to solve this we know what we're trying to find let's assign a variable let's let x be equal to since jessica is involved in both comparisons let's let x be equal to jessica's h now what can we say about albert's age and what can we say about jamie's age well we're told here specifically that jessica is three times as old as albert so if jessica is three times as old as albert that means albert is one-third the age of jessica so then x which is jessica's age over three would be equal to albert's h now we're also told that she's half as old as jamie so if jessica is half the age of jaime that means jaime is two times as old as jessica so i can say then 2 times x again x is jessica's age so 2 times that would be jamie's age okay so this is jamie's age now we have everything set up here we have everything modeled how do we get an equation going so we can get a solution well essentially we have a combined age of 50 years so i can take jessica's age which is x add that to albert's age which is x over 3 and add that to jamie's age which is 2x and i can set that equal to 50 right their combined h so let's go ahead and set that up so we would have x plus x over 3 plus 2x is equal to 50. what i can do to clear this denominator is multiply everything by three so i would have three x plus if i multiply x over three by three i would just have x and then plus two x times three is six x and this equals fifty times three is 150 okay so basically i just multiplied both sides of the equation by 3 again so i can clear this denominator here so now 3x plus x is 4x plus 6x is 10x so you get 10x is equal to 150. let's divide both sides of the equation by 10 and we get that x is equal to 15. so if x was jessica's age that tells me that jessica is 15 years old so let's go back up so again x is 15 so jessica is 15 and then we know that albert is a third of the age of jessica right it's 15 over three which is five and then jamie is twice the age of jessica so it's 2 times 15 or 30. so to answer this question we can say that jessica is 15 comma say albert is five and we'll say jamie is 30. okay now you can check this to make sure it makes sense so again albert jamie and jessica are siblings and have a combined age of 50 years so is 15 plus 5 plus 30 50 yes 15 plus 5 is 20 20 plus 30 is 50 so you're good to go on that part then it says jessica is three times as old as albert so jessica is 15 albert is five 15 is five times three so we're good to go there and then half as old as jamie 30 is jamie's age again jessica's 15 15 is half of 30 so you're good to go there so this checks out jessica again is 15 albert is 5 and jamie is 30. all right let's take a look at one that's a little bit more challenging so we have that nine years ago steve was one third of larry's current age the difference between larry's age and steve's age is the same as max's age 11 years ago if their combined age is 131 years how old is steve larry and max so we want to find out how old is steve larry and again max so i'm going to jump right in and let a variable like x represent the current age of larry okay so let me go down here so we have lots of room so let's let x be equal to larry's current h now what can we say about steve well it says specifically that steve was one-third of larry's current age nine years ago okay so i have to think about the fact that one-third of larry's current age is one-third x but this is steve's age nine years ago so i've gotta add 9 on top of that to get back to our present day and time so for steve's age we could say then one third x so one third of a liar's current age then plus nine right because this part right here is steve's age nine years ago so i've gotta add nine to that to get his current h so this would be steve's current h now let's think about max now so it tells us that the difference between larry's age and steve's age is the same as max's age 11 years ago okay so the difference between larry's current age which is x and steve's current age which again is this one-third x plus nine notice that i'm using brackets here because i've got to subtract the whole thing away this is the same as max's age 11 years ago so to get his current age i've got to add 11 to this amount okay so this would be then we'll say this guy right here let me just kind of slide this down we'll say this is max's current h okay and i might run out of room so let me make that a little bit better so max's current h okay so now it also tells us in the problem that their combined age is 131 years so essentially just like i did in the last problem i can sum the ages and set it to 131 and i'll get a solution so i would have x which is larry's age plus one third x plus nine which is steve's current age and then plus you have x minus you have this inside of brackets 1 3x plus 9 and then plus 11 which again is max's current age this is all equal to 131. so let me copy this we're going to go down to a fresh sheet of paper so let me just kind of clean this up we have x plus one third x plus nine plus x use your distributive property here so you'll have minus one third x then minus 9 then plus 11 and this equals 131. now what do you notice here 1 3x minus the third x that's gone 9 minus 9 that's gone x plus x is 2x and then plus 11 equals 131. if i subtract 11 away from each side of the equation i'm going to get that 2x is equal to 120 divide both sides of the equation by 2 we're going to get that x is equal to 60. so remember x was larry's current age so larry is 60. so larry is 60. we can say that steve again he's 1 3 times 60 which is 20 plus 9 which is 29 and then for max we basically have 60 minus 29 which is 31 and then plus 11 which is 42. so his is going to be 42. all right let's go back up so let's write our answer here again larry is 60. so larry is 60. then we can say steve is 29 and max is 42. again we can check this so it says nine years ago steve was one third of larry's current age so if we go nine years ago larry's current age is 60 so a third of that is 20 okay so steve is 29 nine years ago he would have been 20 and that is one-third of liar's current age which again is 60. then it says the difference between larry's age and steve's age the difference between 60 and 29 is 31 and it says this is the same as max's age 11 years ago well again max is 42 11 years ago he would have been 31 so that checks out as well then it says their combined age is 131 well 60 plus 29 is 89 then if i add 42 i do get 131. so everything checks out here we can say that larry is 60 steve is 29 and again max is 42. all right let's take a look at one more so we have three friends john clyde and beth all attend the local school the school accepts students from age 5 to age 20. this week the school set up a fair in which the minimum age is 12 years old to attend so now we're told in one year clyde will be three times as old as john is today and then one year ago beth was one half as old as clyde is today if the difference of clyde's age and john's age is the same as beth's age in two years find which of the friends can attend the fair okay so we're trying to find which of the friends so we have beth clyde and john can attend the fair remember it tells us that you can only attend the fair if you're at least 12 years old right you need to be 12 years old to attend so we're trying to find out which of these individuals are at least 12 right because if you're 12 you can go to the fair if you're older than 12 you go to the fair and if you're less than 12 you can't go to the fair so let's think about a variable to represent one of their ages and then we can go through a model the rest of the ages based on that variable so it tells us that again in one year clyde will be three times as old as john is today and it tells us one year ago beth was one half as old as clyde is today so since clyde is involved in both comparisons there let's just go ahead and let x be clyde's current h let's do this on a fresh sheet so let's let x be equal to clyde's current h okay so what can we say about john's age well it says in one year clyde will be three times as old as john is today so that means if i take clyde's current h which is x and i add one to it right because it's in one year that amount x plus one would be three times the age of john right so x plus one x plus one is three times the age of john so i can just divide by three to get john's current age so we can say then x plus one over three is john's current h okay so what can i say about beth's current age so if we think about this one year ago beth was half as old as clyde is today so half of clyde's age today is one half times again clyde's age is represented with x so it's one half x now this is beth's age not today but a year ago okay a year ago she was half as old as clyde so if i think about one half x that's best age one year ago if i add 1 to that that's best current h so we would say that then x over 2 which again is her age a year ago then plus 1 would be her current h so this would be beth's current h so in most of these problems you're given a sum right of all the ages but in this case you're not you have to use something else to get your equation and it's this sentence right here so if the difference of clyde's age and john's age is the same as best age in two years so the difference of clyde's age and john's age would be clyde's age minus john's age so let's set that up first so clyde's age is x so this is x and then minus john's age is x plus 1 over 3. now i'm subtracting this whole thing away so let me just put it in brackets so we don't make a sign mistake and then we're saying is the same as so that's equals here we're told beth's age in two years so her current age is x over two plus one so x over two plus one in two more years you would just add another two to this amount so let's copy this we're just going to take it to a fresh sheet of paper and we can solve this guy so we have x and then you have your minus here you can distribute this you can go ahead and set this up as x over 3 plus 1 over 3 and then if i had a minus out in front of that then i would have what i would have x minus x over 3 and then minus 1 3 okay then this would be equal to let me just kind of drag this back up you have your x over 2 1 plus 2 is 3 so plus 3. now if i want i can multiply both sides of the equation by 6 right that would be the lcd so if i did that i would have 6x minus if i have 6 times x over 3 the 6 would cancel with 3 and give me a 2 so this would be 2x then minus if i have 6 times 1 3 that's 2 this equals if i have x over 2 times 6 again 6 divided by 2 would be 3. 3 times x is 3x and then plus 6 times 3 is 18. okay so now here on the left i have 6x minus 2x which is 4x and then minus 2 equals 3x plus 18. and scroll down and get some room going so now i can subtract 3x away from each side of the equation so this is going to cancel i can add 2 to both sides of the equation this is going to cancel 4x minus 3x is x this is equal to 18 plus 2 which is 20. so x is equal to 20. now x was clyde's current h so he's 20. john's current age was what it's 20 plus 1 which is 21 divided by 3 which is 7. best current age is 20 divided by 2 which is 10 plus 1 which is 11. so it doesn't ask for their ages it wants us to find out who can attend the fair so we know that clyde is the only one that can attend the fair because the minimum age is 12. right beth can't go because she's 11 and john can't go because he is 7. so only clyde can attend so that's the only information we need to know it's again find which of the friends can attend the fair and once more you can only attend if you're 12 or older right so let's go back down and say that only clyde can attend the fair okay now if we think about this we can go back and check so in one year clyde will be three times as old as john is today we said that clyde was 20 and we said john was seven so in one year clyde will be what he'll be 21 and that is three times as old as john is today because john is seven so that checks out then it says one year ago beth was one half as old as clyde is today so a year ago beth would have been 10 years old and that is half as old as clyde is today because he is 20. okay so that checks out then it says if the difference of clyde's age and john's age so the difference between 20 and 7 cly's age and john's age is 13 that is the same as best age in two years she's currently 11 in two years she's going to be 13. so that all checks itself out so our answer here is correct only clyde can attend the fair because again he's the only one that's at least 12 years old in this lesson we want to review applications of linear equations and specifically we're going to look at motion word problems all right so let's just go through our procedure real quick on how to solve a word problem this is just a general guideline something you can use as you're solving your problems so the first thing is just to read the problem and determine the question or in some cases questions to be answered the next thing is to assign a variable to represent the unknown express other unknowns in terms of this variable then we want to write an equation then we're going to solve the equation then we're going to write the answer in terms of the question or the questions being asked and then lastly we're going to check the answer using the words of the problem all right so for motion word problems a lot of you who took lower level algebra courses got a lot of experience with these it's a very common type of word problem they rely on the understanding of the formula d or distance is equal to r for rate of speed times t the time traveled this is a very intuitive formula think about being on a road trip and let's say your rate of speed is 80 miles per hour so let's say this is 80 miles per hour and i'll just write 80. and then you do this for let's say 10 hours how far did you go in terms of miles well you just multiply the rate of speed is 80 the amount of time is 10 80 times 10 is 800 so your distance is 800 miles okay so it's a very intuitive formula something you can kind of work out in your head if you just kind of can't remember the distance formula think about being on a road trip and saying okay if i'm going at this rate of speed for this amount of time well i can multiply those two numbers together to get my distance all right let's look at our first example so we have that heather left school and drove toward the ocean daniel left one hour later and drove at a speed that was 15 miles per hour faster than heather after two hours daniel caught up with heather what was heather's average speed so this is what we need to answer here what was heather's average speed that's the key take the second to look at the problem and understand the question that you're being asked that's where a lot of students really you know they read the problem they don't really understand what they're being asked so they're like what do i do okay that's what you've got to understand what was heather's average speed so now we're ready to kind of go through the problem and the first thing that i'm going to do is just let a variable you can let it be x or it could be y or z whatever you want to use doesn't matter so let x be equal to what am i looking for i'm looking for heather's average speed so let's let x be equal to heather's average speed okay and this is going to be in terms of miles per hour because that's what we're working with in the problem now we need to model other unknowns in terms of this variable x so we have heather's average speed modeled but we don't have daniel's average speed model well daniel it tells us in the problem that he's driving 15 miles per hour faster than heather okay well if heather's average speed is x then we can say that x plus 15 okay that amount there is going to be daniel's average speed okay so this is daniel's average speed okay let me make those a little bit better all right so with this information we can take this to a table okay and i'm going to show you this table real quick this is something you can use to kind of organize your thoughts okay so in this case i'm going to put the names on the left so i can just kind of put it as the names if i want and that's going to go down like this so let's put h for heather and let's put d for daniel to the right here in the top i'm going to put d for the distance this is equal to my r which is the rate of speed multiplied by my t which is the amount of time traveled so we already know for heather her rate is x and for daniel his rate is x plus 15. okay what about the time let's go back up and see if we can figure that out so it tells us in the problem specifically that daniel left one hour later okay so he drives for an hour less than heather does when they catch up okay or when they're at the same point so it says after two hours daniel caught up and it also says he left one hour later okay so that means that daniel was driving for two hours because it says specifically after two hours daniel caught up and it tells us also that again because daniel left one hour later than heather that heather was driving for an hour more so she was driving for three hours so i can fill this out and say heather was driving for three hours daniel was driving for two now i can get the distance by multiplying these two together okay so in other words the distance that heather travels is the amount of time she travels for times the rate of speed that she travels at so 3 times x is 3x that's my distance for heather 2 times the quantity x plus 15 is my distance for daniel now let's go back up and see how we can figure out an equation if we think about this remember they leave from the same point heather left school and drove toward the ocean daniel left one hour later okay so he's leaving from the same point now they catch up meaning they go to the same point when they're caught up okay so they leave from the same point they go to the same point so that means that heather and daniel drive the same distance they do it in a different amount of time but they drive the same distance so what i can say here is that the distance for heather which is 3x is going to be equal or the same as the distance for daniel which is 2 times the quantity x plus 15. so once we've set this up it's basically a breeze from here we would use our distributive property on the right this is 2x plus 30. on the left we still just have 3x let me go ahead and subtract 2x away from each side of the equation this is going to cancel i'll have that x is equal to 30. okay so this is another big problem a lot of students get x equals 30 and they stop right we're used to just saying hey this is my answer x equals 30. with word problems you've got to go back up and make sense of your answer so x equals 30 tells me that heather's average speed was 30 miles per hour okay and it also tells me that daniel's average speed was 45 miles per hour because his is expressed as x which is 30 plus 15 30 plus 15 is 45. so let's erase this and say what was heather's average speed we can say heather's average speed was 30 miles per hour okay and you can check this pretty easily you can say are the distances equal well if heather drives at 30 miles per hour for three hours she's going to go 90 miles right again distance equals rate of speed which is 30 times time traveled which is 3. 30 times 3 is 90. so when you think about the units it's going to be 90 miles for daniel he drives for only 2 hours but he does it at 45 miles per hour 2 times 45 is also 90 so you can say this is the correct answer heather's average speed was 30 miles per hour and again although it doesn't ask for it in the problem we could say daniel's average speed was 45 miles per hour all right let's look at another one so we have that i left the mall and traveled toward her beach home at an average speed of 50 miles per hour jeff left the mall at the same time and traveled in the opposite direction with an average speed of 60 miles per hour how many hours before they are 330 miles apart so this is what we need to answer how many hours before they are 330 miles apart so let's label our kind of table down here so we have our names going down let me make that a little bit better so we have aya and we have jeff okay and then for this part right here we have our distance is equal to rate times time okay so what are we given in the problem we're told that i as average speed is 50 miles per hour and jeff's average speed is 60 miles per hour so for aya the rate of speed is 50. for jeff it's 60. do we know the time in each case no we're not given that okay and in fact what we need to find is how many hours before they are 330 miles apart meaning they start at one point and one drives in one direction another drives in the opposite direction and we want to know how long before kind of the distance between them is 330 miles okay so let me make that better so what we want to do is let a variable like x be equal to the amount of hours that they're each traveling for before they are 330 miles apart so time in hours we could say to get to 330 miles apart okay and make that a little bit better so each of them are going to drive for that amount of time so we can go down here and we could say that this is x and this is x now the distance that aya travels is 50 times x or 50x the distance that jeff travels is 60 times x or 60x in the last problem we set the distances equal to each other but that's not going to work here because what we're saying is that the sum of the distances the amount that aya travels plus the amount that jeff travels is going to be 330 miles right 50x plus 60x needs to be 330 miles okay so if we solve this 50x plus 60x is 110x and this equals 330. we divide both sides of the equation by 110 and we're going to get that x is equal to 3. all right so let's erase this and let's say since x was 3 and x represents the amount of hours for them to get to 330 miles apart we'll just say it takes three hours for them meaning aya and jeff for them to be 330 miles apart okay and that would be your answer there and you can think about why that makes sense if aya drives at a speed of 50 miles per hour for three hours she's basically going let's say from this point to here she's going 150 miles okay if jeff goes the opposite direction he starts here and goes this way he's going to go a little bit further because he's going faster 60 miles an hour times 3 hours would be 180 miles okay so the distance between them now from here where jeff is to here where aya is is 180 plus 150 which is going to be 330. all right let's look at one more so jason can drive to farmwood in two hours if he takes the bus it takes twice as long if the average speed of the bus is 25 miles per hour slower than jason's car how far away is farmwood okay so what are we asking we're asking for how far away is farmwood meaning what is the distance from wherever jason starts to farmwood okay well we know that there's two situations there's one where he takes a car and there's another where he takes a bus so let's go down here so to start filling in the table we have one scenario which is a car and another which is a bus so this just represents the vehicle so just put v for that again we're still going to have our d for distance is equal to r for rate of speed times t for time traveled so let's go back up we're told that jason can drive to farmwood in two hours so by car the trip takes two hours so the time by car we're gonna put two there and then if he takes the bus it takes twice as long so two times two would be four so if he takes the bus it's four hours now in terms of the rate of speed we're not given that we're only told that the speed of the bus is 25 miles per hour slower than jason's car okay so let's let a variable like x be equal to the car's average speed and again this is in miles per hour now we can model the speed of the bus based on this right if x is the car's average speed since the bus is going 25 miles per hour slower than the car we can say x minus 25 can represent the bus's average speed so we'll say the average speed of the bus okay all right so x is the car's average speed x minus 25 is the average speed of the bus so rate of speed for the car is x rate of speed for the bus is x minus 25. now we know for distance we multiply so 2 times x is 2x and over here we have 4 times the quantity x minus 25 okay now we know that the distance is equal right jason can drive to farmwood in 2 hours if he takes the bus it takes twice as long so the distance is going to be the same so to solve this one i can just set this distance that he travels by car of 2x equal to the distance of the bus which is 4 times the quantity x minus 25. all right so let's scroll down get some room going so we have 2x on the left is equal to 4 times x is 4x and then minus 4 times 25 is 100. let's go ahead and subtract 4x away from each side of the equation that cancels 2x minus 4x is negative 2x this equals negative 100. let's go ahead and divide both sides by negative 2 and we're going to get that x is equal to 50. okay so what does that mean x equals 50 we don't stop there we've got to go back up so x is the car's average speed so that means the car was traveling at 50 miles per hour okay the average speed of the bus is 50-25 or 25 miles per hour but that still doesn't answer the question it didn't ask what was the speed of jason's car it asked how far away is farmwood well all we have to do now is say since jason can drive to farmwood in two hours and he's driving at 50 miles per hour 50 times 2 is 100. that means farmwood is 100 miles away so farmwood is 100 miles away let me erase all this okay you can check that pretty easily again if he can drive to farmwood in two hours and he's going at 50 miles per hour that's 100 miles if the bus is 25 miles per hour slower than the car 50 minus 25 is 25 it takes him twice as long it takes him four hours 25 times 4 is also 100 so either way he's going 100 miles so that's consistent with our problem right farmwood is 100 miles away in this lesson we want to review applications of linear equations we're going to look at mixture word problems so again when we're working with applications of linear equations we have a little six step procedure that we can use so for solving a word problem we have as our first step we want to read the problem and determine the question or again questions to be answered then we want to assign a variable to represent the unknown we're going to express other unknowns in terms of this variable then we're going to write an equation okay based on the scenario we're given then we're going to solve the equation then we want to write the answer in terms of the question or again questions being asked then lastly we're going to check the answer using the words of the problem this one right here number six the last step is very important you want to make sure your answer is reasonable so all the word problem types that we encounter in elementary algebra courses usually mixture problems give students the most trouble right when you're in that section on applications of linear equations the key to understanding mixture problems is understanding how to find the pure amount of a substance in a mixture to accomplish this task we can multiply the rate of concentration as a decimal by the amount of the solution okay this will give us the pure amount of a substance in a mixture so let's look at a simple example so a local bar serves pebble beer which is nine percent alcohol so i'm just going to highlight this because they gave us a percentage so nine percent alcohol if one serving is 20 ounces let me just highlight that so it's 20 ounces how much pure alcohol is in each serving so obviously this is what we want to answer here how much pure alcohol is in each serving so all i need to do is take the amount of one serving which is 20 ounces forget about the ounces part for a minute just the number 20. and it's 9 alcohol okay so all i need to do is multiply the 20 by 9 percent as a decimal which is .09 if i do that calculation i'm going to get 1.8 okay so what this tells me is that this pebble beer here is going to be 1.8 ounces of pure alcohol so let me erase this and i'll write a nice little sentence and we'll say that pebble beer has 1.8 ounces of pure alcohol per serving okay so a very simple example just something we need to understand how to find the pure amount of something in a mixture or substance so in this case we have a beer it's nine percent alcohol we want to know the pure amount of alcohol in the full serving which is 20 ounces again we take the rate of concentration which is .09 as a decimal we multiply by that serving which is 20 ounces and that tells us we have 1.8 ounces of pure alcohol in each serving all right let's look at a typical word problem now so we have that mary is throwing a party and wants to serve fruit punch which is 22 percent pure fruit juice she currently has two gallons of cherry smash fruit punch which is 50 pure fruit juice how many gallons of bubble berry fruit juice which is 14 pure fruit juice must be added to create her desired fruit punch so understand what we're trying to find here we're trying to find out how many gallons of this bubble berry fruit juice which again is 14 pure fruit juice that's got to be added to create her desired fruit punch and again her desired fruit punch is 22 percent pure fruit juice okay that's what we want so let's get a little idea of what's going on she has these two fruit punches that she's mixing together one is the cherry smash and the other is the bubble berry okay so you can see that little image here so she's mixing them together to get this final mixture here and the final mixture is going to be 22 percent pure fruit juice okay so that's the goal so let's let a variable like x be equal to the amount of the bubble berry fruit punch in gallons okay so that's the amount she's going to use in gallons so we'll say the amount let me kind of erase that so we'll say the amount of the bubble berry fruit punch used and this is going to be in gallons okay in gallons all right so once we know that x is representing the amount of bubble berry fruit punch used in gallons what else do we need well we already know the amount of cherry smash it tells us that in the problem it tells us that she has two gallons of cherry smash okay so let's go down to a little table and organize this information so for our table up here we're going to put the gallons so this is going to be the gallons this is going to be our concentration percentage as a decimal and this amount right here is going to be our pure fruit juice okay then for our products over here we're gonna have the cherry smash and we're gonna have the bubble berry okay so those two products now it tells us in the problem the number of gallons of the cherry smash is going to be two and then we said the number of gallons of bubble berry that we're going to use is going to be represented with our variable x what is the percentage of pure fruit juice in cherry smash so let's go back up so cherry smash it tells us is 50 pure fruit juice 50 pure fruit juice whereas bubble berry is 14 pure fruit juice okay so let's fill this part in so cherry smash again is 50 fruit juice so as a decimal that's 0.5 okay bubble berry is 14 pure fruit juice so as a decimal that's 0.14 now as we just saw in our simple example a minute ago all we need to do to get the amount of pure fruit juice here is multiply across so for cherry smash if i have two gallons of my fruit punch and it's 50 pure fruit juice the amount of pure fruit juice in gallons there is going to be two times point five or one okay then for the bubble berry i just got to multiply across again so i have x gallons it's 0.14 or 14 percent pure fruit juice so the amount of pure fruit juice there is 0.14 times x okay and again this is in gallons now we have this final row here that represents our mixture our desired mixture so we'll say this is our final mixture okay so the number of gallons will be what well we only have gallons coming from two sources right we have the cherry smash and the bubble berry so i can just add these i would have x gallons coming from the bubble berry then plus two gallons coming from the cherry smash now the percentage is given to us in the problem it tells us that the final mixture again is 22 pure fruit juice so this has to be 0.22 and so the pure fruit juice in gallons would be what it would be 0.22 times this quantity of x plus 2. all right so let's stop here from this point it's very easy to set up an equation we know that the amount of pure fruit juice in gallons in the final mixture comes from two sources it comes from this source right here you've got one gallon coming from your cherry smash and then you've got .14 x gallons coming from your bubble berry that has to be equal to the pure fruit juice in gallons in the final mixture this point two two times the quantity x plus two so that's where we're going to get our equation we're just going to set this up so we will have one again the amount of pure fruit juice in gallons from jerry smash plus you're going to have your 0.14 x which again is the amount of gallons of pure fruit juice in the bubble berry we're going to set this equal to the amount of pure fruit juice in the final mixture which is 0.22 times the quantity x plus 2 okay so once we've set up an equation it's very easy so you have your 1 plus 0.14 x is equal to use our distributive property we'll have 0.22 x and then plus 2 times 0.22 is 0.44 and then to solve this guy i'm just going to go ahead and subtract 0.14 x away from each side of the equation and i'm going to subtract .44 away from each side of the equation and what's that going to give me so we know that this would cancel we know that this would cancel so 1 minus 0.44 is 0.56 so the left side is 0.56 and this is equal to 0.22 minus 0.14 is .08 and then times x if we finish this up by dividing both sides by .08 we're going to get 7 okay so x here is going to be equal to 7. now let's make sense of this we said that x was the amount of bubble berry fruit punch used in gallons so we know that x is equal to seven so she needs to use seven gallons of bubble berry fruit punch okay so let's go back up and let's answer our question now so we'll say that she needs to use [Music] seven gallons of bubble berry fruit punch now let's check this let's make sure the answer is reasonable she's throwing a party and wants to serve fruit punch which again is 22 percent pure fruit juice she currently has two gallons of cherry smash fruit punch which is 50 pure fruit juice and again it asks how many gallons of bubble berry fruit juice which is 14 pure fruit juice must be added to create her desired fruit punch we know if she's using seven gallons of bubble berry fruit punch and it's 14 percent pure fruit juice then essentially she's getting 0.98 gallons of pure fruit juice from the bubble berry so this comes from the bubble berry now we're told again in the problem directly that she's using two gallons of the cherry smash and it's 50 pure fruit juice so two gallons times .5 is again one so add that to one this is from the cherry smash so this amount should be equal this one plus 0.98 which is 1.98 should be equal to the amount of pure fruit juice in the final mixture so in the final mixture she's going to have seven gallons of bubble berry plus two gallons of cherry smash so that's nine gallons and again it's 22 percent pure fruit juice and 9 times 0.22 is 1.98 so we're good to go right this is your final mixture you basically have said that the amount of pure fruit juice coming from the bubble berry and the cherry smash is equal to the amount of pure fruit juice in the final mixture and it is 22 percent pure fruit juice so we're good to go she needs to use seven gallons of bubble berry fruit punch to make her desired fruit punch mixture all right let's take a look at another one so double yield farmer's market sells 24 ounce bags of mixed nuts that contain 43 peanuts to make this product they combine raw it mixed nuts which contain 30 peanuts and true nut mixed nuts which contain 50 peanuts how much of each should be used to make their product so again another straightforward kind of question how much of each so we're saying how much of the raw it mixed nuts and how much of the true nut mixed nuts should be used to make their product which again is a 24 ounce bag of mixed nuts that contains 43 peanuts okay so in this case we don't know an amount for either right so we have to kind of get that we're working with ounces so the total amount in the final mixture that they sell is 24 ounces so let's use that information to create a variable so let's let x be equal to the amount in ounces of we'll just go ahead and say roll it mix nuts used okay now because the total mixture in the end is 24 ounces we know that if we subtract away x again the amount in ounces of raw at mixed nuts used what's going to happen is 24 minus x can represent the amount of the true nuts mixed nuts that's going to be used right so if we put then we'll say 24 again the total amount of ounces of the final mixture minus x which is the amount in ounces of raw mixed nuts used so this is going to be equal to the amount in ounces of the true nut the true nut mix nuts used okay so once we've kind of modeled this we can go down to the little table and set things up just like we did in the last example so on the left let me put the name so you have your rollets you have your rollet mixed nuts you have your true nuts and we're going to have a final product so we'll have a final you could say final mixture or final product okay let me make that a little better so again final mixture so up here we'll have our ounces of mixed nuts and then we're going to have our concentration percentage of peanuts so this is the percentage as the decimal and then we're going to have pure peanuts and ounces so this is pure peanuts and again this is in ounces because we're working with ounces all right for the ounces of mixed nuts we know for raw we represented this with x and for true nut we represented this with 24 minus x for the final mixture we know it's 24 okay that's given to us in the problem now the percentages let's go back up we're told that the final mixture is 43 so this would be 43 or 0.43 as a decimal then we're told that raw lit mixed nuts has 30 peanuts and true nut has 50 peanuts so let's go back down so raw it is 30 so 0.3 as a decimal true nut is 50 so 0.5 as a decimal and again to get the amount of pure peanuts we just multiply across if it's 30 peanuts and it's x number of ounces the ounces of pure peanuts would be 0.3 times x again if it's 50 peanuts and there's 24 minus x ounces the pure peanuts would be 0.5 times the quantity 24 minus x and then for the final mixture if it's 43 peanuts and there's 24 ounces of mixed nuts well i can multiply 0.43 times 24 to give me 10.32 this is the amount in ounces of pure peanuts in the final mixture now again we can set up an equation in the same way that we did in the last problem we basically think about the fact that the amount of pure peanuts coming from the raw at mixed nuts plus the amount of pure peanuts coming from the true nut mixed nuts has to be equal to the amount of pure peanuts in the final mixture so all i've got to do is take this point three x and add it to my point five times the quantity twenty four minus x and then set this equal to my 10.32 so let's solve this and we'll be good to go so we have our point three x plus .5 times 24 is 12 then minus you have 0.5 times x which is 0.5 x this equals 10.32 0.3 x minus 0.5 x is going to give us a negative 0.2 x then you have your plus 12 equals 10.32 let's go ahead and subtract 12 away from each side of the equation and this is going to cancel you have your negative 0.2 x is equal to if i do 10.32 minus 12 i get negative 1.68 so negative 1.68 as a final step let's divide both sides by negative 0.2 and then this cancels with this we'll have that let me kind of make a border here we'll have that x is equal to negative 1.68 divided by negative 0.2 is going to be 8.4 okay so 8.4 so let's go back up so this is going to be 8.4 and then 24 minus 8.4 or 15.6 is going to be the amount in ounces of true nut mixed nuts used so let's go back up we'll make a nice little sentence here we'll say that they should use 8.4 ounces of raw it make that better again mixed nuts and you have 15.6 ounces of true nut mixed nuts so true nut mixed nuts okay so again they should use 8.4 ounces of raw mixed nuts and 15.6 ounces of true nut mixed nuts all right so to check this the first thing you want to do is make sure that the number of ounces coming from roll it mixed nuts and the number of ounces coming from the truenut mixed nuts match up so you've got 8.4 ounces coming from raw mixed nuts and 15.6 ounces coming from the true nut mix nuts that does give you 24 ounces okay 15.6 plus 8.4 would give you 24. so you're good to go there if again we have a 24 ounce bag that's 43 peanuts we know there's 10.32 ounces of pure peanuts in that product now this comes from what it comes from the raw mixed nuts which is 30 peanuts so 8.4 times 0.3 is going to give us 2.52 and then it also comes from 15.6 ounces of true nut mixed nuts which is 50 peanuts so 15.6 times 0.5 is 7.8 so if you take the 7.8 the amount of pure peanuts coming from true not mixed nuts and you add that to the 2.52 again the number of ounces of pure peanuts from the raw mixed nuts you do get your 10.32 or again the number of ounces of pure peanuts in your final product for the double yield farmers market so we can say our answer here is correct again they should use 8.4 ounces of raw mixed nuts and 15.6 ounces of true nut mixed nuts all right let's take a look at one more of these so jacob wants to make 17.5 gallons of a 44 acid solution to do this he will mix a 60 acid solution and a 25 acid solution how much of each solution must he use again very simple problem the key to understanding this here again is to think about the final mixture it's 17.5 gallons of a 44 acid solution so basically we have 17.5 gallons 44 of his pure acid now this can only come from what two sources here we've got a certain number of gallons which is unknown at this point of a 60 acid solution and then a certain number of gallons again unknown at this point of a 25 acid solution now if we let a variable like x be equal to the amount in gallons of the 60 percent acid solution that we're going to use then what can we say about the amount of the 25 acid solution we're going to use again the total amount in gallons is 17.5 so we could say then 17.5 again the total amount in gallons of the final solution minus x again the amount in gallons of the 60 acid solution we're going to use this is going to be equal to the amount in gallons of the 25 acid solution let me just drag this to another page so it's nice and clean so let's go down to a little table real quick and let's kind of set this up so we know that we want to have gallons of the acid solution so this is gallons right here we know we want a percentage concentration basically of acid and then we're going to have the amount of pure acid and of course this is going to be in gallons now we have our 60 acid solution we have our 25 acid solution and then we have our final mixture okay so the number of gallons for the final mixture is given to us as 17.5 in the problem we assign our variable x to represent the amount of gallons for the 60 acid solution let me make that clear because it looks like 66 and then we said that 17.5 again the amount in gallons of the final mixture minus x the amount of gallons for the 60 acid solution that we're using is going to be the amount of the 25 acid solution that we're going to use okay so we have the number of gallons modeled for each the percentage is very easy because it's off to the left right so this one right here is going to be 0.6 as a decimal right that's 60 percent and this one right here would be 0.25 as a decimal that's 25 so the final mixture we are told is 44 again jacob wants to make 17.5 gallons of a 44 acid solution so as a decimal 44 is 0.44 so to get the amount of pure acid in gallons for each i would just multiply across so i have x gallons and it's 60 percent pure acid so 0.6 times x is 0.6 x i have 17.5 minus x gallons it's 25 pure acid so 0.25 times the quantity 17.5 minus x and then my final mixture is 44 acid and it's 17.5 gallons if we multiply those together you get 7.7 so this is the amount of pure acid in the final mixture it's 7.7 gallons now for our equation again we use a similar format we think about the amount of pure acid coming from the 60 acid solution which is 0.6 x plus the amount of pure acid coming from the 25 acid solution which is 0.25 times the quantity 17.5 minus x this has to be equal to the amount of pure acid in the final mixture right so 7.7 okay so let's solve this equation and we'll be good to go so we have that 0.6 x plus 0.25 times 17.5 is 4.375 and then minus 0.25 times x is 0.25 x and this equals 7.7 okay so let's clean up on the left 0.6 x minus 0.25 x is 0.35 x so 0.35 x and then if i subtract 4.375 away from each side what will i get well we know this would cancel so we'll say this is equal to 7.7 minus 4.375 is 3.325 all right so to finish this up let's divide both sides of the equation by 0.35 so we'll get x is equal to 9.5 okay so let's go back up so again we said that x was the amount in gallons of the 60 acid solution so this is going to be 9.5 and then 17.5 minus 9.5 is going to give me eight so there's eight gallons of the 25 acid that's used so let's go back up and to answer this we'll say he should use 9.5 gallons of the sixty percent acid solution along with eight gallons of the 25 acid solution again very easy to check first you would say is 9.5 plus 8 the gallons coming from each source if you add those together is that 17.5 right the number of gallons in the final make sure that he has well yes it is right 9.5 plus 8 is 17.5 so that part checks out all right now next remember he has 7.7 gallons of pure acid in the final solution that came from multiplying 17.5 the number of gallons in the final mixture times 0.44 again 44 as a decimal that gives us 7.7 gallons of pure acid in the final mixture so that comes from two sources from one source he has 9.5 gallons that's 60 acid so that would be 5.7 5.7 gallons of pure acid then lastly it has 8 gallons that's 25 percent acid so 8 times 0.25 is 2. so if you add 2 to 5.7 you do get 7.7 right so you've checked the amount of pure acid in the final mixture that matches up and you've checked the number of gallons of the final mixture that matches up so we can say our solution here is correct he should use 9.5 gallons of the 60 acid solution along with eight gallons of the 25 acid solution in this lesson we want to review applications of linear equations and look at bill and coin problems all right so again if we're solving application problems that involve a linear equation in one variable we can use our little six step procedure so again for solving a word problem we want to read the problem and determine the question or questions to be answered then we want to assign a variable to represent the unknown we want to express other unknowns in terms of this variable then we want to write an equation again based on the situation given in our problem then we're going to solve the equation then we're going to write the answer in terms of the question or again questions asked and lastly we're going to check the answer using the words of the problem we just want to make sure our answer is reasonable so another very common problem type that occurs when we're solving applications of linear equations involves denominations of money okay so these might be referred to as dollar word problems you might hear them called coin word problems just based on the situation given okay and essentially all you're doing here you're normally given a value for all of the coins or all of the bills and you need to find the individual amounts so how many of you know this type of coin do you have or how much of this type of bill do you have you know so on and so forth so let's jump in and look at the first example so we have heather a store clerk has a total of nine hundred ten dollars in her cash register this amount comes from only three types of bills twenty dollar bills five dollar bills and one dollar bills now it says the clerk has doubled the amount of twenty dollar bills as she has five dollar bills and half the number of one dollar bills as she has five dollar bills find the amount of each type of bill in her register so this is a very very straightforward problem we just need to find as it says here the amount of each type of bill in her register so how many five dollar bills how many one dollar bills and how many twenty dollar bills and again this is based on the fact that there's a total value of nine hundred ten dollars in the register okay so let's let a variable be equal to one of our unknowns since the five dollar bill is involved in both comparisons let's let x be equal to that right because it says the clerk has double the amount of 20 bills that she has five dollar bills and half the number of one dollar bills and she has five dollar bills so let's let x be equal to the amount or you could say the number of five dollar bills in her register all right so what can we say about the number of 20 bills and the number of one dollar bills so if we go back up it says here specifically the clerk has double the amount of twenty dollar bills as she has five dollar bills so if i take the number of five dollar bills which we modeled as x and i double that or if i multiply it by two that would give me two x and that would be the number of twenty dollar bills that she's going to have let's put then we'll put our 2x is equal to the number of 20 dollar bills okay so we go back up for the number of one dollar bills it says half the number of one dollar bills as she has five dollar bills so if i take half of x or just one half x that's going to represent the number of one dollar bills so we'll say then again we're going to do one half times x equal to x over two whatever you want to do is equal to the number of one dollar let me make that better again one dollar bills okay so how do we get an equation going we've modeled all of our kind of unknowns but we need an equation to get a solution so let's go back up now it says a store clerk has a total of 910 dollars in her cash register where the problem comes in a lot of students at this point they're used to kind of summing what they have and setting it equal to that value so you see a lot of people do this they'll go okay x plus 2x plus one half x is equal to nine hundred ten this is not the correct equation why is that the case well what i have on the left here is going to be the total amount of bills that she has right between the five dollar bills the 20 bills and the one dollar bills what i have on the right is a value okay so it's not the same these two would not be equal so suppose you had six 20 bills in your wallet or purse how would you find the value there you wouldn't say you just have six bills you would say that you have six times twenty or a hundred twenty dollars okay it's in other words going to be found by multiplying the quantity of bills which is six by the value of each bill which is 20 which gives us a 120 right that's how we got the value so we're going to basically do the same thing to get our 910 total value we need to take the number of five dollar bills and multiply it by five plus the number of 20 bills multiplied by 20 plus the number of one dollar bills and multiply it by one so let's do this with a nice little table let's go down here and essentially what i'm going to have here is the quantity of bills and what i'm going to have here is the value per bill and what i'm going to have here is the total value okay pretty straightforward so we have our 20 bills let's just label this as 20 and we have our five dollar bills we'll label that as five and we have our one dollar bills just label that as one so the quantity of bills again if i go back up i know that the five dollar bills are labeled with x the 20 bills are labeled with 2x and the one dollar bills are labeled with one half x so the quantity of bills for the five dollar bills it's x for the 20 bills it's 2x and for the one dollar bills it's one half x or again you can do x over 2. now the value per bill that's pretty easy just look all the way to the left this one's going to be 20. this one's going to be 5 and this one's going to be 1. okay that's pretty straightforward then to get the total value i just multiply across right this times this will be equal to this okay so 2x times 20 would be 40x x times 5 would be 5x and then 1 half x times 1 would be one half x so now i have something that i can set up an equation with i have the total value of the 20s which is 40x plus the total value of the fives which is 5x plus the total value of the ones which is one half x and this is equal to again the 910 that amount of dollars which is the total value in her cash register okay so you've got to go through and make this set up here otherwise you're not going to get the right answer by just going through and saying okay i have 2x plus x plus 1 half x and setting that to 9 10 because you're not setting a value equal to a value okay that doesn't work so let's go ahead and solve this guy i'm just going to copy this and go to a fresh sheet all right so what i can do 40x plus 5x is 45x then plus one-half x equals 910. if we want to clear that denominator let's multiply both sides of the equation by 2. so the 45x if i multiply that by 2 i get 90x and plus one half x times two would be just x and this equals nine hundred ten times two would be eighteen hundred twenty and now ninety x plus x would just be 91x so let's write this as 91x to get x by itself i divide both sides of the equation by 91 and we're going to find that x is equal to 20. okay so again we're not done we have to go back and make sense of this so x equals 20. so x was the number of five dollar bills so that's going to be 20. then 2x is the number of 20 bills 2 times 20 is 40. then one half x is going to be the number of one dollar bills so one half times 20 is 10. okay so let's go back up and write our answer so what we're going to say is that there are there are 20 five dollar bills and then you have 40 20 bills and you're going to have 10 one dollar bills in her let me make that better register all right so let's check this and make sure it makes logical sense so it tells us that again she has a total value of 910 dollars in her cash register so 25 bills is a hundred dollars let me just line this up and put this as a hundred forty twenty dollar bills is eight hundred dollars and ten one dollar bills is ten dollars so sum those amounts a hundred plus eight hundred plus ten is nine hundred ten so this part's a check okay that works out it says this amount comes from three types of bills twenty dollar bills five dollar bills and one dollar bills good to go there the clerk has double the amount of 20 bills she has five dollar bills again she has 25s and she has 40 20s so we're good there and then it says she has half the number of one dollar bills and she has five dollar bills you've got again 25s and you've got 10 ones so good to go there so everything checks itself out we can say our answer here is correct we'll go ahead and just state this again there are 25 bills 40 20 bills and 10 one dollar bills in her register all right let's take a look at another one again these are pretty easy overall so james a local vendor has a total of fifty six hundred dollars in his carry bag this amount comes from twenty dollar and one hundred dollar bills only if he has a total of eighty bills in his bag how much of each type of bill does he have so in other words what we're trying to find here it says how much of each bill does he have we want to know how much of the 20 bills does he have and how much of the 100 bills does he have because again he only has those two types of bills in the bag and again another thing to highlight here is that he has a total of fifty six hundred dollars and there's only eighty bills in the bag okay so let's let a variable represent one of the unknowns it could be the twenty dollar bills or it could be the hundred dollar bills the key to understanding this problem is knowing that if he has a total of 80 bills in the bag whatever i let x represent then 80 minus x is going to represent the other one so in other words if i let x be equal to the number of 20 dollar bills then because again there's only 80 bills in the bag it has to be true that we could say 80 the total number of bills minus x the number of 20 bills has to be equal to the number of 100 dollar bills okay let me make this a little bit better okay so now that we have both of our unknowns kind of modeled let's think about how we can get an equation going so again just like with the last problem we're given a value okay but we've modeled things as kind of an amount of bills that we have so we have to again go through set up a little table and relate the values so let's go down here and again i'm going to say i have quantity of bills i'm going to say value per bill and i'm going to say total value again if i take the quantity of bills times the value per bill i will get the total value so here i'm going to have my 20 bills and here i'm going to have my hundred dollar bills so we said that the quantity of bills for 20s we represented that with x and the quantity of bills for hundreds we represented that with 80 minus x now the value per bill for 20 we know is 20 we know the value per bill for 100 is a hundred and we just multiply across so to get the total value for the 20 bills it would be x times 20 or 20 x to get the total value for the 100 bills it would be 100 times 80 minus x so 100 times the quantity remember you're multiplying by the whole thing 80 minus x okay so now we have the total value of the 20s and the total value of the hundreds and we can sum those amounts and set it equal to the total amount that he has which is 5600 so in other words i would have 20x plus 100 times the quantity 80 minus x and again this is set equal to the total value of his carry bag which is 5600 dollars so let's copy this and go to a fresh sheet so on the left i have my 20x and then plus you've got 100 times 80 which is going to be 8 000 then minus 100 times x is 100x and this equals 5600. so 20x minus 100x is negative 80x and then again you have plus 8000 this equals 5600 and now i'm going to subtract 8000 away from each side of the equation so i'm going to have negative 80x is equal to if i have five thousand six hundred minus eight thousand that's going to be negative two thousand four hundred to get x by itself i'm gonna divide both sides of the equation by negative eighty so we know these cancel and you're gonna get that x is equal to 30. okay so let's take this result here back up again we said x was the number of 20 bills so this is going to be 30. we said 80 minus x or 80 minus 30 or 50 is the number of hundred dollar bills okay so we have our solution let's go back up and now let's answer this guy so we can say there are again 30 20 bills and 50 100 bills to check this is very easy you'd want to make sure the value works out so 30 times 20 is 600. so the value from the 20s is 600 and then 50 times 100 is 5 000. okay so the value from the hundred dollar bills is 5 000. thousand five thousand plus six hundred is five thousand six hundred so you're good to go there now the next thing you wanna check is does the amount of bills from the twenties plus the amount of bills from the hundreds give me eighty bills which is what's in his back so there's 30 20s and 50 hundreds 30 plus 50 is 80. so you're good to go there as well so again our answer is that there are 30 20 bills and 50 100 bills all right let's take a look at one more problem and this is one that's going to be a little bit tedious so we have becky a math teacher at a local elementary school has a jar of mixed change in her classroom the jar only contains pennies nickels dimes and quarters and has a total value of 89.50 she tells her class that the first person to correctly state the quantity of each type of coin present in the jar can take the jar home now she states that there are four times as many quarters as pennies and six times as many dimes as pennies additionally if the number of pennies was increased by 10 there would be three times more nickels than pennies find the quantity of each type of coin in the jar so this one is tedious because you have four coins involved you have pennies nickels dimes and quarters so that's the unknown here again we're trying to find the quantity of each type of coin in the jar so how many pennies how many nickels how many dimes how many quarters so let's start out by just modeling our unknowns so let's let x be equal to the number of pennies and we're going to do that because it's involved in all the comparisons so it's just easier if you do that so what can we say about the number of nickels the number of quarters and the number of dimes so it tells us to start that there are four times as many quarters as pennies and six times as many dimes as pennies so if there's four times as many quarters as pennies that means we could say then four x or four times the number of pennies which is x is equal to the number of quarters then we can also say again since it told us that there are six times as many dimes as pennies well if there's x number of pennies six times that amount would be 6x so that's going to be my number of dimes so 6x is equal to the number of dimes okay so what can we say about the nickels so what can we say about the nickels so then it tells us if the number of pennies was increased by 10. so i have x number of pennies if i increased it by 10 that would be x plus 10 there would be three times more nickels than pennies so in other words if i increase the number of pennies by 10 again if i take x the number of pennies i add 10 to it if i multiply that amount by 3 that's how many nickels i would have so again if i multiplied 3 times the quantity x plus 10 this would be equal to the number of nickels okay so now that we have everything modeled again we're told the total value of all the coins in our jar at the start of the problem that's going to be 89.50 so again we got to think about the value of each coin we're going to sum the values of all the coins and that should be equal to 89.50 so let's use a little table all right so let's set this up in the normal way so we have our quantity of coins and then we have our value per coin and then we have our total value okay let me make that better and again if i multiply the quantity of coins by the value per coin i'm going to get the total value for that set of coins so again i have pennies which i'm just going to put a p for where you could put .01 i have nickels which i'm going to put an n and i'll just put .05 i have dimes which i'll put a d for and then .1 or 0.10 if you want that and then i have quarters which i'll put a q for this is 0.25 so quantity of coins if we go back up again x is the number of pennies 4x is the number of quarters 6x is the number of dimes and 3 times the quantity x plus 10 is the number of nickels so for pennies this was x for nickels this was 3 times the quantity x plus 10 4 dimes this was 6x and for quarters this was 4x now the value per coin is pretty simple for pennies is 0.01 for nickels it's 0.05 for dimes it's 0.1 or you put 0.10 whatever you want to do and then for quarters it's 0.25 so again the total value i just multiply across so for pennies it's x times .01 or 0.01 x for nickels it's 0.05 times 3 which is 0.15 then times the quantity x plus 10. now you can simplify this here or you can wait till later let's just simplify it here 0.15 times x is 0.15 x and then .15 times 10 is 1.5 so let's put 0.15 x plus 1.5 and then for dimes it's 6x times 0.1 so that's 0.6 x and then for quarters it's 4x times 0.25 which is 1x or just x okay so if we sum these total values for each then what we're going to get is a value that is equal to 89.50 so in other words if i take point zero one x the value of all pennies and if i add it to 0.15 x plus 1.5 the value of all nickels and i add it to 0.6x the value of all dimes and i add it to 1x the value of all quarters i'm just going to write x this is equal to again the total value of the jar which is 89.50 okay so let's copy this and go to a fresh sheet so 0.01 plus 0.15 is 0.16 if i add 0.6 to that i get 0.76 and then if i add another one again this has a implied coefficient of 1 then i get 1.76 so the left side will be 1.76 x and then don't forget about this guy so plus 1.5 this equals 89.50 which is 89.5 if you want now if i subtract this 1.5 away from each side of the equation this will cancel here you have 1.76x is equal to 89.5 minus 1.5 is 88. now to get x by itself i'm going to divide both sides by 1.76 so we know that this is going to cancel and you'll have that x is equal to 88 divided by 1.76 is 50. so this is our answer here but again we've got to go back and make sense of it all right so we know that x represents the number of pennies and again we found that x was 50 so there are 50 pennies then 4 times x is the number of quarters again x is 50. 4 times 50 is 200 so there's going to be 200 quarters then 6x is the number of dimes again x is 50. 6 times 50 is 300 so there's 300 dimes and then lastly 3 times the quantity x plus 10 is the number of nickels so x again is 50 50 plus 10 is 60 60 times 3 is 180 so there's 180 nickels so let's go ahead and write our answer so find the quantity of each type of coin in the jar we can say that there are 50 pennies 180 nickels 300 dimes and then lastly we'll say and there's 200 quarters 200 quarters we could say in her jar okay so there are 50 pennies 180 nickels 300 dimes and 200 quarters in her jar so let's check this so it tells us there are four times as many quarters as pennies you've got 50 pennies in 200 quarters so that checks out so i'll just put a check there then we're told there's six times as many dimes as pennies you've got 300 dimes and 50 pennies so that checks out right 300 is 6 times 50. and then additionally if the number of pennies was increased by 10. so if we went from 50 to 60 there would be three times more nickels than pennies well yeah if i had 60 pennies i have 180 nickels 60 times 3 is 180 so we're good to go there as well now the last thing to check is the value if i have 50 pennies that's 0.5 or 50 cents if i have 180 nickels that's 180 times .05 it's basically nine dollars if i have 300 dimes it's basically 300 times 0.1 that's going to be 30 and then if she has 200 quarters that's 200 times 0.25 which is going to be 50. okay so 50 plus 30 is 80 80 plus 9 is 89 plus 50 cents is gonna be 89.50 so we're good to go on our value right if we add everything up we do get 89.50 which again is what it told us in the problem so we could say our answer here is correct there are 50 pennies 180 nickels 300 dimes and 200 quarters in her jar in this lesson we want to review applications of linear equations and look at some simple interest problems so again when we're solving applications of linear equation problems we can use this little six step procedure to help us out so for solving a word problem we have that we want to read the problem and determine the question or again questions to be answered we want to assign a variable to represent the unknown and then with these type of problems you're going to be able to express other unknowns in terms of this variable then we're going to write an equation we're going to solve the equation we're going to write the answer in terms of the question or questions asked and then we're going to check the answer using the words of the problem so this part right here the last part just make sure your answer is reasonable all right so for today's problems we're going to be using the simple interest formula so the simple interest formula goes like this we have i which is the amount of simple interest earned is equal to p which is the principal or amount invested times r which is the rate as a decimal times t which is the time which is generally given to us in years but it could be months it could be weeks it could be whatever they give to okay but generally it's going to be years now one thing we need to understand before we move on is kind of the difference between simple interest and compound interest we will look at examples that deal with compound interest later on in the course we need to kind of build some concepts before we deal with that for right now we'll be dealing with just simple interest so with simple interest i only earn interest on the principal or the original amount invested with compound interest i earn interest on interest so let's look at a very simple example let's say that my original amount that i invest is a thousand dollars okay so this is my account balance with each of these okay so one is a savings account with simple interest one's a savings account with compound interest and let's say each of them pays me 10 okay so it's 10 in each case so no difference at the end of year one let's just assume they hold the money for a year and at the end of the year they give us our interest payment so at the end of year one then this guy right here ten percent of a thousand is a hundred so my balance is one thousand one hundred dollars same thing when it's compounding i didn't get any interest dropped in there yet so it's still one thousand one hundred dollars now at the end of year two what's going to happen is i'm gonna have a difference right because this guy i just get paid on the original amount that i invested which is a thousand so i get my ten percent again which is just a hundred dollars so this guy's going to go to 1200 okay so that's where it is with simple interest with compound interest they're now going to say okay you get ten percent of the balance so the balance is eleven hundred dollars so i'm going to get ten percent of that which is a hundred ten dollars so i'm gonna get one thousand two hundred ten dollars as a balance okay so the difference here is that i have twelve hundred from simple interest or one thousand two hundred ten dollars with compound interest because my interest that was earned which was a hundred dollars at that point earned interest of ten percent okay a hundred dollars ten percent of that is ten bucks so that's where you're going to get the difference in the account balance okay so simple interest you only earn interest on the original amount invested compound interest you earn interest on the full balance or you could say you earn interest on your interest all right so let's look at a simple example here just to kind of get our feet wet and then we'll look at two kind of typical problems that you'd see so nathan invests ten thousand dollars in a savings account which pays three percent annual simple interest what is the total amount of simple interest earned at the end of a two-year period so again for this guy right here you can almost do this mentally you can say what this is i simple interest earned equals p which is the principal times r the rate as a decimal times t the time in this case in years mentally i can say okay ten thousand dollars times three percent as a decimal is point zero three so if you do that ten thousand times point zero three is 300 and so it's basically 300 a year for two years 300 times 2 is 600 so he would earn 600 for the two year period but to plug it into the formula to kind of make it official you would substitute in a 10 000 for p because that's your principal or amount invested you would substitute in a .03 for your rate okay it's always as a decimal and you would substitute in a 2 for the time period in this case that stands for two years okay so again make sure that the rate is in terms of years if you're dealing with years okay so we have 10 000 times .03 which is 300 times 2 which is 600 so you can say that your i or your simple interest earned is 600 and again we can answer this let me just erase everything we have our answer now we'll make a nice little sentence and say that nathan earned 600 at the end of two years okay and lots of different sentences that you can construct to answer this you're just looking for something that addresses what they asked okay so it says what is the total amount of simple interest earned at the end of a two-year period and we just said nathan earned 600 at the end of two years perfectly acceptable okay let's look at a typical problem now one where we have to set up kind of our full array of steps so jason has 70 000 to invest and wants to earn an overall rate of nine percent annual simple interest so he has seventy thousand dollars to invest and he wants to earn an overall rate of nine percent annual simple interest okay let's go down and look at the next part so he wants to invest part of the money in a savings account which pays eight percent annual simple interest the rest of the money will be placed in a corporate bond fund which pays 12 percent annual simple interest then here's the question to meet his investment goal which again was up here he has 70 000 to invest and wants to earn an overall rate of 9 annual simple interest that's the investment goal so to meet his investment goal how much should be invested at each rate so to earn nine percent on seventy thousand dollars how much should he put in the savings account and how much should he put in the corporate bond fund all right so we've read the problem we understand what we need to do let's let a variable like x be equal to the amount he invests in one of these guys doesn't matter which one you choose you can say x is the amount he invests in the savings account or you can say x is the amount he invests in the corporate bond fund it makes no difference so i'm going to come down here and say we're going to let x be equal to the amount invested in the bond fund let me put the in the bond fund okay so then what can we say about the amount invested in the savings account well we're told that jason has a total of seventy thousand dollars to invest and he only puts it in either the savings account or the bond fund so if x is representing the amount that's put in the bond fund then 70 000 minus x has to be the amount that he put in the savings account so 70 000 which is the total amount minus x which is the amount invested in the bond fund is going to be equal to the amount invested in the savings account [Music] okay nice and simple let me make that better okay so now that we have kind of our variable set up and the other unknown is expressed in terms of that variable we're going to move into our next step which is to basically set up an equation to do that i'm going to make a little table a lot of times i think it's a little bit helpful if you're unfamiliar with solving word problems to set up a table some people don't do it they feel like it's a waste of time but it can be helpful so let's just go ahead and do it so we're going to write our simple interest formula up here on the top so i the simple interest earned is equal to p the principal or amount invested times r the rate as a decimal times t the time then for this guy right here this is going to be the investments so i'll just put inv for investments so you have the savings account and you have the corporate bond fund so i'll just kind of initial those so we know that for the amount invested or the principal the bond fund we said was x so let's put an x here for the savings account we said it was 70 000 minus x now the rate is given to us in the problem it says the savings account pays 8 annual simple interest and it says the corporate bond fund pays 12 percent annual simple interest so the savings account again is eight percent or .08 as a decimal the bond fund is 12 or 0.12 as a decimal now what about the time well basically we're dealing with a scenario where the time is just going to be one so if we go back up you can see that it basically says in the first part of the problem jason has 70 000 to invest and wants to earn an overall rate of 9 annual which is yearly simple interest so what are we doing for one year right one year period so we can go back down and just put a one for each time period you can basically get rid of it because multiplying by one just leaves something unchanged right but we'll just leave it there for kind of completeness so we know that i the simple interest earned is the product of the principal times the rate times the time so at the bottom it's pretty easy i would just have 0.12 x for this guy right here it's kind of hard to fit this but we're going to have .08 times the quantity 70 000 minus x so let me kind of scooch this part up so it's out of the way so i'm going to put 70 000 minus x just going to bleed into the next column i can extend the column but it's going to kind of make everything go crazy so let's just leave it like that and we can now kind of think about how we can use this information so i have the simple interest earned from the savings account which is .08 times the quantity 70 000 minus x and then i have the simple interest earned from the corporate bond fund which is 0.12 x so if i sum these amounts together this needs to be equal to what well again it tells us that he has 70 000 to invest and he wants to earn an overall rate of return of 9 so the sum of the simple interest from those two investments has got to be equal to 70 000 which is the principal times the rate which is .09 times the time which is one okay so you can just leave that off so let's go through we've set up our equation now and let's just solve this guy so .08 times 70 000 is 5600 then minus you'd have 0.08 times x then plus 0.12 x and this equals 70 000 times .09 is 6 300. so 6 300. so negative 0.08 x plus 0.12 x is going to be .04 x so you'd have 5600 plus .04 x is equal to 6300 then i could subtract 5 600 away from each side of the equation that'll cancel if i have zero four x is equal to sixty three hundred minus fifty six hundred is seven hundred okay so the last step here to get x by itself is to divide both sides by point zero four i'll get x is equal to 17500. okay so let's take this back up so this is where you really need to again think about a sentence you can write because you don't just put x equals 17500 and turn in the test okay you've got to make sense of it in terms of the problem so x was the amount invested in the bond fund so we know that's seventeen thousand five hundred dollars then seventy thousand minus x so seventy thousand minus seventeen thousand five hundred or fifty two thousand five 500 is the amount invested in the savings account so this is 52 500. so now we're ready to go back up and answer this in terms of the question being asked how much should be invested at each rate well we can say that jason should invest fifty two thousand five hundred dollars in the savings account [Music] and seventeen thousand five hundred dollars in the bond fund okay so jason should invest fifty two thousand five hundred dollars in the savings account and seventeen thousand five hundred dollars in the bond fund okay so how can we check this to make sure that it's accurate well again if we go back to the beginning of the problem jason has seventy thousand dollars to invest so let's check that first is the amount that he put in the bond fund plus the amount that he put in the savings account equal to seventy thousand fifty two thousand five hundred plus seventeen thousand five hundred does equal seventy thousand so that part is a check so this is good to go now he wants to earn an overall rate of return of nine percent annual simple interest so seventy thousand earning nine percent annual simple interest would mean that he's earning sixty three hundred dollars a year in annual simple interest okay so let's check that against what he's going to earn with the two investments so he's going to earn 8 with the savings account and he's going to earn 12 with the corporate bond fund so let's go back up we again we have 6 300 as the amount that he's earning and we're going to check so we have 52 500 in the savings account which is earning 8 percent so that would give me 4 200 a year and then he put 17 500 in the bond fund which is earning 12 percent so that would give him 2100 okay so if you sum these two amounts together the 4200 he gets from the savings account and the twenty one hundred dollars he gets from the bond fund he's going to get the sixty three hundred dollars she's looking for in annual simple interest so we can say this problem does check out and again jason should invest fifty two thousand five hundred in the savings account and 17 500 in the bond fund all right let's take a look at one more of these so jennifer recently retired and needs six thousand dollars per year in supplemental income from investments so she has fifty thousand dollars that will be invested in both junk bonds which pays 15 annual simple interest and commercial paper which pays seven percent annual simple interest how much money should be invested in each so again let's go back up again she retired and need six thousand dollars six thousand dollars per year in supplemental income so she needs six thousand dollars she has fifty thousand dollars so that's going to be her principal and that will be invested in again both junk bonds the junk bonds are going to pay 15 annual simple interest and the commercial paper is going to pay seven percent annual simple interest so how much money should be invested in each okay that's what we're trying to answer so let's let a variable like x be the amount that she invests in let's just say the commercial paper you can put it as either but i'm just going to go with the commercial paper so let's go ahead and do that real quick so let's let x be equal to the amount invested and commercial and let me just abbreviate that because it's kind of hard to write all that so i'm just going to put commercial paper c period p period so then what can we say about the amount that's put into junk bonds well again the total amount she's investing is 50 000 so we can say then 50 000 minus x is going to be the amount invested in junk bonds so i'll put j period b period okay for junk bonds so let's take this information to a table just like we did in the last example so for our investments these are going to go down we have your junk bonds and you have your commercial paper okay and again we have our simple interest formula i the simple interest earned is equal to p the principal times r the rate times t the time okay now in this case we're talking about an annual basis so time is just going to be one for each of these okay so very very simple for the rate we know that for the junk bonds they pay 15 percent so 0.15 we know for the commercial paper it pays 7 percent so 0.07 again you've always got to express the rate as a decimal and then the principal or the amount invested in each of these we just modeled that we said x is the amount invested in the commercial paper and 50 000 minus x is the amount invested in the junk bonds so the principle again the amount invested in the commercial paper is x and the amount invested in the junk bonds is going to be 50 000 minus x okay so my simple interest earned again all i've got to do is multiply in the rows so i would have principal 50 000 minus x times a rate of 0.15 times 1 which really doesn't change anything so you basically have 0.15 times the quantity 50 000 minus x again hard to kind of fit that in the column and then over here it's pretty easy for the commercial paper it's just going to be .07 times x again you can multiply by one but it doesn't really change anything now we are told that she wants to earn six thousand dollars per year in supplemental income from investments okay so what do we know we know that the interest she's going to earn from the junk bonds is 0.15 times the quantity 50 000 minus x then if we add that to the interest she's going to earn from the commercial paper which is .07 x this has to be equal to 6 000 okay because that's what she needs to earn so let's go ahead and solve this equation very very easy 0.15 times 50 000 is 7 500 and then minus 0.15 times x is 0.15 x then plus .07 x again is equal to 6 000. if we combine like terms here you'd end up with negative .08 x so 7 500 minus 0.08 x is equal to six thousand i'm going to subtract 7 500 away from each side of the equation that's going to cancel let me get some room going so on the left i'm just going to have my negative .08 x on the right i'm going to have negative fifteen hundred we'll finish this up by dividing both sides of the equation by negative point zero eight so this is going to cancel with this and i'll have that x is equal to eighteen thousand seven hundred so 18750 okay so that's our answer for x but again we're not done we've got to make sense of that so again x was the amount invested in commercial paper so we know that's 18 750 and then we said that 50 000 minus x which is thirty one thousand two hundred fifty dollars is going to be the amount invested in the junk bonds so thirty one thousand two hundred fifty okay so let's go back up and make a nice little clean sentence so jennifer should invest thirty one thousand two hundred fifty dollars and junk bonds junk bonds i'll just abbreviate it and eighteen thousand seven hundred fifty dollars in commercial paper okay so you can check this pretty easily by going back to the original statement where it said jennifer recently retired and needs six thousand dollars per year okay so that's what she has to earn so just check the simple interest from the two investments again from the junk bonds she's got thirty one thousand two hundred fifty dollars invested and she's getting fifteen percent on that so times point one five that's gonna give you what that's gonna give you four thousand six hundred eighty seven dollars and 50 cents okay so this is what she earns from the junk box i'm just going to put junk bonds okay this is what she earns in yearly simple interest then from the commercial paper she's going to earn 7 eighteen thousand seven hundred fifty so if i had eighteen thousand seven hundred fifty times point zero seven i would get one thousand three hundred twelve dollars and fifty cents so one thousand three hundred twelve dollars and fifty cents so all you have to do to check this problem is sum these two amounts together does it give you six thousand the answer to that is yes right if i add four thousand six hundred eighty seven dollars and fifty cents to one thousand three hundred twelve dollars and 50 cents i will get 6 000 so that does tell me my solution is correct that jennifer should invest 31 250 in the junk bonds and 18 750 in the commercial paper in this lesson we want to review applications of linear equations and look at percent problems so when we're in our section on applications of linear equations again we can use a little six step procedure to get a solution for our word problem so for solving a word problem again the first step is just to read the problem and determine the question or questions to be answered then the next step is to assign a variable to represent the unknown again with these type of problems you're going to express other unknowns in terms of this variable then we're going to write an equation that expresses the situation we're given right based on the scenario in the problem then we're going to solve that equation then we're going to write the answer in terms of the question or again questions asked and then lastly and most importantly we're going to check the answer using the words of the problem we always want to make sure that our answer is reasonable all right so for solving percent word problems it's a very easy case scenario it's one of the easiest types of problems that you deal with when you're talking about applications of linear equations okay so essentially as long as you understand how to convert between a decimal and a percentage you're basically good to go okay so we have that kristen works as a cashier for a local clothing store after a busy shift she had a total of 16 277 dollars and 17 cents in receipts so this amount included the 8.5 state and local sales taxes now if the sales tax rate had been 10 how much would she have collected in state and local sales taxes so this is what we need to figure out again if the sales tax rate had been 10 percent in this problem it was 8.5 so if it had been 10 how much would she have collected in state and local sales taxes so that's what we need to find out so to attack this problem it's very very simple you have two amounts that you're going to work with the first amount is the sixteen thousand two hundred seventy seven dollars and seventeen cents in receipts and the second amount is the eight point five percent state and local sales taxes so think about what this means she's got this amount in total receipts so it included the state and local sales tax okay so think about this you have a pre-tax amount of goods sold so you have some pre-tax amount of the goods sold plus you have the taxes that are collected and when we sum these two amounts together we get this value here this sixteen thousand two hundred seventy seven dollars and seventeen cents so what i'm going to do is i'm going to let a variable like x equal the pre-tax amount of all the goods sold so let's go down and i'm just going to say we're going to let x be equal to the pre-tax amount of goods sold okay so again if i go back to what i wrote there i said that the pre-tax amount which we labeled as x right this will now be x plus the tax amount which how can we get that well if she sold x amount of goods and the tax is 8.5 percent well then all i have to do is multiply eight point five percent times that pre-tax amount right which is x now you do that as a decimal so eight point five percent as a decimal is point zero eight five so plus point .085 times x that would give me my 16277.17 so it's very easy to get our equation let me kind of arrow that up there and i'm just going to copy this because it's kind of busy here so let me remove that from here and let me just kind of throw that down here so we can just get some thought processes going and we see that again the pre-tax amount is x plus the tax amount which is .085 which again is 8.5 percent as a decimal times x which is the pre-tax amount that gives me the tax you sum those two amounts together and again you get our total receipts which is sixteen thousand two hundred seventy seven dollars and seventeen cents so on the left side i can just sum this right so this is basically 1x plus .085 x so this would be 1.085 x and let me make that 8 a little better again this is equal to the 16277. and 17 cents now if i want x by itself just divide both sides by 1.085 again over here 1.085 i'm going to do that with a calculator and i'm going to end up with x is equal to fifteen thousand two okay so let me erase everything other than my x okay i don't need any information anymore because i basically found what i need the pre-tax amount of goods sold so this is fifteen thousand two dollars okay now it doesn't ask us for this information in the problem we go back up it specifically says if the sales tax rate had been ten percent how much would she have collected in stated local sales taxes so they don't care that it's fifteen thousand two dollars in sales pre-tax right they don't care about that at all they wanna know again if the sales tax rate had been ten percent what would she have collected in stayed in local sales taxes so i would just multiply this by .10 that's 10 percent as a decimal that's very easy to do i just moved my decimal point one place to the left so you would have 1 500 and 20 cents okay so 1 500 and 20 cents that's going to be your answer so i'll say she [Music] would have collected so it's going to be 1 500 and then 20 cents okay so she would have collected 1 500 and 20 cents you could put in i'm just going to put sales taxes so sales taxes okay so to check this is pretty easy if she had 16 277 dollars and 17 cents in total receipts and again this included the 8.5 percent state and local sales taxes well if i started with again the fifteen thousand two dollars and pre-tax goods sold if i added in the eight point five percent tax on that one way that i could do that is i could multiply fifteen thousand two by one point zero eight five okay and that would give me the sixteen thousand two hundred seventy seven point one seven so that part would check out now i know that it's fifteen thousand two dollars okay that's pre-tax goods sold again if the sales tax rate had been 10 percent all you got to do is move the decimal point one place to the left okay to multiply by 0.1 10 as a decimal and you would end up with this and twenty 1500.20 again in sales taxes okay so our answer here is correct again she would have collected one thousand five hundred dollars and twenty cents in sales taxes all right let's take a look at another one so we have here that james took a few friends to lunch and paid a total of 141 dollars and 45 cents including sales tax and tip if the state and local sales taxes were 8 percent and he gave a 15 tip on the pre-tax amount what was the price of his lunch before tax and tip so this problem sets up in a similar way but what we want to find out is what was the price of his lunch before tax and tip so before he gave the tax and the tip what would the amount have been so again just like the last problem we have a pre-tax pre-tip amount where we say the original amount of the food so the original amount then you have plus the tax and then plus the tip and this equals this 141.45 so let's bring this down to a fresh sheet so the original amount i'm just going to call that x so let x be the original amount of the food okay this is before tax or tip so if i have x plus my tip and plus my tax again i'm going to get this 141.45 so we said that the tax if we go back up was 8 percent and the tip is 15 so the tax again is eight percent so that's point zero eight times x right that's eight percent as a decimal point zero eight again multiply that by x that's going to give you the tax okay and again this is the original amount or you could say pre-tax pre-tip amount whatever you want to do then plus i've got to get my tip and the tip is 15 of the original amount so that's 0.15 which is 15 as a decimal times x so this is your tip so this equals you've got your 141.45 if i sum everything on the left just combine like terms what's going to happen is i'll have x plus point zero eight x plus point one five x so one plus point zero eight plus point one five is going to be 1.23 and then times x and again this equals 141.45 divide both sides of the equation by 1.23 and what are we going to get we're going to get that x is equal to 115 okay so that was the price of his food before again the tax and the tip so we'll say that james is lunch was 115 dollars before tax and tip okay and again pretty easy to check this if you think about the 115 before taxing tips so if he had 115 dollars his tip would have been what it would have been 15 percent of that so 15 of 115 is 17.25 so you'd have 115 plus 17.25 that's your tip and then plus your sales tax that's eight percent if you did eight percent of 115 you would get 9.2 so if you add those three amounts together you will get the 141.45 okay so that tells us our solution here is correct james's lunch again was a hundred fifteen dollars before tax and tip all right let's take a look at one more problem again these are very very easy problems so evelyn works as a sales floor associate at a local hardware store she receives a weekly salary of 800 along with an 11 percent commission on all of her sales so last week she earned thirteen hundred fifty dollars from her salary and commission what is the dollar amount of the merchandise that she sold so this is what we're trying to find out what is the dollar amount of the merchandise that she sold okay pretty easy so again she receives a weekly salary of 800 and she gets an 11 commission on all of her sales okay again last week this is what she earned sharing thirteen hundred fifty dollars so the key to understanding this is that she's gonna get that eight hundred dollars as a salary no matter if she sells zero dollars or a million dollars okay it doesn't matter she's getting a weekly salary plus commission so if she got thirteen hundred fifty dollars i need to subtract the eight hundred dollars away to find out what our commission was so i would do one thousand three hundred fifty again the total amount she got minus the eight hundred dollars which is kind of the salary or the set amount she gets no matter what so if we subtract this away this comes down five borrow here 13 minus eight is five so she's got 550 dollars from commission so 550 dollars from commission okay so how much was her commission rate again it was 11 okay so it's 11 for the commission rate so what this means is that she has some amount of goods sold let's just say it's x so let x be equal to the amount of goods sold okay by her so if i take x and i multiply it by 11 percent that would give me the amount of commission she made which we know is 550 dollars so as a decimal 11 percent is 0.11 so point 11 then times x again the amount of good she sold should be equal to 550 dollars okay that's again the amount of commission she received if i divide both sides of the equation by 0.11 i'm going to get that x is equal to five thousand okay so this is what she sold in merchandise okay if she sold five thousand dollars worth of stuff and she got an eleven percent commission on it she would get five hundred fifty dollars in commission you could check that in your head five thousand times eleven percent is five hundred fifty again that's what we said she got in commission so let's go ahead and write our solution as a sentence just change my marker here and i'll say that she sold five thousand dollars worth of merchandise if you want to check this it's pretty easy you can check it almost in your head if you wanted to again she got a total of 1 350 from her salary and commission it says right here that her salary was 800 okay so again take the salary out that's going to give you 550 that she got from commission now if she sold five thousand dollars worth of merchandise and she gets an 11 commission again you can check that really quickly and say okay well 5 000 times 0.11 would give you 550 so our answer here is correct again she sold five thousand dollars worth of merchandise in this lesson we want to review the imaginary unit i in elementary algebra courses we typically don't deal with the square root of a negative number if we see something like the square root of negative 25 we know that using the real number system we can't get a solution right so a lot of times we would see this and we would stop and say no real number or no real solution or something like that but as we progress through algebra we learn that there is a way to kind of deal with this scenario so the problem lies in the fact that we cannot square a real number and get a negative if i take something like x and this can be any real number and i square it the result is always greater than or equal to 0 okay for any real number if i square 0 i get 0. if i square a negative i get a positive if i square a positive i get a positive okay so it's not possible to square a real number and get a negative so in order to kind of progress and deal with this scenario or we're taking the square root of something like negative 25 we introduce the imaginary unit i so the definition of i is usually given as i squared is equal to negative 1. so this is now a number that when squared gives us a negative and specifically a negative one okay you may also see that i is defined to be equal as the square root of negative one and this just comes from taking the square root of both sides here right if i took the square root of this side i'd have just i if i took the square root of this side i'd have the square root of negative 1. so using this definition and specifically we're going to focus on this one right here we can rewrite the square root of negative 25 using the product rule for radicals we'll say this is the square root of negative 1 multiplied by the square root of 25 we know the square root of negative 1 is defined right here to be i okay this imaginary units so i'm just going to replace that with an i so i will have i times the square root of 25 we know the square root of 25 or at least the principal square root of 25 is 5. so i'm going to say this is 5 times i okay so in this particular case by using an imaginary unit i i'm able to kind of push past my restrictions with real numbers and get a solution okay so the square root of negative 25 we simplify that to 5 i so in general we can use this rule here we have the square root of negative a and we say this is equal to i okay the imaginary unit times the square root of a all we're doing here we're first assuming that a is a positive real number we're saying the square root of negative a is equal to the square root of negative 1 right i'm just dealing with this negative right here times the square root of a so this is equal to this by definition is i so i put the i here and then times the square root of a now you typically see the i written in front of the square root symbol why is that the case well a lot of times if you put it like this and you say the square root of a and then multiplied by i like that you might make a mistake and then extend the radical a little too far it may be confusing in terms of where it is so to clarify things we just put the i out in front okay so that's why you see that but it is not incorrect to do the square root of a times i like that but you will not typically see that all right let's start out with the square root of negative 100. so kind of the long way to do this is to break it up again using the product rule for radicals i can say this is the square root of negative 1 times the square root of 100 again we know by definition that the square root of negative 1 is i so i'm going to replace that with just i and then times the square root of 100 the square root of 100 or at least again the principal squared of 100 is 10. so i can say this is i times 10 or 10i and that's simplified now how can i speed up this process so let's say that i saw the square root of negative 100 we know that this negative here is basically negative 1. so if i have the square root of negative 1 i know i can pull that out and say that's i so anytime you see a negative involved and it's underneath a square root symbol just go ahead and pull that part out and say okay this is i times this number here is 100 so i times the square root of 100 and then i'll go ahead and just say this is i times 10 or 10i okay you don't need to go through this long process each time this is just to demonstrate where things are coming from all right let's take a look at another one so we have the negative square root of negative 64. so this negative out in front is just going to carry itself over here once we've done that we don't need to worry about it anymore this negative inside of the square root symbol is different right that is representing the square root of negative 1. we're going to replace that with just i now i would think about this as times the square root of 64. square root of 64 is just 8. so i can wrap up this problem and say that i have 8 times negative i or negative 8 i as my simplified answer what about something like the square root of negative 24. so again the first thing i'm going to do is realize that i can pull that negative out right i can think about that as again the square root of negative 1 which is i and then times the square root of 24. the square root of 24 you think about 24 as not being a perfect square right there's no rational number that you can square to get 24 but one of the factors of 24 is a perfect square right 24 is 4 times 6 4 is a perfect square so i can say this is i times the square root of 4 times the square root of 6. square root of 4 is 2. so i'm going to go ahead and say this is 2i times the square root of 6. all right let's take a look at kind of a trap question that you're going to get so suppose we have the square root of negative 9 times the square root of negative 16. what would you do to simplify this well a lot of people would look at this and say okay i'm going to use my product rule for radicals i'm going to have a square root and i'm just going to take these radicands here and i'm going to combine them under this square root symbol so negative 9 times negative 16 we know negative 9 times negative 16 is 144 right so this would be the square root of 144 and if we take the square root of 144 we end up with 12. okay so you might put that down as your answer but the problem is this is absolutely wrong okay this is wrong the reason is is that when we use the product rule for radicals with two negative radicands it doesn't give us the correct answer okay so we can't use our product rule for radicals with two negative radicands we can use it with two positive radicands so if these were both positive we could do this right this would be square root of 144 which would be 12 so that would be correct there or we could do it with one negative and one positive okay so there can be alternating signs it will work in that scenario for this case what i would end up doing is pulling the negative out of this part and saying i have an i times the square root of you'd have 9 times 16 which is again 144. so this equals square root of 144 is 12. so this would be 12i okay that would be the correct solution that's fine but again if you use two negative radicands you cannot use that rule so what you need to do in this scenario is convert each of these first so i'm going to pull the negative out of each part and say this is i times the square root of 9 times you'll have i times the square root of 16. once you've done that then you can use your product rule for radicals it's fine now because if you have square root of 9 9 is a positive number you have square root of 16 16 is a positive number so there's nothing else that's stopping you there so i times i is i squared and then we know that the square root of 9 times the square root of 16 we could put the square root of 144 so this is going to equal i squared times the square root of 144 is 12. now most people will stop and write this as 12 i squared but we should know from the beginning of the lesson that i squared is defined to be equal to negative 1. so anywhere i see i squared i can just replace it with negative 1. so if i had 12 times negative 1 this would be negative 12 okay and that would be my answer so notice how i got negative 12 here whereas if i went with the product rule for radicals i would have ended up with a positive 12 okay so it's a very different answer in terms of the sign so you've got to make sure that if you have two negative radicands you simplify using this method right pull the negative one out of each one of those guys and then you can use the product rule for radicals so let's look at another example suppose we have the square root of negative 6 times the square root of negative 5. again all i'm going to do because i have two negative radicands here i'm going to pull this out of each so i'll have i times the square root of 6 multiplied by i times the square root of 5. so i times i again is i squared and then times square root of 6 times square root of 5 really you can just write that as a square root of 30. we know that i squared is negative 1. so i can go ahead and say this is negative square root of 30. all right let's look at another one so we have the square root of negative 8 times the square root of negative 11 times the square root of negative 3. so again to simplify this i want to change i want to change each of these okay so i'm going to have i times square root of 8 times i times square root of 11 times i times square root of 3. okay what's that going to give me well i know i can simplify the square root of 8 8 is 4 times 2 so i can say this is i times square root of 4 times square root of 2 times i times square root of 11 times i times square root of 3. okay so essentially we know at this point that the square root of 4 is 2 so i'm going to say i have 2 i times i times i is i cubed and then you're going to have square root of 2 times square root of 11 is square root of 22 square root of 22 times square root of 3 is square root of 66. now i can simplify i cubed and we'll talk more about this in another lesson but essentially using my rules for exponents i cubed is what i can say this is i squared which i know is negative 1 times i to the first power or just i so i can replace this part right here with a negative 1 and then say times i which is just negative i so i'm going to erase this i'm going to put the negative part of that out in front and just put an i right there so this would simplify to negative 2i times the square root of 66. so when we work with dividing radicals we're going to use the same process if we have two negatives involved we need to change each form first okay so the square root of negative 15 we're going to write as i times the square root of 15 and then the square root of negative 3 we're going to write as i times the square root of 3. and i can go ahead and simplify this further square root of 15 is what it's the square root of 5 times the square root of 3 this is over the square root of 3 here and then i've got i over i now i know these are going to cancel i over i doesn't matter that's an imaginary unit same thing over itself as long as it's not 0 that's allowed and it cancels to be 1. over here square root of 3 over square root of 3 again same non-zero number over itself that cancels and it's 1. so here i'm just left with the square root of 5. in this lesson we want to review operations with complex numbers so in our last lesson we reviewed the concept of the imaginary unit i again the definition of i is given as i squared is equal to negative one or alternatively you could say that i is equal to the square root of negative one so if we look at something like a plus bi where we would say that a and b these two guys are real numbers and again i represents the imaginary unit when we see a plus bi this is known as a complex number okay so this whole thing together we're going to say is a complex number now in your textbook when you see something written as a some real number plus b another real number multiplied by i again the imaginary unit this is known as standard form for a complex number so something like 3 plus 7 i or something like negative 2 minus 9 i right so on and so forth this is in standard form you want the real part which is this part right here so this is the real part because it only consists of a real number and then this part right here this bi part is last this is your imaginary part this is the imaginary part now some textbooks will differ on this they're only called be the imaginary parts so you just need to be aware that from textbook to textbook from teacher to teacher they might say something different but generally just go with whatever your teacher or your textbook is saying okay so it's either that bi is the imaginary part or just that b is the imaginary part now when we work with complex numbers we need to understand that the real numbers are a subset of the complex numbers what do i mean by that well i can take any real number that i want and write it as a complex number let's say i had the number three i could put this as three plus for my b there i'd use a zero and then times i zero times anything is always zero it doesn't matter if i'm multiplying it by i it's the same so this would be just 3 right i have 3 plus 0 which is just 3. so i took a real number 3 and i wrote it as a complex number 3 plus 0i so again that tells us that the real numbers are a subset of the complex numbers right they're part of them as another example i could take something like negative 16 and i could write it as negative 16 plus 0i and again this is just negative 16. this is the same thing as when we took integers and we showed that we could write them as rational numbers because we could say all integers are rational numbers right so if i took let's say negative 12 i can always write this as negative 12 over 1 and this shows it as a rational number but not all rational numbers are integers i can't take something like two-thirds and write it as an integer okay so it's the same concept here all real numbers are complex numbers you can write them as plus 0i so 3 plus 0i is 3 negative 16 plus 0i is negative 16. so on and so forth but not all complex numbers are real numbers okay the complex numbers include more numbers okay they're a larger set so i could do something like 2 plus let's say 5i this is a complex number i cannot write this as a real number so when we work with complex numbers we use the same properties to simplify things so if i saw something like the quantity negative 3 plus i then plus the quantity 2 plus 5i i would just arrange the real and the imaginary parts together so in other words i would use my associative and my commutative properties to kind of reorder this i could say this is negative 3 plus 2 those are the real parts then plus you have i plus 5i so i plus 5i those are the imaginary parts so we could say negative 3 plus 2 is negative 1 and then plus you have i plus 5i which is 6i so you end up with negative 1 plus 6i as your answer here and notice how we give our answer in the format of a the real part plus your b which is 6 in this case times i okay this is the imaginary part this is the real part so this is your standard form right some people will kind of switch this around and say this is 6i minus 1. this is not standard form okay you want the real part first that's why we have negative 1 out in front then plus your imaginary part the 6i second all right let's take a look at another one so suppose we have the quantity 8 minus 3i then minus we have the quantity 2 plus 4i then minus we have the quantity 2 plus 6i so what i want to do here let me just distribute the negative to each term i'm going to put plus negative 1 here and plus negative 1 here so i'm going to write this as 8 minus 3i then negative 1 times 2 is minus 2 negative 1 times 4i is minus 4i negative 1 times 2 is minus 2 negative 1 times 6i is minus 6i so we can regroup these if you want at this point you should understand that you can add the real parts together and the imaginary parts together so in other words i can say eight minus two minus two eight minus two is six six minus two is four but if it helps you can group it like this 8 minus 2 minus 2 and then plus you've got your imaginary part so you've got your negative 3i you've got your minus 4i and you've got your minus 6i if you need to group it like this for a while that's fine too it's just a little bit slower so again 8 minus 2 minus 2 is 4 so that's your real part then plus negative 3i minus 4i is negative 7i negative 7i minus 6i is negative 13i so let me just put minus 13 i and again it's written in standard form where your real part is first your 4 this is your a and then you have plus in this case you'll have negative 13 that represents your b and then you have i so this part right here is going to be your imaginary part so the real part comes first imaginary part is second all right let's look at some multiplication problems pretty simple to do this we're just going to use our foil method because we have the product of two binomials so we have the quantity 6 plus 6i times the quantity 4 plus 6i all i'm going to do here first term 6 times 4 is 24. my outer terms 6 times 6i is plus 36i make that a little better then plus 6i times 4 is plus 24i and then last 6i times 6i is plus 36i squared okay so what can we combine the 24 i can't do anything with that for right now we'll be able to do that with something within in a second but for right now nothing 36i plus 24i 36 plus 24 is 60. so this would be plus 60i and then 36 i squared remember i squared is defined to be negative one so i can go ahead and say this is plus 36 times negative one which is basically minus 36 or negative 36. so i can erase this and put minus 36 and now i can combine 24 negative 36 as negative 12. so i want the negative 12 out in front because that's the real part then plus i want the 60i in the back because this is the imaginary part so negative 12 plus 60i is your answer in standard form all right let's take a look at a longer one so we have the product of three binomials here so i'm going to use foil with the first two and then i'll take that result and i'll multiply it by the final binomial so we'll just start with these guys right here so what does that give us so 5 times negative 6 would be the first terms that's negative 30. the outer 5 times 6i is plus 30i the inner 3i times negative 6 is minus 18i the last 3i times 6i is plus 18i squared now we know i squared is negative 1. so let's go ahead and erase this and put a minus 18. so to combine things here negative 30 minus 18 is negative 48. so negative 48. then we know that 30i minus 18i is going to give us 12i okay so let's erase this so we're going to put this in parentheses and we'll drag it up and now it's times this guy negative 1 minus 3i okay so let's do the foil again so negative 48 times negative 1 is positive 48 then negative 48 times negative 3i is plus 48 times 3 is 144 and then times i then you'd have 12i times negative 1 which is minus 12i and then lastly you would have 12i times negative 3i that's minus 36i squared and make that a little better all right so let's wrap up the problem now do some simplifying we know that i squared is negative one so if i had a negative one times this negative here this would become positive okay so 48 plus 36 is 84. so this is 84 plus 144i minus 12i is 132 i so again our answer here in standard form is 84 plus 132 i all right so let's talk a little bit about dividing with complex numbers you need to be aware that again the definition of i is that it's equal to the square root of negative 1. so i have negative 7 over 5i this is really like saying i have negative 7 over 5 times the square root of negative 1. we know that a simplified radical expression does not contain a radical in the denominator so we want to rationalize this guy and so we can multiply both the numerator and denominator by the square root of negative one but typically what we'll do is just keep it in this format of i so i'll do i over i right so in the numerator i'll have negative seven i in the denominator 5 times i times i is 5 i squared we know i squared is negative 1 so this would basically be what negative 7 i over negative 5 negative over negative is positive so you end up with 7 i over 5. now it's okay that i is in the numerator because again you can have a radical the square root of negative 1 in the numerator you just don't want in the denominator because it's not considered simplified let's take a look at another one so we have 9 plus 4i over negative 5i again i've got i here in the denominator so to get rid of that i'm going to put i over i here and multiply by that so i've got to use my distributive property i times 9 is 9i then plus 4i times i is 4i squared and we know that's negative 1. so you can go ahead and just replace that and say this is negative okay anytime you get i squared just immediately replace it with negative 1 and see what that gives you because you don't want to have to worry about simplifying it later it might be something you just forget so then down here negative 5 times i times i i times i is i squared so the negative 5 times negative 1 right that's i squared would give me positive 5. so there's really nothing else i can do to simplify here you just be left with 9i minus 4 over 5. if you wanted to for the sake of writing this in standard form you could reorder this first you could break it up as 9 i over 5 then plus negative 4 over 5 and then you want the real part out in front so you would go ahead and say this is negative four-fifths that real number then plus your imaginary part which is going to be the 9i over 5. so again if you wanted to write this in standard form you would want to break it up like this you've got your real part of negative four-fifths plus your imaginary part of nine-fifths times i alright so the last concept we're going to go over is basically rationalizing with a binomial denominator that includes a complex number all right so we previously learned that conjugates were the sum and difference of the same two terms so something like the quantity a plus b times the quantity a minus b so you've got a and a you've got b and b different signs right plus and minus if we were to foil this out the middle two terms drop out you've got a squared minus b squared okay so nothing new there when you do this and you have i involved like a plus bi multiplied by a minus bi the sine is going to change it's because you're going to end up with an i squared and the i squared is negative 1 and so it flips the sign so first terms here a times a is a squared your outer a times negative bi is minus a b i your inner a times b i is plus a b i you can see those would cancel right it's they're basically opposites you've got minus a b i and plus a b i you can erase that you don't need it then we've got positive bi times negative bi we know that's a negative b times b is b squared and i times i is i squared okay we know i squared is negative one so if i replace this with negative one here again you can think about this as plus a negative here negative times negative is positive so this is going to be a squared plus b squared okay so that's where that comes from so we're going to use this part right here to help us simplify a two-term denominator when i is involved so let's look at an example we already know how to deal with so if we saw something like 8 over you have 3 minus the square root of 5 we should know at this point that if we have a binomial denominator here we can multiply the numerator and denominator by the conjugate of the denominator so i would just multiply by 3 plus the square root of 5 over 3 plus the square root of 5. right this is going to clear my radical from the denominator in the numerator 8 times 3 is 24 then plus you'd have 8 times square root of 5. in the denominator 3 times 3 is going to give me 9. so then the outer and inner would cancel you'd have 3 times square root of 5 and then negative 3 times square root of 5. so those would cancel and then lastly you would have minus we know the square root of 5 times square root of 5 would be 5. so let's finish this up we would have 24 plus 8 times square root of 5 over 9 minus 5 is 4. now you can simplify this further because you can factor out a 4 from the numerator so if a four got factored out inside you would have a six plus two times square root of five and this is over four i can cancel this with this and i'd be left with a simplified answer of six plus 2 times square root of 5. now we're going to use the same thought process for this example here so we have 9i over you have negative 5 minus 7i so i want to get rid of this guy right here because it represents a radical of square root of negative 1 in the denominator okay so what i'm going to do is i'm going to multiply the numerator and denominator by the complex conjugate so it's the same concept so i'm going to take this guy right here the negative 5 i'm going to take this guy right here the 7i i'm just going to flip the sign so it's negative it's going to become a positive right if it's subtraction it becomes addition if it's addition it becomes subtraction okay that's all you're doing so over here i have the same thing negative 5 plus 7i and what happens is when i multiply in the denominator i don't need to go through foil i just showed you that if you have a minus bi times a plus bi it's a squared plus b squared okay so a is negative 5. if i square that i get 25 then plus b is 7 okay and 7 only if i square 7 i get 49. and you can pause the video and verify that for yourself but that is what you get remember i times i is i squared and that represents negative 1. okay so that's why you get a different sign in the numerator 9i times negative 5 is negative 45 i then we have 9i times 7i that would be plus 63i squared okay let's see what we can do to simplify well we know i squared is negative 1. so i'm going to start out by just writing this as negative 63. and then minus 45 i and this is over 25 plus 49 is 74. now is there anything at this point that i could cancel so i would have to be able to pull something out from this and this that could cancel with this okay so there's really nothing else you can do but we do want to split this up and say this is negative 63 over 74 and then i'm going to say plus you have negative 45 over 74 times i again this part right here your negative 63 over 74 is the a or the real part and then this part right here this is your b right your negative 45 over 74 and then you have your i so you've got your real part and your imaginary part again the real part goes first that's how you put it in standard form so negative 63 over 74 plus negative 45 over 74 times i all right let's take a look at one more of these so we have 6 minus 6i over you have 2 plus 3i again the idea is i want to get rid of this i in the denominator because it represents a radical in the denominator so this is not currently simplified all i need to do is multiply the numerator and denominator by the complex conjugate so 2 and 3i just change the sign right now it's a positive or a plus so you could say this is going to be a negative or a minus okay so 2 minus 3i all right so now we just do our multiplication again just use the formula down here it's faster if i had 2 plus 3i multiplied by 2 minus 3i again it's a plus bi times a minus bi okay so it's a squared plus b squared okay 2 is a 2 squared is 4 then plus 3 is b 3 squared is not we know 4 plus 9 is 13 so let's just go ahead and write that so the denominator denominator's done we don't need to do anything else in the numerator we've got to use our foil okay so we've got 6 times 2 that's 12. the outer would be minus 18i the inner would be minus 12i so let's go ahead and combine that we know that will be minus 30i and then for my last i would have a negative 6i times a negative 3i we're going to go ahead and say that's a plus 6 times 3 is 18 and then we would have i times i which is i squared now we know that i squared is negative 1. so that would be negative 18. so we can basically say this is minus 18 and then 12 minus 18 is negative 6. so let's go ahead and erase that and that and let me erase this and say this is just negative 6 minus 30 i over 13. again to put this in standard form i'm going to break it up i'm going to say this is negative 6 over 13 then i'll put plus you'll have negative 30 over 13 and then times i okay so this is your real part this is the a this is the b part okay this is what's multiplying i but the whole thing the negative 30 over 13 times i is the imaginary part but again some books go back and forth on this so standard form i'm just going to say is negative 6 over 13 plus you have your negative 30 over 13 times i in this lesson we want to review simplifying powers of i so over the course of the last two lessons we talked about the imaginary unit i and we talked about complex numbers in general and how to perform operations with complex numbers so one more topic that comes up when you first start looking at imaginary units you're going to think about how to simplify powers of i so if you've got something like i raised to the power of 503 you are expected to write that in a simpler form okay so it's very easy to kind of perform this simplification you just need to know your rules of exponents and you need to know the first four so i to the first power i squared i cubed i to the fourth power so the first four powers of i okay with those two things the rules of exponents and the first four powers of i you can simplify any power of i that you'd want to all right so the first thing i have listed here is i to the power of 0 is equal to 1. i just put this here for reference sake we're not going to use it in the lesson a lot of students will ask is i to the power of 0 equal to 1 the answer to that is yes so any nonzero number raised to the power of 0 is always 1. okay when we think about the definition of i we normally get i squared is equal to negative 1 given to us so this guy right here and we know that i to the first power is just equal to i right any number raised to the power of 1 is just itself so that's where those two come from now when i get to i cubed i start using my rules for exponents so i break this up and i say it's equal to i squared times i okay you can say this is i to the first power here if you want to my rules of exponents if i multiply here i stays the same and i add 2 plus 1 to get me back to 3. okay so i haven't done anything illegal then i take the fact that i squared by definition is negative 1 and i just replace it so now i have negative 1 times i which is just negative i okay so right now we know that i the first power is just i i squared is negative 1 and i cubed is negative i now what about i to the fourth power we're going to break this up as i squared times i squared again rules of exponents i would stay the same 2 plus 2 is 4 so nothing illegal there and again we know that i squared by definition is negative 1. so we're just replacing it in each case negative one times negative one is positive one okay so we need to write this down in your notebook or whatever you're using to follow along i to the first power is i i squared is negative one i cubed is negative i and i to the fourth power is one so we just need to know those four okay that's all we need so let's look at our first example so suppose we see i to the power of 48 so the first thing you're going to look for if you're dealing with whole number exponents okay whole number exponent so if you have something like 48 which is a whole number the first thing you want to do is check for divisibility by four if it's divisible by four it's equal to one okay so 48 is divisible by 4. 48 divided by 4 is 12. so i can rewrite this using the power of power rule as i to the fourth power raised to the 12th power okay if i know that i to the fourth power by definition is 1 then i could say this is 1 to the 12th power which is 1. so if you have i raised to a whole number exponent that is divisible by 4 it simplifies to just 1. so if i had something like i to the power of 60 as an example we all know that 60 divided by 4 is 15. so i could write this as i to the fourth power raised to the 15th power again this is just 1 to the 15th power which is 1. 1 raised to any power is always 1 okay so it doesn't matter what you're raising it to once you have this i to the 4th power here being raised to something it's just 1 raised to whatever which is always going to give you 1. okay if you had something like i raised to the power of 200 as an example we know that 200 divided by 4 is 50. so again i could just say that this is i to the fourth power raised to the 50th power this is 1 raised to the 50th power which is just 1. okay so if you have i raised to a whole number exponent and that whole number happens to be divisible by four it's just equal to one okay you don't need to do all this you can just stop and say it's one now what if it's not divisible by four what do we do so suppose we saw i raised to the power of 61. well what you want to do is you want to look for the next number going down that's going to be divisible by 4 okay so in this case we know that if you're at 61 if i go down to 60 60 divisible by 4 60 divided by 4 is 15. so i'm just going to break this up and say this is i to the power of 60. let me make that a little better so again i to the power of 60 times i to the first power now we know i to the power of 60 is what it's 1 right i could write it as i to the power of 4 raised to the power of 15 this is 1. so i can replace this right here with a 1. that's gone don't even think about it anymore so it's 1 times i to the first power which is just i okay very very easy process all right what about i to the power of 202 again is it divisible by 4 no it's not right a number 4 to be divisible by 4 the final two digits in this case that would be two because there's a zero in front of the two has to form a number that's divisible by four two is not divisible by four right if i divide two by four i get point five so what i have to do is again look for a number going down that's divisible by 4. 201 wouldn't be but 200 is so i can break this up and say this is 1. it's i to the power of 200 times i squared okay we know that this guy is going to be a 1 right so we'll say 1 times i squared and we know that i squared by definition is negative 1. all right what about i raised to the power of 1003 okay we know that again i look at the final two digits of the number it's a three it's not divisible by four if that was a four it would be but we're not that high up so we need to go down to two which isn't and then to one which isn't and then to zero which is so the number one thousand is divisible by four so i could write this as i raised to the power of 1000 times i raised to the power of 3. we know this is 1. okay so 1 times i cubed again we know by definition i cubed is negative i okay so this is negative i all right what about i raised to the power of negative 22 don't get scared if you see a negative exponent remember if you get a negative exponent you take the reciprocal of the base okay so it would be one over i and you make the exponent positive so it's to the 22nd power so again i'm just using my same process here so i have 1 over when i think about i to the power of 22 is 22 divisible by 4 no is 21 no is 20 yes so i could write this as 1 over i raised to the power of 20 times i raised to the power of 2. okay so 1 over i know i raised to the power of 20 is 1 times i squared is negative 1. so times negative 1. so you basically have 1 over 1 times negative 1 is just negative 1 so 1 over negative 1 which is negative 1. all right so let's wrap up the lesson by looking at an example where we need to rationalize the denominator so we have i raised to the power of negative 119 and what i'm going to do here is i'm going to use my rule for again negative exponents so i'm going to say i have 1 over i raised to the power of 119. now just ask yourself the question is 119 divisible by 4 right just look at the last two digits so is 19 division by 4 no 18 no 17 no 16 yes so if you did that on a calculator if you did 116 divided by 4 you would get 29 okay so if you broke this up the long way you could say this is 1 over you could do i raised to the power of 116 again 116 there is going to be divisible by 4 and then times you have i cubed so at this point you could further break this up or you could just realize that this right here is 1 either way it doesn't matter let's just go through it the long way one last time so i'm going to go one over i'll go ahead and say this is i raised to the power of 4 and then this is raised to the power of 29 again 4 times 29 is just 116. okay so this right here is going to end up giving me 1. and then i'll say times i cubed okay so i know i cubed by definition is negative i so i can just say this is 1 over negative i now here's where you run into trouble okay and sometimes i forget this myself you want to make sure that when you have i in the denominator you take the step to rationalize the denominator because technically i is going to be the square root of negative one right we don't want radicals in a simplified denominator okay so what i want to do here just take the extra step and multiply this by i over i remember you can do this because i over i technically is 1 multiplying something by 1 does not change the value so this is going to give me 1. it's going to give me i over you could say you have negative let's just treat this as negative 1 times i times i is i squared okay so what is this now it's i over we'll say negative 1 times i squared by definition is negative 1. well negative 1 times negative 1 is 1 and i over 1 would just be i okay so after we rationalize the denominator our simplified result here is just going to be i in this lesson we want to review solving quadratic equations by factoring so in this section of the course we're going to review how to solve quadratic equations so for most of you that took lower level algebra courses you already understand that there are multiple methods to kind of achieve this task the easiest case scenario is when we can solve a quadratic equation using factoring okay but we cannot use this method every time so we've got to have other tools like completing the square and the quadratic formula so we'll talk about those other methods over the course of the next few videos in this lesson we want to focus on just solving quadratic equations by factoring so first and foremost let's explain standard form for a quadratic equation so in your textbook when you first start talking about quadratic equations you'll see something like this we have ax squared plus bx plus c is equal to zero and we basically restrict a the coefficient of x squared and we say it can't be equal to zero and the reason for that lies in the definition of a quadratic equation with a quadratic equation you have a variable like x that's squared and you don't have a variable that's raised to a higher power so you don't have like x cubed or x to the fourth power or anything like that so x squared the 2 is the highest exponential power on the variable when you work with a quadratic equation it's also known as a second degree equation for that reason right because again the highest exponential power is a two when we think about a polynomial if the highest power on the variable is a two it's known as a second degree polynomial when we look at these guys we could say a b and c could be any real number with that one exception that a is not equal to zero so something like five x squared plus two x plus three equals zero or something like let's say 3x squared plus you could have 0x plus 2 equals 0. and this guy right here this plus 0x i can get rid of that and just say this is plus 2 equals 0. okay so this is still a quadratic equation because i've still got my variable raised to the second power so it still fits that definition okay and similarly i could just put something like plus 5x equals zero i don't have to have a constant term that could just be zero the main thing here is to understand how to write this in standard form so this is in standard form this is in standard form and this is obviously in standard form here so let me kind of notate this this is standard 4. okay so you want it as a some coefficient of x squared plus b some coefficient of x plus c the constant is equal to zero okay zero is going to be on one side everything else is on the other in order to solve a quadratic equation by factoring we have to understand something known as the zero product property aka the zero factor property okay so those are the same thing this tells us that if two real numbers are multiplied together and the result is zero so something like a multiplied by b equals zero then essentially one of the following possibilities has to be true we could have that a is equal to zero and b does not equal zero so something like let's say a is zero and b is let's say two zero times two is obviously zero right zero times anything is zero it could also be true that a is not equal to zero and b is zero so let's say a is negative five and b is zero this still gives us zero lastly you could have that a is equal to zero and b is equal to zero so it could be we have zero times zero and that is obviously zero okay so this is the kind of math behind what we're going to do today this is in its simplest form we're going to look at it in a more complex form and to kind of see this let's first start out by working through this problem so we have x squared plus 4x equals x plus 28. so we want to solve this quadratic equation by factoring meaning we want to find the values for x that make it into a true statement okay so the first thing i want to do is i want to write it in standard form so my ax squared plus bx plus c is equal to 0. now it doesn't matter which side 0 is on but typically you'll see 0 on the right side and everything else on the left so let's just kind of follow that so to accomplish this i'm going to use my addition property of equality i'm going to subtract 28 away from each side of the equation i'm also going to subtract x away from each side of the equation so on the right side this is all going to cancel and just become 0. on the left side i have my x squared and then i have 4x minus x so that's plus 3x and then i have my minus 28 and of course this is equal to we already said the right side is 0. so now it's in the format of ax squared in this case a is just 1 plus bx in this case b is 3 plus c in this case c is negative 28 okay so this equals 0. all right so now that we have it in standard form we want to factor it okay we want to factor it so let's drag this up here and let's see how we can factor this if it's a trinomial we're trying to factor this into the product of two binomials this is our easiest case scenario so we have x squared here so this would be x and this would be x and i just need to work out what is this and what is this so give me two integers whose sum is 3 and whose product is negative 28. well for positive 28 it's only 1 times 28 or 4 times 7 okay if you think about 1 and 28 there's no way to play with the signs and make that work you can't get a sum of positive 3. but if you play with the signs with 7 and 4 you can make it work if i had a positive 7 and a negative 4 7 times negative 4 is negative 28 7 minus 4 is positive 3. okay so now that we have factored the left side of this guy what i want to do is think about again my zero product property if a times b equals zero then a can be zero and b cannot b could be zero and a cannot or a and b could both be zero well it's the same thing here you've got to realize that these guys are factors the quantity x plus seven is multiplying the quantity x minus 4. so these are factors these are factors okay so make sure you understand that so if i find a value for x that when plugged in here and here let's just say makes this guy into 0 well zero times whatever this is will give me zero same thing over here if i find a value for x that makes this guy into zero zero times whatever this is would also give me zero so essentially i just wanna set each factor equal to zero and i want to solve and find out what that value is those values will be solutions for my equation okay so i'm going to start with x plus 7 that's one of my factors and i'm just going to set that equal to 0. very easy to solve this i just subtract 7 away from each side of the equation i get that x is equal to negative 7. so let me just kind of put this over here we'll say that x is equal to negative 7. that's one of the solutions i'll put a comma since i have a second solution the other scenario is that x minus 4 equals 0. so x minus 4 equals 0. so i can add 4 to each side of the equation and i'll get that x is equal to 4. okay so let me put a 4 right there now when we check this i want to show you in this kind of factored form what happens let's start out by checking x equals negative 7. if i plugged in a negative 7 for x there what happens you're going to have a negative 7 plus a 7 and then over here you're going to have a negative 7 minus a 4 and then of course this is equal to 0. negative 7 plus 7 is 0. so this is where the magic happens you have 0 times doesn't matter what this is over here it's going to work right because 0 times anything is just 0. negative 7 minus 4 is negative 11 0 times negative 11 is 0. so you get 0 equals 0. so we can go ahead and check this guy off and say it does work right negative 7 is the solution and you can also check that in the original equation if you'd like i'm not going to go through that just for the sake of time but you can pause the video and plug in a negative 7 here here and here and you'll see that you do get the same value on the left as you get on the right let's also check 4 in the factored form so if i plugged in a 4 there and there what would i get well on the left side i get 4 plus 7 which is 11. on the right side i get 4 minus 4 which is 0. 11 times 0 is again 0. so you get 0 equals 0. so this results in a true statement so this one works out as well so our two solutions here are that x equals negative 7 or x equals 4. all right let's take a look at another one so we have 77x is equal to you've got negative 15x squared minus 3x minus 105. so again i want to write this in standard form so we want ax squared plus bx plus c is equal to 0. so essentially i want one side of the equation to be 0 and to accomplish this i can just subtract 77x away from each side of the equation on the left i would just have 0. so 0 is equal to you'd have your negative 15x squared negative 3x minus 77x is minus 80x and then you'd have your minus 105. so let's scroll down a little bit now there's two things here most people like their zero to be on the right there's nothing wrong with just kind of flipping this around and saying this is negative 15x squared minus 80x minus 105 is equal to zero okay that's perfectly legal the other thing is you have negatives for each term here if you don't want to deal with that when you're factoring you can use your multiplication property of equality here you can multiply both sides of the equation by negative one multiplying zero by negative one will still be zero so if i multiply zero by negative one i still have zero if i multiply each term here by negative one i'm gonna just change the sign so i would have 15x squared plus 80x plus 105 okay so now this will be a little bit easier to factor so now that we've written in standard form we're ready to factor the left-hand side here so the first thing i notice is that everything is divisible by 5 on the left so we can go ahead and pull that out that would give me a 3x squared plus a 16x plus 21 and this is equal to 0. all right so we need to keep factoring here so i would have 5 times i'll set up my parentheses i know that this number here is prime this 3 that's the coefficient for x squared i'll do 3x and x and now i just need to work out these two parts right here okay the outer and the inner have to sum to 16x and the last have to give me a product of 21. so i need two of those integers that's going to fit the criteria there well if i think about 21 this kind of final term here it's only 1 times 21 or 7 times 3. so there's no way to make it work with 1 and 21 right 1 and 21 no matter what i do it would just be too large of a sum but with 7 and 3 i might be able to make that work i can't put my 7 here because 7 times 3 would be 21 that's already larger than 16. so that doesn't work i'd have to put a 7 here and a 3 here so does that work well yeah it does because my outer here would be 9x and my inner here would be 7x 9x plus 7x is 16x so this is the correct factorization all right so once we have it in a factored form remember what we did last time we took each factor and we set it equal to zero well here we have a factor of five we don't need to set that equal to zero because that would just be nonsense so we would kind of be more specific here and say we're going to set each factor with a variable in it equal to 0 and we're going to solve so let's go ahead and take this left one first so 3x plus 7 is equal to 0 we would subtract seven away from each side of the equation we'll get that three x is equal to negative seven we will divide both sides by three and we'll find that x is equal to negative seven thirds okay so let me just kind of write that over here we'll say that x equals negative seven thirds and then we'll have another value for x that works so let's work on this guy over here on the right so now we have x plus three we're going to set that equal to zero and we're just going to subtract 3 away from each side of the equation we'll get that x is equal to negative 3. okay so x could be equal to negative 7 thirds or it could also be equal to negative 3. each of those would work as a solution to the original equation in the interest of time i'm not going to check it but again you can pause the video go back to the original equation plug in a negative seven-thirds for each occurrence of x and also a negative three for each occurrence of x and verify that the left and the right sides are equal all right let's take a look at another so we have negative 1 plus 7x is equal to negative 2x squared minus 4. so again i want this in standard form so ax squared ax squared plus bx plus c is equal to zero okay it helps if you write it out and you can kind of look at and see what you need to do so i am going to basically follow this format and have zero on the right hand side so i'm just going to add 2x squared to both sides of the equation and i'm going to add 4 to both sides of the equation so this is going to cancel so the right hand side is just going to be a 0. okay on the left-hand side i'll have my 2x squared and then you have 7x so plus 7x and you've got negative 1 plus 4 which is going to be plus 3 and again this is equal to 0. so now that i have it in standard form i can go ahead and factor the left hand side and this guy again is pretty easy to factor because this part right here is a prime number so i'll go ahead and set this up we'll have our 2x and our x so now i just need to work out this and this right again we want this times this plus this times this to give me this middle term this 7x so what are the factors of 3 well it's only 1 and 3. so it's not very hard here you can either put plus 3 here and plus 1 here that obviously doesn't work you'd end up with 2x plus 3x which is 5x so it has to be true if it's factorable that a 3 would go here and a 1 would go here and that does work right 2x times 3 is 6x and then 1 times x is 1x so this ends up being 7x which is the correct middle term when you do that sum okay so we factored this guy and now we're going to set it equal to zero and again we're just going to use our zero product property so this guy right here we're going to set this equal to zero let's start with that so two x plus one is equal to zero let's subtract one away from each side of the equation we will get that two x is equal to negative one we will take this two x is equal to negative one and divide both sides by two we'll get that x is equal to negative one-half okay so let's say x equals negative one-half and then the other solution comes from this guy on the right so this factor here this x plus three if i set that equal to 0 so x plus 3 equals 0 subtract 3 away from each side of the equation you get x is equal to negative 3. okay so negative 3 goes here as well so my two solutions here we have that x equals negative one half and then also x can be negative three all right let's look at one that's not a quadratic equation and the reason it's not a quadratic equation is because we have x raised to the third power here so that would violate our definition but it's important to know that you can use this technique for any equation if it's factored okay so it doesn't have to be a quadratic equation so we have x cubed minus 3x squared minus 4x plus 12 equals zero and we know if we have a four term polynomial we want to try to factor this using the grouping method so from the first group i can see that i can pull out an x squared that would give me an x minus a 3. from this group here to match this i would pull out a negative 4 and that would give me an x minus a 3 and this equals 0. of course my common binomial factor here is the x minus 3 okay so this ends up giving me my x minus 3 and then times the quantity x squared minus 4 and this equals 0. so at this point we're not actually done factoring because the x squared minus 4 parts is the difference of two squares so we want to factor this all the way so let's put x minus 3 and then times this guy would factor into x plus 2 times x minus 2 and again this equals 0. so all we're doing now is we're taking each factor and setting it equal to 0. so x minus three is equal to zero we add three to each side of the equation we get x is equal to three okay so that's one solution so x is equal to three another solution we have x plus 2 is equal to 0. subtract 2 away from each side of the equation this cancels you get x is equal to negative 2. so that's another solution and then a lot of you can guess that the final solution is x equals 2 but for the sake of completeness let's just do it so x minus 2 equals 0 add 2 to each side of the equation this cancels you get x is equal to 2. so that's our last solution so here x can be 3 it can be negative 2 and it can be 2. in this lesson we want to review the square root property so in our last lesson we reviewed how to solve a quadratic equation using factoring although factoring is the simplest method that we can use to solve a quadratic equation it's not going to work in every scenario so we need other tools that will basically work for every quadratic equation and in this lesson we're going to lay the groundwork for our methods known as completing the square and also the quadratic formula so we're going to start out today by talking about the square root property so essentially if we have a variable like x and it's being squared and it's set equal to some number k then we can solve this equation for x and say that x is equal to the square root of k or x is equal to the negative square root of k so it's very important to understand why we have these two square roots here right the principal and the negative so let's say we had something like x squared is equal to just the number not well if i want to solve this guy for x again we're always looking to undo what's being done to x here x is being squared so the opposite operation of squaring something is taking the square root so i'm going to take the square root of each side of the equation and if i don't go plus or minus here i'm not going to get my full solution we know that this index here of 2 will cancel with this exponent of 2 and i get that x is equal to the principal square root of 9 is just 3. now think about this format did i get the correct full solution if x squared is equal to 9 well 3 squared is equal to 9 but also negative 3 squared is equal to 9. so in order to get the full solution here i've got to have plus or minus the square root of 9. so this leads to x equals 3 or x equals negative 3. again 3 squared is 9 and also negative 3 squared is 9. so because of the squaring operation and the fact that it takes a negative and creates a positive and also takes the positive and gives us a positive we've got to include the plus or minus symbol when we take the square root over here on the right hand side let's take a look at an example so suppose we had x squared equals 121. so let me rewrite the problem over here so we have x squared is equal to 121. again if i want to solve this guy for x i would take the square root of this side let me kind of make that a little better and i would go plus or minus the square root of this side okay so we know that this guy right here is just going to be x this is equal to plus or minus the square root of 121 is 11. so we get that x equals 11 or x equals negative 11. okay so you can leave it in this format this is more compact or you can write it like this x equals 11 comma negative 11 or you could use your or keyword you could say that x equals 11 or x equals negative 11. okay all of these are kind of different ways to notate the same thing but when you work with these equations you don't need to go through and take the square root each time you can just use your property right i could have started this by just saying okay i have x squared is equal to some number let's just call it k so this leads to x is equal to the square root of k or x is equal to the negative square root of k so if i have this format i can say x is equal to the square root of 121 or x is equal to the negative square root of 121. again this would just lead to x is equal to 11 and negative 11. okay because the principal squared of 121 is 11 and again the negative squared of 121 is negative 11. all right let's look at another so we have x squared equals 54. so again if i want to solve this for x again i can use my property i can say that x is equal to plus or minus the square root of 54. okay so what is 54 it's not a perfect square so we can break this down 54 is basically 27 times 2 we know 27 is 9 times 3 9 is a perfect square because it's 3 times 3 all right so you have kind of those prime factors there so i can say that x is going to be equal to plus or minus the square root of 9 is 3. so go ahead and pull that out you would have 3 times the square root of 3 times 2 which is 6. and essentially you can leave your answer like that that would be perfectly acceptable or again you could write it like this you could say that x is equal to 3 times square root of 6 and then comma negative 3 times square root of 6. all right let's look at one that's a little bit more challenging so we have 2x squared plus 10 is equal to 152. so how do i solve this for my variable x again just like when we worked with any other type of equation we want to start by getting the variable term with x involved isolated on one side so how do i do that i just want to subtract 10 away from each side of the equation so i would have that 2x squared is equal to 152 minus 10 is 142. now again i want to get the x squared by itself so i need to divide both sides of the equation by 2 so i can isolate the x squared part so we know that this is going to cancel and i'm going to have x squared is equal to 142 divided by 2 is 71 and now that i have it in the format of x squared equals a number again i can just use my property i can say that x is going to be equal to the principal and negative square root of 71. now 71 is going to be a prime number so i can't really do anything else with this this is just how i'm going to list my solution or again i can say that x is equal to the square root of 71 and then comma the negative square root of 71. again any kind of way that you want to notate that all right let's look at another one so we have 5x squared plus 9 is equal to negative 7. so again i want to isolate x i want to isolate x and to start this off i want to isolate the term with x involved okay so i'm just going to subtract 9 away from each side of the equation this is going to give me that 5 x squared is equal to negative 7 minus 9 is negative 16. again now i want to isolate the x squared so i'm going to divide both sides by 5 and this is going to give me that x squared is equal to you've got negative 16 fifths okay so now we have x squared equals some number so now i'm going to use my square root property let me kind of divide this up we're going to say that x is equal to plus or minus the square root of negative 16 fifths now there's a few things here we have the square root of a negative so we're going to use our imaginary unit i to kind of simplify this so i'm going to say that x is equal to plus or minus again we know how to use i at this point this guy right here is just going to come out as i so i'm going to put that right here and then times i'm going to break up the square root into the square root of 16 over the square root of 5 okay now i can simplify this further because the square root of 16 is 4. so let me scroll down a little bit so we're going to say that we have x is equal to plus or minus again the square root of 16 is 4 so i'd have 4 times i so 4i over the square root of 5. now we're not done yet because again we have to rationalize the denominator here so i can do that pretty quickly i would say the square root of 5 over the square root of 5 is what i need to multiply by in the denominator square root of 5 times square root of 5 is 5. so we'll just put x is equal to plus or minus in my denominator i have 5 in my numerator i'm going to have 4 i times square root of 5 okay so this is basically your answer x equals positive or negative 4i times square root of 5 over 5. and again you can break that up into two different solutions you have 4i times square root of 5 over 5 or you have negative 4i times square root of 5 over 5. all right let's take a look at another one so suppose we had the quantity x minus 7 that's being squared and it's equal to 64. well we can extend our property to this scenario as well in fact this is what we're going to end up doing when we use our completing the square method so what i'm going to do let me kind of show this the long way so we have x minus 7 this quantity squared is equal to 64. i'm going to take the square root of this side and we all know that what's going to happen is this 2 right here is going to cancel with this 2 right here so on the left side i'm just left with my x minus 7. but on the right side when i take the square root i've got to go plus or minus the square root of 64. okay so this equals the principal square root of 64 is 8 and the negative square root of 64 is negative 8. so you could say plus or minus 8 there now we're not done so in this particular case i've got to kind of break this up and look at both solutions so i've got to say that x minus 7 could be equal to positive 8 then or x minus 7 could be equal to negative 8 okay so i'm going to solve each equation here so i add 7 to each side of the equation i get x is equal to 15. then or i add 7 to each side of the equation and i get x is equal to negative 1. so let's erase everything we're going to go back up and i'm going to check these two solutions and we're going to think about why they work so again we get that x equals 15 or x equals negative 1 and we can write that in a more compact manner by just saying x equals 15 comma negative 1 like that let's think about why these solutions work if i started out with 15 then i would have 15 minus 7 inside of parentheses being squared and this equals 64. essentially what i'm saying is 15 minus 7 is 8 i'm saying that positive 8 squared is equal to 64. so another thing would be that i need negative 8 squared to give me 64 as well and that's what negative 1 is going to give me if i put negative 1 in there what's going to happen is negative 1 minus 7 is negative 8 so negative 8 squared is also 64. so again that's why you have those two solutions you don't know whether it's 8 that's being squared that gives you 64 or whether it's negative 8 that's being squared that gives you 64. so that's why you've got to say that x minus 7 could be equal to the positive or the negative square root of 64. okay that's why you have to have those two scenarios to give you all of the possible solutions all right let's look at another one so suppose we have the quantity 2x minus 1 squared is equal to 18. so same thing here i'm going to take the square root of the left side so the square root of this side and again we know that this 2 cancels with this 2. then on the right i've got to do the positive and negative square root of 18. okay so we know on the left that we're just going to have our 2x minus 1 this is equal to on the right if i think about 18 i know it's 9 times 2. 9 is a perfect square it's 3 times 3. so i can pull that 9 out and say we have plus or minus 3 times the square root of what's left which would be 2. okay so now i've got two different solutions that i've got to work out i've got 2x minus 1 is equal to 3 times square root of 2 and then i've got 2x minus 1 is equal to the negative of 3 times square root of 2. so let's set those up and solve them individually so i have 2 x minus 1 is equal to 3 times square root of 2 or i've got that 2x minus 1 is equal to negative 3 times square root of 2. okay so let's solve these guys so over here i've got where i can add 1 to each side of the equation so that's going to cancel i'll have 2x is equal to 3 times square root of 2 plus 1. if i want to solve for x i just divide both sides by 2 and i'm going to get that x is equal to 3 times square root of 2 plus 1 over 2. and i really can't simplify that any further so i just leave it in that format so over here if i add 1 to each side this cancels i'm going to have that 2x is equal to negative 3 times square root of two plus one then what i can do is divide both sides of the equation by two and i get that x is equal to negative three times square root of two plus one over two so when we notate this we can just put or between these two because x could be three times square root of two plus one over two or x could be negative three times square root of two plus one over two the other method would be to use a shorthand so i could say x is equal to plus or minus 3 times square root of 2 plus 1 over 2. okay so both of these ways to notate this would be acceptable and i just want to show you real quick another way to do this so let's erase all this i'm going to show you kind of a faster method so you don't need to split this up into two different equations you can solve this without doing that so from here i can just add one to each side of the equation i would say this is 2x is equal to plus or minus 3 times square root of 2 plus 1. so now if i want x by itself i can just divide both sides by 2 and what do i get i get that x is equal to plus or minus 3 times square root of 2 plus 1 all over 2. so this is a more compact form and it's also a quicker way to get the solution right again x is equal to the positive or negative of 3 times square root of 2 plus 1 over 2. so it's two solutions x is going to give me 3 times square root of 2 plus 1 over 2 and x is also going to give me the negative 3 times square root of 2 plus 1 over 2. all right let's take a look at one more of these and this problem is going to lead us into our next lesson where we talk about completing the square so here we have 16x squared minus 56x plus 49 equals negative 27. how can i solve this for x this currently doesn't look like x squared equals a number or even kind of something like this where we have the quantity ax plus b squared is equal to some number k right we know how to solve this one and we know how to solve this one but we don't know how to solve this one yet so essentially what i want to think about is the fact that the left side of this guy is a perfect square trinomial when we have a perfect square trinomial we can factor it into a binomial squared okay so think about the fact that you could factor the left side here into 4x minus 7 that quantity squared okay so if we set this equal to negative 27 now we have something that we can solve right so essentially what i'm going to do here is i'm just going to take the square root of this side and i'm going to just kind of slide this down i'm going to take the positive or negative square root of this side okay so we know that this cancels with this and i have 4x minus 7 on the left and this is equal to plus or minus the square root of negative 27. now to simplify here i can pull this negative part out and let's go ahead and say this is 4x minus 7 is equal to plus or minus if i pull that out i get i if i think about 27 i know it's 9 times 3 9 is a perfect square right so i can pull the square root of 9 out as 3 so i'll say this is the 3i right there then times the square root of 3. all right so now we can solve this guy for x and again we can do it in a quicker way by just not splitting this up so i'm just going to add 7 to each side of the equation here and i'll say this is 4x is equal to plus or minus you've got your 3i times square root of 3 plus 7 and then to finish this up and to solve for x i'm going to divide both sides of the equation by 4 and that's going to give me a final answer here of x is equal to plus or minus 3i times square root of 3 plus 7 all over 4 okay so again this is two solutions we've got x equals 3i times square root of 3 plus 7 over 4 and then we've got x equals negative 3i times square root of 3 plus 7 over 4. in this lesson we want to review solving quadratic equations by completing the square so in our last lesson we talked about the square root property we said that if we had a variable like x that was squared and it was set equal to some number let's just call it k then our solutions for this equation we could say that x is equal to the principal square root of k where x is equal to the negative square root of k and we can abbreviate this and say that x is equal to plus or minus the square root of k remember when you write it like this this is two solutions the principal square root of k and then also the negative square root of k and remember why we have to kind of set it up this way if i had something like let's say x squared and let's say this was equal to the number 16. well if i want to solve this for x if i take the square root of this side to undo the squaring operation well then over here i've got to put plus or minus the square root of 16. if i just took the principal square root of 16 i wouldn't get all the answers okay and if we go through and kind of cancel here we know that this index would cancel with this exponent that isolates x on the left side on the right side i have plus or minus the square root of 16 which is plus or minus 4. now think about the fact that in the original equation if i kind of erase all this you had x squared equals 16. so 4 squared is 16 but also negative 4 squared is 16. so when you go through and you solve these things make sure that on the side over here with your number part you go plus or minus and then the square root of that okay that's going to give you all of your solutions again in this case you just get that x is equal to plus or minus the square root of 16 is 4. so let's look at two examples real quick and then we'll go into completing the square so x squared equals negative 36. so we have some variable x that's squared and it's equal to some number negative 36. so using the square root property i know that this would just be x is equal to the positive or negative square root of negative 36 again where does this come from well if i have x squared equals negative 36 i can take the square root of this side so that i can get x by itself but then over here when i take the square root of this number negative 36 i've got to go plus or minus the square root of that guy to get all my solutions so that's how i get plus or minus the square root of negative 36 but of course we can kind of simplify this let me erase this real fast we can simplify this we know that this guy right here if i have a negative under a square root symbol i can pull that out as i okay so i can say this is x is equal to plus or minus i times the square root of 36 the square root of 36 is 6 so i can say x is equal to plus or minus 6 i now you can check this if you want plug in a 6i for x and square it then plug in a negative 6i for x and square in each case you will get negative 36. now let's talk about this other scenario this quantity x minus 5 squared equals 29. so this is what we want to focus on today because this part right here is what we need to understand we can also use the square root property there so if i had the quantity x minus 5 and it's being squared and it's equal to 29 well what happens is i can take the square root of this side and then we kind of slide this down i can go plus or minus the square root of this side it's the same thing i'm taking the square root over here i need the plus or the minus so if i write this with a visible index of a 2 this cancels with this and i'm left with just the x minus 5 that's what was being squared and this equals plus or minus the square root of 20 knot okay so i'm going to split this up into two different equations so i would have that x minus 5 is equal to the square root of 29 then or x minus 5 is equal to the negative square root of 29. now if you wanted to you could also solve this just kind of by doing this you can add 5 to each side of the equation there that's fine too in this particular case because you can't simplify the square root of 29 but for right now i'm just going to do this the long way so over here i'm going to add 5 to each side of the equation i would get that x is equal to i'll just put my 5 in the front and then plus the square root of 29 and then or over here i'm going to add 5 to each side of the equation i'm going to say x is equal to 5 minus the square root of 29. so obviously if you kind of look at it this way you can say okay the only difference between this is the positive or principal squared of 29 and this one's the negative square root of 29 so you can say this is x equals 5 plus or minus the square root of 29. this is the most compact way to write it this is also correct or again you could write x equals 5 plus the square root of 29 comma x equals 5 minus the square root of 29. any way you want to notate it it's all the same thing all right so let's talk a little bit about completing the square if you've never done this before the process is very tedious and you end up working with a lot of fractions and just a lot of people really don't like completing the square it's not that it's hard it's just a lot of like kind of scratch work that you have to do so it's something you do need to understand but in the next lesson when we talk about the quadratic formula you'll be able to solve any quadratic equation that you want without much effort okay so this is just something you need to understand for your kind of future in math so completing the square so the first two steps can kind of go back and forth based on the situation you're in and i'll explain that as we get into the examples so the first step is to make sure the squared term has a coefficient of 1. so if i had something like let's say 6x squared plus let's say x minus 2 equals 0. in this particular case i don't have a coefficient of 1 on the squared variable okay i have a 6. so what can i do remember i can multiply or divide both sides of the equation by the same non-zero number and not change the solution so i can legally divide both sides of the equation by 6 and i can transform this into x squared plus you could say 1 6 x and then 2 over 6 is 1 3 so minus 1 3 is equal to zero okay so in some cases you'll do that first in other cases you'll do it second all right so the second step or in some cases it could be the first you're going to move all variable terms to one side and the constant to the other so let's say i had something like x squared plus 3x minus 7 equals 0. what i want is i want this part right here on one side by itself and i want the number on the other side so i would just add 7 to both sides of the equation okay that way over here i would say this is equal to 7. all right now this part right here is where it gets a little bit confusing and we'll look at this in the example so we're going to multiply the coefficient of the x to the first power term by one-half then square and some people say divide by 2 if you divide by 2 or you multiply by half it's the same thing okay so however you want to remember that but remember you're going to take it and you're going to cut it in half and then square it okay and we'll see that in a minute when we look at an example all right then we're going to add the result to both sides of the equation this will create a perfect square trinomial on one side of the equation so a perfect square trinomial on one side of the equation so in case you forgot a perfect square trinomial is a trinomial that we can factor into a binomial squared so something like x squared plus 2 x y plus y squared we previously learned in special factoring that this factors into the quantity x plus y squared we want to put it in this format so that we can use our square root property okay here i have that we're going to factor the perfect square trinomial into a binomial squared okay and then you're just going to solve the equation using the square root property so it's a lot of steps that i gave you you probably want to go back and just write all the steps down and we're just going to look at a lot of examples and this is one thing that you just have to practice and after you get enough practice on it it becomes very very easy all right let's go ahead and look at the first example this one's very very easy i kind of set everything up for you already so we have x squared minus 13x equals 14. okay so the first two steps that i gave you one of them was to make sure that the coefficient on the squared variable is a one you already have that in this case this is a one the second step is to move all the variable terms to one side and all the numbers to the other here you have a variable term x squared and a variable term negative 13x on one side and you have a constant 14 on the other so you don't need to do any of the first two steps here they're already done for you okay so we're going to move into the more complicated step which is to basically complete the square so again what am i trying to do here i'm trying to make the left side this x squared minus 13x into a perfect square trinomial so i want something that will factor into a binomial squared that's going to allow me to use my square root property and get a solution for x so how can we do that again we want to look at the coefficient for the x to the first power we want to cut it in half and then we want to square it so if i took negative 13 and i multiplied it by a half and then i squared that what would i get you would get negative 13 halves squared negative 13 squared is 169. 2 squared is 4. okay so i'm going to add this value to the left hand side but to make it legal i've also got to add it to the right hand side so let's go ahead and set that up so on the left my equation will become x squared minus 13x plus 169 fourths is equal to again to make it legal i've got to do it to the right side so 14 plus 169 fourths okay so let's scroll down get some room going all right so the first thing i'm going to do now is just factor the left side of this so this is going to factor into x minus 13 halves that quantity squared okay then this equals here comes a part that's pretty tedious you've gotta simplify over here and you need to get a common denominator so sometimes this can take a little bit of time again that's why students don't like this method because it's just very very tedious so let me kind of move this down so plus 169 fourths i'm going to multiply this by 4 over 4 so i can have a common denominator 14 times 4 is 56 and if i added 56 and 169 i get 225. so this right side here would be 225 fourths okay so let's put 225 fourths and now i have it into a format where i can solve using the square root property you've got something over here that's squared this quantity x minus 13 halves that's squared and it's just equal to some number well just like we did earlier in the lesson if i take the square root of this side remember this has a index of 2 here this cancels with this and i've got x minus 13 halves but again you got to remember this number here on the right i've got to go plus or minus the square root of this okay so this equals you've got your plus or minus you can split this up legally so you can say the square root of 225 which is 15 over the square root of 4 which is 2. okay so i've got my two different equations to work on you're going to have x minus 13 halves is equal to 15 halves and you've got x minus 13 halves is equal to negative 15 halves okay so let's set those up so x minus 13 halves is equal to 15 halves and you could do or x minus 13 halves is equal to negative 15 halves okay so let's solve these guys and we'll have our solution okay so all i'm going to do here is just add 13 halves to each side of the equation that's going to give me that x is equal to 15 plus 13 would be 28 28 over 2 is 14 okay so that's one solution my other solution if i add 13 halves to both sides of the equation what am i going to get well negative 15 plus 13 is negative 2 negative 2 over 2 is negative 1 so then x would be equal to negative 1. so there's your two solutions you get that x equals 14 or x equals negative 1. you could leave it like this of course you could write it as x equals 14 comma negative 1. again many different ways to kind of notate this all right let's kind of turn the heat up just a little bit and look at one that's a little bit harder so we have x squared plus 18 x plus 85 equals zero so i still have a coefficient on x squared of one so i can kind of skip that step but in this particular case i have this constant term on the left hand side right i'm adding 85. so i want to get rid of that before i do anything else so i'm just going to subtract 85 away from each side of the equation again you want all the variable terms on one side all the numbers on the other so i'm going to get x squared plus 18x is equal to negative 85. let me kind of move this down all right so i want to complete the square on the left hand side again i'm trying to get a perfect square trinomial over here i want this to end up being a binomial squared okay so how can i do that again all i want to do is look at the coefficient of x to the first power in this case it's 18. so what is one half of 18 squared again take 18 multiply it by half or as i say cut it in half and then square it 18 times a half or 18 divided by 2 is 9 9 squared is 81. so this is what i need to add to both sides of the equation to complete the square so let me kind of make some room here so i add 81 on the left side that completes my square this equals to make it legal i've got to add 81 on the right side okay on the left i can factor this now as x plus 9 that quantity squared on the right negative 85 plus 81 is negative 4. okay so now i can solve this using my square root property so i can take square root of this side and then again you've got to do plus or minus the square root of this side okay so this quantity here x plus 9 is what we're going to have on the left this equals plus or minus the square root of negative 4. we know that the square root of negative 4 could be written as 2i so let's go ahead and write this as x plus 9 is equal to plus or minus 2i okay nice and easy from here i'm just going to go ahead and solve the two equations that i would get so let's set this up and say that x plus 9 is equal to 2i or x plus 9 is equal to negative 2i okay now in this particular case because you really can't do anything you'll see what i mean by that you really can't do anything once you subtract 9 away from each side of the equation you can kind of just write it in a compact form so let me do it as or and then i'll kind of bring it back together so i'm going to subtract 9 away from each side i'm going to say x is equal to i'm going to write my negative 9 first and then plus my 2i then or over here i'm going to subtract 9 away from each side of the equation so x is equal to again negative 9 now minus 2i so the only difference between these two solutions is the plus here and the minus here so when this happens you can use that more compact form that's typically what you're going to see so you can just say that x is equal to negative 9 which is the same in each case then plus or minus the 2i okay this is just more convenient to do it that way but this is fine too if you want to notate it this way perfectly valid they say the same thing they mean the same thing don't get caught up on the notation all right let's look at a harder example so we have negative 105 plus 23x equals negative 10 minus 6x squared so in this particular case you kind of have a mess right you have terms all over the place you definitely are not going to have a 1 as the coefficient for x squared so if you get something that looks like this the first thing you need to do is clean it up so what i would say is go to your second step first which is to put all the variable terms on one side all the numbers on the other so what i'm going to do i like my variables to be on the left i'm going to add 6x squared to both sides of the equation and that's going to cancel this over here then what i'm going to do is i'm going to add 105 to each side of the equation and that's going to cancel that over there so on the left i'm going to have my 6x squared plus my 23x and this is equal to on the right negative 10 plus 105 is 95. so from here i have all the variable terms on one side and a number on the other don't forget to make this coefficient into one if you don't do that you're not going to get the right answer a lot of students they kind of set this up and then they start the process and they go okay 23 cut it in half and square it no okay don't do that you need to make sure that this coefficient here is a one so divide both sides of the equation by 6. again that's legal so this cancels with this i now have x squared plus i have 23 6 times x and this is equal to 95 over 6. okay all right so what do i want to do now i'm going to complete the square on the left side so that 23 6 that i'm working with i want to cut that in half so multiply by half square the result okay so you're going to end up with 23 12. you're going to square that we know that 12 squared is 144 that's easy but does anybody know what 23 squared is well it happens to be 529 okay so this is the value here that you're going to add to both sides of the equation so that on the left side you end up with a perfect square trinomial so on the left i've got my x squared plus my 23 over 6 times x plus my 529 over 144 and this equals on the right i'm going to have 95 over 6 and then let me kind of just drag this down so it's going to be plus this value okay and i'll scooch it out so i can have room for my common denominator all right so on the left side i can factor this into a binomial squared so this is x plus this guy is going to be 23 12 and this is squared and it's going to be equal to over here to get a common denominator i need to multiply this by 144 divided by 6 is 24. so this would be multiplied by 24 over 24. so 95 times 24 is 2280 and then 2280 plus 529 is 2809. so this would be 2809 over 144. again this is why students don't like completing the square because you end up with just kind of a big mess like this these are kind of larger numbers to work with i know you can use a calculator but still when you see the quadratic formula we'll see how much easier that is all right so we can solve this using our square root property square root of this side plus or minus square root of this side again don't forget this so over here i have x plus 23 12 is equal to you've got plus or minus the square root of 2809 is 53 square root of 144 is 12. okay so let's scroll down and get some room going so you're going to have that x plus 23 12 is equal to 53 12 or let me kind of make that two a little bit better you'll have that x plus twenty-three twelfths is equal to negative fifty-three twelfths okay let me make that equal sound a little better okay so let me scroll down and get some room going so we can see what's going on on the left equation here let's subtract 23 12 away from each side of the equation so that cancels you've got x is equal to 53 minus 23 is 30 so you'd have 30 over 12 30 is divisible by 6 and so is 12 30 divided by 6 is 5 12 divided by 6 is 2. so 5 halves would be one solution for x another solution if i go over here if i subtract 23 12 away from each side of the equation then this cancels you're going to get that x is equal to i have negative 53 minus another 23 that would give me negative 76. so let me write this down here and i'm just going to put or instead of a comma so x equals you've got negative 76 over 12. i know that 76 divided by 4 is 19. so i could write this as negative 19 and 12 divided by 4 is 3. so i have two solutions x equals 5 halves or x equals negative 19 thirds all right let's take a look at another one so we have 4x squared plus 8 equals negative 1 plus 2x squared minus 7x again if it's a mess like this go ahead and do the second step first clean everything up put all the variable terms on one side all the numbers on the other you can see that you have like terms here and here so they're on two different sides of the equation so you got to clean that up first before you can really do anything so let's subtract 2x squared away from each side and let's subtract 8 away from each side and let's add 7x to each side okay so that cancelled and that cancels and on the left i'm going to have negative 2x squared plus 4x squared which is 2x squared and then plus 7x is equal to negative 1 minus 8 is negative 9. okay so from here i can now think about the coefficient of x squared we've got to make that a 1. let's scroll down and i'm going to divide each side by 2. okay make sure you do this step again it's very crucial so you have your x squared plus your 7 halves x is equal to negative 9 halves okay so now i want to complete the square on the left-hand side so what do i want to do i want to take 7 halves the coefficient for x to the first power and i want to cut it in half mean i'm going to multiply by half or divide by 2 whatever you want to do and then square it okay so 7 halves times a half square it seven halves times a half is 7 4. if i squared that i would get 49 okay so that's what i need to add to both sides of the equation so on the right side let me just give myself some room so 49 16 on the left side let me kind of scooch this down just a little bit so plus 49 16. okay so at this point you could probably do this in your sleep factor the left hand side it's x plus 7 4 squared on the right side i've got to get a common denominator so multiply this by 8 over 8. we know negative 9 times 8 is negative 72 if i add that to 49 i get negative 23. so this is negative 23 16. so i can solve this with my square root property so square root of this side plus or minus square root of this side kind of scroll down here so here i'm going to have x plus 7 4 is equal to plus or minus pull this negative out and just say this is i then times i'm going to do the square root of 23 and this is over square root of 16 would be 4 okay now you can't do a whole lot to simplify this so i'm going to kind of do a quicker method i'm just going to subtract 7 4 away from each side of the equation and i'm going to write this as x is equal to i'm going to put negative 7 plus or minus i times square root of 23 over the common denominator of 4 okay notice if you had negative 7 4 and then plus or minus this whole thing here you'd have a common denominator of 4 okay so that's why i wrote it like this you can break it up if you want you can say x is equal to negative 7 over 4 plus or minus i times square root of 23 over 4. okay you can do it that way as well it's two different solutions either way it's negative 7 plus i times square root of 23 over 4 and negative 7 minus i times square root of 23 over 4. over here it's negative 7 4 plus i times square root of 23 over 4 and negative 7 force minus i times square root of 23 over 4. so however you want to notate it again i get a lot of questions about notation this is the same as this it does not matter okay please don't get tripped up on something that's very very inconsequential okay doesn't matter all right let's look at one more of these and i think we'll pretty much have it so again we have a situation where it's pretty messy i mean it's not as messy as the other ones but i have a 9x squared here on the left and a four x square here on the right so i want to clean that up first so let me just move all the variable terms to one side all the numbers to the other as the first step and then we'll get the coefficient of x squared to be a 1 in the second step so for this one i'm going to subtract 4x squared away from each side of the equation i'll have 5x squared minus 6x subtract 23 away from each side of the equation so this is going to be equal to negative 23. so again in order to complete the square this coefficient has to be a 1. so we can divide both sides of the equation by 5 to accomplish that so i'll have that x squared minus 6 fifths x is equal to negative 23 fifths okay let me make that a little better scroll down get some room going so now i want to complete the square on the left so again i want to take this guy right here the coefficient for x to the first power i want to cut it in half so just multiply by half and then square it okay so negative six fifths times a half is going to be negative six tenths so negative six tenths squared this is going to be what it's going to be 36 over 100 so on this side i'm going to add 36 over 100 on this side i'm going to add 36 over 100 okay so from this point it's pretty easy on the left side i'm gonna factor this don't forget that you have a minus there so this is x minus you're going to have six tenths here and then this is squared and this equals on this side i need a common denominator so let me kind of slide this down so negative 23 fifths times to get a denominator of 100 i need to multiply by 20 over 20 okay so negative 23 times 20 is negative 460. if i added 36 to that i would get negative 424 so equals negative 424 over 100. now if you want to you can make this simpler at this point i know that each is divisible by two i also know that each is divisible by four right because 424 forget about the fact that's negative it doesn't affect the visibility the 24 there tells me that's divisible by 4. so 424 divided by 4 is going to give me 106. so go ahead and write this as negative 106 and then 100 divided by 4 is 25. now 25 is 5 times 5. 106 is not divisible by 5 so there's nothing else you can do there so what i'm going to do is solve this with my square root property plus or minus square root over here so i've got x minus 6 10 on the left and this equals let me get some room so i've got plus or minus the square root of negative 106 over square root of 25 is 5. so i know that i can pull out the negative part as i so let's go ahead and write x minus 6 10 is equal to plus or minus i times the square root of 106 and this is over 5. all right so to solve this i'm going to add 6 tenths to each side of the equation i'm going to get that x is equal to i'm going to have 6 10 and then plus or minus i times the square root of 106. over 5. now i want you to realize that 6 over 10 can be reduced 6 is 3 times 2 10 is 5 times 2. so if i divide this by 2 i get 3 if i divide this by 2 i get 5. so now i have a common denominator so i can write this as x is equal to i'll have 3 plus or minus i times the square root of 106 over that common denominator of 5. again this is two solutions you've got x equals 3 plus i times square root of 106 over 5 and x equals 3 minus i times square root of 106 over 5. in this lesson we want to review solving quadratic equations using the quadratic formula so in our last lesson we reviewed how to solve any quadratic equation by completing the square now most students hate this method because it's just so tedious so fortunately for us there's an easier method known as the quadratic formula so this basically allows us to take any quadratic equation which is written in standard form and just plug in the values for a the coefficient of x squared b the coefficient of x to the first power and see the constant and immediately obtain a solution so first and foremost the quadratic formula is derived from taking this ax squared plus bx plus c equals 0 and essentially just solving it for x by completing the square so i'm going to go ahead and show you that process just so you know where this is coming from and then from here on out we can just use the quadratic formula it's very very easy so the first thing i want to do if i was completing the square on this forget about the fact that you have no numbers involved other than zero you have a that's the coefficient of x squared we want to make sure this is a 1. so to ensure that we're just going to divide each side of the equation by a so that's kind of your first step so we know that this cancels with this and we would have that x squared plus b over a times x plus c over a is equal to 0. now the next thing is to make sure that all the variable terms are on one side and all the numbers on the other so what i'm going to do is i'm going to subtract c over a away from each side of the equation and that's going to give me that x squared plus b over a times x is equal to negative c over a all right so now i'm at the step where i want to complete the square so essentially i want to transform the left side into a perfect square trinomial so i want to take the coefficient of x to the first power which in this case is b over a and i want to cut it in half so that means i want to multiply it by a half and then i want to square the result so this just becomes b over 2a that amount squared so this becomes b squared over 2 squared is 4 and then a squared is just a squared okay so we're going to add this right here to both sides of the equation so let me erase all this we're going to have that x squared plus b over a times x plus b squared over 4 a squared is equal to negative c over a again plus this quantity so plus your b squared over 4 a squared all right so let's factor the left side as x plus remember we squared b to get b squared so this is b over we square 2a to get 4a squared so 2a and this whole thing is squared and this equals if i get a common denominator going over here i would need to multiply this negative c over a by 4 a over 4a okay so that would give me b squared minus 4ac over the common denominator of 4a squared okay so now i can use my square root property and i can take the square root of this side and then plus or minus the square root of this side let's get some room going so that would give me x plus b over 2a is equal to plus or minus the square root of b squared minus 4ac if i take the square root of 4a squared i just get 2 a okay so that's going to be my denominator so if i want to solve for x all i've got to do is subtract b over 2a away from each side of the equation and notice how you have a common denominator here so i can say that x is equal to i'm just going to put the negative b out in front then plus or minus my square root of b squared minus 4ac this is going to be over the common denominator of 2a so this is the quadratic formula this is what we can use to solve any quadratic equation so if it's in the format of ax squared plus bx plus c equals 0 i just record the values for a b and c and i plug them in so i'm going to plug in here for b and here for b i'm going to plug in here for a and here for a and i'm going to plug in here for c okay so that's all i've got to do so let me erase this and we'll take this down to the next page and we'll see an example so let me paste our quadratic formula here so this is what we're going to use to get our solution so the very first thing you need to do is you need to make sure this is in standard form it's not in standard form because you have 14x squared over here and 10x squared over here you want ax squared plus bx plus c is equal to zero so let's move this down for one second i'm just going to subtract 10x squared away from each side of the equation and this would cancel give me 0. so essentially what i'd have is 14x squared minus 10x squared is 4x squared and then minus 5x and then plus 1 equals 0. all right so let me erase this all i need to do i don't need to do anything fancy with this equation i just need to record the values for a b and c so a is the coefficient of x squared so a here is going to be 4. b is the coefficient for x to the first power so here that's negative 5. and c here is my constant term in this case this is 1. i want to caution you against something you need to make sure this is in this format before you record the values okay if your equation is all jumbled up like it is here you're not going to get the right answer okay so make sure you put it in standard form first before you start this process so now that i have the values for a b and c i can just plug into my quadratic formula and i'll have my solution so x equals you've got negative b b is negative five so the negative of negative five is five then plus or minus the square root of b squared negative 5 squared is 25 then minus 4 times for a we have 4 and then for c we have 1. then this is over 2 times a a is 4 in this case so 2 times 4 would give me 8. so now i just simplify this and i have my answers so x equals 5 plus or minus 25 minus basically 4 times 4 times 1 is 16. so 25 minus 16 is 9. so you'd have 5 plus or minus the square root of 9. we know the square root of 9 is 3. so just go ahead and write that as 3. and then this is over 8. so we can go ahead and say that we have x is equal to five plus three is eight so you'd have eight over eight which is one and then five minus three is two two over eight is one fourth so you'd have one fourth there as well okay so just make sure you break that up and do both solutions especially in this case because you want to say you have five plus three over eight so that's one solution for x and this leads to eight over eight which is going to give us one okay so that's how we got this guy right here then the other scenario is 5 minus 3 over 8 so that gives us 2 over 8 which is 1 4 so that's how we got this guy right here so x basically is going to be equal to 1 or 1 4 okay and look how much quicker that was versus completing the square or even if we were to use factoring okay using the quadratic formula is generally the fastest method all right for the next one we're going to look at negative 5 minus 2x equals negative x squared again i always want it in the format of ax squared plus bx plus c equals 0. so i'm going to add x squared to both sides of the equation i get x squared minus 2x minus 5 is equal to 0. so now all i need to do is record the values for a b and c and plug into the quadratic form that's all i need to do so for a here it's the coefficient of x squared it's just going to be 1. so a is 1 b is what it's the coefficient for x so it's a negative 2 and c is going to be a negative 5 right it's the constant okay so let's erase this and let's just kind of drag this up a little bit and in case you forgot the quadratic formula you can just keep writing it you'll remember it pretty quickly so it's x is equal to you've got negative b plus or minus the square root of b squared minus 4ac all over 2 a so i'm going to plug in for b i've got a negative 2 so i'd have the negative of negative 2 which is plus 2. so x equals you'll have two plus or minus the square root of b squared you'd have negative two squared which is four then minus four times a times c a is one c is negative five so you'd have negative four times one which is negative four then times negative 5 which is positive 20. so we'll put plus 20 here let's get some room going then i have it over 2a a is 1 so this is just all over 2. so this is just all over 2. so x is equal to you've got 2 plus or minus 4 plus 20 is 24 so the square root of 24 which we know we could simplify into what square root of 24 is the square root of 4 which is 2 times square root of 6. so this is 2 times square root of 6 and then over 2. now you could leave it like this but you want to always simplify if you can you want to think about the fact that you have a 2 here a 2 here and a 2 here so really what i could say is x is equal to factor out the 2 from the numerator you would have 1 plus or minus the square root of 6 over 2. this guy would cancel with this guy and i have that x is equal to one plus or minus the square root of six and again this is two different solutions this is x equals one plus the square root of six and also x equals one minus the square root of six all right let's look at another one so we have seven minus five 5x is equal to we've got negative 4x minus 3x squared so again clean this up put it in standard form you want your ax squared plus your bx plus your c is equal to 0. so what i'm going to do is i'm going to add 3x squared to both sides of the equation and i'm going to add 4x to each side of the equation and that's going to give me what these cancel this is 0 on the right so i would have 3x squared and then 4x minus 5x is negative x and then you'll have your plus seven okay so this is equal to zero so now all i need to do is record my values for a b and c and i'm basically set right so a here is going to be 3. a here is 3. b here is negative 1 okay b here is negative 1 and c here is going to be 7. so i plug into my quadratic formula so x equals negative b plus or minus the square root of b squared minus 4ac all over 2a just keep writing it and it's going to become second nature for you you'll remember it in your sleep so i'm going to go ahead and say that x is equal to the negative of b b here is negative one so the negative of negative one is one plus or minus the square root of again b is negative one so b squared is one so you'll have one minus four times a a is 3 here so 4 times 3 is 12 times c c is 7 and 12 times 7 is 84 so you'd have 1 minus 84 which is negative 83 and this is all over 2 times a a in this case is 3 so 2 times 3 is 6. all right so let me erase all this and we'll kind of look at these two different solutions here and see what we can do so we know we would have x equals 1 plus the square root of negative 83 over 6 and then x equals 1 minus the square of negative 83 over 6. what i can do to simplify here say that x is equal to 1 plus or minus pull out the negative part and say this is i times the square root of 83 over 6. all right let's talk a little bit about the discriminant so the discriminant is basically the b squared minus 4ac it's the part of the quadratic formula that's underneath the square root symbol so essentially by looking at the discriminant we can get a lot of information we'll know the type of solution we're going to get whether it's real or imaginary and we'll also know the number of solutions we're going to get so in the case where b squared minus ac is greater than zero we're going to have two real solutions now if this guy right here happens to be a perfect square then you're gonna get two rational solutions now if b squared minus 4ac is equal to 0 then you're going to get exactly one rational solution and the reason for that is the plus or minus the square root of b squared minus 4ac if this guy is zero you have plus or minus zero which is just zero okay so it kind of renders the plus or minus part ineffective so if b squared minus four i say is equal to zero then you're just going to have one rational solution so then the last kind of case is that b squared minus 4ac is less than zero so in this case you're going to have two complex solutions all right let's take a look at another one so we have 10x squared plus x plus 1 equals 2x i just want to subtract 2x away from each side of the equation just to put this in standard form so i'm going to have 10x squared x minus 2x is negative x and then plus 1 equals 0. so again i want to record my values for a which is 10 in this case b which is negative 1 and c which is positive 1. so let's think about the quadratic formula so x equals you've got negative b in this case b is negative 1. so the negative of negative 1 plus or minus the square root of b squared which is negative 1 squared minus 4 times a a is 10 times c c in this case is 1. okay now this guy is going to be over 2 times a so 2 times in this case a is 10. now what is this part right here again the discriminant this will predict the number of solutions and the type of solutions we're going to have so negative 1 squared is 1 and then minus 4 times 10 is 40 times 1 is still 40. so you'd have 1 minus 40 which is negative 39. so if this is negative 39 i know that i'm going to have two complex solutions okay so let's go through this the negative of negative 1 is just 1 and then if i think about plus or minus the square root of negative 39 39 is 13 times 3 so there's nothing i can really do other than i pull this negative out so i can put this is i i times square root of 39 and then 2 times 10 is just 20. so essentially you can leave your solution in this format this is perfectly fine or you could write that x is equal to 1 plus i times square root of 39 over 20 and then also x is equal to 1 minus i times square root of 39 over 20. all right let's look at one more so we have 8x squared minus 4 is equal to negative 8x plus 12x squared so again clean this up put it in the format of ax squared plus bx plus c equals 0. so i'm just going to subtract 12x squared away from each side of the equation and i'm going to add 8x to both sides of the equation so this is going to cancel and become 0. i'm going to have that 8x squared minus 12x squared which is negative 4x squared and then we have plus 8x and then minus 4 is equal to 0. a little trick to make this simpler whenever you work with stuff and you see that this is divisible by 4 this is divisible by 4 this is divisible by 4 and 0 is divisible by 4. i can just divide both sides of the equation by 4 and make this easier i can also divide both sides of the equation by negative 4 if i want it does not change the solution okay it just makes it easier to work with so this becomes x squared this becomes negative 2x this becomes plus 1 and this equals 0. notice how i'm going to be working with smaller values for a b and c it's just something you can do very quickly to make your life a lot easier so let's go ahead and erase all this let's drag this up and again let's record the values for a which in this case is a 1 for b which in this case is negative 2 and for c which in this case is 1. so we get x is equal to negative b in this case b is negative 2 so the negative of negative 2 plus or minus the square root of b squared so you're going to have negative 2 squared minus 4 times a which is 1 times c which is 1. okay this is all over 2 times a a is 1. so 2 times 1. okay so let's think about our discriminant and predict our solutions so we have negative 2 squared which is 4 okay and then we're going to subtract away 4 times 1 times 1 which is 4 4 minus 4 is 0. so this is going to be taking the square root of 0 which is 0. so since the discriminant is 0 i know i'm going to have one rational solution right because you can basically get rid of this part here and you can say that the negative of negative two is two and two over two would just be one so x here would just be one okay that would be your only solution in this lesson we want to review solving equations that are quadratic in form so in some cases we're going to come across an equation that is not a quadratic equation but can be made into a quadratic equation with a simple substitution so these equations are known as quadratic in form and we can solve them using the same methods we used when we worked with quadratic equations so we can use completing the square we can use the quadratic formula in some cases we can use factoring so let's take a look at the first example so we have x to the fourth power minus 25 x squared plus 144 is equal to zero so i want you to notice here how the smaller exponent on x is a 2 and the larger exponent on x is a 4. generally when we work with a quadratic equation you have ax squared plus bx and i'm just going to say bx to the first power plus c is equal to 0. so notice how this guy right here this larger exponent is double that of the smaller this is the same thing here the larger exponent is double that of the smaller or you could say the smaller exponent is half of the larger okay so we see that pattern and we also see that we have a constant here and we have a constant here so we can say that this equation is quadratic in form so what we can do is make a little substitution okay so we're going to grab our variable that's raised to the smaller power and we're going to set that equal to some other variable it could be q it could be z a lot of people like to use u okay so i'm going to stick with that i'm going to say let's let our variable u be equal to x squared now in order to do the substitution in order to show it to you i'm going to use my rules of exponents i'm just going to rewrite my equation so x to the fourth power i'm going to write that as x squared being squared okay i haven't done anything illegal power to power rule that would still be x to the fourth power then minus 25 x squared then plus 144 equals zero let me make that one a little better so everywhere i see an x squared i'm just going to plug in a u okay because u is equal to x squared all right let me get a little room going here what i'm going to say is that i have u that's being squared again just plugged in a u there then minus 25 u i'm plugging in the u there it's not squared because i'm plugging in a u for the x squared okay then plus 144 is equal to zero now if i saw this initially i would have a quadratic equation and i could solve that either by completing the square or the quadratic formula or again in some cases with factoring so let's go ahead and solve this using the quadratic formula because i think that's usually the easiest scenario so let's go ahead and put this as a visible 1 here i'm going to say that my a my coefficient for u squared in this case normally it's x but here it's u squared is a one b the coefficient for u to the first power is a negative 25 and c my constant is 144. so a is 1 b is negative 25 and c is 144. okay so let's plug into the quadratic formula so we would have that u is equal to you've got the negative of b plus or minus the square root of b squared minus 4ac and the whole thing is over 2a so let's go through and just replace so you've got the negative of b b is negative 25 the negative of negative 25 is 25 plus or minus the square root of you've got b squared negative 25 squared is positive 625 so 625 then minus you've got 4 times a times c a is 1 so we can just get rid of it you can put times 1 if you want but it's not going to change anything and then c is 144. so c is 144. so what is 4 times 144 well that's 576. so 576. what is 625 minus 576 well that's going to give us 49. so we've got a 49 here and of course if we took the square root of 49 we would get 7. so let's go ahead and put a 7 there now we have 2 times a in the denominator a is 1 so we basically just have 2. so what we have here is that u is equal to you've got 25 plus 7 which is 32 over 2 that would give me 16 and then you also have 25 minus 7 okay 25 minus 7 which would be 18 over 2 which would be 9. so u equals 16 and also not so here's the problem a lot of students get to this step and they go i've got my answer i'm done and they move on to the next problem you're not finished because the original equation was in terms of x not in terms of u okay so you've got to go back and you've got to substitute again so let's copy this all right so initially we said that u is equal to x squared so essentially what i can say here since u equals 16 or 9 i can say that x squared is equal to 16 or 9 because u is just x squared so now i can erase this and this i don't need any more now i want to set x squared equal to 16 and then i also want to set x squared equal to 9 and i want to get a solution okay so if x squared equals 16 using my square root property i know that x would be equal to plus or minus the square root of 16 which is 4. okay so over here if i use my square root property x squared equals nine i know x would be equal to plus or minus three okay so we have four solutions here we have that x could be equal to plus or minus 4 so positive 4 or negative 4 and then also plus or minus 3 right positive 3 or negative 3. all right let's look at another one so we have 2 times the quantity 3x minus 1 squared plus 5 times the quantity 3x minus 1 and this is equal to negative 2. so this one might not jump out at you as being quadratic in form but realize that you have the same quantity here as you have here okay you've got that 3x minus 1 you've got the 3x minus 1. this is typical see how this is squared and this is basically raised to the first power if i substitute this quantity with a variable like again u i can solve my equation using the quadratic form okay so let's let u be equal to this quantity 3x minus 1. so i'm going to plug in a u there and there so what would that give me i would have 2 u squared plus 5 u is equal to negative 2. and let's scroll down and get some room going we'll come back up to this in a minute if i add 2 to both sides of the equation to put this in standard form i'm going to have 2 u squared plus 5 u plus 2 is equal to 0. so if i want to use my quadratic formula i'm going to record the value for a which is 2 the value for b which is 5 and the value for c which is 2. okay so let's use our quadratic formula to get a solution here so we have that u is equal to the negative of b plus or minus the square root of you've got b squared minus 4ac all over 2a so let's go through and replace some stuff so negative b b is 5 so negative 5. plus or minus the square root of b squared again b is 5 5 squared is 25 then minus you've got 4 times a times c a is 2 and c is 2 so you have 4 times 2 which is 8 times 2 again which is 16. if i did 25 minus 16 i'd get 9 and the square root of 9 is 3. okay down here you have 2 times a a again is 2 so this is 4. so i've got two solutions for you i've got that u is equal to you've got negative 5 plus 3 which is going to give you negative 2 and then over 4. so negative 2 over 4 is negative 1 half then another solution is negative five minus three which is negative eight over four which would give us negative two so u is equal to negative one half or negative two okay so we have that again u equals negative one half or negative two and we say that u is equal to three x minus one so we've gotta set three x minus one which is again u so we're putting that there and say this is equal to each solution so it's equal to negative one half or three x minus one is equal to negative two all right so let's erase all of this we don't need that anymore and we'll get our solutions for x okay so on the left hand side here this equation 3x minus 1 equals negative one-half i would add one to each side of the equation so i would have that 3x is equal to you've got 1 minus one-half now you can multiply both sides of the equation by two if you don't like working with fractions or you can just multiply this guy by two over two it's a little quicker and you would basically have that two minus one which is one over two is your right hand side so three x equals one half we can divide both sides by three or you can also multiply by a third same thing makes it easier if you're working with fractions to do that so you have x is equal to one half times 1 3 which is 1 6 okay so that's one answer all right so over here i'm just going to add 1 to each side of the equation and we know that this is going to cancel i'll have that 3x is equal to negative 2 plus 1 is going to be negative 1 divide both sides by 3 and i get that x is equal to negative 1 3. so we get our two solutions here x is equal to 1 6 or x is equal to negative 1 3 of course you can write it as x equals 1 6 comma negative 1 3 any way you want to notate it is fine all right let's look at another one so suppose we had x to the power of negative 2 minus x to the power of negative 1 minus 6 equals 0. so again look at the fact that this guy right here it's a negative 1 but this guy right here is a negative 2. so it follows that same format right if i doubled the negative 1 i would get to negative 2 right so it follows the same format as a quadratic equation so i'm going to let u be equal to my x raised to the power of negative 1. okay so in this particular case i'm going to rewrite this using the rules of exponents i'm going to say this is x to the power of negative 1 squared okay again negative 1 times 2 is negative 2. i haven't done anything illegal here then minus x to the power of negative 1 then minus 6 equals 0. so to substitute here again this guy is going to be u and this guy's going to be u so i'm going to have u squared minus u minus 6 is equal to 0. okay so let's erase this we don't need it anymore and let's kind of drag this up a little bit so to put in some visible coefficients here let's put a 1 here this is going to be my a this will be a negative one here for my b and this will be a negative six for my c okay so for my quadratic formula i'm gonna have that u is equal to again you've got negative b plus or minus the square root of b squared minus 4ac all over 2a and again you keep working with this over and over again you're going to memorize it okay this is something you're going to recite in your sleep after a while so for the negative of b i know b is negative one the negative of negative one is positive one plus or minus the square root of you've got b squared negative one squared is one minus four times a times c a in this case is one c is negative six so if you did one times anything is just itself so really you get negative four times negative six which is twenty-four so you can write this as plus twenty-four we know one plus twenty-four is twenty-five and we also know that the square root of 25 is 5. okay down here we have 2 times a a is 1 so this is just 2. so i'm going to get two solutions for you i'm going to find that u is equal to 1 plus five which is six over two so that's three and then also u is equal to one minus five which is negative four over two which is negative two so let's copy this all right let's paste this right here so again u is equal to 3 and negative 2 and we said that u is equal to x to the power of negative 1. so if x to the power of negative 1 is u and u is equal to 3 or negative 2 then x to the power of negative 1 is equal to 3 or x to the power of negative 1 is also equal to negative 2. now how do we solve something like this remember what an exponent of negative 1 means this is basically like saying that we have 1 over x now you can solve it that way or if you don't want to work with that you can raise this side to the power of negative 1 and this side to the power of negative 1. you can also do that over here and what's going to happen is power to power rule negative 1 times negative 1 is 1 so you'd have x by itself and this is equal to if you have 3 to the power of negative 1 it's 1 3. so now i'm going to say or over here same thing x raised to the power of negative 1 raised to the power of negative 1 x is just going to be raised to the power of 1 right negative 1 times negative 1 is 1. this is equal to you've got negative 2 being raised to the power of negative 1 so this is basically 1 over negative 2 okay so you could write that in a simpler format and just say this is negative one-half so x here is one-third or x is negative one-half again i could write this as x equals one-third comma negative one half all right let's take a look at another one so we have three times the quantity x over x plus one squared plus seven times the quantity x over x plus one then minus six equals zero so again it's the same thing you have this guy right here and this guy right here that are the same in one case it's squared in the other case it's raised to the first power so this is another scenario where you can use your kind of quadratic in form technique to solve so what i'm going to do is just use a little substitution so i'm going to let u be equal to this x over x plus 1. so i'm going to sub in there i'm going to have 3 u squared plus 7 u minus six is equal to zero so a quick note here we notice that we have a variable in the denominator here so we wanna make sure that our solution is not negative one right if x ends up being negative 1 if i plugged in a negative 1 there or there notice how you would get 0 in the denominator and that is undefined okay so you want to guard against that if you end up with a final answer where x equals negative 1 you've got to throw that solution away because it does not work okay so again let's kind of return to the problem we've got this set up here and let's think about this for a second using our quadratic formula this is my a this is my b and this is my c right so a is three b is seven c is negative six so i've got that u is equal to we've got negative b in this case b is seven so i'm just gonna write it out negative seven plus or minus the square root of we've got b squared 7 squared is 49 then minus 4 times a a is 3. so 4 times 3 is 12 times c c in this case is negative 6. so that would give me negative 72. so basically you'd have minus a negative 72 which is plus 72 49 plus 72 is 121. if i took the square root of 121 i would get 11. okay so this is 11 here and this is over 2 times a a is 3 so 2 times 3 is 6. so u will have again two solutions uke will be equal to you've got negative 7 plus 11 which is 4 over 6 so 4 6 we know is two-thirds right you can divide each one of those by two you would get two thirds there then alternatively you've got negative seven minus eleven which is going to give me negative eighteen over six which is negative three okay so those are my two solutions for you so now if u was set equal to the quantity x over x plus one then we know that since u equals two-thirds and also at equals negative three i can set x over x plus one equal to these amounts so two-thirds then or x over x plus one could be equal to negative three so let's solve these two equations and basically i'm gonna drop this down here i have x over x plus 1 equals 2 3. what i want to do is multiply both sides of the equation by the lcd so i've got an x plus 1 and i've got a 3. so i'm going to multiply this side by 3 times the quantity x plus 1. i'm going to do the same thing over here so 3 let me kind of slide this down so 3 times the quantity x plus 1 and you'll see that this will cancel and this will cancel so on the left you've got 3x is equal to on the right you've got 2 times the quantity x plus 1 so that's 2x plus 2. all right now this is pretty easy to solve we'll subtract 2x away from each side of the equation and on the left hand side now i just have x and this is equal to just 2. so x equals 2 is one solution and again one of the solutions is that x is equal to 2. i can erase this because i don't need it anymore let's put our or here let's solve this guy so i need to multiply both sides by the lcd in this case i just have the x plus 1 i don't have a denominator here i just have negative 3. so i'm going to multiply this side by x plus 1 and this side by x plus 1 and this will cancel with this on the left i'll have x and this is equal to on the right you've got negative 3 times this is a quantity here so it's negative 3 times x which is negative 3x and then negative 3 times 1 which is minus 3. okay so let's go ahead and scroll down and get a little bit of room going we want to add 3x to both sides of the equation this gives me 4x on the left and it's equal to negative 3. i can divide both sides by four and this will give me x is equal to negative three fourths so let me erase this and we'll say that x equals two or x equals negative three fourths so x equals two comma negative 3 4. and notice how we did not get a solution for x that violated our domain okay remember we said initially that x couldn't be negative 1 because if we plug in a negative 1 for here or here right we would get a 0 in the denominator that is undefined so you've got to guard against that so if you ever have a variable in the denominator and you're working with an equation you've got to look at where would this guy be restricted where is it undefined and make sure you say that specifically here x cannot be negative 1. it's not in this case so you really don't even need to put that you just put your solutions here as x equals 2 or x equals negative 3 4. so at this point i think we pretty much have this down and we're just gonna look at one more problem so we have x to the power of two thirds minus x to the power of one third minus six equals zero again look how this guy right here is doubled out of this guy if i multiplied one third by two i would get two thirds so let's let u be equal to the variable raised to the smaller power in this case that's x to the power of one-third so i'm going to again rewrite this using my rules of exponents i'm going to say this is x to the one-third power squared minus x to the one third power minus six equals zero everywhere i see x to the one third power i'm plugging in a u so i'm gonna have u squared minus u minus six equals zero now for this particular guy we can solve this using factoring it'll be a little bit quicker we know how to use the quadratic formula so let's just solve it using factoring so i'm going to set this up as u and u put my 0 over here give me two integers whose sum is negative 1 and whose product is negative 6 well you can do a negative 3 and a positive 2. a negative 3 and a positive 2. all right so now we can use our zero product property and set each factor with a variable equal to 0 and solve so in other words i would do u minus 3 equals 0 or u plus 2 equals 0. so i'm just going to solve those this one i would just add 3 to both sides of the equation i would get that u is equal to 3. okay so one solution is u equals 3. then over here i would just subtract 2 away from each side of the equation i would get that u is equal to negative 2. so or u equals negative 2. all right so let's get rid of all this we don't need it anymore and let me make that 2 a little bit better again since u is equal to x to the power of 1 3 i can basically say that x to the power of 1 3 is equal to 3 or x to the power of 1 3 is equal to negative 2. let's erase all this and let me make that 2 a little bit better all right so how can we solve these equations well i can cube both sides to clear the 1 3 power right because if i have 1 3 raised to the third power it's 1 right power to power rule so if i cube this side and i cube this side i'm going to get that x is equal to 3 cubed is 27. so x is equal to 27 then or over here if i cube this side and i cube this side i'm going to get that x is equal to negative 2 cubed is negative 8. so or x equals negative 8. all right so let's erase this and we can basically state our solution as x equals 27 again or negative 8. in this lesson we want to think about applications of quadratic equations so throughout our study of algebra we often come across word problems or application problems so so far in our course we've only discussed word problems that involve solving a linear equation in one variable now we're going to go one step further and take a look at some word problems that involve quadratic equations so i want you to recall that we have a little six step procedure that we can use for solving word problems and although we've introduced this method for use with applications of linear equations we can also use it in this section as well so let's just kind of recap this six step procedure i think you'll find it useful if you write it down and kind of go through the steps when you're solving the problem so solving a word problem we have that our first step is to read the problem and determine the question or perhaps questions to be answered you always want to make sure that you understand the question or again the questions that you're being asked to kind of answer it helps you kind of format things in a way well what am i trying to find then the second step is to assign a variable to represent the unknown and in some cases we're going to have to express other unknowns in terms of this variable then third we want to write an equation that describes the scenario we're given and then fourth we're just going to solve that equation then fifth we want to write the answer in terms of the question or again perhaps questions asked and then the sixth thing probably the most important we want to check the answer using the words the problem we want to make sure that the answer is reasonable okay you're going to see in some of the examples we get today we're going to have some solutions to our equation that we set up that don't quite work in the context of the problem we're given okay we'll get to that later on all alright so the first few problems we're going to look at are going to involve the pythagorean formula i know that most of you already know about the pythagorean formula but just in case you don't it doesn't hurt for a little bit of a review so the pythagorean formula basically shows the relationship between the lengths of the sides of a right triangle so again we know we have a right triangle because of this signature mark here this is telling us we have a 90 degree angle or a right angle okay so if you have a right angle in a triangle it's known as a right triangle now with a right triangle we have this special relationship between the sides okay so you have these two shorter sides known as legs you have this here leg a which is this kind of vertical measure here and then you have this leg b which is this horizontal measure here okay and then you have c which is the hypotenuse okay the hypotenuse is always going to be opposite of the right angle so opposite of this and it's the longest side so most of you already know this the pythagorean formula tells you that a squared right which is basically the measure of leg a that guy squared plus b squared again the measure of b leg b squared if i sum those two amounts together it's equal to c which is the hypotenuse that measure squared okay this is your pythagorean formula and it's true for any right triangle that you come across so using this pythagorean formula basically if we don't know one of the sides of a right triangle but we know the other two we can solve for the unknown okay so let's take a look at our first sample problem all right so jason needs to wash a window in a building that is 12 feet from the ground to avoid a fence he decides to rest the ladder against the building for stability jason decides he should place the ladder nine feet away from the building how long of a ladder will jason need okay so the first thing we need to understand again is what am i being asked to find and in this case it's this little question here right so how long of a ladder will jason need so we'll just highlight that that's what we're trying to find so when we go to assigning a variable we're just going to let x represent exactly what we're trying to find which is the measure of jason's ladder and this is going to be in feet because we're working with feet in the problem so let's go down here to a fresh sheet we're going to say that we're going to let x be equal to the measure of jason's ladder and of course this is going to be in feet okay so in feet all right so i'm not going to write feet every time i'm just going to work with the number parts we're just going to understand that we're working with feet all right so to help us understand what's going on we have this little illustration here it's not perfect but it'll just give you an idea of what's kind of happening in our problem so when we started out we said that jason wanted to wash a window in a building that is 12 feet from the ground so this guy right here this represents our window okay that's in our drawing and basically you could say that it's 12 feet from the ground so from right here to right here that measurement let me kind of draw a little vertical line okay not perfect but it'll have to do so this guy's going to be 12 feet again i'm just going to put 12. and then also it tells us that for stability jason decides he should place the ladder 9 feet away from the building so the ladder we can't see because it's hidden in the grass but let's just say it's about right there so from right there to right here let me use a different color so from right there to right here that horizontal measure is going to be nine feet so let me again just draw a little horizontal line okay again not perfect but you get the idea so this guy is nine feet i'm just going to put nine and i can put my little symbol here this is going to be a right angle so i've got that symbol there so we have a right triangle right we have that this guy here this horizontal leg is 9. this guy right here this vertical leg is 12. and i'm solving for this guy right here which is the length of his ladder this is exactly what i need to find i could just put this as x right here this guy is going to be the hypotenuse okay so essentially i can say that again for my pythagorean formula a squared plus b squared equals c squared where a and b are the lengths of the two shorter sides or the legs right so we know that's 9 and 12. so we just plug in and say that we have 9 squared plus 12 squared and it doesn't matter the order that you do that n is equal to instead of c i just have x so i'm going to say this is x squared okay so 9 squared we know is 81 plus we know 12 squared is 144 this equals x squared and 81 plus 144 is of course 225 so we get that x squared okay x squared is equal to 225 so we can solve this with our square root property we know that x would be 15 or negative 15. right take the square root of each side so square root of this side is equal to plus or minus the square root of this side and let me kind of scroll down and get a little bit of room going and we can basically say again that x is equal to plus or minus 15. now does it make sense okay does it make sense for the measure of this ladder to be negative 15. no it's nonsense okay so that's where you know checking your answer when you think about kind of the steps we gave makes a lot of sense so if you incorrectly said that the ladder could be 15 feet or negative 15 feet when you go to check that you can say okay that doesn't make sense right a ladder can't measure negative 15 feet that that's complete nonsense so you just have to throw that solution out and say that x or the kind of length of the measurement of that ladder is going to be 15 feet okay so i'm just going to write that in there and say this is 15 or again 15 feet so let's go back up and again we can just kind of write a little sentence to describe our answer so how long of a ladder will jason need well we'll say that jason needs a ladder jason needs a ladder that is 15 feet long okay so checking this is really really easy again we just go through our pythagorean formula and it's easier to check something like this by looking at the picture again just so you get an idea what's going on again we said that with jason he needs to wash a window in a building that's 12 feet from the ground so this is where the window is and this is where the ground is so the distance from here to here that's 12 feet okay so that's where we got the 12 from and then to avoid a fence he decides to rest the ladder against the building so that's why this ladder is drawn in this manner okay and then basically it says that for stability jason decides he should place the ladder nine feet away from the building so from here to here is nine feet that's where we got that nine from okay so this part all checks out so then we're asked how long of a ladder will we need well again we see that we have a right triangle and so we can use our pythagorean formula we plugged in we got 9 squared plus 12 squared equals x squared x was the length of the ladder and so we got 81 plus 144 on the left side of this which is 225 so we ended up with x squared equals 225 we took the square root of each side right and we found that we had x equals plus or minus 15. we threw out negative 15 because it didn't make any sense so that's how we got our solution where we said that jc needs a ladder that is 15 feet long okay so this solution is correct all right let's go ahead and take a look at another one so we have that larry and steven were both parked in the same parking garage larry leaves first and drives north at 50 miles per hour one hour later steven leaves and drives east at 70 miles per hour how long after larry starts driving does it take for the two cars to be 290 miles apart so again we want to understand the main question and here it's just to find how long after larry starts driving does it take for the two cars to be 290 miles apart so for this problem we're going to end up using our pythagorean formula and we're also going to use the distance formula that we see with motion word problems right the distance equals the rate of speed times the amount of time traveled all right so before we kind of get into that though we need to assign a variable to represent the unknown in this case we're going to end up having kind of two unknowns because we don't know how long larry's going to drive for and we also don't know how long stephen's going to drive for okay so what we're going to do is since larry leaves first we're going to say that x is the amount of time that larry drives for before the two are 290 miles apart okay so let's go down here to a fresh sheet and i'm just going to say we're going to let x be equal to the amount of time and i'm just going to put for larry okay and we know what that means it's going to be specifically in hours and it's going to be the amount of time that larry drives for before again the two cars are 290 miles apart now for stephen we can kind of base his time on larry's right we know larry's time is x well for steven we know that one hour later steven leaves so he's been driving for an hour less at this point when they are 290 miles apart so i can just go x minus one to represent stephen's top okay so we're going to say then i'll do that in a different color so we'll say then x minus 1 this quantity is the amount again of time for steven okay we'll just make it nice and simple like that you could write out something that's really long and say it's the amount of time for steven you know in hours before they're 290 miles apart but you kind of get the idea and we want to make it as short as possible all right so the next thing we want to do is just come down to a little table just organize our information so that we know what's going on so our distance formula in case you forgot the distance traveled which i'm going to say is d is equal to the rate of speed which is r multiplied by the amount of time traveled which is t okay so the rate of speed is in miles per hour the time is in hours and the distance is going to be in miles okay and then we have two participants here we have larry and then we have stephen okay so those two guys are involved in the problem so what information do we have well we know the time because we said that larry's going to be driving for x hours when the two are 290 miles apart and we say that stephen since he drives an hour less is going to be driving for x minus 1 hours again when they're 290 miles apart now the rate we can get directly from the problem it tells us specifically that larry leaves first and drives north at 50 miles per hour so this 50 miles per hour is going to be his rate of speed and then for stephen he leaves and drives east at 70 miles per hour so 70 is going to be his rate of speed again in terms of miles per hour so let's just fill this in for a larry it's going to be 50 miles per hour for steven it's going to be 70. now if i have a rate of speed and i have a time i have a distance right because i can just multiply r times t gives me d so 50 times x is 50x and that's my distance that's larry's going to travel again when they're 290 miles apart so for steven he's going to have traveled 70 times the quantity remember this is a quantity and this is multiplication so put parentheses so times the quantity x minus 1. and you can go ahead and simplify this now because you're going to have to later you can just say this is 70x minus 70. so we'll write this as 70x minus 70 just so that we have it ready to go okay so before we kind of set up our equation i want you to think about something because usually at this point people make a drastic mistake they're used to kind of solving these in another scenario kind of with the opposite direction travel with that one they end up saying okay well the distance larry travels which is 50x plus the distance steven travels which is 70x minus 70. they sum those two amounts and they say okay well that's got to be equal to 290 okay this is wrong this is what we saw kind of earlier in the course this is not the current setup okay what we have here and just to kind of illustrate what's going on i drew a little picture a little illustration and basically what we have again is another application of the pythagorean formula because we end up with something that is a right triangle okay so we have that larry is driving north so you can picture him starting from here let's just say that's the parking garage and he drives north which on every map that you look at that's basically going up okay so this is going to be a vertical leg for the right triangle and we know that measure the distance if i go back is going to be what it's 50x so let's go back here and i'm just going to write this in as 50x and we know that's for larry so i'm just going to put a little arrow here and say that's l for larry okay that's his distance that's again from here to here that's the point he's going to be at when these two guys are 290 miles apart from each other now for stephen let me put a little s here and go this way he's got this horizontal leg so this guy to this guy that distance there again if we go back up that's 70x minus 70. so let me put this here as 70x minus 70. okay so you can obviously see what's going on here we've got our two legs for this right triangle and again the formula is that a squared plus b squared okay we square each of the legs and we sum those amounts together this is equal to the hypotenuse squared so c squared again normally this is c but in this case we have kind of expressions involved so we just kind of plug those in and work with those so we end up with let's just say 50x that quantity squared then plus this guy is the quantity 70x minus 70. the whole thing is squared so make sure you wrap it in parenthesis and this equals our c or our hypotenuse is going to be 290. so that amount that's going to be squared okay so we're going to just copy this and go to a fresh sheet because we'd run out of room here because it's going to be quite tedious so let's just paste this in here and let's get started all right so for 50x squared if we wanted to simplify that i would need to square 50 and also x so that would give me 50 squared times x squared now 50 squared is 2500 so let me just replace that and say it's 2500 okay so then plus now don't make the mistake of doing this a lot of students again i've said this throughout the course they do what they say this is 70x squared minus 70 squared okay that is wrong this is wrong don't do that remember if you have something in here with addition or subtraction you've got to expand it okay remember our special products formula you can use that here to do this pretty quickly you're going to square the first guy which is 70x that whole thing squared right so the whole thing is squared so it's 70 squared which is 4900 times x squared okay then you're going to have minus you're going to have 2 times this guy times this guy so it's 2 times 70x times 70. okay we'll just go ahead and crank that out real quick 2 times 70 is 140 then times 70 again would be 9 800. so basically what you'd have is minus 9800x here so minus 9800x you could say 9800x whatever you want to say and then lastly you need to take this 70 the last term there and square it so we're going to go plus 70 squared we already know is 4 900. so this is 4 900 okay so that's this side right here and this equals 290 squared is 84 100. let me slide this down so i can fit this all on one one line so 84 100. okay so we have kind of some big numbers involved here so let's do some simplifying and then i'll show you what we can do to kind of make this as easy as possible all right so 2 500 x squared plus 4 900 x squared those are like terms so we can combine them so if i had 2 500 and 4 900 i get 7 400. so this is 7 400 x squared then minus 9 800 x i can't do anything with that as of yet so we'll just leave that as it is and then what i can do over here i'm going to subtract 84 100 away from each side of the equation okay and what that's going to do is that's going to give me 0 over here on the right side okay because i'm solving a quadratic and i want it in the format of ax squared plus bx plus c equals zero so i can use my quadratic form okay so 4900 minus 84 100 is going to give me minus or negative 79 200 again this equals 0. now if you want to plug these into the quadratic formula as they are you can most of you i know pretty much all of you are going to be using calculator i'm using a calculator if you're working with bigger numbers or smaller numbers with calculator it doesn't really matter right it does the calculation for you but just to make this a little easier a little simpler so what we'll notice here is that each of these numbers ends with two zeros this guy over here is a zero so it's not going to matter but if they all ended two zeros they're divisible by a hundred right so if i divided every part of this equation by a hundred on the right zero divided by 100 is still zero so that's fine basically i would just knock two zeros off from everything right so that makes it a lot simpler right you have 74x squared minus 98x minus and this comma would disappear you'd have 792 and this equals zero okay perfectly mathematically legal to do that and you would notice that you could also still divide each part by two so if you wanted to make it even simpler you could say that 74 divided by 2 is 37 so 37x squared and then 98 divided by 2 is 49 so minus 49x and then minus if you take 792 and divide by 2 you would get 396. and we can say this is equal to zero okay so this is a lot simpler a lot easier to work with so let's kind of drag this up here and see what we got now so we're going to use our quadratic formula so again this is going to be a this is going to be b okay it's the whole negative 49 and this is going to be c again the negative 396 okay and i circled the negative ones because sometimes you get confused and you try to just take the number part without the sign when it's positive obviously it doesn't matter right you're not bringing a sign with you it's just positive so it doesn't have a sign in front of it all right so with the quadratic formula again in case you have forgotten it x is going to be equal to the negative of b plus or minus the square root of you've got b squared minus 4 times a times c this whole thing is over 2a so all i'm going to do is just plug in okay very very easy so for negative b again b is negative 49 the negative of negative 49 is 49 plus or minus you've got b squared so you'd have negative 49 squared so let's just go ahead and do that real quick if i square negative 49 i get 2401. so let me write that in 2401 that's kind of hard to fit in there so let me redo this so square root of 2 401 again minus 4 times for a i have 37 and then for c i have my negative 396. and we can crank this out now or come back in a minute let's just come back in a minute for a we have what we have 37 so let's plug that in there all right let's do our calculation so negative 4 times 37 is negative 148 then times negative 396 is going to give me 58 608 so let's just put plus 58 000 again 608 if we sum two thousand four hundred one and fifty eight thousand six hundred eight we get sixty one thousand none okay so let's write that there so sixty one thousand nine then down here two times thirty seven we know from earlier is seventy four okay all right so what is the square root of sixty one thousand nine well if you punch that up on a calculator you need 247. so let's erase this and put 247. now we have two cases here but you're going to see that one of the cases ends up being nonsense right remember we're solving for x which is the amount of time that in this particular case larry is going to travel for a drive for before the two cars are 290 miles apart so you know that you can't have a negative amount of time that wouldn't make sense in this scenario i can't say well he drove for negative 18 hours as an example okay what does that mean it doesn't mean it's nonsense so you have to throw it out so the one of these guys that's going to make it negative we would just immediately exclude that okay so the first scenario would be that x is equal to 49 plus 247 over 74. so we keep that one the other scenario is what it's x is equal to 49 minus 247 over 74. this is going to end up being negative so you can just throw this out okay you don't need to worry about that solution at all so focusing on this one 49 plus 247 is 296. so if you had 296 over 74 you would have four okay so nice and simple so x equals four again that tells me that larry drove for four hours before the two cars were 290 miles apart again x was the amount of time for larry x minus one was the amount of time for steven so this guy is four and this guy is three right because four minus one is three so again officially answer this how long after larry starts driving does it take for the two cars to be 200 90 miles apart we'll say larry drives for four hours before the two cars are 290 miles apart okay and if you want to check this again it's easy to go to the diagram and kind of visually see what's going on you want to mark down some information that larry's going at 50 miles per hour and steven is going at 70 miles per hour so let's go to the little drawing we have and let's see if we can make sense of this we can erase all this we can actually leave the a squared plus b squared equals c squared we'll actually use that so if i plug in a 4 for x i have 200 so that's for larry we know that he drives 200 miles from here to here and that's basically over the span of four hours and that's how far he's going to have traveled when he gets to this point here when they're 290 miles apart and then for steven you can plug in a 4 which is what x is you get 70 times 4 is 280 280 minus 70 is 210 or you could again do it kind of manually and say okay well he was going 70 miles per hour he only drove for 3 hours 70 times 3 is 210 so either way you get 210 as his distance from here to here okay so again we know about the right triangle if this is 200 and this is 210 plugging in if one is 200 and the other is 210 if we sum those two squares it should be this guy right here squared so is that true 200 squared is what we'll punch that up on a calculator you're going to get 40 000. okay 210 squared punch that up on a calculator you're going to get 44 100 and if you sum those two amounts together you are going to get 290 squared because that's going to be 84 100 okay you can easily see that 40 thousand plus forty four thousand one hundred is eighty four thousand one hundred so we can say here our answer is correct larry drove for four hours before the two cars were 290 miles apart all right so another common type of problem in this section involves finding an unknown area so generally speaking these problems are going to deal with some type of rectangular shape this could be a pool it could be a garden it could be a quilt the example we're going to see next is going to be a rug okay so something rectangular in nature so what i want to cover really quickly is just the formula for the area of a rectangle so we have a rectangle on our screen we have our length and we have our width okay so the area of a rectangle we'll put a here is equal to the length which i'm going to put capital l times the width which is capital w okay so that's the formula for the area of a rectangle you might want to write that down because we're going to use it immediately in the next problem so just as a quick example suppose i said that the length here was eight inches and suppose i said the width here was six inches okay and i said well what's the area of this rectangle given this information well the area also put capital a is equal to what it's 8 times 6 which is 48 you're just multiplying and then inches times inches you multiply your units that's going to be inches squared or you could say square inches so we're going to say this is 48 square inches okay and let me make this q a little bit better okay so again 48 square inches when we have a length of 8 inches and a width of 6 inches all right so again let's put this formula to immediate use so we have here for our sample problem that jamie wants to put a rug in a bedroom that is 15 feet long and 12 feet wide she wants to leave a strip of uncovered floor that has a uniform width around the rug after looking at pricing information she decides that she can afford to buy a rug with an area of 108 square feet okay so what are the dimensions of the rug so again what are we being asked to find well given the fact that she can afford to buy a rug that has a maximum area of 108 square feet we want to know what are the dimensions of the rug okay well to solve this problem if you haven't seen it before it might be a little challenging for you to kind of set it up okay so usually what you do with this is you draw a little diagram and i have that already set up and what you'll see here is that the entire rectangle is meant to represent the bedroom okay and then you have this kind of smaller rectangle which represents your rug and then the brown you see that's the strip of uncovered floor okay that's around the rook okay and that's actually the key to understanding this problem so when we go back up i just want to read this part one more time she wants to leave a strip of uncovered floor that has a uniform width around the rock this is very important so this uniform width around the rug okay let me go back down and we're just going to label a few things so what do we know we know the length of the bedroom or the measurement from here to here is 15 feet okay that's given in the problem we know the width the measurement from here to here is 12 feet that's given to us in the problem we also know that this kind of strip of uncovered floor has a uniform width so in other words i know from here to here and then from here to here and from here to here and from here to here this is all going to be the same all the way around the width is all the same okay so the key to this problem is letting a variable like x or whatever you want to use represent that uniform width for that strip of uncovered floor okay so i'm just going to say this is x this is x this is x and this is x okay so given the information that we have now can you see how you would set up and find the dimensions of the rock okay so for the rug we want to know what's the length and what's the width well for the length i know the length of the bedroom it's given to me it's 15 feet so that's from here to here if i want the length of the rug well i need to subtract out kind of from here to here and from here to here okay so what's that measure and i kind of went over there so let me draw that a little bit better so what's that measure well it's going to be 15 the total amount minus this which is x okay this is x and then minus this which is x so 15 minus x minus x which is going to be 15 minus 2x okay so that's going to be my length of the rug now for the width it's a similar thought process okay the width is going to be from here to here and again i just want to think about the width of the whole room which is 12 feet so i'm just going to put 12 there and i want to subtract out this guy right here which is x and i will subtract out this guy right here which is also x so 12 minus x minus x is 12 minus 2x okay so now i have a length as an expression 15 minus 2x and i have a width as an expression 12 minus 2x so let's copy that real quick and i'm just going to paste that there and what we know from our problem it tells us that she can afford to buy a rug with an area of 108 square feet so we're going to assume that that's what she buys so my length times my width gives me my area so what i'm going to do is i'm going to say 15 minus 2x that quantity that's my length multiplied by the quantity 12 minus 2x that's my width this should equal my area and we're told that the area is 108 square feet so i'm just going to put 108 okay so i'm going to solve this equation for x and that's going to allow me to find out what is the length and what is the width for my rug that jamie has okay so 15 minus 2x times 12 minus 2x we're going to use 4 for that so 15 times 12 is 180 and then 15 times negative 2x is minus 30x negative 2x times 12 is minus 24x if i combine like terms i'm going to have negative 54x so minus 54x all right so then we have negative 2x times negative 2x which is going to be plus 4x squared and then this is just equal to 108. okay so let me kind of rearrange things here let me put this as 4x squared minus 54x plus 180 is equal to 108. now what i want to do is subtract 108 away from each side of the equation so that i can have 0 on the right side and i can go ahead and use my quadratic formula or you could factor whatever you want to do okay so in this case i'm going to have 4x squared minus 54x and then plus 180 minus 108 is going to be 72 this equals zero now if you want to do this this is an optional step you don't need to you can divide each part here by two okay that's going to give you smaller numbers to work with so if i divide this by two this by 2 this by 2 and this by 2 i'm going to get 2x squared minus 27x plus 36 equals 0. okay that's just something i did to give us smaller numbers to work with again completely optional all right so now let's use our quadratic formula again you can solve this one with factoring but i'm going to use the quadratic formula so this guy right here this is going to be a this guy right here this is going to be b this guy right here is going to be c okay so i'm going to plug into x equals the negative of b plus or minus the square root of you've got b squared minus 4ac the whole thing's over 2a okay so the negative of b b is negative 27. so just plug that in there then b squared you've got negative 27 squared 4 times a times c a is 2 c is 36 and then a again is 2. okay so we filled all this in now we can just kind of crank it out the negative of negative 27 is 27 okay so that's easy enough negative 27 squared is 729 so this would be 729 negative 4 times 2 is negative 8 and then negative 8 times 36 is negative 288 so this is basically minus 288 if i do that calculation real quick i'm going to get 441 right 729 minus 288 is 441. if i took the square root of 441 i would get 21. okay so this is 21 here and then 2 times 2 is 4. okay so we're going to get two answers you're going to find that one of them doesn't work but we'll get to that in a minute so 27 plus 21 is going to be 48 48 divided by 4 is 12. so x equals 12 and then also 27 minus 21 is six six over four you could say that's three halves or one point five so let's take these back to our illustration and we're gonna see if they make sense okay so let me paste this up here so can x be 12. remember i'm plugging in for x in each case so i'm plugging in here and here well 2 times 12 is 24. 15 minus 24 is negative okay it's going to be negative 9. does it make sense for this rug to have a length of negative 9 okay no it can't be negative 9 feet that doesn't make any sense it's nonsense so you get that when you solve these problems you just have to reject one of the solutions so go ahead and throw that out okay you can't use that can we use three halves yes we can three halves is one point five if you multiply two by one point five you get three okay so what you'd have in each case you could just replace this with three and this with three fifteen minus three is twelve so that's the length of the rug and twelve minus 3 is 9 so that's the width of the root okay so we basically found our answer for our problem so what are the dimensions of the rug we'll say that jamie's rug okay jamie's rug has a length of 12 feet and a width of 9 feet okay is that consistent with what we're told in the problem well 12 times 9 is 108 feet times feet is feet squared or square feet so that does match up with this okay but you can also check to make sure that it is consistent with the rest of the problem so when we come back to our little illustration again we just want to check that this 1.5 this figure matches up with what we got right so this 1.5 is the width of this strip of uncovered floor which has a uniform width it's supposed to be the same all the way around so it's 1.5 all the way around let me write that like this okay so does that make sense well if the length or the measurement from here to here is 12 and i add 1.5 for this and 1.5 for this well yeah 12 plus 1.5 plus 1.5 is 12 plus 3 that's 15. that's this guy right here the length of the bedroom okay so that makes sense same thing is going to happen for the width so the measurement from here to here we know that's 9 then plus another 1.5 plus another 1.5 9 plus 3 is 12. again that's given to us in the problem so that checks out and we already checked the fact that the rug matches up and it gives us 108 square feet so we can basically say we're good to go this rug has a measurement of 12 feet for the length and nine feet for the width all right so let's wrap up the lesson and talk about another common problem you see in this section we're going to be dealing with the height of a propelled object in feet so when we get to this section we basically have a little formula that we work with if air resistance is disregarded the height of an object that's propelled directly upward can be thought about with this formula so we have h is equal to negative 16 t squared plus v sub zero times t plus s sub zero so when you're in this section in your algebra course you're going to be given this formula they're going to give you stuff to plug in and then you're going to have to solve for something okay so h is the height of the projectile in feet t is the number of seconds v sub 0 is the initial velocity in feet per second and s sub 0 is the initial height in feet so when we're talking about height if i get a height of 0 that means i'm on the ground okay if i get a height that's positive let's say it's 50 let's say for h again it's 50 that means i'm at a height of 50 feet in the air okay if i get an initial height in feet it's the same thing if i got 50 that means i'm launching from 50 feet all right let's look at a sample problem i think you'll find this to be very very straightforward a toy rocket is launched vertically upward from an initial height of 500 feet the initial velocity is 20 feet per second find out how long before the rocket will hit the ground okay so obviously we're just answering this part right here where it says to find out how long before the rocket will hit the ground we're just going to take the formula let's go back for a second and we're going to grab this guy and i'm just going to rewrite it right here real quick so h is equal to you've got negative 16 t squared you've got to be careful here because the t gets confused with the plus sign so i'm going to put that in a different color so plus you've got your v sub 0 times t or you could say t times v sub 0. so let's put v sub 0 times t and again you've got another plus sign so let's do that in a different color and then you have your s sub zero okay so this is your formula that you're working with now we can plug in a lot of information here it says that the initial height is 500 feet s sub 0 is your initial height so you can just go ahead and erase this and put 500 there and then it says the initial velocity is 20 feet per second so that's this v sub 0 here so i'll erase that and put a 20 there okay so now we have two variables involved i have h and i have t but it says find out how long before the rocket will hit the ground we already said that when the rocket hits the ground h is zero right that's the height so i can just replace h with 0 and all i have here is a standard quadratic equation right so i have negative 16 t squared plus 20t plus 500 if i replace t with x it's going to look very familiar for you okay and i'm not going to keep it this way i'm going to keep it with t because that's consistent with our formula but if you were struggling and you couldn't couldn't kind of grasp what's going on just replace it with x solve it and then go back and replace that for t okay not a big deal so i can solve this using the quadratic formula i could also solve this particular one by factoring what i'm going to do to make this a little bit simpler i'm going to divide everything by 4 just so we have smaller numbers to work with so divide this by 4 this by 4 this by 4 and this by 4 i'm going to get and i'm also going to kind of rearrange things so i'm going to get negative 4 t squared then plus i'm going to have 20 over 4 is 5. so 5 t and then plus 500 divided by 4 is 125 this equals 0. okay let's go ahead and plug into the quadratic formula so we know that what we know that this guy right here this negative 4 would be a we know this guy right here this 5 would be b and this guy would be c okay this 125. so instead of x we have t so we're going to say t is equal to you've got the negative of b b is 5 so negative 5 plus or minus the square root of you've got b squared again b is 5 5 squared is 25 minus 4 times a a is negative 4 times c c is 125 and i'll do this in a minute this is over 2a a is negative 4 so 2 times negative 4 is negative 8. okay so negative 4 times negative 4 is plus 16. so let's just put plus you'd have 16 times 125 that's going to be 2000 okay so 25 plus 2 000 you do this all in one shot this is 2025. if you punch up square root of 2025 on a calculator you're going to get 45. okay so this guy's 45. now one of these solutions you're going to throw out okay because we're solving for t which is time so it's got to be a positive value or zero okay it's basically got to be non-negative if you end up with something like negative 800 seconds what does that mean right it doesn't mean anything again it's nonsense we've seen that throughout this kind of lesson if you get something that doesn't make sense just reject it throw it out okay so the first one would be negative 5 plus 45 so let me kind of do that over here so t equals so negative 5 plus 45 which would be positive 40 over negative 8 this would give me negative 5 seconds which again doesn't make sense so we're just going to throw that solution out okay we don't need it then t is equal to the other one is negative 5 minus 45 which would be negative 50 and then divided by negative 8 negative over negative is positive i can also simplify by dividing each by 2 so this would end up being 25 4. now in decimal form that's 6.25 and that's what we're going to use so let me erase all this we don't need any of it anymore so find out how long before the rock will hit the ground well we found that it's going to happen at 6.25 seconds right so we'll say that the rocket will hit the ground at 6.25 seconds okay and again a very very easy problem very very easy if you get this question on homework or test you just plug into the formula and you're basically just have to set up your quadratic equation and solve okay so how about checking something like this what would we do well again we just go through the formula again it's h is equal to we have your negative 16 times t squared for t i've got 6.25 again i can plug in a fractional form here to make it a little simpler so i'm going to put 25 fourths this guy squared then plus v sub zero the initial velocity was 20. so we're going to put 20 times you've got your t which again i'm going to put 25 fourths for that and then you've got plus your s sub 0 which is 500 that was the initial height in feet so that's 500 feet okay so h is supposed to be zero so we just check this make sure we got the right answer so let's go back down so 25 squared is 625 4 squared is 16. so this would be 625 i'm going to write that a little better so 625 over 16. this guy's going to cancel with this guy and give me a negative 1. so this would basically be negative 625 here and most of you can already see this is going to work out this cancels with this and gives me a five five times 25 is 125 so basically what you have is 125 plus 500 which is 625 and then that's being added to negative 625 so that does give you zero right you end up with zero equals 0. so this is consistent with what we're told in the problem so we can say here that our answer or solution is correct the rocket will hit the ground at 6.25 seconds in this lesson we want to review solving quadratic inequalities so a quadratic inequality is of the form we have ax squared plus bx plus c is less than zero where we can say that a the coefficient of x squared b the coefficient of x to the first power and c the constant term represent any real number with one exception we can't have a allowed to be equal to zero because if a is 0 our x squared term would disappear and we'd no longer have a quadratic inequality now additionally in this definition we can say that our less than here could be a less than or equal to a greater than or greater than or equal to all right so how can we solve a quadratic inequality well there's quite a few different methods that you can use we'll talk about a method that relies on graphing kind of later on in the course graphing is good to conceptualize what's going on but it's not necessarily the quickest method in most cases it's not going to be quickest so we're going to focus on a method that relies on test values so for solving quadratic inequalities the first thing we want to do is we want to write the inequality as an equation and solve so what this is going to do is it's going to give us our boundaries okay or some people will call them endpoints some people will call them critical values now we're going to use the solutions from the equation to set up intervals on the number line and again we'll see that in a moment when we look at an example then this is the important part here we're going to substitute a test value from each interval to determine the solution set now the problems we're going to look at today are going to be inequalities such that when we transform them into equations these quadratic equations are going to have real number solutions and more specifically they're going to have rational number solutions so these are going to be kind of easier examples just to get the concepts down you can work with these when the equations have complex solutions and also when they have irrational solutions but we'll talk about that later on again when we revisit this topic using graphing all right so let's go ahead and take a look at the first example so we have x squared minus 6x is greater than 16. so the first thing i'm going to do is kind of write this where my quadratic expression is on the left side and 0 is on the other side okay so i can do that by just subtracting 16 away from each side of the inequality i'll just write this over here this would cancel become 0. i would have that x squared minus 6x minus 16 is greater than 0. okay so now what i want to do is i want to change this inequality into inequality so all i got to do is change this greater than symbol to an equal symbol so i'm going to make this into x squared minus 6x minus 16 is equal to 0. so i can solve this using factoring i can solve this with my quadratic formula whatever you want to do it's pretty easy to solve this with factoring so let's just do that so this would be an x and this would be an x so give me two integers whose sum is negative 6 and whose product is negative 16 we know that would be negative 8 and positive 2. so now i can just set each factor equal to zero and solve so i would have x minus eight is equal to zero or i would have that x plus two is equal to zero for this one i can add eight to each side of the equation and i'll get that x is equal to for this one i can subtract 2 away from each side of the equation i'll get that x is equal to negative 2. so let's erase all of this we don't need any of it anymore we see these solutions so at x equals 8 or at x equals negative 2 we're basically saying the left side of this quadratic inequality is going to be 0. so here's 8 and here's negative 2. so what i'm going to do is i'm going to put a vertical line at each to kind of separate the number line into intervals so anything less than negative 2 or anything to the left of negative 2 let's just call this interval a for reference sake anything between negative 2 and positive 8 let's call this interval b for reference sake and anything that's greater than 8 or you could say anything to the right of 8 let's call that interval c again for reference sake so three intervals on the number line what we're going to do is test a value in each interval and see if it works in our inequality and the reason this works is that again at 8 or negative 2 the left side here is going to be 0. now all other numbers other than 8 or negative 2 will make the left side either negative or positive and the sign of the quadratic expression can only change from positive to negative or negative to positive at a number that makes it 0. so this means that if a number in the interval that we test so if a number in interval a or a number on interval b or a number on interval c if a number in one of those intervals satisfies the inequality then every number in that interval will satisfy the inequality the reverse is true if a number in that interval does not satisfy the inequality then no number in that interval will satisfy the inequality so in interval a let's just go ahead and choose something like negative 4 that would be a good test point and we can try that in this guy right here so let's plug in a negative 4 everywhere we see an x and so we would have negative 4 squared which is positive 16 then negative 6 times negative 4 is plus 24 then you'd have minus 16 and this is greater than zero well we know that 16 minus 16 is zero so you have that 24 is greater than zero which is true so i'm just going to put for interval a that this is true okay so every number in interval a would work now let's choose something in interval b using 0 is always pretty quick so let's choose that let's plug in a 0 for each x you would have that negative 16 is greater than 0 which is false so for interval b we'll say this is false and then for interval c let's choose a value of 10 because that would be pretty easy to work with so 10 squared is 100 minus 6 times 10 which is going to be 60 then minus 16 is greater than zero 100 minus 60 is 40. 40 minus 16 is 24. 24 is greater than zero this is true so for interval c we'll say it's true okay so let's erase all of this we don't need this anymore and you can label this visibly you can say that interval a is true interval b is false and interval c is true in most of the situations you're going to look at when your quadratic inequality sets up as a quadratic equation with real number solutions you're going to basically have three intervals where either the left interval and the right interval are going to work in the middle doesn't or where the middle works and the 2 on the outside so the leftmost interval and the rightmost interval don't work okay so in this situation we have that anything to the left of negative 2 would work so x is less than negative 2 or we have that anything to the right of 8 would work so x is greater than 8. now in this particular case because we have a strict inequality we see that our endpoints don't work right if i plug in a negative 2 or an 8 again the left side would be 0 0 is not greater than 0 so negative 2 is not a solution and 8 is not a solution so let me erase this and kind of drag this over here we can write our solution also in interval notation we could say that from negative infinity up to but not including negative two and the union width we have anything larger than eight okay so this is our interval notation this is using x we can also kind of graph this on a number line so we could say again anything less than negative two i'll put a parenthesis at negative two to show it's not included i'll just shade everything to the left we also said anything greater than eight i'll put a parenthesis at eight shade everything to the right all right so that one was pretty easy this one's also going to be easy but just a little bit more tedious so we have 25x squared plus 71x plus 16 is less than or equal to negative 7x squared minus x so again the first thing i'm going to do is write this as a quadratic expression on the left 0 on the right so i can add 7x squared to both sides of the inequality and i can add x to both sides of the inequality so this is going to cancel and become 0. now 25 plus 7 is 32 so this will be 32 x squared and then plus 71 plus 1 is 72 so 72x and then plus 16 is less than or equal to 0. let's erase this and we'll just kind of slide this up okay so let's replace the inequality symbol with equals and solve that equation so we're going to have 32x squared plus 72x plus 16 is equal to 0. now each number in this equation the 32 the 72 and the 16 are each divisible by 8. so 32 divided by 8 is 4 so this would be 4x squared 72 divided by 8 is 9 so plus 9x 16 divided by 8 is 2 so plus 2 and this equals 0. right obviously 0 divided by 8 is still 0. so let's erase this anytime you can make the equation simpler and work with smaller values it's always a little bit better so for this guy i could solve it using factoring or i can use my quadratic formula because this number right here is not prime it's probably going to be faster to use my quadratic formula so the 4 here represents a the 9 here represents b and the 2 here represents c so my quadratic formula in case you forgot is x equals the negative of b plus or minus the square root of b squared minus 4ac all over 2a okay so negative b we have negative 9 plus or minus b squared 9 squared would be 81 minus 4 times a times c well a here is 4 c here is 2 so negative 4 times 4 is negative 16 negative 16 times 2 is negative 32. so this would be minus 32 here then we have 2 times a in the denominator a is 4 so 2 times 4 is going to be 8. so then under our square root we have 81 minus 32 which is 49 and the square root of 49 is 7. so we have negative 9 plus 7 which is going to be what is going to be negative 2 and then over 8. so we'll say that one solution is x equals negative 2 over 8 which we know would simplify to negative 1 over 4 so negative 1 4. then the other solution would be negative 9 minus 7 and that's going to be negative 16 and then negative 16 over eight would be negative two all right so let's erase all this we don't need this anymore so we can see our two kind of critical values or boundaries are negative one fourth and negative two so at negative two i could draw a kind of vertical line there and then negative 1 4. we don't have a notch for it but let's say it's about right there so i'm just going to label this as negative 1 4 and let's just put a kind of vertical line there as well so i'm splitting this up into three intervals anything to the left of negative 2 is interval a anything between negative 2 and negative 1 4 is interval b and anything to the right of negative 1 4 is interval c all right so let's just pick a test value for each of these and let me write my 0 there again so let me divide this by 8 so this would have been 4x squared and this would have been 9x and this would have been 2. again just doing that so i have smaller values to work with so i'm going to pick an interval a a value of negative 4 so let's plug that in there and also there so negative 4 squared is 16 4 times 16 is 64. and then 9 times negative 4 is going to be negative 36 so minus 36 and then plus 2 we don't even need to do the kind of operations here we know that this would not be less than or equal to 0. so anything in interval a would be false okay this would be false so let's erase this and now let's look at something in interval b so again the way most of these work if something in interval a doesn't work then b will work and c won't work okay so for b let's pick negative one so let's plug that in here and here so negative one squared is one one times four is four so nine times negative one is minus nine and then we have plus two and this is less than or equal to zero we can tell that this is going to be negative on the left right so it will be less than zero so we can say for interval b it's true and then again in most cases okay not every case but in most cases if a works the leftmost region then b will not work and c the rightmost region will work so you have a case where kind of the leftmost region works the middle region doesn't and the rightmost region will work the other case that happens a lot will be that the middle region works and the one on the left and the one on the right don't work now this isn't true every single time but in most cases it follows this pattern okay so for region c let's pick a value of 0 because that's going to be available it's real easy to work with so i'm going to plug in a 0 there and there so we know this would be gone and this would be gone you would have that 2 is less than or equal to 0 which is false okay so let's erase this we have our solution region now integral b does work anything between negative two and negative one fourth so x is greater than or equal to because we have a non-strict inequality negative 2 and less than or equal to again because we have a non-strict inequality negative 1 4. whenever you have that non-strict inequality the endpoints are going to be solutions okay so we can show this in interval notation put a bracket here we'll have negative 2 because that's included comma you'll have your negative 1 4 and then another bracket so graphically let me kind of keep that notch there for negative 1 4. let's put that right there what we're going to do is we're going to put a bracket at negative 2 and a bracket at negative 1 4 just shade everything in between so this is our graphical solution this is interval notation and then again we can write this with our variable x and just say that x is greater than or equal to negative 2 and it's less than or equal to negative 1 4. all right let's take a look at another one so we have negative 2x squared minus x plus 3 is greater than 0. so in this case i have my quadratic expression on the left and a zero on the right so i can just change this greater than here to equals and solve the resulting equation so we have negative two x squared minus x plus three is equal to zero i can solve this using factoring pretty quickly so i know that i can first divide everything by negative 1 so that the leading coefficient is going to be positive that just makes it easier for me to factor so i'm going to say this is positive 2x squared and then we're going to have plus x minus 3 is equal to 0. okay so let's just erase this and i can factor this on the left side here pretty quickly because this 2 is a prime number so this would be 2x and this would be x i just got to work out the outer and the inner such that i get a plus 1x as my middle term right so the last term here is a negative 3 that can only come from a positive 1 times a negative 3 or a negative 1 times a positive 3. so if i want a positive 1x as the middle term i would want to put a plus 3 here and a minus 1 here right because the outer would be negative 2x and the inner would be plus 3x if you sum those you get plus 1x okay so i'm not going to set each factor equal to zero so 2x plus 3 equals 0 then or i'm going to set x minus 1 equal to 0. so for this guy i'm just going to subtract 3 away from each side of the equation i'll get that 2x is equal to negative 3 divide both sides by two and i'll get that x is equal to negative three halves for this guy i would add one to each side of the equation and i would get that x equals positive one okay so i'm going to erase all this i don't need any of this information anymore i'm just going to kind of drag this up here these are our critical values or again the boundaries or the endpoints it's basically going to be the values that make this left side 0. okay so negative 3 halves we can just say that's again halfway between negative one and negative two in decimal form it's negative 1.5 so let's just say that's right there so this is negative three halves and then we have positive one so i'm just going to put a vertical line at each just to separate the number line into kind of intervals so anything to the left of negative three halves i'm going to call that interval a anything between negative three halves and positive one i'll call that interval b and anything larger than positive one i'll call that interval c okay so in interval a let me just choose negative four so i'd have negative two times negative four squared minus negative four and then plus three is this greater than 0. so minus a negative 4 is plus 4 so we see that we know that negative 4 squared is going to be 16 and negative 2 times 16 is negative 32. so we can see that this is not going to be greater than zero so region a we know that's false again for most of these not all of them but for most of these it sets up to where the one on the left if that's going to be false then the middle part is going to be true and the part on the right is going to be false again but again you've got to really test these because you have some special case scenarios so let's look at 0 for region b so if i plugged in a 0 for x this would be gone and this would be gone i'd have that three is greater than zero which is true and then for c let's just choose four so if i square four i get sixteen sixteen times negative two is negative thirty-two and then minus four plus three is greater than zero we can tell that's going to be false okay let's erase this so basically the region that works is region b and we have a strict inequality so the end points or the boundaries or the critical values will not work so i can say that x is greater than negative three halves and less than positive one in interval notation i would have my negative three halves next to a parenthesis and then comma a 1 next to a parenthesis then kind of graphically let me erase all this i'm going to keep my notch there for negative 3 halves so graphically i'm just going to put a parenthesis at negative 3 halves and a parenthesis at positive one and then just shade everything in between so this is my graphical solution this is my interval notation and then again using x x is greater than negative three halves and less than one so pretty easy overall to solve a quadratic inequality as we move throughout our course we'll see some more challenging scenarios and we'll get to this when we start talking about graphing parabolas okay so this would be the graph of a quadratic equation we'll look at some more challenging situations with quadratic inequalities at that time so we'll just wrap up our lesson by looking at some special case scenarios and then as the final problem here we'll talk about something that's not a quadratic inequality but that we can use the same thought process to solve okay so if we saw something like x squared plus 2ax plus a squared is greater than negative k the idea here is that we know this is generic but we can show that the left side here is a perfect square trinomial so this can be factored into x plus a that quantity squared right and this is greater than negative k so don't get kind of tripped up by the fact that this is so generic right we have a negative k we have an a these are kind of foreign things when we see all these letters involved other than x but again don't let that kind of rattle you all we're trying to say here is that in this particular case i can plug in anything i want for x and i would get a true statement why is that this guy over here this negative k just represents some negative real number so it could be negative 3 or negative 10 or negative 2 billion whatever you want it to be think about the fact that x can be anything and a can be anything and i'll always get a true statement x plus a can be whatever it could be 0 it could be negative it could be positive once i square this guy if i square a negative i get a positive right a positive would always be greater than a negative if i square 0 i get 0 0 would always be greater than a negative if i square a positive i'm going to get a positive and that's always going to be greater than a negative so there's nothing you can plug in for x that won't work so if you see this scenario you're basically going to have a solution where x is any real number okay so you can say all real numbers all right let's look at an example so we have 9x squared minus 48x plus 64 is greater than negative 5. so again if you look at the left side and you have an inkling that this guy's going to be a perfect square trinomial you check the n's right so 9x squared that comes from 3x being squared and then 64 comes from 8 being squared so is the middle minus 2 times this guy the 3x times this guy the 8 well 2 times 3 is 6 6 times 8 is 48 48 times x is 48x and we've got the minus there so this is minus 48x right so this is a perfect square trinomial that means i can factor it into a binomial square right this would be 3x minus right because you have the minus there the 8 that quantity squared let me make that a little better and this is greater than negative 5. well as we just said this is going to be true for all real numbers right it doesn't matter what i plug in for x here i'm going to multiply it by 3 i'm going to subtract away 8 no matter what it is if it's negative or if it's 0 or if it's positive when i square it it's either going to be 0 or some positive number so it's always going to be greater than negative 5 so this is true for all real numbers so for this i can just write all real numbers now another thing i can do you can kind of erase this you don't need this anymore you can write this in interval notation from negative infinity to positive infinity you might want to do that or lastly you can just shade the entire number right so shade this arrow and then all the way through and then shade this arrow okay so any real number would work as a solution for x in this scenario all right let's talk about the other scenario you're going to come across so we have x squared plus 2ax plus a squared is less than negative k again the left side is a perfect square trinomial so i can factor this into x plus a that quantity squared if i say it's less than the negative of k so this is a negative number here on the right so what happens is if i choose a value for x it doesn't matter what it is i add some number a and then i square that result i'm always going to get something that's non-negative so the left side here would be 0 or something positive and that's never going to be less than a negative value so when you see this there's not going to be a solution now if you get something like x squared plus 2ax plus a squared and that's less than zero again this guy right here this is x plus a quantity squared is less than 0. this would carry no solution as well but if we change this and say it's less than or equal to 0 well what happens is you would have one solution there okay in this particular case because you could set this part right here equal to zero and you could solve that that would give you one solution okay one value that would satisfy the inequality the less than part would never work but the or equal to would work for one value okay so let's look at an example so we have 49x squared minus 140x plus 100 is less than negative 2. so the left side here i know i can factor this into a binomial squared this guy here on the left is 7 x being squared this guy here on the right is 10 being squared and the middle is minus 2 times 7x which is 14x times this guy 10 which is 140x so you have minus 140x so this would work itself out to be factored into 7x you have your minus there so minus the 10 being squared and this is less than negative 2. this will not have a solution okay there's no solution here so there's no solution okay so you can put the symbol for the null or empty set but again no value here plugged in for x would give you a true statement all right so let's look at one more example here and so what we have is the quantity x minus 1 times the quantity x plus 2 times the quantity x minus 4 is less than or equal to 0. so this if i multiplied it together on the left would not be a quadratic expression it would be a cubic expression right so i wanted to give you one where it's not quadratic and you can see that you can use pretty much the same process to solve it so if you multiply this together on the left you would get x cubed minus three x squared minus six x plus eight just went ahead and factored that already because we're not at a level where we can factor something like that yet we'll learn how to do that later on in the course so if i replace this inequality symbol with equals i can solve the resulting equation because it's already in factored form this will be pretty quick for us so we know that our solutions here would be what you'd have x minus one equals zero so x would be one x plus two equals zero so x could be negative two and then x minus four would be equal to zero so x could also be four so these are our critical values here so we have that x could be 1 so let's put a visible border there x could be negative 2 and then x could be positive 4. so we're going to have 4 intervals here so interval a will be anything to the left of negative two interval b will be anything between negative two and positive one interval c will be anything between positive one and positive four and interval d will be anything larger than positive 4. so let's erase this i don't need any of this anymore let's just choose a point from each interval and see if it works so for interval a let's just choose negative 4 so negative 4 minus 1 is negative 5. and then negative 4 plus 2 would be negative 2 so times negative 2 and then negative 4 minus 4 would be negative 8. so is this less than or equal to 0 i don't even need to go through here and multiply i know that a negative times a negative times a negative basically three negatives would give me a negative a negative is always less than zero so this part is true so then for interval b i can use zero right that's easy to work with so zero minus one is negative one zero plus two is two zero minus four is negative four so this guy because there's two negatives being multiplied together that would be a positive so for interval b that would be false right a positive is not going to be equal to 0 and a positive is not going to be less than 0. all right for interval z let's choose 2 so 2 minus 1 is 1 and then 2 plus 2 is going to be 4 and then 2 minus 4 is going to be negative 2. so this guy would be negative so this is going to work this is true and then for interval d we can choose let's say 5. 5 minus 1 is 4 5 plus 2 is 7 so it'd be times 7 and then 5 minus 4 is going to be 1. so everything here is going to be positive again a positive is not less than 0 and it's not equal to 0. so for interval d it's going to be false okay so our solution here again because we have a non-strict inequality the endpoints will be included anything in interval a so anything to the left of negative 2 and also including negative 2. so x can be less than or equal to negative 2 then or anything in interval c so anything between one and four with one and four included so x can be greater than or equal to one and less than or equal to four okay so those two kind of regions on the number line so in interval notation i can say from negative infinity up to and including negative two and it's the union with will have anything from one to four with them both being included so graphically we can say that at negative two we'd put a bracket we'd shade everything to the left and then at one we'd put a bracket and at four we'd put a bracket let me make that a little bit better and then we'd shade everything in between here okay so this is our solution graphically this is our solution using interval notation and then again using x we can say x is less than or equal to negative two or x is greater than or equal to one and less than or equal to four in this lesson we're going to review solving radical equations so when we talk about solving radical equations this is an equation with a radical involved somewhere so you might have a square root somewhere a cube root a fourth root a fifth root so on and so forth so the way we undo a square root is by kind of squaring both sides the way we'd undo a cube root is by cubing both sides you know a fourth root we'd raise both sides to the fourth power so on and so forth but there's a kind of problem that comes up when we work with radical equations to see what i'm talking about here let's look at a simple example suppose you have something like x equals a number in this case we're just going to let that number be 10. if i squared both sides so that would give me x squared is equal to 100 notice how this new equation has more solutions than the original so what do i mean by that well we know from the last section where we talked about solving quadratic equations that x squared equals 100 can be solved by the square root problem this would be x is equal to plus or minus the square root of 100 which is 10. so this basically is x is equal to 10 or x is equal to negative 10. okay well 10 works as a solution to the original equation if i plug in a 10 for x there i get 10 is equal to 10 which is true that works but if i use negative 10 let's say i plugged in a negative 10 there negative 10 does not equal 10. this is false okay this is false so what happens here is that when we squared both sides we lost a little bit of information okay we don't know whether it was 10 that was squared or negative 10 that was squared because 10 squared is 100 but also negative 10 squared is 100 so we don't know which was squared so this guy right here this x equals negative 10 is an extraneous solution okay that's what we call a solution to a new equation okay that was formed that doesn't work in the original okay so that guy is extraneous so to guard against this what happens is we have to check all of our proposed solutions when we're working with radical equations all right so for solving radical equations the first thing you want to do is isolate the radical on one side of the equation then you're going to apply the power rule so that just means if it's a square root you isolated you raise both sides to the second power if it's a cube root you raise both sides to the third power you know so on and so forth so the next step is if a radical remains you're going to repeat steps 1 and 2. so you're going to have to isolate again and then again if it's a square root square both sides if it's cube root cube both sides so on and so forth so once we've kind of dealt with all the radicals we're going to solve the equation then we're going to check all potential solutions so this guy is very important because again we want to guard against extraneous solutions all right let's take a look at the first example so we have the square root of 10x plus 1 is equal to 9. all right so in this particular case my square root operation is already isolated on one side and i just have a number on the other so essentially i just want to square both sides so i can clear that square root operation okay so we have square root of 10x plus one is equal to nine i'm just going to square both sides we know that this index of 2 here cancels with this exponent of 2. so we know that the square root operation is basically gone and i just have my radicand which is 10x plus 1 and this equals i've got to square this side as well so this is 81 okay so another way to think about this before i go further remember you can always write this in exponential notation so if you're struggling you can say this is 10x plus 1 raised to the one-half power and this is equal to 9. okay i haven't done anything illegal i just wrote the square root of 10x plus 1 as 10x plus 1 that quantity raised to the one-half power so now if i use the power to power rule here it becomes very clear one half times two is basically one so this guy right here becomes an exponent of one so that's how i know i'm getting my 10x plus 1. okay so if it's not clear for you by doing it in that way you can always use exponents to make it a little bit cleaner all right let's scooch this up and let's just solve this equation so let's subtract 1 away from each side this gives me that 10x is equal to 80. i want to divide both sides by 10 so that i can isolate my variable x and so we know that this would cancel with this and i get x is equal to 80 over 10 is 8. okay so this is my proposed solution here x is equal to 8. we need to check again to guard against errors so if i plugged in an 8 there i would get 10 times 8 which is 80 plus 1 which is 81. so we would have the principal square root of 81 is equal to 9 that's 9 equals 9. so this guy does check out so x equals 8 is the correct solution all right let's take a look at another one so suppose we saw x is equal to 2 plus the square root of x minus 2. so again the first thing i want to do is isolate the radical on one side of the equation so that square root of x minus 2 i've got to isolate that how can i do that well i've got this 2 that's being added to it so i'm just going to subtract 2 away from each side of the equation we know this would cancel on the left side i would have my x minus 2 this is equal to i'm going to have my square root of x minus 2. okay so now that i've isolated the radical on the right hand side i can square both sides to clear the radical okay but i've got to be careful here on the left this is the quantity x minus 2 squared okay we haven't talked about this in a while but again a common error is for students to just say this is x squared minus 2 squared that is wrong please don't do that remember this is a quantity x minus 2 squared so i have to go x minus 2 that quantity times x minus 2 right but again we all know at this point we can use our special products formula so let's square the first guy which is x so that's x squared then minus two times this guy which is x times this guy which is two so that would be minus four x and then plus this guy two being squared which is four then this is equal to if i squared the square root of x minus 2 i would just get that x minus 2. okay let's scroll down get some room going we'll come back up so now i want to solve this guy so let's go ahead and clean it up let's subtract x away from each side of the equation and that's going to cancel let's add 2 to both sides of the equation and that's going to cancel so on the right i know i'm going to have 0 right on the left i'm going to have that x squared minus 5x right negative 4x minus x is negative 5x and then plus 4 plus 2 is 6. okay so this is equal to 0. now i can solve this with the quadratic formula i can solve it with factoring whatever you want to do factoring is a little bit easier here so let's go ahead and set this up as x and x this equals zero over here so what are my two missing kind of items here i need one there and one there well you need two integers whose sum is negative five and whose product is positive six so that would be negative two and negative three all right so now i can use my zero product property and i can set each factor equal to zero so i'm gonna set x minus 2 equal to 0 then or i'm going to set x minus 3 equal to 0. and to solve this guy on the left i would just add 2 to both sides of the equation and we get x is equal to 2. to solve this guy on the right i would add 3 to both sides of the equation and we get x is equal to 3. so these are our two proposed solutions so let's copy this and we're going to check all right so if i plugged in a 2 for x here and here what would i get i would get that 2 is equal to 2 plus the square root of 2 minus 2 is 0. so basically this is just 2 equals 2 so this guy is going to check out if i plugged in a 3 for x so there and there you would get that 3 is equal to you have your 2 plus the square root of 3 minus 2 is 1. square root of 1 is 1. 2 plus one is three so three equals three that does check itself out all right let's look at a harder example so suppose we have the square root of two x minus two minus the square root of eight x plus one and this is equal to negative three all right so the idea here is that i've got two square roots i've got to isolate one of them first square both sides then i'm still going to have a radical so i'm going to have to do it again so this is just a more tedious problem all right so let's go ahead and add the square root of 8x plus 1 to both sides and what i'm going to have here is that the square root of 2x minus 2 is going to be equal to you'll have the square root of 8x plus 1 and then minus 3. okay so now i'm going to square both sides i'm going to square both sides this guy right here i'm just going to have my 2x minus 2 this is equal to again you've got to remember when you square this guy here it's got to be expanded okay so you could use foil or your special products formula it doesn't matter but you can't just distribute the exponent to each term there okay don't do that so if i squared the square root of 8x plus 1 i would get 8x plus 1. so that would be the first guy there then you have a minus here so this would be minus you're going to have 2 times this square root of 8x plus 1 and then times the 3. so 2 times 3 is 6. so let's just put a 6 there then plus the last guy here 3 squared okay let's scroll down get some room going all right a lot to clean up here so i know that 1 plus 9 is 10. so let me just write this as 10. and let me go ahead and subtract 10 away from each side of the equation now so that's going to cancel and let me subtract 8x away from each side of the equation so that's going to cancel on the left side 2x minus 8x is going to be negative 6x then negative 2 minus 10 is minus 12 and this is equal to you've got your negative 6 times the square root of 8x plus 1. now you have a number which is negative 6 that is multiplying your radical at this point you can square both sides as it is but to make it simpler we're going to divide both sides by negative 6 okay all right so we know that if we divide like this we can split this up and say this is negative 6x over negative 6 and i could say plus negative 12 over negative 6 just to make that a little simpler so on the left hand side i would have x and then plus 2 right negative 12 over negative 6 is 2 and this is equal to this cancels so i just have my square root of 8x plus 1. so now to clear this radical i'm just going to square both sides of the equation so i'm going to square this side and i'm going to square this side so the quantity x plus 2 squared again i've got to use my special products formula or use foil or whatever you want to do just don't distribute the exponent to each term okay that's wrong so this would be x squared plus 2 times x times 2 so 4x and then plus 2 squared which is 4. this is equal to if i square this guy we get 8x plus 1. okay so now let's clean this up let's subtract 1 away from each side of the equation that's going to cancel let's subtract 8x away from each side of the equation and that's going to cancel so i'm going to have that x squared minus 4x plus 3 is equal to 0. so again i can solve this guy with factoring or i can solve it with the quadratic formula whatever you want to use it's probably easier to solve it just with factoring so let's go ahead and put this as x here and x here and two items that i'm missing here those numbers would sum to negative four and give me a product of positive three so that would be negative three and negative one okay so now i can use my zero product property to get a solution we'd have that x minus 3 is equal to 0 or x minus 1 is equal to 0. so if i solve this guy here i add 3 to both sides of the equation i get x is equal to 3. if i solve this guy right here i add 1 to both sides of the equation i get that x is equal to 1. so x equals 3 are x equals 1. so we're not done we need to check these so let's check x equals 3 first so i'm going to plug in a 3 there and there so 2 times 3 is 6 6 minus 2 is 4 square root of 4 is 2 then minus 8 times 3 is 24 24 plus 1 is 25 so the square root of 25 is 5. so is 2 minus 5 equal to negative 3 yes it is so that checks out if i plugged in a 1 for each occurrence of x what would i get 2 times 1 is 2 2 minus 2 is 0. so that's gone 8 times 1 is 8 8 plus 1 is 9 so you would have the negative square root of 9 square root of 9 is 3. if i put a negative out in front it's negative 3. so negative 3 equals negative 3 so x equals 1 checks out as well all right now let's look at one with an extraneous solution okay so this guy right here you're going to have a problem so we have the square root of 7x plus 2 is equal to negative 3 minus the square root of 3 minus x okay so on the left hand side this square root operation is already isolated so let's get things started by just squaring both sides so i have the square root of 7x plus 2 i'm going to square this side and then i'm going to square this side so negative 3 minus the square root of 3 minus x again squared on the left i'm just going to have my radicand which is 7x plus 2. on the right again i've got to use my special products formula so i'm gonna go ahead and square this guy right here this negative three so that's going to give me a positive nine then minus you've got two times you've got your negative three and then times the square root of 3 minus x so negative 2 times negative 3 is 6. so let's put plus 6 there and then times again that square root of 3 minus x now you have to be careful when you use this when the first term is negative okay we always account for this sign right here if it's positive then you get a positive there in the middle term if it's negative you get a negative there in the middle term okay but in this case we have a negative as the first term and you don't usually see that when you're working with these kind of sample problems okay so you have to account for this negative so that's why we have a positive here because we ended up with a negative times a negative that gave us a pause okay so make sure you understand all right then plus we've got this last guy squared if i squared the square root of 3 minus x i would get 3 minus x okay so let's clean this up now we would have 9 plus 3 which is 12 okay so this is 12. if i subtract 12 away from each side of the equation i would have 7x plus 2 minus 12 is negative 10. so you put plus negative 10 or minus 10 whatever you want to do this is equal to then i'm going to add x to both sides of the equation and we know this is going to cancel over here 7x plus x is going to be 8x so let me erase this and just put 8x here so then i have on the right hand side just my 6 times the square root of 3 minus x now at this point i can square both sides but it might be a little easier if i kind of divide both sides by six you're not going to get whole numbers but i find it to be easier to work with smaller numbers even if they're fractions okay so i'm going to divide both sides by 6 on the right side i'm going to have the square root of 3 minus x on the left side you think about 8 over 6 that's 4 3 times x and then i'm going to put plus you have negative 10 over 6 let's put minus 5 thirds okay now i'm going to square both sides so i'm going to square both sides so use my formula i'd square 4 3 x so that would be 16 9 times x squared then this guy would really be a minus here right i wrote plus negative but it's really a minus so let's put a minus out in front you have your two times the four x times the five-thirds okay so if i go through and multiply that you have two times four which is eight eight times five which is forty so this is forty in the numerator over 3 times 3 which is 9 in the denominator then plus you're lastly going to have 5 3 squared 5 squared is 25 and then 3 squared is 9. okay so this is equal to over here the square root of 3 minus x being squared is just 3 minus x so now what i want to do to clean this up i'm going to multiply both sides of the equation by 9. so if i multiplied these guys by 9 this guy right here this guy right here and this guy right here the denominators would just be cleared right so you'd have 16x squared minus 40x plus 25. this equals i've got to multiply each term here by 9. 3 times 9 is 27 then minus 9 times x is 9x okay so let's clean this up if i add 9x to both sides of the equation that will cancel and then if i subtract 27 from both sides of the equation that'll cancel okay so on the left hand side i just have my 16x squared and then negative 40x plus 9x would give me negative 31x and then 25 minus 27 is -2 this equals 0. okay so i'm going to solve this using my quadratic formula and so basically let me scroll down a little bit i know that this 16 is my value for a the negative 31 is my value for b and the negative 2 is my value for c so x is equal to the negative of b b here is negative 31 so that's 31 plus or minus the square root of b squared if i squared negative 31 i would get 961 then minus you've got 4 times a a is 16 times c c is negative 2. so negative 4 times 16 is going to be negative 64 then times negative 2 is positive 128. so you basically have a plus 128 there so 961 plus 128 is 1089. so this is 1089. this is over you've got 2 times a a is 16 so this is 2 times 16 which is 32. now the square root of 1089 is 33 okay so this is 33 here so this gives me two solutions we've got that x is equal to 31 plus 33 is 64 and 64 over 32 is 2. the other solution is 31 minus 33 which is negative 2 over 32 so negative 2 over 32 each is divisible by 2 there this would be a 1 and this would be a 16. so let's go ahead and check these okay so let's start off by checking x equals 2. so if i plugged in a 2 there and there what would i get well 7 times 2 is 14 14 plus 2 is 16. the principal square root of 16 is 4. this equals over here we'd have a 2 for x so 3 minus 2 is 1 square root of 1 is 1. so you'd have negative 3 minus 1 negative 3 minus 1 is negative 4. that does not work out 4 does not equal negative 4 this is false so go ahead and mark this one off okay that's not a solution that one is extraneous then let's try negative 160. okay so we're going to plug that in so here and also here so this one's going to be a bit messy so 7 times negative 1 16 is negative 7 16 then plus 2 i get a common denominator here by multiplying 2 by 16 over 16 so that would be 32 over 16. so negative 7 plus 32 is going to be 25 so what i'd have here is i'd have the square root of 25 over 16 which basically is 5 4 okay so that's my left side on the right side here if i have 3 let me kind of slide this over if i have 3 minus again negative 1 16 so i can put plus 1 16. well what's going to happen is i need to get a common denominator so i'll multiply this by 16 over 16. 3 times 16 is 48 so you'd have 48 plus 1 which is 49 so this is 49 over 16. so the square root of 49 over 16 is 7 4. okay so you'd have negative 3 minus 7 4. to get a common denominator here let me multiply this by 4 over 4 so this would be negative 12 fourths so end up with negative 12 minus 7 which is negative 19 over 4. so we can clearly see that this one doesn't work out either okay this is also an extraneous solution because 5 4 does not equal negative 19 force right this is false okay let me make that a a little bit better so in this particular case you do not have a solution for your equation you can go ahead and just mark this as no solution no solution [Music] in this lesson we want to continue to review solving radical equations all right let's go ahead and take a look at an example with some cube roots involved so we have the cube root of 4x plus 3 is equal to the cube root of 2x minus 1. again the idea here is the same you want the radical isolated on one side of the equation in this case we have a radical on one side and a radical on the other so they're both isolated so all i need to do is use my power property and that just tells me i can raise both sides of the equation to the third power since i have a cube root in each case so let me just rewrite this i have the third root of 4x plus 3. i'm going to raise this to the third power this is equal to the third root of 2x minus 1 i'm going to raise this to the third power okay so we know that the index here would cancel with the exponent here i'm just left with the radicand so 4x plus 3 is equal to again index with exponents just left with the radicand so you have your 2x minus 1. so from this point it's really easy to solve i would subtract 3 away from each side of the equation and i would subtract 2x away from each side of the equation and this is going to cancel and this is going to cancel 4x minus 2x is going to give me 2x and this is equal to negative 1 minus 3 is going to be negative 4. divide both sides of the equation by 2 so that we can isolate our variable x this cancels with this i get x is equal to negative two okay so again we said x was equal to negative two let's plug in a negative two there and then also there so we would have the cube root of four times negative two is negative 8 negative 8 plus 3 is negative 5. so the cube root of negative 5 is equal to the cube root of 2 times negative 2 is negative 4 negative 4 minus 1 is also negative 5. so you get the cube root of negative 5 is equal to the cube root of negative 5. it's the same value on both sides of the equation so we can go ahead and check this off and say x equals negative 2 is the correct solution all right let's go ahead and look at another one so we have 4 times the cube root of x minus 4 plus 36 is equal to 28. again the idea is to isolate the radical on one side of the equation we can then use the power property to get rid of the radical so what i'm going to do here is i'm going to subtract 36 away from each side of the equation and this will cancel so i'll have 4 times the cube root of we've got our x minus 4 there this is equal to 28 minus 36 is going to give me negative 8. all right so at this point since i have 4 times the cube root of x minus 4 i can cube both sides of the equation but again it's always easier if you get rid of this part right here that's multiplying the radical so i'm going to divide both sides by 4. sometimes when you do this the number on the other side of the radical is not going to be an integer or whole number if you want to think about it that way you'll sometimes have to deal with a fraction i prefer to do that but you might not okay and a lot of books will not do that if they get a fraction on the other side so in this case it turned out to be an integer so this is going to cancel with this we get the cube root of x minus 4 is equal to negative 8 over 4 is negative 2. now we can cube both sides to get rid of that radical so i know that this index here will cancel with this exponent i'm just going to have my x minus 4 that radicand this is equal to negative 2 cubed is negative 8. okay so very easy to solve this equation i just add 4 to both sides of the equation and i get x is equal to let's go ahead and visibly cancel this negative 8 plus 4 is going to be negative 4. so that's my solution let's go ahead and check it all right so if i plugged in a negative 4 there for x what i get so you would have 4 times the cube root of negative 4 minus 4 which is negative 8. and let me make that 3 a little bit better then plus 36 is equal to 28. okay so the cube root of negative 8 is negative 2. so i can say i'd have 4 times negative 2 which would be negative 8 plus 36 equals 28 and this is a true statement negative 8 plus 36 is 28 so you get that 28 is equal to 28 so i can go up here and say that x equals negative 4 is the correct solution all right let's take a look at one with a fourth root involved so we have the fourth root of x squared minus 16 is equal to 2. so very easy to solve this if i want to get rid of a radical with kind of an index of a 4 i'm just going to raise both sides of the equation to the 4th power right this guy's already isolated for me so again i've got the fourth root of x squared minus 16 is equal to two i'm going to raise this side to the fourth power and this side to the fourth power so i know that again this index cancels with this exponent and i just have my radicand so i have x squared minus 16 is equal to 2 to the fourth power is 16. so let me scroll down and get some room going here to solve this i'm going to add 16 to both sides of the equation that's going to cancel i'll have x squared is equal to 32 i can solve this using my square root property so i know this would be x is equal to plus or minus the square root of 32. now 32 is 16 times 2 16 is a perfect square right the square root of 16 is 4. so let me kind of get some more room going here i'll say this is x is equal to plus or minus again the square root of 16 is 4 and times the square root of 2 okay so this is my solution you could say that x is equal to 4 times square root of 2 or x is equal to negative 4 times square root of 2. all right we have the fourth root of x squared so if i squared 4 times the square root of 2. let's do that off the side so 4 times square root of 2 if i squared this i would end up with what square 4 and you get 16 you square the square root of 2 and you get 2. so 16 times 2 would be 32. so this guy would be 32 minus 16 and this should equal 2 over here so 32 minus 16 is 16 and the fourth root of 16 is 2. okay so this guy right here does check itself out now you're going to get the same thing when you plug in this guy right here because the negative there gets squared okay so the negative 4 times square root of 2 if you square that whole thing you would again end up with 32 32 minus 16 is 16 the fourth root of 16 is 2 so this guy would check itself out as well all right let's look at a more challenging problem so we have the square root of 2 times the square root of 7x plus 2 is equal to the square root of 3x plus 2. so the easiest way to kind of solve this problem is to use exponents so i'm going to rewrite this and say that i have 2 times the quantity seven x plus two so this guy right here if you notice is just underneath a square root symbol so that means it's raised to the one half power so that parts raised to the one half power and then this whole thing here is raised to the one half power okay think about that the two times the square root of seven x plus two is underneath this radical here okay so the whole thing is raised to the one half power but you've got another one half power that's going to surround the 7x plus 2. okay so make sure you understand how to set that up so once we have this we'll use our power to power rule and we'll do that in a minute this is equal to this guy's easy it's just 3x plus 2 inside of parentheses again raised to the one-half power okay let's scroll down get some room going so now because this is multiplication here i can bring this exponent inside to kind of each guy that's involved in this multiplication so 2 will be raised to the one half power and then we have times the 7x plus 2 raised to the one half power is now raised to the one half power again so i'm going to use my power to power rule and say i have 7x plus 2 raised to the power of one-half times one-half which is one-fourth and this equals you've got your 3x plus 2 raised to the one-half pop okay let's scroll down and get some room going so how can we solve this equation well we're going to look for the lcm of the denominators and basically raise each side to this power in this case it's a 4 okay so i want to raise both sides to the 4th power that's going to basically clear all those fractional exponents so i'm going to raise this side to the fourth power and i'm going to raise this side to the fourth power okay so again power to power rule i would have 2 raised to the power of one half which is raised to the fourth power so four times one half is two so this is two squared times you've got the quantity seven x plus two raised to the one fourth power raised to the fourth power so the seven x plus two part stays the same one fourth times four is one okay and this equals over here i've got the quantity three x plus two raised to the one half power and then raised to the fourth power again three x plus two that quantity stays the same then i multiply one half times four which is two okay let's go down and get some more room all right so we know that two squared is four and we basically have four times seven x which is 28 28x and then plus 4 times 2 which is 8. then this is equal to you would have this guy right here that you need to expand again this is applying to something that has addition involved this is not multiplication so you can't distribute the 2 to each kind of term in there you have to foil this out or again use the special products formula so to use the special products formula i would square 3x which is 9x squared then plus you have 2 times this term 3x times this term which is 2 then plus your final term 2 squared which is 4. now 2 times 3 is 6 6 times 2 is 12. so this would be 12x in the middle okay so let's see what we got going here we can go ahead and subtract 8 away from each side of the equation and subtract 28 x away from each side of the equation so this is going to cancel over here and here and i'll just say this is a 0. so i'll have that 9x squared and then 12x minus 28x is negative 16x so we'll put minus 16x and then 4 minus 8 is negative 4 and this is equal to 0. so let's copy this and bring this to a new page okay so we can use our quadratic formula on this so x would be equal to the negative of b b in this case is what it's negative 16 right because this is a this is your b this is your c so the negative of negative 16 is 16 plus or minus the square root of b squared negative 16 squared is 256 then minus 4 times a a is 9 times c c is negative 4. so what's this going to give me negative 4 times 9 is negative 36 then times negative 4 is 144. so 256 plus 144 is 400 if i have the square root of 400 that would be 20. okay so write this as a 20 here then this is over 2 times a again a is 9 so this is 18. so i've got two solutions here i've got x is equal to 16 plus 20 which is 36 over 18 which is 2 and then x is equal to 16 minus 20 which is going to be negative 4 over your 18. now this one can be reduced right if i divide both of these by 2 i would get negative 2 over 9. all right so let's go ahead and check these solutions so if i plugged in a 2 here and here what would i get well on the right side 3 of 2 is 6 6 plus 2 is 8. so this would be the square root of 8. we know that simplifies into square root of 4 is 2 so this would be 2 times square root of 2. 2 times square root of 2. what do i have over here 7 times 2 is 14 14 plus 2 is 16. so the square root of 16 is 4 so i would have the square root of 2 times 4 which we know is 8 right the 4 can come out there and you would have 2 times square root of 2. so 2 times square root of 2 on each side of the equation so we can check this one off all right let's look at one that's a little bit more tedious to check so now we're going to look at the negative 2 9. so on the right side 3 times negative 2 9 so 3 times negative 2 9 we can see that this cancels with this and gives me a 3 there so you would have negative two-thirds you would have negative two-thirds plus two so get a common denominator going here i'd multiply this by three over three two times three is six so you would have six plus negative two which is four so you basically have the square root of four thirds here so the square root of four thirds if you split that up you would get 2 you would get 2 over the square root of 3. now you can rationalize the denominator there but it's not necessary because this isn't your answer right you don't need to present a simplified answer here you're just checking to make sure you get 2 over square root of 3 on the left side and you'll be good to go all right so let me kind of slide this out of the way and put this over here for now so on the left 7 times negative 2 9 is what you would get negative 14 9 you would get negative 14 9. then plus get a common denominator here if i'm adding 2 i can write that as 18 9. so essentially what i would have here is negative 14 plus 18 which is 4 so you get 4 9 there okay so 4 9. so the square root of 4 9 square root of 4 is 2 square root of 9 is 3. so you get two thirds so you would have two times two thirds underneath a square root symbol we know that would be the square root of four thirds which is exactly what we had over there square root of four is two square root of 4 is 2 over the square root of 3. again i don't need to rationalize this because i'm just checking to make sure i get the same value on the left as i get on the right and i did so 2 is a valid solution and so is negative 2 9. all right let's take a look at one more of these and what we're going to find here is that we have a mix of indexes okay so we have the square root of x plus 3 is equal to the cube root of 2x plus 5. so what do we do in this situation i can't just raise both sides to the second power because that's not going to clear all the radicals i can't just raise both sides to the third power because that's not going to clear all the radicals so what you need is you need the lcm between the two indexes you need to raise both sides to that power that's how you're going to clear both radicals so since this index is a 2 and this one's a 3 i need to raise both sides to the 6th power okay so i would say this is the square root of x plus 3 i'm going to raise this to the 6th power and this over here is the cube root of 2x plus five i'm going to raise this to the sixth power now if it's easier for you to kind of conceptualize this with exponents that's fine you could write this as x plus 3 to the one-half power and then raise that to the sixth power and you can write this as 2x plus 5 to the one-third power and then raise that to the sixth power perfectly fine okay it doesn't matter which way you do it now the power to power rule if i have x plus three raised to the one half power and then it's raised to the sixth power then my x plus three will be raised to the power of one six times a half is basically going to be three right and this is equal to over here my two x plus five raised to the one third power is raised to the sixth power again six times a third would give me two so this is two x plus five squared okay so let me erase this we're not going to need it anymore and let me kind of scooch this up give myself a little bit of room because this guy will be kind of messy okay so we know the formula for a binomial cubed it's going to be the first guy cubed so x cubed plus you've got three times this guy squared so x squared times this guy to the first power so this would be a nine here then plus you've got three times this guy is now raised to the first power and this guy's going to be squared so 9 times 3 is 27 this would be 27x and then lastly plus this last guy 3 cubed which is 27. okay so this is equal to if i squared this guy i would have 4x squared plus 2 times 2x times 5 so 2 times 5 is 10 10 times 2 is 20 and then times x and then lastly plus 5 squared is 25. so kind of hard to fit that all on the screen but let's go through and see what we can do so we've got our x cube nothing to combine with that i can subtract 4x squared away from each side of the equation so that's gone i can subtract 20x away from each side of the equation so that's gone and i can subtract 25 away from each side of the equation so that's gone so i'm going to have that x cubed plus 5 x squared plus seven x plus two is equal to zero so at this point most of us don't know how to factor this yet we haven't learned about the rational root theorem we'll talk about that later on in the course so i'm just going to go ahead and tell you that this will factor into x plus 2 times over here you would have x squared plus 3 x plus 1 this equals 0. so i'll just give you this later on in the course we'll learn how to factor polynomials on the left side like this for right now we'll just take this as a given and we're going to solve this by using our zero product property so i could set x plus two equal to zero and i know that if i subtract 2 away from each side i would get x is equal to negative 2. so that's going to be one solution then my other solution i would set x squared plus 3x plus 1 equal to 0. i can solve that with my quadratic formula because i can't factor this guy let's go ahead and label this 1 as the coefficient of x squared as a let's label this 3 as the coefficient of x to the first power is b let's label this one out here this constant is c so x is equal to the negative of b b is 3. so the negative of 3 plus or minus the square root of b squared so 3 squared is 9 and then minus 4 times a a is 1 times c c is 1 also so 9 minus 4 is 5. let's just write this as square root of 5 here and this is over 2 times a a is 1. so this is as simple as you can make this you have negative 3 plus square root of 5 over 2 and then negative 3 minus square root of 5 over 2. so let's copy these and we'll go back up so for x equals negative 2 let's just plug in a negative 2 here and here very straightforward negative 2 plus 3 would be one square root of one is one this is equal to two times negative two is negative four negative four plus five is one the cube root of one is one so this guy definitely checks out this is a solution right x equals negative two now for this guy right here it's very very tedious especially since you have different kind of indexes here so what i'm going to do is i'm going to approximate each root here okay so i'm going to say that x is approximately if i did negative 3 plus square root of 5 over 2 i'm going to go ahead and approximate that to negative 0.382 okay then another one i'm going to go x is approximately we've got negative 3 minus the square root of 5 over 2. i'm going to say this is about negative 2.618 okay so it's just easier if you use approximations here you're going to see that they're very close to being equal right if they work and then they won't even be close at all if they don't work okay so let's go ahead and plug this guy in for x so negative 0.3 let's plug that in for x as well all right so on the left side 3 minus 0.382 is 2.618 so i would have the square root of 2.618 and if i approximate this i can say it's about 1.62 okay about that so let me kind of erase this and i'll say on this side it's approximately 1.62 so i'm looking for something on this side that's very close to that okay so 2 times negative 0.382 would be negative 0.764 so let's say this is negative 0.764 then plus 5 that would give me 4.236 so i would have the cube root of 4.236 and i can approximate this to 1.62 as well so these sides are approximately the same when we use again these approximations so that's enough for me to say that x equals negative 3 plus the square root of 5 over 2 is a solution so i can check this where i can kind of break these up which would be better i'll say x equals negative 3 plus square root of 5 over 2 let me kind of erase this this is a valid solution so let me kind of spring this up and let me check this one now so let's erase this so you've got negative 2.618 plus 3 that's going to give you a 0.382 if you take the square root you can say that's about 0.618 okay that's approximately what that's going to be on this side if i plug this in for x you're going to see that you have a problem 2 times this guy right here is going to give you negative 5.236 well when i add that to 5 i'm going to get a negative the cube root of a negative is going to be a negative this side's a positive so it's not possible for this one to work so you can reject the solution of x equals negative 3 minus the square root of 5 over 2. this guy is going to be extraneous okay so if you get in a situation like that don't be afraid to use approximations you can use that to see if your solution is going to work or not right we can immediately see that this guy doesn't work this guy was very close so you know if you plugged in the actual number it would work so this is an acceptable answer and so is this one so x equals negative 2 or x equals negative 3 plus square root of 5 over 2. in this lesson we want to review solving radical inequalities so in our last lesson we review the process used to solve radical equations in this lesson we're just going to build on that knowledge and review how to solve radical inequalities so solving radical inequalities is not something that's hard but it can be very very tedious depending on the complexity of the problem so before we kind of jump in and start looking at problems there's a few things that we need to remember so first and foremost when we're working in the real number system a radical with an even index so a square root a fourth root a sixth root you know so on and so forth it's got to have a non-negative radicand so something like the square root of negative 25 again this is not real okay this is not real and i know we can use the imaginary unit i here to kind of get a solution but we're not talking about the complex number system here we're talking about just real numbers okay so in the real number system there's no number that we can square that would give us a result of this radicand here of negative 25 so this is not real so to guard against this taking the square root of a negative when we have a variable involved let's say we had something like the square root of x we've got to set that variable as greater than or equal to 0 right to guard against a square root of a negative okay or if you had something like the fourth root of x well again x has to be greater than or equal to zero because again if you plug in a negative here or here you can't do that right you don't want the square root of a negative you don't want the fourth root of a negative you don't want the even root of a negative now if you get something more complex let's say you had the square root of something like x minus seven well here you need to make sure that this radicand ends up as a non-negative value so what you do is you take your expression and you're going to set it as greater than or equal to 0. this is going to give you any domain restrictions that you would have you're going to find the allowable values for x by doing this so i'm going to add 7 to both sides of the inequality and i'm going to find that x must be greater than or equal to 7 okay so if x is 7 we're okay because 7 minus 7 is 0 square root of 0 that's fine that's just 0. but if x is something less than 7 well then we have a problem right if i plugged in 6 there 6 minus 7 would be negative 1 square root of negative 1 again is not real so when you're working with again square roots 4th roots 6 roots you know so on and so forth you need to make sure that if it's an expression under there with a variable you set that guy as greater than or equal to 0 and you solve it and that's going to give you any domain restrictions that you would have again for this example we found that x must be greater than or equal to 7. now if we're working with a radical with an odd index you don't have this issue right we know that if we have something like the cube root of negative 8 this is a real number this is negative 2 right because i can raise a negative number to an odd power and get a negative okay so the cube root of negative eight is negative two or the fifth root of let's say negative 243 would be negative three right because negative three to the fifth power is negative 243. so if i got something like let's say the cube root of x this is valid for any real number right so i can plug in anything for x you can say x is any real number okay so with an odd index you don't have those restrictions all right let's jump in and look at a problem so we have that 3 times the square root of x plus 3 plus 13 is greater than or equal to 16. so the first thing i would do with this type of problem is find the domain restriction so we would take this expression here for our radicand this x plus 3 and we would say it has to be greater than or equal to 0 right this is what we're going to accept as values for x so let's subtract 3 away from both sides of the inequality we're going to find that x must be greater than or equal to negative 3. again that makes sense because if i plug in a negative 3 here negative 3 plus 3 is 0 so you're good there right square to 0 you're fine that's 0. if you plug in something less though you've got a problem if you plugged in a negative 4 negative 4 plus 3 is negative 1 square root of negative 1 again is not real so let's state this again as our domain restriction x must be greater than or equal to negative 3. so if we solve this guy and we get something that violates the domain we've got to kind of exclude that okay we'll get to that in a minute so now let's talk about solving the inequality so the way i'm going to do this is i'm going to replace this inequality symbol with equals and i'm going to solve the resulting equation so i'm going to have 3 times the square root of x plus 3 then plus 13 is equal to 60. okay so i'm going to solve this i'm going to get my boundary okay and then i'm going to look at my number line and i'm going to test each side of the boundary okay so how do we solve this we know that our goal is to isolate the radical right to start so i'm going to subtract 13 away from each side of the equation and we know that this is going to cancel we'll have 3 times the square root of x plus 3 is equal to 16 minus 13 is 3. so now to isolate the radical all i want to do is divide both sides of the equation by 3 and that's going to give me what let me kind of scroll down and get some room going that's going to give me that the square root of x plus 3 is equal to 1. now that i have my radical isolated i can square both sides to get rid of it so i'm going to square this side and i'm going to square this side now remember when you square both sides of an equation you might end up with extraneous solutions okay so we've got to make sure we check so this guy is going to end up as x plus 3 is equal to 1. okay so to solve this guy i subtract 3 away from each side of the equation and end up with x is equal to negative 2. so this is my boundary okay if i plugged in a negative 2 for x the left side is going to end up as 16. so let's see that real quick so let me write this down here as my boundary x is equal to negative 2. so again if i plug that in here negative 2 plus 3 is 1 square root of 1 is 1. so you basically have 3 times 1 which is 3 plus 13 which is 16 so you get that 16 is greater than or equal to 60. again if this was an equation you'd see that you have a true statement right so we know that this solution here is not extraneous okay but consider the fact that you get 16 is greater than or equal to 16. so we know that a value of negative 2 works as a solution for the inequality so it's going to be included now what you need to do is test values on each side of negative 2 to see if they work all right so we come down to the number line and essentially we know that x equals negative 2 is our boundary okay let me just label this for the first problem so this is our boundary again some people call them critical values critical points whatever term you want to use so at negative 2 i'm just going to put a vertical line here so to the left of negative 2 we have values that are less than negative 2 to the right of negative 2 we have values that are greater than negative 2. so we want to see what's going to work in the original inequality so we have 3 times the square root of x plus 3 then plus 13 is greater than or equal to 16. now this is subject to our constraint right our domain restriction that specifically told us that x had to be greater than or equal to negative three okay so really you could make another interval here if you wanted to you don't have to but you could make another one just to visibly say that hey anything to the left of negative 3 is not going to work so if it shows up as a solution you've got to reject it so i'm just going to label this as interval a i'm going to label this as interval b and i'll label this as interval c just to keep track of what's going on so what we're going to do here is we're going to test a value in interval b and c only we don't need to test stuff in interval a that's an automatic rejection zone we'll put a false there and f for that because again it violates the domain so between negative 2 and negative 3 so i'm going to pick a value of negative 2.75 so negative 2.75 plus 3 would be 0.25 so i would have 3 times the square root of 0.25 plus 13 is greater than or equal to 16. so what's the square root of 0.25 or what's the square root of 1 fourth well it's going to be one half so this would be three times point five or three times one half plus 13 is greater than or equal to 16. and again three times a half is just three halves so you'd have three halves plus 13 is greater than or equal to 16. without going through and getting a common denominator and all that stuff you can see that's not going to work right this is basically going to be 13 plus 1.5 which is 14.5 that's not greater than 16 and that's not equal to 16. so things that interval b are not going to work we'll put this as false also so what about things in interval c what about things that are greater than negative 2 we already know that negative 2 works because it's a non-strict inequality and we tested it earlier but what about things that are greater than negative 2. does that work so for interval c well let's go ahead and try to make this easy on ourselves a value of 1. so we'll plug that in 1 plus 3 is 4 square root of 4 is 2 so you'd have 3 times 2 which is 6 plus 13 which is 19 so you'd have 19 is greater than or equal to 16 which is true so things in interval c will work so that's where we're gonna have our solution region so integral c is anything greater than negative two and again because this is a non-tricked inequality here and we tested it right if we plug in a value of negative 2 negative 2 gives us a value of 16 on the left we'd have 16 on the right 16 is greater than or equal to 16. that's true right so negative 2 does work so we would say that our solution here is x is greater than or equal to negative two and the great thing here is that it doesn't violate our domain right our domain says that it has to be greater than or equal to negative three or greater than or equal to negative two so we're basically good to go so in interval notation we put a bracket a negative 2 a comma an infinity symbol and a parenthesis and then graphically you can kind of follow that same pattern we'll put a bracket at negative 2 and we'll just shade everything to the right okay and that'll be our solution all right let's take a look at another one so we have the square root of x plus 12 minus the square root of x is greater than 2. so since we have two square roots involved we're going to think about two different domain restrictions and we got to go with the stricter one okay so for this one i would set x plus 12 and say that's got to be greater than or equal to zero and then for this one i would say x has to be greater than or equal to zero so for this one i subtract 12 away from each side i find that x has to be greater than or equal to negative 12. for this one we know that x just has to be greater than or equal to zero this one is more restrictive okay so i'm going to go with this one as my domain because if i got a value that's between negative 12 and 0 it wouldn't work right it might work in this one but it's not going to work in that one okay so you've got to go with the stricter one so what we're going to say is that x has to be greater than or equal to 0. so let's just write that over here we'll say x has to be greater than or equal to 0. all right now to solve this i'm just going to replace this with equals so i'm going to have the square root of x plus 12 minus the square root of x is equal to 2. all right so i'm going to isolate one of the radicals i have two involved so you can isolate either one first doesn't matter it's probably going to be easier to isolate this one so what i'm going to do is i'm going to add the square root of x to both sides of the equation and so on the left i'm going to have the square root of x plus 12 and this is going to be equal to on the right i'm going to have the square root of x and then plus 2. so now i'm going to square both sides let me get some room going i'm going to square this side and i'm going to square this side so on the left i'll just have x plus 12. on the right remember i have to expand this so you can use foil or you can use your special products formula so i would square this first guy so the square root of x squared is x and then plus you'd have 2 times this guy times this guy so 2 times 2 is 4 then times your square root of x then plus the last guy 2 squared which is 4. so i still have a radical involved so i need to make sure that i clear that guy out so i want to isolate that again so what i'm going to do first is just kind of simplify or clean things up i have an x on each side if i subtracted that away from each side it would cancel everywhere and basically what i'm going to do is subtract 4 away from each side so i'm going to have that 8 is equal to 4 times the square root of x again if i want to isolate the radical let's divide both sides of the equation by 4 and let me kind of get some more room going so this cancels with this and gives me a 2. this cancels with this and gives me a 1. you've got 2 is equal to the square root of x if we square both sides we're going to get that 4 is equal to x okay or x equals 4. so let me erase everything all right so we found that our boundary was x equals 4. again you can check this if you plug in a 4 there and there 4 plus 12 is 16 square root of 16 is 4. then minus the square root of 4 is 2. again i replace this with equals so 4 minus 2 is 2 2 equals 2 so that works out right we don't have an extraneous solution now this is a strict inequality so 4 does not work as a solution okay so we know that what we want to do is go to the number line and we want to test values on each side of 4 to see what works and what doesn't now we know that we have a domain restriction here that x has to be greater than or equal to zero all right so we found that our boundary was at x equals four all right so here's four on the number line so i want to test values on each side of this remember x needs to be greater than or equal to 0. this is our domain restriction so if i test values on the left side of 4 what would i get so let's go ahead and rewrite our inequality so we have the square root of x plus 12. and then minus the square root of x and again this was greater than 2. so i can choose anything i want with the exception that it has to be greater than or equal to 0 right so it's got to be kind of to the right or equal to that 0 there so let's just go ahead and pick one so if i plugged in a 1 there and there what would i get well here on the left you'd have square root of 1 plus 12 which is square root of 13. then minus square root of 1 is 1 this should be greater than 2. so the square root of 13 is about 3.6 okay if i subtracted away 1 you'd have about 2.6 so that would work that would be greater than 2. so basically values to the left of 4 work with the exception of they have to be greater than or equal to 0. so we can say that anything let's call this interval a let me kind of divvy this up and let's call this interval b so anything in interval a that's to the left of 0 doesn't work right because it violates our domain then anything in interval b is going to work right because it's greater than or equal to 0 but less than 4 so that's going to be good to go what about anything in interval c so that's going to be anything that's larger than 4 so let's test that out let's erase this and i can check a value over here let's just choose the value 9. plug that in there and there so 9 plus 12 is 21. so you have the square root of 21 then minus the square root of 9 is 3. is that greater than 2. well the square root of 21 is about 4.58 okay that's an approximation if i subtracted away 3 i'd have about 1.58 so that's not greater than 2 so this would be false so we have our solution here we found that values in interval b will satisfy the inequality so essentially it's anything that includes zero up to but not including four so x can be greater than or equal to zero and less than four okay anything in that range in interval notation again you put a 0 next to a bracket and then a 4 next to a parenthesis and then graphically okay graphically we would put a bracket at 0 and we put a parenthesis at 4 and we would shade everything in between all right let's take a look at the next example so we have the square root of 4x minus 1 is greater than the square root of x minus 5. so for this guy again i've got two domain restrictions that i need to consider i've got that 4x minus 1 has to be greater than or equal to 0 and let's do that first so let's add 1 to each side of the inequality you'd find that 4x is greater than or equal to 1. divide both sides of the inequality by 4 you'd find that x has to be greater than or equal to 1 4. so here this one's going to be more restrictive right so here if i set x minus 5 is greater than or equal to 0 i add 5 to both sides of the inequality i find that x has to be greater than or equal to 5. so again this one's more restrictive so that's going to be our domain so x has to be greater than or equal to 5. okay let's erase all this now that we know what we're looking for and let's replace this inequality with an equals so we would have the square root of 4x minus 1 is equal to the square root of x minus 5. let me make that a little better so both radicals are isolated here so let's just square both sides and i'll have my 4x minus 1 is equal to x minus 5. let's add 1 to both sides of the equation you'll get that 4x is equal to x and then negative 5 plus 1 is going to be negative 4. let's subtract x away from each side of the equation and kind of do this over here we'd end up with 3x is equal to negative 4 divide both sides of the equation by 3 you get x is equal to negative 4 thirds okay so let's erase this so what happens is this guy is extraneous and you can immediately see that it's not going to work if you plug it in here because it's going to violate your domain right x has to be greater than or equal to 5. so you can kind of get rid of this at this point a lot of people would stop and say okay well there's no solution but that's not actually true either okay so here's where you've got to kind of go down to the number line and consider things so we know that this guy is only valid for values of x that are greater than or equal to 5. but i want you to think about something you have the square root of something and you have the square root of something you're saying where is the square root of this expression greater than the square root of this expression well if i'm doing the same thing to each side again you can just consider where is 4x minus 1 greater than x minus 5. if you solve this you're going to end up with the same thing you've got when you solve the equation you're going to end up with x is greater than negative four-thirds right when we solve it we get x is equal to four-thirds but essentially as long as x is a value that's larger than negative four-thirds then the right side here is less than the left side we can say the left side is greater than the right side okay so that means that for values where it's valid in the domain where x is greater than or equal to 5 the left side should be greater than the right side so we can check that let's go down to the number line and again the domain here is that x is greater than or equal to 5. so with 5 i'm just going to put a little border and i'm going to say that we can test something to the right of 5. we know anything to the left of 5 doesn't work because it violates the domain so i'll call this interval a and this interval b and i'll say that in interval a nothing works okay because it violates the domain so for interval b i can just choose something let's just choose something like 6 and we can check that so remember we had that the square root of 4x minus 1 is greater than the square root of x minus 5. if i plug in a 6 for x i'd have 4 times 6 which is 24 24 minus 1 is 23 so the square root of 23 is greater than 6 minus 5 is 1 square root of 1 is 1. okay so what's the square root of 23 if i approximate that well it's going to be about 4.79 so that's clearly greater than one so in this interval b things are going to work okay so this is going to be our solution it's basically that x is going to be greater than or equal to 5. okay so we'll say x is greater than or equal to 5 that's both our domain restriction and our solution for the inequality and graphically i can just put a bracket at five and just shade everything to the right and in interval notation i can put a brackets a five comma and then infinity okay and a parenthesis all right let's look at one with a cube root involved so these are a lot easier when you have a cube root you don't have a domain restriction so i don't have to start by saying x minus seven is greater than or equal to zero this value here this radicand can end up being whatever okay it can be negative it can be zero it can be positive remember in the real number system you can take the cube root of a negative number okay so you're good to go there so all i'm going to do is just solve this guy so let's go ahead and replace this guy with an equals and solve so 2 times the cube root of x minus 7 plus 6 is equal to 2. let me again isolate the radical so let me subtract 6 away from each side of the equation that will cancel you have 2 times the cube root of x minus 7 is equal to negative 4. let me divide both sides by 2 so i can isolate the radical we kind of move this up here you'll have the cube root of x minus 7 is equal to negative 4 over 2 is negative 2. so to get rid of the radical i'm going to cube both sides of the equation i'm going to find that x minus 7 is equal to negative 8. let me add 7 to both sides of the equation and i'll find that x is equal to negative 1. so again that's the boundary so let me erase all this x equals negative one again that's the boundary and i don't have a domain restriction so let me go down to the number line so here's negative one so basically you're going to have two scenarios you're gonna have this interval a and this interval b okay you don't have any domain restrictions so just pick something in interval a and test it so again we have 2 times the cube root of x minus 7 plus 6 and this should be greater than 2. okay so something less than negative 1 we can pick let's say negative 20. so negative 20 minus 7 is negative 27 so the cube root of negative 27 is going to be negative 3. so 2 times negative 3 is negative 6. so you'd have negative 6 plus 6 is greater than 2 which is false so i know values in interval a don't work so this is going to be false then what about values in interval b so let's choose 8 so i'll plug in an 8 there 8 minus 7 is 1 cube root of 1 is 1. so 2 times 1 is 2 plus 6 is greater than 2 you get 8 is greater than 2 which is true so for interval b it's true so anything larger than negative 1 would work so we can say our solution here is that x is greater than negative 1. all right so in interval notation you've got negative 1 comma infinity again with a parenthesis next to each let me erase this graphically i'm going to put a parenthesis at negative 1. i'm just going to shade everything going to the right in this lesson we want to continue to talk about solving radical inequalities so in the last lesson we started talking about solving radical inequalities in this lesson we're just going to have a follow-up to that we're going to have basically a part two we're just going to look at some more challenging or you could say more tedious problems so the first one we're going to start with is pretty easy we have the fourth root of 2x minus 5 is less than or equal to 1. now in our first lesson we didn't see anything with a fourth root we just worked with a square root and a cube root okay but with the fourth root again it's an even number here for the index so you've got to remember in the real number system if i have an even index of my radical my radicand or the number under the radical symbol cannot be negative okay it can be 0 or something positive so when we set this up the first thing we want to do is we want to take our radicand which in this case is 2x minus 5. and we want to say that that needs to be greater than or equal to 0 okay so i'm going to add 5 to both sides of this inequality and we're going to find that i have 2x is greater than or equal to 5 let me divide both sides by 2 and i'm going to find that x needs to be greater than or equal to 5 halves so in other words if i plug in 5 halves for x i would get 0 there right for my radicand i can take the 4th root of 0 that's just 0. that's ok but if i plug in something less than 5 halves i'm going to have a problem right because this guy the radicand would end up being negative i can't in the real number system have the fourth root of a negative value because it doesn't exist okay there's no number that i can raise to the fourth power and end up with a negative value because again an even number of negative factors will always give you a positive result okay so let me erase this and i'm just going to write this up here so we have that x is greater than or equal to 5 halves again that's my domain restriction so the idea with this is if i get a solution for my inequality and it violates the domain i've got to reject it okay so how do we solve the inequality well one way that we can do this we can replace the inequality symbol with equals and we could solve that to get our boundary okay so we can say that we have the fourth root of 2x minus 5 and that's equal to 1. very easy to solve this equation my radical is already isolated on the left so to get rid of the radical since my index is a 4 i'm going to raise this to the 4th power and then to make it legal i'm going to raise this to the 4th power so this is going to cancel here and then i'm going to have my radicand right so just my 2x minus 5 this will be equal to 1 to the fourth power is obviously 1. now it's very easy to solve from this point we just add 5 to both sides of the equation so that's going to cancel and we'll get that 2x is equal to 1 plus 5 is 6. one last step we want to isolate x so i'm going to divide both sides by 2 and i'm going to find that x is equal to 6 over 2 is 3. so i'm going to put x equals 3 here and that's going to be my boundary okay so let's take these two values here down to the number line and we're going to test the value on each side of 3. okay 3 is the boundary so i'm going to test the value on the left side of 3 something less than 3 and i'm going to test something on the right side of 3 something greater than 3. but you have this restriction here remember x has to be greater than or equal to 5 halves so we got to keep that in mind so let's erase all this and before i go down there i want to show you one thing real quick the first thing is that three works is a solution in this inequality this is non-strict here if i plug a 3 in 2 times 3 is 6 6 minus 5 is 1 the 4th root of 1 is 1 so you get that one is less than or equal to one which is true okay so this guy right here works as a solution so let's go down to the number line so again our domain restriction is that x has to be greater than or equal to five halves and then our boundary is at x equals 3 which again works as a solution so here's three let me put a vertical line there and then five halves as a decimal is 2.5 so let's put something right there and say this is 5 halves so i can also put a vertical line there now you can split this into regions okay this is very easy to do for reference sake so anything to the left of five halves i'm just going to call that region a or interval a anything between five halves and three and it's a tight window i'm gonna call that interval b okay i'm just gonna put an arrow there because i can't really fit my letter there and then anything to the right of three i'm gonna call that interval c so for interval a anything to the left of 5 halves we know that's an automatic rejection right because it violates the domain so i'm going to put an f there for false now for interval b i need to test the value so my original inequality was the fourth root of 2x minus 5 is less than or equal to 1. so something between 2.5 and 3 i could do 2.75 so let me get some room i'll come back up so if i plugged in a 2.75 there for x what would i have i would have the fourth root of 2 times 2.75 which is 5.5 so you'd have 5.5 minus 5 is less than or equal to 1 so then i have the fourth root of 0.5 is less than or equal to 1. now if i took the 4th root of 0.5 basically i could approximate that to be about 0.84 okay about .84 so if we had about .84 that would be less than one okay so that would be a true statement so let me erase everything and i'm going to say that things in interval b work okay let me slide this down just a little bit and i'm gonna put interval b right here and we'll say that's true okay so anything between five halves and three including both okay three is included and five halves included because essentially there's anything that's less than or equal to three so it's going down to and including five halves because that's part of your domain but anything to the left of five halves isn't going to work because it's not included in the domain so what about things in interval c so things that are greater than 3 well you can test anything you want so let's just pick 4 and plug that in again i'm just going to slide down for a minute so if i plugged in a 4 there what would i get well 2 times 4 is 8 8 minus 5 is 3. so you'd have the fourth root of 3 is less than or equal to 1. so if i approximate the 4th root of 3 i'm going to get about 1.32 which is larger than 1. okay so this is false so this is false so things in interval c don't work and we'll say that for interval c it's false so i have my solution again x is going to be any value that's five halves and including five halves going up to and including the number three so let me erase all of this i don't need it anymore so again x is greater than or equal to five halves and less than or equal to three all right so in interval notation i'm going to put a bracket next to five halves then comma i'm going to put a bracket next to three and then graphically again let me put my 2.5 or my five halves back on the number line so it's going to be right there again this is my five halves this is going to be a tight window but i'm going to put a bracket here i'm going to put a bracket here i'm just going to shade in between so again it's from five halves or 2.5 up to and including three so the five halves is included and the three is included and all the values in between all right let's take a look at another one so for this one we have the square root of 2x plus 6 is greater than x minus 1. remember if you have a square root involved or again a fourth root like we just saw or sixth root any even root then you need to figure out what the domain restriction is going to be so in this case two x plus six has to be set as greater than or equal to zero so two x plus six my radicand there again has to be greater than or equal to 0 let me subtract 6 away from each side of the inequality that's going to give me that 2x needs to be greater than or equal to negative 6. we'll divide both sides by 2 and we'll find that x needs to be greater than or equal to negative 3. so that's my domain restriction again if i get something that's less than negative 3 i've got to reject it okay so let me go ahead and erase this and i'll slide this up now let's go about solving our inequality again the method i'm going to use i'm just going to replace this with an equals and i'm going to solve the resulting equation so we're going to have the square root of 2x plus 6 is equal to x minus 1. so again the radical on the left side is already isolated so i can go ahead and just square both sides so i'm going to square this side and i'm going to square this side so what's going to happen over here is this is going to cancel with this i'll just have my radicand so the left is 2x plus 6. on the right side remember i have to expand this okay so x minus 1 that quantity squared you can use your formula it's the first chi squared so x squared then you have a minus there so it's going to be minus 2 times this guy times this guy so 2 times x times 1 is 2x then plus the last guy 1 squared so plus 1. okay so what i'm going to do since i'm going to end up with a quadratic i'm just going to go ahead and move everything on the left over to the right so my left is going to be 0 okay so what i'm going to do i'm going to subtract 6 away from each side of the equation so that's going to cancel and i'm going to subtract 2x away from each side of the equation so that's going to cancel so the left side is now 0. so i'm going to have that 0 is equal to x squared negative 2x minus 2x is minus 4x and then 1 minus 6 is going to be minus 5. so you can flip this around if you want i know most of you including myself like the variable parts on the left so you can say x squared minus 4x minus 5 is equal to 0. that's perfectly legal so let me scroll down a little bit more this is one that we can solve using factoring so that's what i'm going to do for the left side i can factor that by putting an x here and an x here and they give me two integers whose sum is negative four and whose product is negative five well negative five can only come from one times negative five or negative one times positive five so what i'm looking for is positive one and negative five so if we solve this using our zero product property we would set x plus one equal to zero of course that would give us a solution of x is equal to negative one and we would set x minus five equal to zero and that would give us a solution of positive five now one of these is going to be extraneous okay so let me erase everything and we're going to check both of these and i'll show you which one is extraneous all right so again we found that the solutions for this equation are x equals negative 1 and 5. if you plug in a 5 you're fine so if i plugged in a 5 there and also there i'd be okay 2 times 5 is 10 10 plus 6 is 16. the square root of 16 is 4. over here 5 minus 1 is 4 so this one is good to go but with negative 1 you're going to have a problem okay with negative 1 you have an issue if you plug that in here and here 2 times negative 1 is negative 2 negative 2 plus 6 is positive 4. so you have the square root of 4 which is 2 and that's equal to negative 1 minus 1 is negative 2. so obviously this is false okay this is false so this guy needs to be rejected okay it's an extraneous solution so your boundary here is 5 and basically your restriction is that x needs to be greater than or equal to negative 3. so let's take this information down to the number line so again my inequality is the square root of 2x plus 6 and again this is greater than x minus 1. my restriction on the domain is that x needs to be greater than or equal to negative three and my boundary occurs at five so x equals five is the boundary so on the number line five is here should draw a vertical line there and negative three is here so let's draw another vertical line here so we know anything to the left of negative three doesn't work so let me go ahead and put this as interval a again anything to the left of negative three and i'm just going to put that this is an automatic rejection right so an f or false then you have anything between negative 3 and 5 and then you have anything that's larger than 5. so this area will be called b and this one will be called c okay so let's start by testing something in interval b so again this is between negative 3 and positive 5. so if i plugged in let's say i don't know how about 0. so let's plug that in there and there and see what we get so we know that 2 times 0 is 0 so you have basically the square root of 6 is greater than 0 minus 1 is negative 1. well we don't need to know what the square root of 6 is we don't need to even approximate it because this is a positive number of course it's going to be greater than a negative 1 because a positive number is always greater than a negative okay so this is true so let me erase this we'll basically say that things in interval b will work so for interval b it's going to be true so it's going to be anything that's less than 5 going down to and including negative 3 because negative 3 is still in the domain okay now what about things in interval c do they work to make this easy on ourself let's choose 29 so i could plug that in for x there 2 times 29 is 58 then if i add 6 i get 64. so what you'd have is the square root of 64 is greater than 29 minus 1 which is 28. i know the square root of 64 is 8. 8 is not greater than 28 that is a false statement so things in interval c will not work so i'll put an f there for false all right so we have our solution now so anything in interval b will work so again we've got to be specific here it's anything less than 5 but because we have this domain restriction that anything less than negative 3 doesn't work you basically say that x has to be greater than or equal to negative 3 but less than 5. okay so negative 3 is included 5 is not all right so in interval notation i can say that again negative 3 is included by putting a bracket there and then comma you have a 5 and then a parenthesis and then graphically you erase this so again at negative 3 i'm going to have a bracket at 5 i'm going to have a parenthesis i'm just going to shade everything in between all right let's take a look at one to wrap up the lesson that's pretty tedious so we have the square root of x squared minus 5x plus 6 is less than x plus 3. so what makes this problem tedious is that you have a quadratic inequality that you're going to need to set up to find your domain restriction right because you have x squared minus 5x plus 6 that's underneath the square root symbol so we need to set this as greater than or equal to zero okay so this guy right here again it's a quadratic inequality at this point we know how to solve this we're going to replace this guy right here with equals and we're going to solve x squared minus 5x plus 6 equals 0. this will give us our boundary or again critical values and then we're going to test values on each side of that okay so we're going to set up intervals and test just like we've been doing so what i'm going to do is solve this by factoring i know that this guy would be x and this guy would be x two integers that sum to negative five and give me a product of positive six would be negative two and negative three right negative two plus negative three is negative 5 negative 2 times negative 3 is positive 6. so if i solve this equation i would find that x is equal to what x minus 2 is equal to 0 that has a solution of x equals positive 2. if x minus 3 is equal to 0 that has a solution of x equals 3. okay so 2 and 3 are going to be my boundaries so let's go down to the number line so again the critical values are the boundaries x equals 2 and 3. so here's 2 here's 3. and again we're going to set up intervals so anything to the left of 2 is a anything between 2 and 3 will be b and anything greater than 3 will be c so again you've got a b and c so let's test some values so again we've got x squared minus five x plus six is greater than or equal to zero so let's start out by testing zero that's something in interval a we know that that would basically mark out this guy and this guy so i would have 6 is greater than or equal to 0 which is true so for interval a things work that's true what about interval b well i can take something like 2.5 that's between 2 and 3. so 2.5 so if i plugged in a 2.5 there and there what would i get well 2.5 squared is 6.25 then if i subtract away 5 times 2.5 5 times 2.5 is 12.5 then if i add 6 is this greater than or equal to 0 well no it's not because 6.25 plus 6 is going to be 12.25 so 12.25 and if i subtract away 12.5 i'm going to get negative i'm going to get negative 0.25 so that's not greater than 0 and it's not equal to 0 because obviously it's a negative and so it's not going to be greater than 0. okay so let's erase this we're going to say things in interval b don't work okay so let's go back up and so we'll say things in interval b again between 2 and 3 those don't work so that's going to be false the last list check something in interval c so let's just pick four that's easy to work with so i'll plug in a four there and there so four squared is sixteen and then minus five times four is twenty and then plus 6 is this greater than or equal to 0 well 16 plus 6 is 22 22 minus 20 is positive 2 so you'd have that 2 is greater than or equal to 0 which is true so for interval c or region c or whatever you want to call it this is going to be true so i've found my domain okay i've found my domain basically we can say that x needs to be what it needs to be greater than or equal to positive 3 and the or equal to is because i had a non-strict inequality so the end points or the boundaries are going to be included then or x needs to be less than or equal to that value of 2. so the values that don't work in our inequality are going to be things between 2 and 3. 2 works 3 works but it's got to be less than 2 or equal to 2 or greater than 3 or equal to 3 but anything in between will not work so i'm going to leave this here because we're going to work on this later on so this will just stay here but i'm also going to copy it and bring it up for reference sake we're done with this information we can erase it so to solve the inequality now i know my domain i'm just going to replace that with equals so i'm going to have that the square root of x squared minus 5x plus 6 is equal to x plus 3. so to solve this my radical is already isolated on the left so i'm just going to square both sides so this is going to cancel with this and i'll have my radicand of x squared minus 5x plus 6 this is equal to again you've got to expand this okay so this is going to be x squared plus 2 times x times 3 which is 6x plus 3 squared which is 9. now i have x squared on each side i can subtract that away from each side again if you see the same thing on each side you can just get rid of it so i can cancel this and this so on the left side i have negative 5 x plus 6. on the right side i have 6x plus 9. so let's move all the variable terms to one side all the numbers to the other so let me subtract 6 away from each side and let me subtract 6x away from each side so negative 5x minus 6x is negative 11x this is equal to over here 9 minus 6 is 3. so to get my solution for x i'm going to divide both sides by the coefficient of x which is negative 11. so i'm going to find that x is equal to negative 3 11 okay negative 3 11. so let me erase everything right so x is negative 3 11 that's our boundary and again the restrictions on the domain x has to be greater than or equal to 3 or less than or equal to 2. so let's go to the number line with this information so again we found that x was equal to negative 3 11. so that's our boundary so negative 3 11 as a decimal is a repeating decimal it's negative 0.27 where that 2 7 repeats forever okay so if we wanted to put this on the number line we could go about right here and just say that this represents negative 3 11. so i'm going to put a vertical line there okay and then again things need to be greater than or equal to 3 so i can put a vertical line here and things need to be less than or equal to 2 so i can put a vertical line here so if i call this to the left of negative 3 11 interval a i could call between negative 3 11 and 2 interval b i could call between 2 and 3 interval c and i could call anything larger than 3 as interval d now we know that things in interval c okay between 2 and 3 don't work because they violate the domain okay so i have a b c and d again this is false it doesn't work so let's choose a value in interval a and see if that works so again our inequality was the square root of x squared minus five x plus six again this had to be less than x plus three so i'm going to pick a value of let's say negative one that's pretty easy to work with so let me scroll down and get some room going we'll come back up so if i plugged in a negative one there and there and there what would i get well you'd have negative 1 squared which is 1 and then negative 5 times negative 1 is plus 5 and then you'd have plus 6. so you'd have the square root of basically 12 right 1 plus 5 is 6 x plus 6 is 12. so the square root of 12 could be simplified as 2 times the square root of 3 right because 12 is 4 times 3 4 is a perfect square so you could pull out the kind of 2 from that so you would have 2 times the square root of 3. okay so this is less than negative 1 plus 3 is 2. well that's going to be false if you do 2 times the square root of 3 you're going to get about 3.46 and that's clearly not less than 2 so things in interval a don't work okay those are going to be false so for interval a i'm going to put an f for false okay what about things in interval b so this is between negative 3 11 and again 2. so let's think about the number zero that's easy to work with so let's go ahead and plug that in for x so zero squared is zero then minus five times zero is zero so you basically just have your square root of 6 there and that's less than if you have 0 plus 3 that's 3. so the square root of 6 is again about 2.45 you could say so that's going to be less than 3. so values in interval b are going to work so let's erase this we'll come back up so we'll put this one as true okay so anything in interval b is going to work so anything larger than negative 3 11 is going to work okay negative 3 11 won't work so that's going to include 2 because again 2 is in the domain so we're good to go there now you can test something in interval d just to be safe but essentially it's going to work right because this is basically telling us that anything larger than negative 3 11 is going to be a solution with the exception of the numbers that violate the domain which again is anything between 2 and 3. but let's take something from interval d let's just take 4 and see if that works just to be safe so i plugged in a 4 there and there and there what would i get all right so what i'd have here is the square root of 4 squared is 16 then minus you have 5 times 4 which is 20 then plus 6 and this is less than 4 plus 3 which is 7. 16 minus 20 is negative 4 negative 4 plus 6 is 2 so you have the square root of 2 is less than 7 that's obviously true so things in interval d will work okay so let me erase all this we're ready to put our solution together okay so basically anything to the left of negative 3 11 doesn't work we know that then anything greater than negative 3 11 will work but you have this restriction here basically where between 2 and 3 it's not part of the domain so you just have to exclude between two and three but two and three itself are in the domain so those are included okay so x has to be greater than and strictly greater than negative three elevenths and less than or equal to two or x could also be greater than or equal to positive 3 okay so again 2 and 3 are included and negative 3 11 is not included so let me erase all of this we don't need it anymore so in interval notation it's the union of these two sets here so i would have a parenthesis next to negative 3 11 and then comma you'd have a 2 and then a brackets and the union width you'd have a bracket next to 3 and this goes out to positive infinity so again use a parenthesis with infinity so let me write negative 3 11 again right there so this is negative 3 11 and again i'm going to put a parenthesis there and at 2 i'm going to put a bracket i'm just going to shade everything in between at 3 i'm going to put a bracket and then i'm going to shade everything to the right in this lesson we want to review solving rational equations so a rational equation is nothing more than just an equation with rational expressions involved so something like x minus five over two x squared is equal to one over four x plus one over x squared so earlier in our course we talked about solving linear equations in one variable with fractions involved in that lesson we saw that we could clear an equation of fractions by multiplying both sides of the equation by the lcd the least common denominator of all fractions involved well it's basically going to be the same process here if i want to solve this equation i would first find the lcd of all the rational expressions involved and then i would just multiply both sides of the equation by the lcd and i would clear all the denominators but there's one thing you have to watch out for when we multiply both sides of an equation by a variable or a variable expression we will sometimes get extraneous solutions okay so we talked about these in the last section where we're solving radical equations extraneous solutions are solutions that work in the new equation that you form but they don't work in the original that you were trying to solve so specifically when you're talking about a rational equation you want to guard against division by zero because in a lot of cases you have variables in your denominator so let's go ahead and try to solve this first guy here we have x minus five over two x squared is equal to one over four x and then plus 1 over x squared first and foremost what's my lcd we'll have a denominator here of 2x squared a denominator here 4x and a denominator here of x squared so for my lcd the number part is what you've got a 2 which doesn't factor and a 4 which is 2 times 2. so the number part is a 4. then for the variable part i've got an x squared an x and an x squared so it's going to be x squared so now that i have my lcd of 4x squared i'm just going to multiply both sides of the equation by the 4x squared so let's set this up we would have 4x squared being multiplied by x minus 5 over 2x squared and this is equal to i'm going to go ahead and multiply the 4x squared by each part here so we'll have 4x squared times we have 1 over 4x i've got to make that a little better then plus i have my 4x squared times my 1 over x squared okay so let's go through and multiply each let me kind of scroll down get some room going we'll come back up to this in a minute so on the left the 4 would cancel with the 2 and give me a 2 and then x squared over x squared is 1. so essentially you have 2 times the quantity x minus 5. so you've got to make sure that you multiply the 2 by each term here don't just put 2x minus 5. that's wrong and it's a common mistake so you want to say this is 2 times the quantity x minus 5 again wrap x minus 5 inside of parentheses then this is equal to this 4 would cancel with this 4 here and then x squared over x the x cancels and i'll take one of these away so i basically have x to the first power times 1 which is x then plus i can cancel this x squared with this x squared i basically have 4 times 1 or just 4. okay very easy equation to solve now 2 times x is 2x then minus 2 times 5 is 10 and this equals x plus 4. let's go ahead and continue now so if i add 10 to both sides of the equation and i subtract x away from each side of the equation what am i going to have this would cancel and this would cancel so 2x minus x is just x and this is equal to 4 plus 10 is 14. so my proposed solution here is x equals 14. and again the reason i say proposed solution is that we might get extraneous solutions when we multiply both sides by again either a variable or variable expression so we need to check so let's copy this we'll go back up to the top so we want to check if x is equal to 14. so i've got to plug in a 14 everywhere there's an x so that's going to be here and here here and also here okay so in the numerator on the left 14 minus 5 is 9 and this is over 2 times what is 14 squared well that's going to be 196. if i just do that multiplication 2 times 196 is 392. so this is 9 over 392 and that is not going to simplify then next i have 1 over 4 times 14 4 times 14 is 56 then plus well i've got 1 over 14 squared which we know is 196. okay so do we have the same value on the left and the right well let's go ahead and combine these fractions here on the right and see what we get we would think about getting a common denominator i know 56 is 8 times 7 right 8 is 2 cubed then times 7 we know that 196 was built from 14 times 14 right 14 is 7 times 2 so i can say 196 is what it's 7 squared times 2 squared and let me kind of reverse that order so let me put 2 squared first times 7 squared so my lcd is going to be 2 cubed 2 cubed which is 8 times 7 squared which is 49. if i multiply 8 by 49 i do get 392. so i know my lcd is 392. so the lcd is 392. okay so what do i need to multiply 1 over 56 by in order to get a denominator here of 392. well 392 divided by 56 is 7. so i'd multiply this by 7 over 7 and i would end up with 7 over 392. so let's erase that and just put 7 over 392. then for this guy right here i'd need to multiply it by two over two so by two over two and i would end up with two over three hundred ninety-two so two over three hundred ninety-two so seven plus two is nine over three hundred ninety-two is exactly what i'm looking for so i end up with 9 over 392 on each side of my equation so i can go back up and i can say that x equals 14 is a valid solution now you always want to check your answer if you have time one thing that would have saved you a little bit of time here if you were on a test and you were kind of pressed for time you can easily see that in each case for the rational expressions for this one this one and this one the only number that would violate the domain is an x value of zero right so if you were pressed for time you could figure out what is the domain for each rational expression and find any value that violates that domain you would exclude that as a possible solution right those would be extraneous so you can kind of do it that way if you're pressed for time but the way i suggest is solve the equation and then go back and plug in the solution and make sure first off that you've got the right answer and second off that it's not extraneous all right let's look at one that's a little bit more tedious so we have 7x plus 4 over x squared plus 11x plus 28 this is equal to we have 4 over x plus 4 minus we have x minus 8 over x plus 7. okay so the first thing i want to do is just find the lcd so for these guys right here this x plus 4 and this x plus 7 those denominators are not going to factor right but this guy right here it looks like i can factor that so what i'm going to do is i'm going to go ahead and write my x here and my x here and if i think about two integers that sum to 11 and give me a product of 28 well i know that's going to be 4 and 7. right so i get x plus 4 and then x plus 7. so if i look at x plus 4 here and x plus 7 here here i have x 4 here i have x plus 7. well this guy right here is going to be my lcd okay so what i'm going to do is i'm going to multiply both sides of the equation by that lcd i'm going to be a little bit pressed for room so i'm just going to do this individually okay so i'm going to multiply this by my 7x plus 4 over my again i'll do this in factored form so x plus 4 times x plus 7 and we can clearly see that this would cancel with this and i'd be left with my seven x plus four so i would just have my seven x plus four and then let's go ahead and do that for this part right here so i'm gonna multiply let by just kind of erase this my 4 over x plus 4 so in this case i would have the x plus 4 cancel with the x plus 4 and i would have 4 times the quantity x plus 7. so let's put equals we'll have 4 times the quantity x plus seven and then lastly let me kind of erase this i'll be multiplying by you have your minus and then this guy right here this x minus eight over x plus 7. so we would have that this guy cancels with this guy and i'm left with my x minus 8 times my x plus 4. now you have to be careful here you have this subtraction here which i've written here to make sure that you don't make a sign mistake put a bracket there and then go ahead and write that so you have x minus 8 times x plus 4 so x minus 8 inside of parentheses times x plus 4 okay and we'll go ahead and foil that out and again that minus sign out in front of the brackets is going to make sure that i get the correct sign okay so let's go ahead and rewrite this so we have 7x plus 4 is equal to i'll go ahead and clean this up 4 times x is 4x and then plus 4 times 7 is 28 i'll leave my minus out in front of these brackets go ahead and foil this so you get x squared the outer would be 4x and the inner would be negative 8x so if i combine those i get negative 4x and then for the last negative 8 times 4 is minus 32. again because i have this minus out in front of these brackets i've got to go in and change the sign of each term here okay so let me go ahead and go to the next line we'll say we have 7x plus 4 is equal to 4x plus 28. again each term in here is going to change so minus x squared then plus 4x then plus 32 okay so now i know i have the correct sign everywhere and then what i can do here since i have a quadratic i have an x squared term i want to get everything on one side all the variables and all the constants everything on one side and zero on the other okay so let's just go ahead and move everything to the right side so first and foremost 4x plus 4x is going to be 8x so let's erase this and just put 8x we know that 28 plus 32 is 60. let's just go ahead and put this as 60 here and nothing to do with the negative x squared on the left side i can subtract 7x away from both sides and i can subtract 4 away from both sides so on the left i'm going to have 0. okay so on the right let me kind of scroll down and get some room going i'm going to have my negative x squared and then 8x minus 7x is just plus x 60 minus 4 is plus 56. now if you've been watching these videos in order you know that i like my variable on the left it's no problem to switch this around and say i have negative x squared plus x plus 56 is equal to zero okay perfectly legal now if you want to make this a little easier to read you can divide both sides by negative 1 and have a leading coefficient that's positive it's not necessary but it's something you can do if you want so if i divide both sides by negative 1 just to kind of clean this up a little bit i would end up with x squared minus x minus 56 is equal to we know zero over anything other than zero is zero okay so this is what i'm working with let me get some more room going i'm going to have my x squared minus x minus 56 equals 0. so you can solve this with factoring you can solve it with the quadratic formula whatever method you want to use i'm just going to solve it using factoring because it's a little bit easier so i'm going to have an x here and an x here and this course equals 0. give me two integers whose sum is negative 1 and we have a product of negative 56 well that's going to be negative 8 and positive 7 okay so now i can solve this using my zero product property so i would say that x minus 8 is equal to 0 or x plus seven is equal to zero very easy to solve these equations for this one i would just add eight to both sides and i'll get x is equal to eight for this one i would subtract seven away from each side and i'll get x is equal to negative seven so let's go ahead and copy these all right so we want to check x equals 8 and then also x equals negative 7. so let's start by checking x equals 8. so i'm going to plug in an 8 here and here and here and here and here and here now if you plug in an 8 here what's going to happen 8 minus 8 is 0 and over here 8 plus 7 is 15. we know 0 over 15 is zero so i can go ahead and mark this guy out i only have to deal with these two okay so in the numerator here on the left 7 times 8 is 56 56 plus 4 is 60. this is over you have 8 squared which is 64 plus 11 times 8 which is 88 plus 28. if i add all those amounts together i get 180 we know that 60 over 180 is one-third okay so the left side here is one-third this guy right here i'd have 4 over 8 plus 4 which is 12 we know that 4 12 is also 1 3 so we can go ahead and say that this solution x equals 8 is a valid solution now let's plug in x equals negative 7 and see what happens so i'm plugging in a negative 7 here here here here here and here now before i do anything do you see a problem yes right here is where your problem is if you plug in a negative 7 for x there negative 7 plus 7 is 0. you can never divide by zero it's undefined so if you see that again one way to do this is to look at the domain of each rational expression first if i look at the domain of this rational expression all the way on the right i find that i have a problem right because x can't be negative 7 in that case so this guy right here has to be rejected okay you can reject this and say it's an extraneous solution so x equals 8 is the only valid solution here all right let's take a look at one more so we have 1 over x cubed minus 11x squared plus 30x and this is equal to we have 1 over x squared minus 6x and then plus we have x plus 1 over x squared minus 5x so again i want to find the lcd first so that means i need to factor everything so for this guy i know i could pull out an x first and then i would have x squared minus 11x plus 30 inside and then i would have to factor this again because it further factors so this would be x and this would be x give me two integers whose sum is negative 11 and whose product is positive 30 well you can go ahead and say this is negative 5 and negative 6 right negative 5 plus negative 6 is negative 11. negative 5 times negative 6 is positive 30. okay so let's erase this and kind of drag this up so this is that factored form then this guy right here is going to factor into x times the quantity x minus 6 and this guy right here is going to factor into x times the quantity x minus 5. so if we look at the denominators we can see that this is going to be the lcd right x times the quantity x minus 5 times the quantity x minus 6. so what i want to do is multiply each side of the equation by that lcd again because i don't have enough room to kind of set this up in that long of a format i'm just going to do them individually so i'm going to multiply this lcd by kind of each part here individually and i'll write the answer below so we can kind of move forward all right so if i have 1 over x times the quantity x minus five times the quantity x minus six let me kind of make that a little better and i multiply this by the lcd x times the quantity x minus five times going to x minus six what's going to happen this denominator will cancel with this and i'll be left with just 1. so on the left side of the equation i'll just have a 1. now let me erase this and i'm going to multiply this guy by this right here now so i would have equals so you have 1 over your x times the quantity x minus 6. so this guy is going to cancel with this and this and i'm left with just the x minus 5 times 1. so x minus 5. you could put that in parentheses but you don't have to you're just multiplying by one so it's not going to really change anything then you have a plus so let me write plus here and let me kind of get rid of this so now we're going to multiply by x plus 1 over you have x times the quantity x minus 5. so what we're going to do is cancel this with this and this and i'll be left with x minus six times x plus one so make sure you wrap each of these in parentheses so you have x plus one that's going to multiply x minus six okay so let's scroll down we've cleared all the denominators and so we have 1 is equal to you've got your x minus 5 plus go ahead and foil this out x times x is x squared the outer would be negative 6x the inner would be plus x so if you combine those that's going to be minus 5x and then for the last you have 1 times negative 6 which is minus 6. if you clean this up on the right side x minus 5x is going to give me minus 4x so i'll write 1 is equal to i'm just going to put minus 4x here and then i have negative 5 minus 6 which is negative 11 and then plus x squared and i'll reorder this in a minute i want to subtract 1 away from each side of the equation so that this side on the left is 0. so i'll have that 0 is equal to i'll write x squared out in front and then i'll write minus 4x and then negative 11 minus 1 is minus 12. okay so recall that i can just flip this around if i want to it's perfectly legal i can say this is x squared minus 4x minus 12 is equal to 0. so again i can solve this using factoring i can use the quadratic formula whatever you want to do since we haven't used it today let's just go ahead and use the quadratic formula get a little practice so my a is a 1. my b is a negative 4 and my c is a negative 12. so for the quadratic formula we kind of just write it over here x is equal to the negative of b b in this case is negative 4 so the negative of negative 4 is plus 4 and then plus or minus the square root of you've got b squared negative 4 squared is 16 minus 4 times a a is 1 times c c is negative 12. so you basically have negative 4 times negative 12 which is going to be positive 48. so 16 plus 48 is 64. so this is 64. if i took the square root of 64 i would get 8. so this is an 8 right here and this is over we would have 2 times a a is 1 so just over 2. okay so i'm going to get two solutions here x is going to be equal to 4 plus 8 which is 12 over 2 which is 6. and then x is going to be equal to 4 minus 8 which is negative 4 over 2 which is negative 2. all right so let's copy these solutions all right so let's check this so if i plugged in a 6 here here here here here here here and here what would happen well i want you to recall that the factored form of this denominator is x times the quantity x minus six if you put a six in right there you would have six minus six which is zero and zero times six which is still zero right you can just kind of do this real fast let's go ahead and say that i replace this with six and i replace this with six so six minus six is zero zero times six is zero so this denominator here would be zero right you can't divide by zero so this guy right here is going to be an extraneous solution you have to reject that okay so x equals six is not a valid solution so let's check x equals negative 2. okay so i'm going to plug that in here and also here and also here and also here and then also here and here and here one more time and right there okay so a lot of stuff to plug in let's kind of get started here so in the denominator here i have negative 2 cubed which is negative 8 and then minus you have 11 times negative 2 squared negative 2 squared is 4 so 11 times 4 is 44 so minus 44 and then you have 30 times negative 2 which is negative 60. so negative 8 minus 44 minus 60 is negative 112. let me kind of move this out of the way it's going to scooch this down a little bit so i'm going to say that this side on the left is negative 1 over 112. okay let's erase this all right so if i look at the right hand side now i'm going to have 1 over you've got negative 2 squared which is 4 and then basically you have negative 6 times negative 2 which is positive 12 or you could say plus 12. so 4 plus 12 is 16. let's put this as 16 here and then we have plus i've got negative 2 plus 1 which is negative 1 over you've got negative 2 squared which is 4 and then you've got negative 5 times negative 2 which is going to be plus 10. so this is going to be 14 down here okay so what's the lcd well 14 is 7 times 2 and 16 is basically 2 to the fourth power so it would be 16 times 7 which is 112 okay so that's my lcd so i would need to multiply this guy right here by 7 over 7 so this would end up being 7 over 112 i would need to multiply this guy right here by eight over eight so this would end up being negative eight over a hundred twelve so negative eight over a hundred twelve if you sum 7 plus negative 8 you do get negative 1 and it's over the common denominator of 112. so you see that you get negative 1 over 112 on each side of the equation so x equals negative 2 is a valid solution in this lesson we want to review solving rational inequalities so when we solve rational inequalities we follow a similar process as when we solved quadratic inequalities so if you've got the process of solving quadratic inequalities down this will be pretty easy for you all right so solving a rational inequality the first thing we want to do is we want to write the inequality so that 0 is on one side and a single rational expression is on the other all right so after you've done that the next step is to create two equations we want to set the numerator and denominator each equal to 0 and solve the resulting equations so the solutions are going to give you your boundaries or again some people call these critical values some people call them endpoints now we're going to use the solutions from each equation to set up intervals on the number line so this is the same thing we did with quadratic inequalities now we're going to use test values to find intervals that satisfy the inequality so the reason this works is that if you pick a value within an interval and it satisfies the inequality then every value in that interval is going to satisfy the inequality okay but the reverse is true if i pick a value in an interval and it does not work if it does not satisfy the inequality then no value in that interval will now when you choose your test values you cannot pick the end points themselves okay those have to be considered separately which is what we're saying here consider the end points separately okay now if you have a strict inequality strictly less than or strictly greater than the end points or the boundaries or the critical values whatever you want to call them are always going to be excluded from the solution if you have a non-strict inequality so a greater than or equal to or less than or equal to then essentially when we set our numerator equal to zero those values that satisfy that equation are going to be part of your solution okay but if it's the part where we set the denominator equal to zero remember if we set the denominator equal to zero those solutions make the denominator 0 and that gives us division by 0 which is undefined so those values those endpoints those critical values those boundaries have to be excluded from the solution all right so let's go ahead and start out with 7 over x plus 5 is greater than or equal to we have 8 over x plus 6. again what i want to do i want to have a rational expression on one side and 0 on the other so if you get something that sets up like this all we're going to do is subtract 8 over x plus 6 away from each side of the inequality again you can do that because of the addition property of inequality so on the right hand side here this is going to cancel and become 0 which is exactly what we want okay on the left hand side i'm going to have 7 over x plus 5 and then minus we have 8 over x plus 6 and again i'm going to have this is greater than or equal to 0. now we know how to add and subtract rational expressions we just need to get a common denominator going so the least common denominator is the product of these two denominators so x plus 5 that quantity times x plus 6 that quantity so let's go ahead and do this we're going to have 7 over x plus 5 multiplied by this denominator here is x plus 6 so it's going to be x plus 6 over x plus 6 and i'm going to wrap these all in parentheses then minus you're going to have your 8 here and this is over your x plus 6 and i'm going to multiply by the denominator here's x plus 5 so x plus 5 over x plus 5. okay all right so we're going to write this as greater than or equal to 0. let's scroll down get some room going so we'll start here on the left we have 7 times x that's 7x and then plus 7 times 6 is 42 and then let's just keep going i'm going to put a minus here and i'm going to put a bracket again i'm subtracting everything away from there so i've got to be very careful about my sign okay so that's why i put the bracket there just to make sure i don't make a silly sign mistake so a times x is 8x and then plus you've got 8 times 5 that's 40. you can close the bracket now and let's write this over we've got our common denominator which is x plus 5 times we've got our x plus 6 again this is greater than or equal to 0. all right let's get some room going all right so in the numerator let's go ahead and distribute this negative to each term and i'm just going to write it as a negative 8x and then a minus 40. okay when you remove the brackets like that and you have a negative out in front of it you've got to change the sign of each term inside okay so in the numerator 7x minus 8x is negative x and then 42 minus 40 is plus 2. again this is over my common denominator of x plus 5 and then times x plus 6. okay and let me extend that a little bit all right so this is greater than or equal to 0. so i'm going to copy this all right so now what we want to do since we have a rational expression on one side and 0 on the other we want to set the numerator equal to 0 solve that equation we want to set the denominator equal to 0 and solve that equation so if i set the numerator equal to zero i get negative x plus two is equal to zero i'm going to subtract two away from each side of the equation and this cancels so i get negative x is equal to negative two so at this point you can multiply both sides by negative 1 divide both sides by negative 1. it doesn't really matter let's just divide both sides by negative 1. and the negative here cancels with the negative 1. so you have x by itself on the left on the right negative 2 over negative 1 is 2. so x equals 2 solves the equation you set up from your numerator so off to the side here i'm going to say we have x equals 2 and i'm going to say that this is from the numerator okay that's important because we have a non-strict inequality here so because 2 makes the numerator into 0 think about that 2 makes the numerator to 0 it won't make the denominator into 0. so 0 divided by something that's not 0 is 0. so if you think about the left side here this inequality is going to say what it's going to say that 0 is greater than or equal to 0 which is true okay so this guy 2 will be a solution for this rational inequality all right so now let's think about the denominator so we have x plus 5 that quantity times x plus 6 that quantity and we're setting this equal to zero now it's in factored form so we can just use our zero products property and we can set each factor equal to zero and solve so i'm going to have that x plus five is equal to 0. i'm going to subtract 5 away from each side of the equation that will cancel i'll have x is equal to negative 5. so one of the solutions will be x is equal to negative 5. the other solution comes from the factor x plus six so x plus six is equal to zero let's subtract six away from each side of the equation and again this is going to cancel right here so we'll have x is equal to negative 6. okay so we specifically want to state that these guys right here are from the denominator okay the reason we want to state that is if i plug in a negative 5 for x here think about this negative 5 plus 5 is 0 0 times whatever this would be is 0. that means i'm dividing by 0 and that's undefined so this negative 5 here and this negative 6 here will not be solutions because again they make the denominator into 0 and division by 0 is undefined okay so this guy right here this 2 is the only endpoint that will satisfy the inequality all right so let's go ahead and copy these kind of endpoints or boundaries or critical values or whatever you want to call them and we're just going to set up some intervals so everywhere i see one of these boundaries i can just put a vertical line to kind of split up the number line into my intervals so at 2 i'm just going to put a vertical line here at negative 5 i'm going to do the same and at negative 6 i'm going to do the same okay doesn't have to be perfect just a visual representation of what's going on and i like to label the intervals with capital letters you don't have to do that you can do whatever you want but it just makes it easier for me to keep track of what's going on so this will be interval a and that will consist of the numbers that are less than negative 6. between negative 6 and negative 5 will be interval b and then greater than negative 5 but less than 2 will be interval c and then anything larger than 2 will be interval d what we want to do now is very very easy but also very very tedious we want to write the original inequality which was 7 over we had x plus 5 and this was greater than or equal to we had 8 over x plus 6. so we want to take a value from each interval plug it in and see if it works okay so let's start with interval a i'm just going to choose negative 8 and if i plug that in for x what would i get so i would have 7 over negative eight plus five is greater than or equal to eight over negative eight plus six so negative eight plus five is negative three so this would be negative seven thirds and then negative eight plus six would be negative 2 okay so this would be negative 2. we know that 8 over negative 2 is negative 4. so ask yourself the question you have negative 7 over 3 is that greater than or equal to negative 4. well one way you can figure this out if you have absolutely no idea remember you can check to see if one fraction is larger than another by cross multiplying so i could write this as negative 7 over 3 question mark i'm going to write this as negative 4 over 1. go ahead and cross multiply so 3 times negative 4 is negative 12 1 times negative 7 is negative 7. now the larger number okay the larger product you could say if you want to be more specific is next to the larger fraction now this is reversed because we're working with negatives so this guy right here negative 12 is larger in terms of absolute value but because we're working with negatives that means the negative four part is the actual smallest number right if these were positives this would be positive seven this would be positive twelve we could say four is larger than seven thirds but because we have negatives the reverse is true because what we have is a larger negative which means we have a smaller number okay so that means negative seven thirds is greater than negative four so this is true okay this is true so that means anything in interval a is going to work okay so we can just write a t in that interval and it's kind of busy there but let's just write it down here let's put a t for true okay so anything in interval a will work you can even write it off to the side and say interval a we'll put a t okay just so we can keep track so interval b i'm between negative 6 and negative 5 so let's use negative 5.5 we'll just do that in decimal form so 7 over we have negative 5.5 plus 5. we know that would be negative 0.5 right negative 0.5 so what am i really asking there 7 divided by negative 0.5 is the same as 7 divided by negative a half we know that would basically be 7 times negative 2 so this is negative 14. okay this is negative 14. all right this is greater than or equal to if i plugged in a negative 5.5 here negative 5.5 plus 6 is 0.5 so 8 divided by 0.5 again is the same thing as 8 divided by 1 half which is the same thing as 8 times 2 which is 16. so is negative 14 greater than or equal to 16 no this is false this is false so we can say that for interval b it's false right nothing's going to work all right let's look at interval c now so we have 0 there 0 is really easy to work with so let's just choose that we have seven over zero plus five is five is greater than or equal to eight over zero plus six is six you can reduce eight six if you want to divide each by two eight divided by two is four six divided by 2 is 3. now again i don't know off the top of my head if seven-fifths is a larger value than four-thirds i know it's not equal to four-thirds but again i can check by cross-multiplying so seven-fifths and thirds put a question mark there five times four is twenty three times seven is twenty one again the larger product is next to the larger fraction so in this case seven fifths is larger because twenty one is larger than twenty so this is a true statement this is true seven-fifths is greater than four-thirds so for interval c we can say that it's true again right so let me put a t here and i forgot to put an f here but let's go ahead and put that in now okay so now we just have interval d left so let's go ahead and check that and i'm just going to pick the number six easy to work with so 7 over 6 plus 5 is 11 so 7 11 is greater than or equal to eight over six plus six is twelve if i have eight twelfths i can reduce that eight divided by four is two twelve divided by four is three okay so that becomes two thirds again if you don't know which one's larger go ahead and set this up and cross multiply 7 11 question mark 2 thirds 11 times 2 is 22 3 times 7 is 21 22 is a bigger number so two thirds is the bigger number so 7 11 is not going to be greater than 2 3 so this is false okay so this is going to be interval d so this is false let's put an f here now recall that this guy right here is going to be a solution for our inequality and the other endpoints will not be right because they make the denominator zero okay so for our solution region remember a works that interval so anything less than negative six but negative six is not included because it makes the denominator zero so x is less than negative six and again strictly less than then or we also have this region here between negative five and 2 where 2 is included but negative 5 is not so we could say that negative 5 is less than x which is less than or equal to positive 2. okay so this is our solution we can also write this in interval notation as the union of the two intervals so we can say from negative infinity up to but not including negative six and the union with you have anything larger than negative five again negative five is not included out 2 and including positive 2 2 is included so you get a bracket there now we can go ahead and graph this on the number line let's erase all this again anything less than negative 6 works so i'm going to put a bracket negative 6 and shade everything to the left and then between negative 5 and 2 with only 2 included so negative 5 and 2 again bracket at 2 parenthesis at negative 5 and just shade everything in between so there's your solution graphically there's a solution in interval notation and here it is just using our variable x so again x is less than negative 6 or x is greater than negative 5 and less than or equal to 2. all right let's take a look at another so we have 2x squared minus 9x minus 5 over x squared minus 9 and this is less than 0. so because we have a strict inequality here none of the endpoints are going to be solutions okay because it's strictly less than so what i want to do here since i have a rational expression on one side and 0 on the other i can move into the second step which is just to set the numerator equal to 0 and solve and the denominator equal to 0 and solve so for the numerator i've got my 2x squared minus my 9x minus 5 is equal 0. this is pretty easy to solve using factoring you can also use the quadratic formula whatever's easiest for you is what you should do for me i can just factor this pretty quickly and the reason i say that is because this guy is a prime number so i know i can kind of get through this really fast so this is a 2x and this is an x i need to figure out this and this but that's pretty easy because this guy right here is a negative five okay so that means that it can only come from negative one times positive five or positive one times negative five okay now because i want a middle term of negative 9x i know that i'm going to put my negative 5 here and my plus 1 here because the outer would be negative 10x and the inner would be plus x okay that would give you the correct middle term again of negative all right so we can solve this using our zero product property i can set 2x plus 1 equal to 0. let me kind of squeeze this down and i can solve this so minus 1 on each side of the equation we know that this cancels you'll get that 2x is equal to negative 1. divide both sides of the equation by 2 and we'll cancel this 2 with this 2 x is equal to negative 1 half okay so let's put this over here we'll put x is equal to negative one-half you don't really need to specify that this comes from the numerator because again you have a strict inequality it's not going to work as a solution so let's erase this let's look at the other factor there so we have x minus 5 equals 0. add 5 to both sides and we get that x is equal to 5. so my solutions here will be that x equals negative 1 half and then also 5. all right let's think about the denominator now which is really easy to work with we have x squared minus nine so that's the difference of two squares if you want you can factor that and solve it or you can also use your square root property whatever is faster so if we have x squared minus 9 equals 0. again you could factor this but it's even quicker to say x squared if i add 9 to both sides we'd say it's equal to 9 by square root property we know that this comes to x is equal to plus or minus the square root of nine which is three okay so for the denominator we get x is equal to positive three and negative three okay those are our solutions so let's grab this and go to our number line and we'll just set up these intervals so negative one-half we don't have a notch for that but that's halfway between zero and negative one so let's just say that's about right there so i'm going to label that as negative one-half so you've got that as a visible boundary so let's put a vertical line there you've got three you've got negative three and you've got positive five okay so what i'm going to do again is i'm going to label these with capital letters so i'm going to say this is a anything less than negative 3 i'm going to say that b is anything greater than negative 3 and less than negative 1 half i'm going to say that c is anything that's greater than negative one-half and less than three i'm going to say that d is anything that's greater than three and less than five and i'm gonna say that e is anything larger than five okay so the original inequality we had our 2x squared minus 9x minus 5 over you've got your x squared minus 9 and this was less than 0. okay so a lot of stuff to check here in interval a let's just grab negative six so i would have we'll have two times you'll have negative six squared which is 36 and then minus nine times negative six so negative nine times negative six you can go and do that now that's 54. so plus 54 and then minus five so two times 36 is 72 so you would have 72 plus 54 which is 126 then if i subtract 5 away i get 121. so this would be 121. then this is over if we squared negative 6 you get 36 and then 36 minus 9 is 27. now it doesn't matter what this value actually is is it less than 0 no it's a positive number okay so in interval a it's false right because this doesn't work this is false so we'll put an f in interval a and then we'll go ahead and say that interval a here this is going to be false all right an interval b let's go ahead and grab negative 2. so i would have 2 times negative 2 squared minus 9 times negative 2 minus 5. this is over negative 2 squared let's just go ahead and do that that's 4. 4 minus 9 is negative 5. now if you go through here negative 2 squared is 4 4 times 2 is 8. so this is 8. you've got a negative 9 times a negative 2 which is 18. so this is plus 18 and then minus 5. so this is supposed to be less than zero now what you can tell here is that this guy is going to be positive in the numerator and it's going to be negative in the denominator so a positive over a negative is going to give you a negative and a negative is always less than zero so interval b here we can say is true for the sake of completeness we can go ahead and do this 8 plus 18 is 26 26 minus 5 is 21 so you would have 21 over negative 5 which again is a negative number and that's going to be less than 0. so for interval b let's put true okay so this works all right for interval c let's pick 0. so if i plugged in a 0 for x this would be gone this would be gone and this would be gone you would have negative five over negative nine well what happens is negative over negative is positive and that's not less than zero so in interval c it's false right it does not work all right for interval d let's plug in a 4 so i would have 2 times 4 squared minus 9 times 4 minus 5. this is over 4 squared is 16 16 minus 9 is 7. so 4 squared is again 16 16 times 2 is 32. so this is 32 here and then we have a negative 9 times 4 which is negative 36 so this is minus 36 and then minus 5. so this is supposed to be less than 0. so 32 minus 36 is negative 4 and negative 4 minus 5 is negative 9. so you would get negative 9 7 which is less than 0. so this is true so for d we'll say this is true okay so let's check something in interval e now and let's just choose 6. so we have 2 times we've got our 6 squared minus 9 times 6 minus 5 over 6 squared is 36 36 minus 9 is 27. so is this less than zero well 6 squared again is 36 36 times 2 is 72 so this is 72 here negative 9 times 6 is negative 54. so this is going to be minus 54 and then you have minus 5. so 72 minus 54 is 18 18 minus 5 is 13. so 13 27. again this is a positive number it's not less than zero so for e this is going to be false okay this is going to be false okay so let's go ahead and write our solution remember this is a strict inequality so none of the endpoints are going to work so we know in interval a it doesn't work in interval b it does so that's between negative 3 and negative 1 half so x is greater than negative three and less than negative one half okay so that's one part of it so then we're gonna say or here we also have this region between three and five so x is going to be greater than 3 and less than 5. so these are the two intervals that work or that satisfy your rational inequality in interval notation i can say that we have from negative 3 to negative 1 half again with neither included and it's the union with you have between 3 and 5 again with neither included so graphically let me erase all this and let me write my notch in for negative one-half again so i'm gonna go ahead and put a parenthesis at negative three and a parenthesis at negative one-half again shade everything in between i'm gonna put a parenthesis at three and a parenthesis at 5 and shade everything in between so this is our graphical solution this is our interval notation again using our variable x we say x is greater than negative 3 and less than negative 1 half or x is greater than 3 and less than 5. all right let's take a look at another so we have 6x squared plus 7x minus 24 over x cubed minus 6x squared plus 12x minus 8 this is greater than or equal to 0. okay so it's already in the format of a rational expression on one side and 0 on the other so that means we can basically move into the second step which is to set the numerator which is 6x squared plus 7x minus 24 equal to 0. we'll also set the denominator equal to 0 but let's just take care of this one first so for this guy right here it's a little bit more challenging to factor it because we have a non-prime number here right and a non-prime number here so 6 and you can think about negative 24 or 24 if you want to think about like that those are both composite numbers right so it takes a little bit more effort to factor there so let's just use our quadratic formula because in this case it'll be a little bit quicker at least it is for me so this is my a this is my b and this is my c so hopefully you remember your quadratic formula it's x is equal to you've got the negative of b plus or minus the square root of b squared minus 4ac all over 2a so i'm just going to plug in so i have negative b so that's negative 7 then i have b squared so that's 7 squared then i have a times c a is 6 and c is negative 24 and then 2a i've got 6 for a so 2 times 6 is 12. i'm going to get rid of that 7 squared is 49. negative 4 times 6 is negative 24. negative 24 times negative 24 is 576. so this is plus 576 and then 49 plus 576 is 625. okay so the square root of 625 is 25. that's pretty easy so we have two solutions here you've got x equals negative 7 plus 25 which is going to be 18 over 12. so one solution is that x equals 18 12. you can of course reduce this each is divisible by six eighteen divided by six is three twelve divided by six is two so one solution is x equals three halves another solution is we have negative seven minus twenty-five which is negative thirty-two and then this is over 12. so i can reduce this by dividing each part by four 32 divided by 4 is 8 so that would be negative 8 there 12 divided by 4 is 3. so you've got 3 halves and negative eight thirds so this is for the numerator and we want to specifically mark that as from the numerator because we have a non-strict inequality here we have this greater than or equal to so these guys will work as solutions okay so let me just kind of move this up here a little bit and we're just going to say that this comes from our numerator okay and let me kind of tighten this down a little bit so we have three halves and again negative eight thirds all right so now we want to set the denominator equal to zero so we have x cubed minus 6x squared plus 12x minus 8 equals 0. so you might pause for a minute and say how in the world are we going to solve this this is not a quadratic equation this is not something we learned how to solve right we'll talk about these types later on in the course when we talk about the rational root theorem but for right now you can't solve this by factoring it by grouping right that's what you might start out by thinking okay i'm going to factor this using grouping it's a fortunate polynomial but in fact you have to remember that this comes from the special factoring lesson this guy right here would factor into a binomial which is going to be x minus 2 and that quantity cubed right if you went through this and expanded it you would get back to this okay so this is equal to 0. now the cool thing about this is is that you have basically x minus two times x minus two times x minus two so if you use the zero product property you only get one solution right you get x minus two you set that equal to zero you add two to both sides of the equation and you get that x is equal to 2 okay so from the denominator we get that x is equal to 2 and again that right there will not satisfy the inequality right you'd be dividing by 0 and that's not allowed so let's copy these and bring them to the number line all right so let's go ahead and set up our intervals on the number line so three halves is basically 1.5 so let's say that's about right there and negative eight thirds is going to be negative two and two thirds so let's say that's about right there okay so that's negative eight thirds okay so let's go ahead and make some vertical lines here so i'm gonna go ahead and put one here and also here and this is gonna be tight but i also have to put one here at two okay so my interval a will be anything that's less than negative eight thirds my interval b will be anything between negative eight thirds and three halves then my interval c this is going to be a tight one is anything between 3 halves and 2 and then interval d will be anything larger than 2. okay so let's pick a test value from each region here and my original inequality again was 6x squared we had plus 7x we had minus 24 and this is over we'll have x cubed minus 6x squared plus 12x minus 8. but again in factored form i'm just going to put this as x minus 2 that amount cubed and this was greater than or equal to 0. now to save time i'm just going to put some parentheses wherever there's an x just so i can quickly substitute things in and let's put some parentheses here okay so in region a i can choose something like negative 4. so let's plug in a negative 4 here and here and here so negative 4 squared is 16. let me kind of scroll down a little bit so negative 4 squared again is 16 16 times 6 is 96 and then you would have 7 times negative 4 is minus 28 and then minus 24. so 96 minus 28 minus 24 is 44. so you would have 44 in the numerator in the denominator negative 4 minus 2 is negative 6 negative 6 cubed is negative 216. so we don't need to go through and reduce this or anything like that we know that this value is negative and a negative is not greater than zero so this guy is going to be false okay so this is false so we can say that interval a is going to be false right nothing's going to work here for interval b i have zero so that's pretty easy to work with so essentially this would be zero this would be zero and this would be zero so you can go ahead and say this is gone this is gone and this is gone so you would end up with negative 24 in the numerator in the denominator you would have negative two cubed which is going to be negative eight so you've got negative over negative which is positive right it ends up being positive three so three is greater than or equal to zero which is true so this guy is true so interval b is true for interval c you've got a real tight window you've got between 3 halves or 1.5 and 2. so what you can do is you can take 1.75 so i'm gonna take six times and i'm gonna plug that in here i don't really have enough room so i'm just gonna do it right here so six times you've got 1.75 squared plus seven times 1.75 and then minus 24. okay then this is over you're going to have 1.75 minus 2. so 1.75 minus 2 this is going to be cubed okay so let's kind of scooch this over here so we have some room we want to say this is greater than or equal to 0. so 1.75 squared is going to be 3.0625 if i multiply that by 6 i get 18.375 so let's put 18.375 then plus 7 times 1.75 is 12.25 so if i add those amounts together i'll get 30.625 so 30.625 and then minus 24. so the main thing here forget about the number you know that it's going to be positive so let's just say the sign in the numerator is positive in the denominator i know this guy is going to be negative right so 1.75 minus 2 is basically negative 0.25 if i cube that i'm still going to have a negative so positive over negative is negative a negative is not greater than zero so this guy's going to be false okay this guy's going to be false so for region c we can put an f for false okay one more region to check we need to look at region d so for region d let's just go ahead and pick 4 so we'll plug that in there and there and there so 4 squared is 16 so you would have 6 times 16 which is 96 plus you'd have 7 times 4 which is 28 and then minus 24. so if i did 96 plus 28 minus 24 i'd get 100. so the numerator here would be a hundred for my denominator 4 minus 2 is 2 2 cubed is 8. so forget about the fact that you could reduce this this is going to be greater than zero it's a positive number so this guy is going to be true okay and i forgot to write false here it's kind of hard to squeeze it in but let's put an f there all right so we know what our solution is now we've gone through all the possibilities we can say that anything in region b so that means x is going to be greater than in this case or equal to negative eight thirds and less than or equal to in this case 3 halves again i included the negative 8 thirds and the 3 halves because those were from the numerator okay then or the other scenario where it's true is anything larger than two so x is greater than two but i don't include this endpoint here because this came from my denominator makes the denominator zero so it's going to be excluded so in interval notation we have between negative eight thirds and three halves with each included so i use a bracket next to each and it's the union with anything larger than two so a parenthesis next to two again this goes out to positive infinity so let's graph this so again we said negative eight thirds was about right here and we said three halves was about right here so we're going to put a bracket for each of these so a bracket there and a bracket there shade everything in between and then we know where 2 is that's right here i'm going to put a parenthesis there because it's not included let me make that a little bit better i'm just going to shade everything to the right so this is my solution graphically then this is my solution interval notation and then using my variable x i can say x is greater than or equal to negative eight thirds and less than or equal to three halves or x is greater than two in this lesson we want to review applications of rational expressions and specifically here we're going to look at some motion and some work rate problems so at this point you should be pretty comfortable with solving word problems and although some problems in this section will not result in linear equations we can still rely on our little six step procedure that we used in our section on applications of linear equations so again for solving a word problem the first thing we want to do is read the problem and determine the question or questions to be answered then the next thing we want to do is assign a variable to represent the unknown and for the scenarios we're going to have today we can express other unknowns in terms of this variable then the next thing we want to do is write an equation and this is based on the scenario we're given then we want to solve the equation then we want to write the answer in terms of the question or again questions asked then lastly we want to check the answer using the words of the problem just make sure that things are reasonable so if you're in the section on applications of rational expressions the most common problems you're going to see are going to be work rate problems or some people call them rate of work problems or you're going to see motion word problems right these involve again the distance formula distance equals rate of speed times time travel okay so we talked about this formula when we looked at applications of linear equations so we see it again okay and we'll see this throughout our course many times so again if you don't know the formula d for distance is equal to r for rate times t for time so if i'm going 40 miles per hour and i do this for 3 hours i multiply 40 times 3 that gives me my distance of 120 miles okay so this would be 120 in that case so pretty easy formula to understand very very intuitive and we're just going to use it to kind of solve this first problem so the conor river has a current of 8 miles per hour find the speed of lauren's boat in still water if it goes 210 miles downstream in the same time it takes to go 98 miles upstream okay so what are we being asked for we're being asked to find the speed of lauren's boat in still water okay this is what we need to find now we're given a lot of information to help us find this it tells us the connor river has a current of eight miles per hour it also tells us that her boat can go 210 miles downstream in the same time it takes to go 98 miles upstream so when you get these type of problems you need to understand the difference between still water upstream and downstream okay so the way these problems work if you're in still water basically there's no car okay if you're going upstream you are fighting the current okay so you have to subtract the speed of the current away from the speed of the boat in still water if you're going downstream the current is pushing on you okay so you're going with the current and so now you need to add the speed of the current to the speed of the boat in still water okay so we'll talk about that more in a second for right now our next step is to what is to assign a variable to represent the unknown our unknown here is the speed of lauren's boat in still water so i'm just going to let x be equal to that so let's let x be equal to the speed of lauren's boat in still water okay so that's water with no current all right so let's take this down to a little table so we can organize our information so the first thing i'm going to do is put my distance formula up here again d stands for distance and this is equal to r which stands for rate of speed multiplied by t which stands for time traveled okay again if i'm moving at 40 miles per hour and i do this for three hours 40 times 3 is 120 so i know it would go 120 miles okay it's a very intuitive formula so i'm just going to put my two scenarios over here on the left so we have upstream and we have downstream okay so what am i told in the problem again my rate in stillwater is unknown that's what we're trying to find but we expressed it as x so i'm going to put x here and i'm going to put x here now in the case where i'm going upstream again i'm fighting the current the current is eight miles per hour so i've got to subtract eight away from the speed of the boat in stillwater so i subtract eight away from x in the case of going downstream it's the opposite right the current's pushing on me so i'm going to add 8 or again the speed of the current to the speed of the boat and still water so for upstream my rate is x minus 8. for downstream my rate is x plus 8. now what about the distance well it specifically tells us that the boat goes 210 miles downstream in the same time it takes to go 98 miles upstream so my distance for downstream will be 210 and my distance for upstream will be 98. so for downstream it's 210 for upstream it's 98. okay so now what about the time well to do the time what we can do we can solve the distance formula for time so again distance is equal to rate of speed times time traveled okay so if i divide both sides by r what's multiplying t there we know that this would cancel with this i basically have that time is equal to distance over rate so right now i have a distance and i have a rate for each scenario so if i divide my distance by my rate i'm going to have my time modeled okay so let me erase this so for upstream my distance is 98 and my rate is x minus 8. so the time again is distance divided by rate so it's going to be 98 divided by x minus 8. for downstream again my distance is 210 and my rate is x plus eight so my time for downstream would be 210 over x plus eight now it tells us in the problem let me go back up that the boat can go 210 miles downstream in the same time it takes to go 98 miles upstream so all we have to do is set the two times equal to each other because again in the problem it tells us that it's the same it says in the same in the same time okay so i can take 98 over x minus 8 so 98 over x minus 8 that's the time for going up stream and set this equal to my time for going downstream which is 210 over x plus 8. so let me just copy this we'll go to a fresh sheet so how can we solve this we already know how to solve rational equations it's very very easy in this particular case i don't need to multiply both sides by the lcd right the lcd would be x minus 8 times x plus 8. i could multiply both sides by that but a quicker method here would be just to cross multiply so you take this guy right here multiply by this you take this guy right here multiply by that so essentially what i'd have is 98 times the quantity x plus 8 is equal to you'd have 210 times the quantity x minus 8. so i could just start by using my distributive property on each side but one thing you want to see here is that 98 and 210 are each divisible by 14 right so to make this a little easier on myself i'm going to divide both sides of the equation by 14. remember this is multiplication here so this is legal to cancel 98 divided by 14 is going to give me 7 so i'd put a 7 here then 210 divided by 14 is going to give me 15. and now i have smaller numbers to work with so it's just going to make my life a lot easier so 7 times x is 7x and then plus 7 times 8 is 56 this equals 15 times x is 15x and then minus 15 times 8 is 120. okay so let me scroll down a little bit get some room going i can subtract 15x away from each side of the equation and i can subtract 56 away from each side of the equation so we know that this is going to cancel and this is going to cancel 7x minus 15x is going to give me negative 8x and this will be equal to negative 120 minus 56 is negative 176. if i finish this up by dividing both sides by negative 8 right the coefficient of x there what i'm going to get is this cancels with this i'll have x is equal to 22. so this is what we're looking for i can't just put x equals 22 obviously i've got to answer my question but x tells me the speed of lauren's boat and still water which is exactly what we want to know right so it's going to be 22 miles an hour so i can say this is 22 here and let me go up and just answer my question so i can just say that lauren's boat has a speed let me make this better of 22 miles per hour in stillwater so how can we check to make sure we got the right answer well again if we go back to the fact that time is equal to distance over rate we can just kind of think about the scenario we're given for the downstream scenario we know that her distance is 210 miles and we know that her rate of speed is the speed of the boat in stillwater which is 22 plus the speed of the current which is eight so that's going to be 30. so 210 divided by 30 would be 7. so this is her downstream scenario it took her seven hours to go 210 miles now the other scenario let me just set this up over here we know that she goes 98 miles going at a rate of speed remember if it's going upstream it's going to be 22 the speed of the boat in still water minus 8 okay the speed of the current which is going to be 14. okay so this is 14 and we know that 98 divided by 14 is 7. so this is true right it took the same amount of time to go 210 miles downstream right seven hours that it takes to go 98 miles upstream against seven hours so our answer here is correct again we can say that lauren's boat has a speed of 22 miles per hour in still water all right now let's move on and talk about rate of work problems otherwise known as work rate problems so essentially these are very easy problems to solve once you understand how to kind of set them up and they just deal with the rate of speed at which a job can be completed given the individual work rates from the people performing the job okay so again very very easy if you understand the distance formula that we just used in the last problem then you could understand this formula i'm going to give you here today so the distance formula again is d for distance is equal to the rate of speed r times t for time travel okay very intuitive again if i go 40 miles per hour for 3 hours 40 times 3 is what it's 120 so i know i've gone 120 miles okay very intuitive very easy to understand the same thing goes when we work with work rate problems or rate of work problems so instead of d you have a which stands for the amount of work completed and we think about this in terms of jobs completed so it could be one jobs completed in pretty much every scenario that i've seen but if you wanted to extend that and say okay we have two jobs completed 11 jobs completed 2 million jobs completed whatever that is okay but a here stands for the amount of work completed this is equal to r which is your rate at which the work is done okay otherwise known as a work rate that's how these problems get their name okay then times t which is our time and that's the amount of time that you're working for so let's say for example my rate of work and this will make more sense when i explain in a minute is 1 over 4 okay and i multiply this by a time of eight what does that give me we know that eight times one fourth is two so what does this mean my work rate tells me how much of the job i can complete in one unit of time so let's say as an example i'm working with hours and let's say i was painting a house so it takes me four hours to paint the house well in one hour or one unit of time i'm done with one fourth of that job if i work at this rate for eight hours okay i continue it eight hours i'm going to paint two houses right okay that's all that's saying if i change this to four then this just becomes one right so if i work at a work rate of one-fourth of the job per hour and i do this for four hours well then i've got one completed job now in almost every case or in every case that i've ever seen a is going to be 1. so basically what you want to focus on here and let me erase this we don't need this information anymore this r here is your rate of work or your work rate that's what you're focused on in the problem so i'm going to divide both sides of the equation by t that's what's multiplying r so i can isolate and i'll say that r my work rate or rate of work is equal to 1 over t so this is very important you probably want to write that down in your notes if you're taking notes this is what we're going to use in almost every scenario to get an answer okay so you've got to know that if you're working on a problem like this and you want your work rate just take a one and divide it by the amount of time it takes to complete the job so again in the case where we could paint the house in four hours while i would get a work rate my r would be equal to one over takes four hours to complete it so one over four and all this is telling me is that in one unit of time or one hour i've completed one fourth of the job okay as another example suppose it takes me 15 minutes to wash my car okay in one unit of time which is now one minute because we're working with minutes then i've done one over again it took me 15 minutes to complete it so 1 over 15 or 1 15 of that job is complete okay so that's the general idea behind these problems and now let's kind of jump in and look at one all right so working alone it takes heather 30 minutes to wash the car jacob can wash the same car in 45 minutes how long would it take jacob and heather to wash the car if they work together all right so the first thing i'm going to do here just like we always do is just find out what are they asking us to find well the question we need to answer here is how long would it take jacob and heather to wash the car if they work together this is a pretty typical scenario so you have their individual rates of work right we know that heather takes 30 minutes to wash the car we know that jacob takes 45 minutes to wash the car but hey if they work together at the same speed that they're working at when they do it individually how long would it take for them to wash the car okay so that's what we're trying to figure out so what i'm going to do is i'm going to let a variable like x be equal to that so i'm going to go to the next page and i'm just going to say that we're going to let x be equal to the amount of time and i'm just going to say to complete the job to complete the job right but if you wanted to be more detailed you could say the amount of time to complete the job when jacob and heather work together okay and of course since we're working with minutes here the units are going to be minutes okay so you got to make sure of that because since some of these problems you have crossovers between minutes and hours you could have minutes and seconds you know so on so forth but for this one it's very easy we're just working with minutes all right so let's go down to a little table and we're going to organize some information and then once you see this in the next few problems you're going to go right through it okay very very simple so of course we have jacob and we have heather now the first thing i'm going to write is the rate of work so the rate of work and before we started this problem we talked about how to get it it's one over the amount of time they take to complete the job so i'm going to put a one over and then blank for each one let's go back up so for heather it takes 30 minutes to wash the car so her rate of work would be 1 over 30. again for every minute that passes she's done 1 30th of that job for jacob his is going to be what it takes him 45 minutes to wash the car so it's going to be one over 45. every minute that passes he's going to be done with one over 45 or 145th of that job so let's go down again for jacob he's got a work rate of 1 over 45 for heather she's got a work rate of 1 over 30. okay so now i'm going to put time working together so time working and then together now we modeled this on the page up as x the amount of time to complete the job again this is working together so i'm just going to put this as x here and x here okay it's the same in each case now the next thing you need is the total contribution of each person how much of the car does jacob get done in this amount of time how much of the car does heather get done in this amount of time so i'm just going to write this as total contribution okay so to get this we can just multiply right because if you want to know the amount of work that jacob performed i just multiply my r times my t right my a the amount of work that's performed is r the rate of work times t the time that you're doing that work for well i've got my rate of work i've got my time that i'm working for so my contribution here or the amount of work that's being performed for jacob is 1 over 45 times x you could write it as 1 over 45 times x or you could write it as x over 45 which is probably a little better same thing goes for heather it's going to be x over 30 right just multiply across very very easy and now what we need to understand is that between jacob and heather they're going to complete one job okay so the equation here is very easy to set up jacob's contribution which is x over 45 plus heather's contribution which is x over 30 is going to be equal to one again that's for one completed job so the first time you see this it might not make a whole lot of sense but again after you solve like two or three of these you're going to find that these are the easiest problems in the world to solve so let's copy this so what i'm going to do to clear these denominators i'm going to multiply both sides of the equation by the lcd the lcd if i look at the denominators i got 45 and 30 it's going to be 90. so i'm going to use my distributive property here 90 times x over 45 90 divided by 45 is 2 2 times x is 2x then plus 90 divided by 30 is going to be 3 then times x is 3x and this equals 90. some on the left so 2x plus 3x if i combine like terms there i get 5x this equals 90. divide both sides of the equation by 5 and we're going to find that x is going to be equal to 18. so 18 is the number of minutes that they each have to work when they're working together to complete the job right so if we go back up again we said amount of time to complete the job that's when they're working together and of course this is in minutes so let me just say that they work for 18 minutes to complete the job okay you can answer a whole lot of different ways there that's just a very simple sentence to say they work for 18 minutes to complete the job now is it correct how can we check this well we can take the fractional amount that heather does and the fractional amount that jacob does and add those together and see if it equals 1. so we know that if we go back down a is equal to r times t i'm going to put two scenarios a equals r times t so how much or what amount of the job does heather do well remember her r her rate of work is 1 over 30. and she works for t minutes and again we found that the time that they're going to work together which we labeled as x was going to be 18. so essentially i'm going to put this as times 18 that's going to give me the amount of work that she does so between 18 and 30 i have a common factor of 6. so if i divide 18 by 6 i get 3 if i divide 30 by 6 i get 5. so i get 3 fifths there so this is going to be three-fifths that's the amount of work that she completes now for jacob he's a little bit slower so his rate of work is 1 over 45 and the time is the same right because they work the same amount of time when they're working together so this is 18 each here is divisible by 9. 45 divided by 9 would be 5 18 divided by 9 would be 2 so this is 2 fifths now if you go ahead and take the amount of work that jacob completes which is two-fifths that's what he does in 18 minutes and you add that to the amount of work that heather completes she's got three fifths of the job that's completed and that's what she does in 18 minutes you do get one right two fifths plus three fifths is five fifths five fifths is one so we can say our answer here is correct they work for eighteen minutes to complete the job all right very easy overall let's take a look at another one so working together james and nathan can mow a yard in 24 minutes when james mows the yard by himself he takes four times longer than nathan how long would it take for each person to mow the yard if they were working alone so in the last problem we knew their individual rates of work right we knew that for jacob he could wash the car in 45 minutes for heather she could wash the car in 30 minutes but we didn't know or we could say we had to find how long it would take if they worked together in this case we know how long it takes when they work together right it tells us right here it's 24 minutes but we don't know how long it takes them individually okay so that's the question here how long would it take for each person to mow the yard if they were working alone so how long does it take for james how long does it take for nathan so we're specifically told that james takes four times longer than nathan so let's let x be equal to the amount of time it takes for nathan to mow the yard by himself so the time for nathan and i'm not going to write the rest you can write the rest if you want but it's just the time for nathan to again mow the yard by himself so then because james takes four times longer again it tells us that right here when james moves the yard by himself he takes four times longer than nathan what i can do is just multiply four times x to get the time for james so i could say then 4x is equal to the time for james okay let's go back to our table and kind of organize so again we have nathan and james so we have nathan and we have james and again we have the rate of work the rate of work and we have the time working together and then we have their total contribution so for the rate of work remember it's one over the amount of time it takes for that person to complete the job so if it takes me four hours to paint a house again the rate of work is one over four that's how much i'm going to do in one unit of time so in an hour i'm going to do one fourth of the job here we know for nathan this is in terms of minutes he can complete 1 over x amount of that job every minute okay so for james it's going to be 1 over 4x every minute so for nathan it's 1 over x for james it's 1 over 4x the time working together it tells us that in the problem it tells us specifically that james and nathan can mow a yard in 24 minutes that's again when they're working together so each person is going to work for 24 minutes in this scenario so what's their contribution again you just multiply this is the rate of work this is the time working so 24 times 1 over x you could do just 24 over x to make it simpler and then here it's 24 over 4x which we could simplify 24 over 4 is 6 so i could just say this is 6 over x now how do we get an equation going well we know that when we sum the amount of work done from both of them over this again 24 minute period they've completed one job so the amount of work that nathan does is 24 over x the amount of work that james does is 6 over x if i add these two amounts together i should get 1 completed job so let's copy this all right so what can i do here well i can multiply both sides of the equation by x right that would be the lcd so i'll multiply this by x this by x and this by x so essentially this is going to cancel this is going to cancel you'll basically just have what 24 plus 6 which is 30 is equal to x so that's really a piece of cake we've immediately found that x is 30 and we know that x was the time for nathan so this is going to be 30 minutes and then for james it's going to be four times that or 120 minutes if you wanted to put that in terms of hours you could that's two hours if you didn't know that it's okay you can put it in terms of minutes right it doesn't matter so i'm just going to say that nathan can mow the yard in 30 minutes so nathan can mow the yard in 30 minutes then i'll say while james can do it and you can say two hours or 120 minutes i'll just say 120 minutes because the problem was given in minutes but again you can put two hours it's the same thing so does this make sense let's kind of scooch up just a little bit so again working together james and nathan can mow a yard in 24 minutes so we know that when james mows the yard by himself he takes four times longer than nathan so according to our answer james does it in 120 minutes and nathan does it in 30 minutes so if this is correct we're good to go right because it is four times 30 times four will give me 120. so that part does check out we just need to make sure that they can work together and get it in 24 minutes so again think about this the amount of work that nathan does is he's going to do one over 30 of that job every minute so he's going to do this again if they work together for 24 minutes then if i add to this how much james does so he's going to do 1 over 120 of that job for every minute and he's going to do that for 24 minutes if i sum these amounts i should get 1. this is the amount that nathan does all the way to the left then we have plus again we have the amount that kind of james does over here again if i sum them i should get one so how can i reduce this here well i can divide both by six that would give me a four here and a five here for this one 120 divided by 24 is 5. so i can cancel this with this and put a 5 here so you've got four fifths of the job which is what nathan is going to do plus one-fifth of the job which is what james is going to do you sum those two amounts together you get five-fifths or one again for one completed job so our answer here is correct nathan can mow the yard in 30 minutes while james can do it in 120 minutes or again two hours okay so pretty easy overall let's look at one that's a little bit more challenging but again it's not so much because these problems are really easy so clear's empty swimming pool can be completely filled from a garden hose in 30 hours her outlet pipe can completely drain a full pool in 120 hours if for some reason both the garden hose and the outlet pipe are turned on at the same time how long would it take to completely fill claire's empty pool so the question we need to answer here is how long would it take to completely fill claire's empty pool given these two competing forces you've got one kind of hose pipe that's filling the pool and then you've got an outlet pipe or you could say a drain pipe that's emptying the pool so they're kind of working against each other you can think about this as if you're filling your kitchen sink okay and you kind of you got the water on full blast but you hold the stopper just up just a little bit right so some of the water can drain so if the amount of water coming in is more than the amount of water going out eventually you're going to fill your sink right but it's just going to take longer than if you had the sink kind of all the way stopped up okay so that's the idea of what's going on here so how can we model this let's just say we're going to let x be equal to the amount of time it would take to completely fill claire's empty pool so i'm just going to say let x be equal to the amount of time to fill the pool okay let's go down to the kind of table let's organize some stuff so we've got our hose that's filling the pool and then we've got our outlet pipe that's trying to drain it and let me be more specific here and just put garden hose so garden hose because that's what it says in the problem now again we're going to have our rate of work we're going to have our time working together and then we're going to have our total contribution so we know one thing so far it's the time working together is x because we just let a variable represent that so this is x and this is x the rate of work let's go back up we've got to find the garden hose and outlet pipe so the garden hose can fill the pool in 30 hours so every hour it's done one over 30 of that job so the rate of work is 1 over 30. so i'll put 1 over 30. now the outlet pipe which is trying to drain the pool that's its goal okay so every hour since it takes 120 hours to do this it's done one over 120 of that so i'll put 1 over 120 so that's the rate of work now the contribution works a little bit differently here because they're competing forces right you can kind of think about the garden hose that's filling it so i'm going to put a positive 1 over 30 times x or you just put x over 30. really you can think about the outlet pipe as fighting against this right you can think about what it's doing as subtracting away so you can put this as a minus x times 1 over 120 which would be x over 120. so if you sum these two amounts together you should set it equal to 1 and solve that equation so i would have x over 30 i'll just put minus or you can put plus negative it doesn't matter i'll put minus x over 120 i'll set that equal to 1. so in other words the total contribution or the total amount of water that goes in from the garden hose minus the total contribution or the total amount of water that gets pulled out from the outlet pipe when that's set equal to one the value for x is going to tell me the amount of time it takes for them to complete that job again working together so let me copy this so what i want to do is just multiply both sides by the lcd which is going to be 120. pretty easy to do so i know that i'm going to distribute this to each term inside the parentheses 120 divided by 30 is 4. right so i'd have 4x minus 120 divided by 120 is 1. so i just have x there this equals 120. 4x minus x is 3x this equals 120. divide both sides of the equation by 3 and you're going to find that x is equal to 40. so that tells us what it tells us that the garden hose and the outlet pipe can both be left on and the pool is going to go from completely empty to completely full in 40 hours let's go back up so it would take 40 hours to fill the pool okay so let's check this so again i go back to my formula of a equals r times t so think about this for each scenario you have another one over here so what i'm going to do is i'm going to say the garden hose how much water does it put in so we think about the rate again it does a full job in 30 hours so the rate is 1 over 30. when they work together again they're going to work for 40 hours so the amount that's going to be done is what i can divide both of these by ten this will be three and this will be four so it's going to be four thirds that's going to be completed so four thirds of the job is completed now i'm subtracting away what gets completed by the outlet pipe what's going to get drained away here so the rate there again it accomplishes the whole thing 120 hours so it's 1 over 120 and then the time again it's going to go for 40 hours so 120 divided by 40 is 3. so this would be a three and this would be a one so basically this is one third so does this make sense yes it does if the garden hose completes four thirds of the job in 40 hours and the outlet pipe completes one third of the job in 40 hours well four-thirds minus one-third gives me three-thirds or one again what's happening here is the garden hose is filling a pool and then another third of a pool okay so four thirds then the outlet pipe is taking a third away right it's taking that extra third away so you're left with three thirds or just one full pool okay so we can say our answer here is correct it would take 40 hours to fill the pool in this lesson we want to talk about solving absolute value equations with quadratic and rational expressions all right so previously in our course we talked about how to solve kind of simple absolute value equations where basically we were only dealing with linear equations okay so we're going to go a little bit deeper into this topic and we're going to look at some additional examples that are slightly more challenging you might say not more challenging but more tedious because we're going to involve some quadratic expressions and some rational expressions so we'll end up having to solve some quadratic equations and also some rational equations okay so we're going to boil this down to a little rule if u is an algebraic expression and a is some positive number so it's greater than zero so it's not zero and it's not negative right because zero and negative would each give us a special case we'll talk about that in a second so the absolute value of this algebraic expression u if that's set equal to this positive number a then what we get is that the algebraic expression u could be equal to the positive number a or the algebraic expression u could be set equal to the negative of that number a okay so as a quick example this is something we already know how to do let's say we have the absolute value of 2x minus 5 and this is equal to 3. the first thing is notice how the absolute value operation is isolated on one side and you have a positive number on the other okay so that's always what we want to set up so then we can use our little rule again it's kind of cut off the screen here but if the absolute value of some algebraic expression u is equal to some positive number a then u is equal to a or u is equal to negative a we're just going to use that here again this expression here 2x minus 5 that's u that's the algebraic expression and then 3 is the positive number that's a so i would say that 2x minus 5 could be equal to 3 or 2x minus 5 could be equal to negative 3. now what we're basically saying is that this guy right here could be set equal to 3 right if i get a value for x that gives me a result of 3 i take the absolute value of that i get 3. also this guy right here can be set equal to negative 3 because if this guy in here evaluates to negative 3 well when i take the absolute value of negative 3 i also get 3 okay so that's why this ends up working itself out so let's just solve this real quick i'm going to add 5 to both sides of the equation i'm going to get that 2x is equal to 8 divide both sides by 2 i'm going to get that x is equal to 4. over here i'm going to add 5 to both sides i'm going to get that 2x is equal to negative 3 plus 5 is 2 divide both sides by 2 i'm going to get that x is equal to 1 okay so we get x equals 4 or x equals 1. now you can easily check that you could plug in a 4 for x there and basically 2 times 4 is 8 8 minus 5 is going to be 3 right so you know with the absolute value of 3 which is 3 so that works itself out okay the other scenario is going to give you the absolute value of negative 3 because that's also going to give you 3. right if i plug in a 1 for x 2 times 1 is 2 2 minus 5 is negative 3 again the absolute value of negative 3 is also 3 so this works as well all right so before we go any further we have to understand the special cases that we might run across and we covered these earlier in the kind of course when we talked about absolute value equations but we need to kind of jog your memory so that you are ready for all the scenarios you're given so let's say you have the absolute value of 4x minus 5 and this is equal to 0. well what happens is if this guy is isolated and it's set equal to 0 well think about the fact that 0 is 0 units away from itself on the number line right so the only number that's going to have an absolute value of 0 is 0 okay so you're not going to have a compound equation with or here you just have to set 4x minus 5 this expression equal to 0 and then just solve okay so this is a little bit easier a little less tedious so to get our solution there we just add 5 to both sides of the equation and we'll get that 4x is equal to 5 and let's divide both sides of the equation by 4 and we'll find that x is equal to 5 4. okay and obviously if you check that if you plugged in a 5 4 there 4 times 5 4 the 4's would cancel you just have a 5 5 minus 5 is 0. the absolute value of 0 is 0 so that checks itself out as well all right last special case scenario is if the absolute value operation is isolated and it's equal to some negative value there's no solution okay and the reason for that is the absolute value operation takes a negative it makes it positive takes zero and spits out 0 takes a positive and spits out the posit okay so you can't plug in anything for x here and then multiply it by 6 that guy subtract away 2 right take the absolute value and end up with a negative 12. right because the absolute value operation is going to kind of stop all that because it will never ever be equal to some negative value so you can't get a negative 12 as an answer for that so this guy right here is no solution okay it's no solution we can use our symbol for the null or empty set all right so now let's jump in and look at some problems where we have some quadratic expressions involved so suppose we have the absolute value of x squared minus 5x and this is set equal to 6. so we have exactly what we talked about earlier if we have the absolute value of some expression u and it's set equal to some positive number a well then u could be a or u could be the negative of a okay so it's the same concept we're just using kind of more complex expressions so we would have x squared minus 5x x squared minus 5x we'd set that equal to 6 and then or we'd have x squared minus 5x would be equal to negative 6. okay all i did was follow this this expression is my u this guy right here this number is my a okay so i'm just following that just like we saw okay so now i can solve each of these they're kind of easy quadratic equations to solve because they're both factorable i can subtract 6 away from each side of the equation here i would have that x squared minus 5x minus 6 equals 0. and i'll go ahead and solve this real quick so as i alluded to this can be factored the left side here if i think about two integers whose sum is negative five and whose product is negative six well you think about negative six and positive one right negative six and positive 1 because negative 6 times positive 1 would give you a negative 6 negative 6 plus 1 would give you negative 5. so this is our factorization here then we would set each factor here equal to 0. so x minus 6 would be equal to zero and then we'd also have x plus one equal to zero and let's go ahead and solve that so we would add six to both sides of the equation here x is equal to six over here i subtract one away from each side of the equation so x is equal to negative 1. so the two solutions here are that x equals 6 or x can be negative 1. so let me kind of copy this because this is going to get a little bit too busy and let me kind of come back up here for a second and i'm just going to erase this we already know the rule and i'll just kind of paste this in here so those are two solutions i'm going to go ahead and get rid of this whole branch we don't need it anymore all right so for this one it's another easy one we can solve with factoring so we have x squared minus 5x i'm going to add 6 to both sides so plus 6 equals 0. so i'm going to factor this guy so this is x and this is x give me two integers whose sum is negative 5 and whose product is positive six well that would be negative two and negative three right negative two times negative three is positive six negative two plus negative three is negative five so we're good to go there all right so what we want to do is just set each factor equal to zero so x minus two equals zero and then x minus three equals zero add two to both sides of the equation here you're gonna get that x is equal to two add three to both sides of the equation here you'll get that x is equal to three okay so x equals 2 or 3 those are another two solutions so let's just write those in right here let me erase all this so let's pop those in here so 2 and then 3. so a little bit more tedious and you actually have more solutions here you have four solutions whereas the problems we looked at before in most cases you had two solutions right and you can go through and plug all these in and verify they work and just for this one we can go ahead and check it just to see what happens for the rest of them i'm not going to check it just in the interest of time but i just want to show you that you're going to have two of these that are going to evaluate here to be the absolute value of 6 so that's going to work out then the other two are going to evaluate to be the absolute value of negative 6 so those are going to work out as well right because in each case the absolute value of six or the absolute value of negative six it produces six so the left and the right side will be equal okay so for this scenario six squared is 36 so you'd have 36 minus 5 times 6 which is 30. 36 minus 30 is 6. so it's this scenario here so this one works then for this one here negative 1 squared is 1 then you have negative 5 times negative 1 which is positive 5. so you sum those and you get 6. so again this scenario here so that one would work then for these two you're going to end up with negative 6 right so 2 squared is 4 minus you have 5 times 2 which is 10. 4 minus 10 is negative 6. so that's the scenario here so that works then the other one plug in a 3 3 squared is 9 minus you have 5 times 3 which is 15 9 minus 15 is also negative 6 so this guy is going to work as well right so in this scenario again we want the inside of this guy right here we want that expression to evaluate to either 6 right that's this scenario here or negative 6 that's this scenario here either way we'd end up with an answer of 6 so 6 equals 6 so our solutions here are correct all right let's take a look at one that's a little bit more challenging it's just more tedious we have 4 times the absolute value of x squared minus 3x plus 7 equals 47. okay so in this particular case our absolute value operation is not isolated okay we know from earlier in the course that we've got to have it isolated we can't just start doing stuff i see a lot of students when they get this thing okay this is 4 times the absolute value of x squared minus 3x you know plus 7 equals 47 and then they do this again and they say okay 4 times the absolute value of x squared minus 3x plus 7 equals negative 47. no remember your rule is that the absolute value of u okay this is isolated if that's equal to a then u equals a or u equals the negative of a okay but this has to be isolated notice how there's nothing else besides that absolute value of u so if it's not isolated you've got to do a little bit of work to kind of get things set up get things started first okay so let's go ahead and subtract 7 away from each side of the equation we'd have 4 times the absolute value of x squared minus 3x and then this is equal to 40. now i can divide both sides of the equation by 4 right because that's what's multiplying this guy so this cancels you'll have the absolute value of x squared minus 3x is equal to 10. okay so this is something we can deal with this is something that matches this format here again just a little bit of extra work but not too bad overall okay so let's go ahead and set up our two cases so the first case is that x squared minus 3x this algebraic expression is equal to 10 okay then or the second case is that x squared minus 3x is equal to the negative 10 or just say negative 10. and now i'm just going to solve these two equations so let's kind of scroll down get some room going this guy right here i'll subtract away 10 from each side of the equation i'm going to have that x squared minus 3x minus 10 is equal to zero and i can solve this guy with factoring so the left side of this guy i can set this up as x here and x here give me two integers that sum to negative 3 and that have a product of negative 10 that's going to be negative 5 and positive 2 right negative 5 times positive 2 would give me negative 10 negative 5 plus 2 would give me a negative 3. so let's set each factor equal to zero so i'll say x minus five equals zero add five to each side of the equation i get x is equal to five then over here i would say x plus two is equal to zero subtract two away from each side of the equation i get x is equal to negative 2. so those are two solutions we've got x equals 5 and x equals negative 2. so let me erase this part we don't need it anymore because we already have the solutions just going to erase this and i'm going to go back to the top and just kind of write my answers up there because the screen's just going to get a little too full okay so i'm just going to write that x is equal to again we have 5 and then negative 2. okay so let me go down and now we're just going to solve this equation here okay so we have x squared minus 3x i'm going to add 10 to both sides of the equation so plus 10 this equals 0. so unfortunately this guy is not going to be factorable this is something we need to use our quadratic formula to get a solution for so with our quadratic formula we're going to label this as a right we have a 1 an understood coefficient there for x squared we're going to label this this negative 3 as b we're going to label this this tan this constant as c so the quadratic formula if you will recall x is equal to the negative of b plus or minus the square root of you've got your b squared minus your 4ac all over 2a so what is b again that's negative 3 so the negative of negative 3 is 3 then negative 3 squared would be 9 then minus 4 times a a is 1. so negative 4 times 1 is just negative 4 then times 10 which is c negative 4 times 10 is negative 40. so you would have 9 minus 40. 9 minus 40 is going to be negative 31. so this is negative 31 here and then a again is 1 so this is just 2 okay so x equals 3 plus or minus the square root of negative 31 over 2. we can clean this up by using our imaginary unit so we'll say x is equal to 3 plus or minus i'm going to pull the negative out as i times the square root of 31 again this is over 2. okay so again this is two solutions it's 3 plus i times square root of 31 over 2 and it's 3 minus i times square root of 31 over 2. okay so let me copy this and i'm going to erase everything else okay so i'll put comma and i'll put this here so we have four solutions x can be 5 x can be negative two x can be three plus i times square root of thirty one over two and then lastly x can be three minus i times square root of thirty one over two all right let's move on and look at one with a rational expression involved so here we have 7 times the absolute value of we have 5x plus 1 over x minus 2 then plus 9 this equals 30. so again it's the same thing what we want to do is isolate this absolute value operation because again i'm just going to write this rule the absolute value of u this algebraic expression u which is this in this case if that's equal to some positive number a then u equals a or u equals negative a okay so what can we do let's just subtract 9 away from each side of the equation we're going to have that 7 times the absolute value of 5 x plus 1 over x minus 2 is equal to 30 minus 9 is going to be 21 okay so 7 is multiplying this absolute value operation so let's divide both sides by 7 and that's going to get us to what this cancels at the absolute value of 5x plus 1 over x minus 2 and this is equal to 21 over 7 is 3 okay so at this point again i have an absolute value operation that's isolated on one side i've got a positive number on the other so i can go ahead and just use my rule and what i'm going to say is that i have 5x plus 1 over x minus 2 that's set equal to 3. then or we have 5x plus 1 over x minus 2 that's equal to negative 3 okay so pretty easy to solve these in each case here it's going to be quickest to just cross multiply so i'm going to just write this as 3 over 1 i'm going to write this as negative 3 over 1. so x minus 2 times 3 would be three times the quantity x minus two and then one times five x plus one is just five x plus one okay so if i go through use my distributive property this would be three x minus six is equal to 5x plus 1. all right so to solve this we can just subtract 5x away from each side of the equation and then add 6 to each side of the equation so we know this cancels and this cancels 3x minus 5x is negative 2x this is equal to 1 plus 6 which is 7 divide both sides by negative 2 and we're going to end up with x is equal to negative 7 halves okay so that's one solution let me kind of erase this because the screen gets just too packed and i'll just say that x is equal to negative seven halves okay that's one solution we're going to have another so let's go back down so now let's work on this one again i'm just going to cross multiply so i would have negative 3 times the quantity x minus 2 this is equal to 1 times the quantity 5x plus 1 which is just 5x plus 1. okay so negative 3 times x is negative 3x negative 3 times negative 2 is plus 6 this equals 5 x plus 1. so to solve this let me subtract 5x away from each side of the equation and subtract 6 away from each side of the equation so we're going to see that this is going to cancel and this is going to cancel so negative 3x minus 5x is negative 8x and this equals 1 minus 6 is going to be negative 5. let's divide both sides by negative 8 and we're going to end up with x is equal to 5 8. okay so 5 8 is a solution and negative 7 halves is a solution so let's go back up and just list 5 8 as another solution okay so this guy again x equals negative 7 halves and x equals 5 8. again you can check it i'm not going to in the interest of time but you'll see that each one of these does give you a true statement when you plug it in for each occurrence of x all right so additionally we might see some problems where we have two absolute value operations set equal to each other we talked about these earlier when we looked at absolute value equations in our course and again for this basically what you have is you have to think about the fact that two numbers can only have the same absolute value if they're the same number meaning they're equal to each other or if they're opposites so for this scenario you're going to set up another compound equation with or for the first case you're just going to drop the absolute value bars so for the first case i'm just going to say that x squared minus 1 is equal to 2x minus 1 okay then for the second case i'm going to drop the absolute value bars so i'm going to say x squared minus 1 is equal to 2x minus 1. well what i'm going to do is i'm going to make one of the expressions into its opposite okay so the way to do this is just to wrap the expression inside of parentheses and just put a negative in front so you can do this one or you can do this one but don't do both right if you did both then you're basically just going back to this because i can divide both sides by negative one okay so make sure you just do one of those okay so i'm going to solve this one first and let me put my or kind of in the middle here and i'll just drag this down so i have enough room to work and so let me start by kind of moving everything to the left side so let me subtract 2x away and let me add 1 to both sides so this is going to be 0 over here on the right so what we're going to have is that x squared minus 2x negative 1 plus 1 is 0. so this is just equal to 0 over here okay so x squared minus 2x equals 0. i can quickly solve that with factoring pull an x out you'd have x minus 2. so basically x could be what it could be 0 or it could be 2 right so x could be equal to 0 and i found that by just setting x equal to zero or x could be two i found that by saying x minus two this factor here set that equal to zero x would be two okay so those are the two solutions from there so x equals zero or 2. okay let's look at the other kind of equation that's involved here so i'd have x squared minus 1 is equal to i got to make that a little better you have this negative that's got to get distributed so it would be negative 2x and then plus 1. okay so again move everything over so you're going to have x squared add 2x to both sides of the equation so plus 2x subtract 1 away from each side of the equation so over here again this is 0 and then negative 1 minus 1 is negative 2 this equals 0. so can we solve this with factoring no this guy isn't going to be factorable so let's go ahead and set up our quadratic formula so let me erase this so we have a little room to work and we already know that this has an implied coefficient of 1 so that's my a this is my b and this negative 2 here is the c so x equals what it's negative b b is 2 so negative 2 plus or minus the square root of b squared again b is 2 2 squared is 4 minus 4 times a a is 1 times c c is negative 2. so negative 4 times negative 2 is positive 8 8 times 1 is still 8 so you would have 4 plus 8 which is 12. okay so this would be 12 here this is over 2 times a a is 1 so just 2. okay so square root of 12 i know that 12 is 4 times 3 so let's write this as 4 times 3 we know the square root of 4 is 2 so i can pull this out and say this is 2 times square root of 3. okay so let me erase this now we know we can simplify this a little bit so to make this simpler let me factor out a 2 from the numerator that would give me a negative 1 plus or minus a 1 times square root of 3. this is over 2 and we can see that this would cancel with this and so what i'm left with i can just erase this i don't need a fraction anymore i basically have negative 1 plus or minus the square root of 3 okay so my other two solutions would be negative 1 plus or minus the square root of 3. so those are my four solutions x could be 0 x could be 2 x could be negative 1 plus square root of 3 or x could be negative 1 minus the square root of 3. all right let's take a look at one more example of these and this is with rational expressions so the absolute value of 3 over x minus 1 is equal to the absolute value of 4 over x plus 2. so for this again the first scenario just drop the absolute value bars they could just be equal so 3 over x minus 1 is equal to 4 over x plus 2. then or for the second scenario drop the absolute value bars so 3 over x minus 1 is equal to 4 over x plus 2 and then make one of the expressions into its opposite so for this guy all i need to do is make one of the numerators negative so i can just say this is negative or i can say this is negative but just don't do both of them okay so let me just solve this one first i'm going to cross multiply so you'd have 4 times the quantity x minus 1 is equal to 3 times the quantity x plus 2 let's kind of scroll down and solve this so you'd have 4x minus 4 is equal to 3x plus 6. so now let me subtract 3x away from each side of the equation let me add 4 to both sides of the equation and this guy is going to cancel this guy's going to cancel 4x minus 3x is x 6 plus 4 is 10. so one solution is that x is equal to 10. then over here let me cross multiply so you could have what you'd have that negative 3 times the quantity x plus 2 is equal to 4 times the quantity x minus 1. negative 3 times x is negative 3x negative 3 times 2 is minus 6. this equals 4 times x which is 4x and then 4 times negative 1 is negative 4 okay so let me add 6 to both sides of the equation let me subtract 4x away from both sides of the equation so this is going to cancel and this is going to cancel you have negative 3x minus 4x which is negative seven x and this equals negative four plus six which is going to be two we could finish this up by just dividing both sides by negative seven we're going to find that x is equal to negative two sevenths okay so here x could be 10 or x could be negative 2 7. so let me erase everything so i'll say that x could be equal to 10 or it could be negative 2 7 okay that's going to be our answer and again you can check this if you want plug in a 10 for each occurrence of x you'll see you get the same value on the left and the right plug in a negative 2 7 for each occurrence of x you'll again see that you get the same value on the left and the right so this guy is correct in this lesson we're going to talk about solving more advanced absolute value equations so up to this point we've worked with some absolute value equations that can be solved pretty easily in pretty much all the cases we just needed to isolate the absolute value operation and then we could set up a compound equation with or now i know we looked at some more challenging scenarios in the last lesson where we involved some quadratic equations and some rational equations but now we're going to go even further and look at some absolute value equations that end up either being challenging to solve or being really tedious to solve okay a lot a lot of extra work so let's start out by talking about how to solve an absolute value equation where one absolute value operation is nested inside of another so this is a common problem you will see so suppose we had something like the absolute value of you have the absolute value of 9x plus 4 minus 10 and this is equal to 21 okay so notice that you have an outer absolute value operation and then let me highlight this in a different color you have an inner absolute value operation okay so the way you kind of attack this problem is you first consider the larger absolute value expression okay the one that surrounds the entire left side of the equation the one that i highlighted in yellow so what we're going to do is we're going to set up an or with this guy so we're going to drop those absolute value bars so that would give me the absolute value of 9x plus 4 and then minus 10 and then i'm going to say this is equal to 21 then i'm going to do my or so now i'm going to have the absolute value of 9x plus 4 okay again minus 10 now it's going to be equal to negative 21 okay so it's the same thought process as what we used in the previous kind of lessons but we're going to have to do this basically twice okay so it's just extra work okay so for this guy on the left let me kind of move down just a little bit we know how to solve this this is basically what we've looked at in the past so let me isolate the absolute value operation so i'm going to isolate this guy now so all i'm going to do is i'm going to add 10 to both sides of the equation i'm going to get that the absolute value of 9x plus 4 is equal to 21 plus 10 is 31 okay so now all i've got to do is set what's inside of the absolute value operation equal to 31 then say or what's inside the absolute value operation is equal to the negative of 31 okay so we get 9x plus 4 is equal to 31 or 9x plus 4 is equal to negative 31. and you can see how tedious this becomes this is just one of those ones where it's not hard once you kind of understand the steps it just takes a long time a lot of scratch paper and sometimes that can involve you making some mistakes if you're not really careful so for this guy on the left i'm going to subtract 4 away from each side of the equation this is going to cancel i'm going to have that 9x is equal to 31 minus 4 is 27 okay so if i divide both sides of the equation by 9 i get one solution and that's that x is equal to 3 okay so that's one solution over here if i subtract 4 away from each side of the equation this cancels i'm going to get that 9x is equal to negative 31 minus 4 is negative 35. again i'm going to finish this up by dividing both sides by 9 and i'm going to find that x could also be equal to negative 35 9. okay so those are two solutions that we have but we've got to go back up and deal with the other scenario so let me just kind of copy this real fast and i'm just going to erase all this work okay and i'm just going to paste that in right here right there okay so that's for this guy right here so i'm just going to erase all of this we don't need that anymore and let me kind of drag this over here and let's work on this guy now so what i'm going to do is i'm going to add 10 to both sides of the equation to kind of get things started i'm going to have the absolute value of 9x plus 4 is equal to negative 21 plus 10 is going to give me negative 11. now you'll notice that there's a problem here we know that if we have the absolute value operation isolated on one side and we have a negative value on the other side there's no solution okay again if i take the absolute value of a negative number it becomes positive if i take the absolute value of zero i get zero if i take the absolute value of a positive it stays positive so there's nothing i can plug in for x here multiply by nine and then take that and add to 4 and take the absolute value and get a negative 11. it's not going to happen because again the absolute value of operation is going to give you a non-negative result and so this guy could never be equal to negative 11. so for this part we would say there's no solution okay there's no solution okay but for the equation as a whole we found one part had a solution so this is going to be the solution for the whole thing this guy you can just throw it away right that part just leads to a no solution type of answer okay so for this guy as a whole the solution is going to be x could be 3 or x could be negative 35 9. all right let's take a look at another one of these and then we'll move on and talk about another type of problem that you'll probably encounter so we have the absolute value of and then inside of that you have the absolute value of 5x minus 2 and then minus 6. this whole thing is equal to 2. so again i'm going to start with the outside absolute value operation and what i'm going to do is i'm going to set up two scenarios i'm going to say that the absolute value of 5x minus 2 minus 6 is equal to 2 then or i'm going to also say that the absolute value of 5x minus 2 minus 6 is equal to what it's equal to the negative of 2 okay so just split it up into two cases and then we're just going to start working so over here i know i want to isolate the absolute value operation so i'm going to do that by adding 6 to both sides of the equation so i'm going to get that the absolute value of 5 x minus 2 is equal to 2 plus 6 is going to be 8 okay so this splits up into two more cases right so we can say that five x minus two this expression inside the absolute value bars is equal to eight or five x minus two is equal to negative eight okay so let's solve those real quick we'll come back up so let me go ahead and add 2 to both sides of the equation we're going to have that 5x is equal to 10. let's divide both sides of the equation by 5. we're going to get that x is equal to 2. okay so that's one solution over here let me add 2 to both sides of the equation and i'm going to get that 5 x is equal to negative 6 divide both sides by 5 and i'm going to get that x is equal to negative 6 fifths okay so we have two solutions we've got x is 2 or x is negative 6 fifths so let's work on the other side and see about getting some more solutions okay so let me just copy these and i'm gonna erase all this work we don't need it anymore all right so i'm just gonna erase this part right here and let me just drag this other one over here because we're gonna need to work with this and i'm just gonna paste these solutions in this is what we have so far and i'm just gonna erase this and put a little comma so x equals two and i'll put comma negative six fifths okay let's take a look at this kind of second case so again we want to isolate the absolute value operation and what we're going to do is just add 6 to both sides of the equation and so this guy's going to cancel we would have that the absolute value of 5x minus 2 is equal to what negative 2 plus 6 is positive 4. so in the other example we saw that in our kind of second case we had something with no solution in this example we don't have that so we can keep going so what we're going to do is again set this up into two more cases so 5x minus 2 the expression inside the absolute value bars could be equal to 4 or 5x minus 2 could be equal to negative 4 okay so we're just going to solve those two scenarios so let me add 2 to both sides of the equation here i'm going to get that 5x is equal to 6 divide both sides by 5 we're going to get x is equal to 6 fifths okay so that's the solution over here i'm going to add 2 to both sides of the equation and i'm going to get that 5x is equal to negative 2. we're going to divide both sides by 5. so i'll put or here x could also be equal to negative two fifths okay so we're gonna have a total of four solutions let me kind of copy this and i'll erase everything okay so we have six-fifths and negative two-fifths so let me put comma six-fifths and comma negative two fifths okay so those are our four solutions i'm not going to check them in the interest of time but you can plug each one in for x and you see that you get a true statement in every case right so x is two negative six fifths six-fifths or negative two-fifths all right so now let's move on and look at something that's even more tedious so we've previously seen examples where two absolute value operations were directly set equal to each other okay so for that scenario it's very easy to solve but what happens when we have an equation with two absolute value operations and then kind of an extra or a loose number okay so this is a very common scenario that you'll come across so something like the absolute value of x minus 1 then plus the absolute value of 3x minus 1 and this is equal to 5 okay so in this case we can't directly set the two absolute value operations equal to each other okay so we can't use that method so we have to kind of turn to a more advanced approach or you could say a more tedious now there's other ways to do this than what i'm going to show you and i'll explain that when i get done with this method so what we're going to do is we're going to think about the expressions that are inside of the absolute value operation so you've got two of those okay so what we want to do is think about where these expressions are 0 okay so where is the expression x minus 1 equal to 0 and where is the expression 3x minus 1 equal to 0 okay because those represent basically our turning points it's where the sign is going to change from positive to negative or from negative to positive okay this can only happen at a value of 0. so for this one i'm just going to solve i'm going to add 1 to both sides of the equation so i'm going to have that x is equal to 1. okay for this one i'm going to add 1 to both sides of the equation i'm going to get that 3x is equal to 1. i'm going to divide both sides of the equation by 3 and i'm going to say that x is equal to 1 3. so this value here of 1 is what makes this expression here equal to 0 right if i plugged in a 1 there 1 minus 1 would be 0. okay this value of 1 3 is what makes this expression equal to 0 3 times a third is 1 1 minus 1 is 0. so we can see that so we're going to use these two values to kind of split the number line up into intervals so let me go down here real quick to a little table so this is a good way to kind of organize the information so one interval would be anything from negative infinity up to but not including 1 3. so this interval here so we have negative infinity and we have one third and again neither of those is going to be included here then the next one if we go back again the values were 1 3 and 1 so i want to think about between 1 3 and 1 what's going to happen so what happens between 1 3 and 1 okay neither of those going to be included and then my last interval would be anything greater than 1. so i'll put a 1 next to a parenthesis and then comma infinity okay so that interval there so what we're doing here in case you're confused we're thinking about the value of the expression in each interval okay so if it's negative we're going to replace the expression with the opposite and if it's positive we just replace it with the expression itself okay so if i go back up again this guy right here i'm just going to copy this down here so we have the absolute value of x minus 1 and then plus the absolute value of 3x minus 1 this is equal to 5. so when we think about this interval anything that's less than one-third we know that both of these guys would be negative okay because this guy right here is zero when x is equal to one and this guy right here is zero when x is one third so if it's less than one third both of those guys would be negative and so you would take each expression okay each expression and you would make it into its negative or it's opposite okay so in other words this guy would be negative of x minus one that quantity this guy would be the negative of you have three x minus one that quant okay so that's the expressions we're going to be dropping the absolute value bars in each case and we're going to be replacing those with these kind of expressions we're setting up here now between these two numbers one third and one remember this guy is made zero by one third so anything larger than one-third means that this guy is going to be positive so i can replace this one with the just expression itself so three x minus one then for this guy right here again one is what makes it zero so anything that's less than 1 is going to result in this being negative so this is negative of the quantity x minus 1. then in this scenario here where we're larger than 1 both of these guys are going to be positive so i can say this is x minus 1 and this is 3x minus 1. okay so what i'm doing here is i'm going to go through let me kind of scroll down a little bit and i'm going to basically replace my absolute value expression in each case with what i came up with for the interval so for this first interval instead of having the absolute value of x minus 1 because this guy is negative i'm going to say it's negative of x minus 1. then we would have plus you'd have the negative of this guy as well so you could put plus negative or you just put minus doesn't matter so minus you have the quantity 3x minus 1 and this is equal to 5. okay so let's scroll down and solve this and then we'll come back up we just want to make sure we have enough room so i'm going to distribute the negatives to each part here so this would be negative x plus one distribute the negative so negative three x plus one this equals five so negative x minus three x would give me negative four x one plus one is two so plus two this equals five and then what i'm going to do here is i'm going to subtract 2 away from each side of the equation and we're going to find that this is going to cancel we're going to have that negative 4x is equal to 3. we're going to divide both sides of the equation by negative 4. so divide this by negative 4 and this by negative 4 we're going to get that x is equal to negative 3 4. okay so i want to erase everything and i want to show you what we do here so let me erase all this so the solution we found in this interval is that x is equal to negative 3 4. now what you end up doing in this process is if your solution lies in the interval you accept it if it lies outside of your interval then you're going to reject it since negative 3 4 is in this interval right from negative infinity of 2 but not including 1 3 we can accept this solution okay so we're going to put that as part of our solution okay now we're going to work in this interval okay so let me go down a little bit so this guy right here is going to be replaced with its negative so the negative of x minus 1 then plus this guy is positive so we just say 3x minus 1 this equals 5. okay so let's go ahead and solve this and then we will come back up so distribute the negative to each part you'd have negative x and then plus 1 then plus 3x and then minus 1 equals 5. so we know that 1 minus 1 those would cancel right that's basically 0. so you'd have negative x plus 3x which is 2x this is equal to 5 okay so what i'm going to do from here is i'm going to divide both sides of the equation by 2 and i'm going to get that x is equal to 5 halves now let me erase everything let me paste this in so we have x equals 5 halves let me kind of do this in a different color so it doesn't get confusing so x equals 5 halves does 5 halves lie in the interval which is between 1 3 and 1 right with neither included no it doesn't because 5 halves is 2.5 so this is the solution we're going to reject right we're going to reject this solution now we're going to move on to this interval right here which is anything larger than 1. so in this interval both of these are positive so what i'm going to do is i'm going to drop the absolute value bars in each case and just write what i have so i'd have x minus 1 plus 3x minus 1 and this is equal to 5. so then i would add x plus 3x that's 4x negative 1 minus 1 is negative 2 this equals 5. we would add 2 to both sides of the equation and so we would have that 4x is equal to 7. we can divide both sides of the equation by 4 and find that x is equal to 7 4. okay so let me erase this once again and we'll say x equals 7 4. okay so is 7 4 in the interval again anything larger than 1 well yes it is 7 divided by 4 in decimal form is 1.75 so that's clearly larger than 1. so your two solutions here let me just kind of check this off you would have that x is equal to negative 3 4 or x is equal to 7 4. so let's go back up here i'm just going to erase this and we'll say that x is equal to negative 3 4 and then also 7 4 right and we rejected 5 halves that didn't work you can check this if you want you can even check 5 halves and prove that to yourself that it doesn't work if you plug in negative three-fourths here so you have the absolute value of negative three-fourths minus one so minus four-fourths i'm just getting a common denominator then plus the absolute value of three times negative three fourths okay then minus one and then this is equal to five negative three minus four is negative seven so this would be negative seven fourths and then if i took the absolute value of that i'd have seven fourths okay so this would be seven fourths so that works okay and then this guy right here 3 times negative 3 is negative 9. so this is negative 9 4. again getting a common denominator just write this as 4 over 4. negative 9 minus 4 is negative 13. so you'd have negative 13 fourths okay so you'd have negative 13 4. take the absolute value of that you get 13 fourths okay so this is 13 4 here and what do you get 7 plus 13 is 20 over the common denominator of 4 you would get 5 right so you know this one would work itself out so this one checks out all right so let's check the other one so the absolute value of you'd have 7 4 minus for 1 i'm going to put 4 4 and then plus the absolute value of three times seven fourths minus again for one i'm gonna put four fourths so what do we have here seven minus four is three so over the common denominator four this would be three fourths absolute value of three fourths is three fourths okay so that's three then this guy here 3 times 7 is 21 so you would have 21 here then 21 minus 4 is 17 so you would have 17 4 okay 17 4. absolute value of 17 4 is 17 fourths and you see that this works right 3 plus 17 is 20 20 over 4 is 5 so you get 5 equals 5 so this works out as well so you can also do 5 halves and prove that it doesn't work so you'd have 5 halves minus in this case i'm gonna write one is two over two and then plus you'd have three times five halves and then minus one so we know that this would be what this would be three halves this would be three halves absolute value of three has us three halves so this would be three halves and then this guy three times five is fifteen so you'd have 15 halves again i'm going to write this as two over two 15 minus two is 13 so you'd have 13 halves absolute value of 13 halves is 13 halves so in this case you get 3 plus 13 which is 16 and so you'd have 16 over 2 which is 8 so you get 8 equals 5 which is false okay that's false so that solution where x equals 5 halves again you've got to reject this this does not work okay and we found that because it wasn't in the interval we knew it wasn't going to work now let's talk a little bit about the alternative method to solve this this is one that i don't recommend because it just takes a lot more time so what you can do is you can consider all the cases that you would be presented with this so in other words you can consider if this expression is positive and this expression is positive you would just have x minus one plus three x minus one equals five okay so that's one scenario that you would encounter then another scenario would be this one's negative and this one's positive so in other words i would put a negative out in front of this and say the negative of x minus 1 plus and then this guy's positive so 3x minus 1 and this equals 5. so i'd solve that equation right and you keep going so this would be positive and this would be negative and this would be negative and this would be negative so you basically have four equations to set up and solve but then you get extraneous solutions so you have to go back and check each one of those okay and with our method we really don't need to check we just need to verify that our solution lies in the interval that we've created and then we're basically good to go okay so once more the solution here is that x equals negative 3 4 or 7 4. all right let's go ahead and take a look at another one of these so we have the absolute value of 5x plus 3 then minus the absolute value of 2x minus 7 and this equals 21. so again it's the same thought process i'm thinking about where this guy is 0 and where this guy is 0. i'm using that to kind of set up intervals on my number line so we'd have 5x plus 3. we'd say where is that equal to 0 so i would subtract 3 away from each side of the equation i would get 5x is equal to negative 3. divide both sides by 5 and i get that x is equal to negative 3 fifths okay so that's going to be one of those guys then for this guy right here where is 2x minus 7 equal to 0 let's add 7 to both sides of the equation we get 2x is equal to 7 divide both sides by 2 we get x is equal to 7 halves okay so basically i'm going to have an interval that is anything less than negative three fifths so let's write that first so we'll put negative infinity here up to but not including negative three fifths then i'm gonna have an interval between negative three fifths and seven halves with neither included so you have negative three fifths and seven halves again neither one's included and then lastly we have an interval where we're greater than seven halves okay so where were greater than seven halves okay so let's rewrite our absolute value equation so we have the absolute value of five x plus three and then minus the absolute value of two x minus 7 this guy is equal to 21. so again i'm just going to look into each interval and decide where each expression is positive and negative again if it's positive i can just kind of drop the absolute value bars if it's negative i can drop the absolute value bars wrap it in parentheses and put a negative out in front right to change it into its opposite so we know that if we're less than negative three-fifths both of these expressions are going to be negative right so 5x plus 3 i'd wrap that in parentheses put a negative out in front 2x minus 7 i'm going to wrap that in parentheses put a negative out in front if i'm between negative three-fifths and seven-halves i know this guy right here is going to be zero at negative three fifths so if i'm greater than negative three fifths that guy's positive right so 5x plus three is positive and i can go ahead and just say it's going to be positive in this interval as well right now this guy right here this 2x minus seven this guy is going to be 0 when x is 7 halves so it's still going to be negative in this interval here so we would wrap this in parentheses and put a negative out in front and again in this interval it's positive because everything's positive once we get larger than 7 halves because 7 halves makes this guy zero so we'll say this is just 2x minus 7. okay so let's go through the possibilities let's scroll down let's think about the first one so all the way to the left i'm going to replace each one of these with their opposites so i'm going to have the negative of 5 x plus 3 and then minus you've got to be very careful because this is a minus here and you're putting a negative out in front so minus a negative is really plus a positive right so minus a negative and then times the quantity 2x minus 7 i'm going to put plus okay so plus 2x minus 7 okay then this equals 21. all right so i'm going to distribute this negative to each term i'll have minus 5x and then minus 3 and then plus 2x and then minus 7 and this equals 21. okay let me scroll down and get some room going we'll come back up so this guy right here so negative 5x plus 2x that's going to be negative 3x and then negative 3 minus 7 is negative 10 this equals 21. let's add 10 to both sides of the equation we're going to get negative 3x is equal to 31. we're going to divide both sides of the equation by negative 3 and we're going to get that x is equal to negative 31 thirds okay so let's copy this so let me just write that in here x equals negative 31 thirds okay negative 31 thirds so is this valid you just need to ask yourself the question does it lie in this interval and the answer to that is yes right negative 31 thirds is more negative or to the left of negative three-fifths so you can clearly see it would be in this interval so this guy will go ahead and check it off and say that's a valid solution all right let's talk about in this interval here between negative three-fifths and seven-halves so this guy's positive and this guy's negative so let's scroll down a little bit and we'll say that we have five x plus three again if i have a minus a negative it's plus a positive so you just say plus there again two x minus seven and i really don't need the parentheses i can just drop them and then this equals 21. so let's just solve this real quick i'm just going to scroll down and get a little room going we'll come back up so 5x plus 2x is 7x 3 minus 7 is negative 4. this equals 21. and let's go ahead and add 4 to both sides of the equation so we have 7x is equal to 25. let's divide both sides of the equation by 7 and we'd find that x is equal to 25 7. all right so let me kind of copy this real quick and i'm just going to erase all this work we're going to go back up so again x is equal to 25 7. so 25 7. so is that in the interval again between negative three fifths and seven halves well negative three fifths is basically negative point six as a decimal seven halves is three point five okay and if i think about 25 sevenths that's a little bit larger than 3.5 it's 3.571 if you want to kind of go to three decimal places so it is larger than 3.5 so it's outside of this interval so we're going to end up rejecting this solution okay it's not going to work so let me do that in red i'm going to reject this and so now we just have to check this interval here so everything's positive so i can just drop the absolute value bars so we kind of go down and do that so we would have 5x plus 3 and then basically you're subtracting away this whole thing okay you've got to be careful there so what i'm going to do is i'm going to say minus i'm going to wrap this in parentheses 2x minus 7 because again this whole thing is being subtracted away don't make the mistake of just putting a minus 2x and then leaving this as negative 7 because this has to be changed and so does this okay you got to be very careful there so then this equals 21. okay so i'm going to go ahead and say 5x minus 2x is 3x and then 3 plus 7 is 10 so plus 10 this equals 21. let's go ahead and subtract 10 away from each side of the equation and we'll say that 3x is equal to 21 minus 10 is 11. divide both sides by 3 we're going to get that x is equal to 11 thirds okay so once again let me erase everything all right so we found that x is equal to again 11 thirds okay so is 11 3 in this interval well 7 halves is 3.5 so it needs to be larger than that and if you punch up 11 divided by 3 on a calculator you get 3.6 where the 6 repeats forever okay so that is going to be larger than 3.5 so this guy is going to work out as a solution so the two solutions are going to be x equals negative 31 thirds and then x equals 11 thirds so let me go back up so we'll say that x is again equal to we have negative 31 thirds and also 11 3. all right so let's wrap up our lesson and look at another common type of problem that you'll see so let's say we ran into something like this we have 5x squared minus 9 times the absolute value of x minus 2 and this is equal to 0. so in this particular case we have this absolute value operation that's kind of in the middle of our equation and we can't really isolate it so there's kind of two different ways to solve this one of them is just to consider the two cases and then just kind of go back and think about if the solutions that we get work right so you just plug them back in the other one would be to kind of make a little substitution and then you can consider if things are valid or not from that so let's try it the first way and then i'll show you the substitution method i prefer the substitution method on this but you can do it either way so there's two cases here one case would be that this value here is going to be positive so in that case we just remove the absolute value bars so you would have 5x squared minus 9x minus 2 equals 0. and then the other case kind of the case over here would be let's say this expression was negative so then i'd remove the absolute value bars and i would make x negative x right so we would have 5x squared minus 9. it's being multiplied by this and i'm multiplying it by the negative of x and then minus 2 and this equals 0. of course we could simplify this because negative times negative is positive so i can now write this as plus 9x okay so plus 9x and let me kind of tighten this down a little bit just move that over there okay so let's start over here on the left so this is something we can factor over here on the left so let me kind of set this up and we'll factor this guy real quick so i know that this is going to be 5x and this is going to be x because this is a prime number here so i need to make sure the outer and the inner work itself out so i'm looking for what i'm looking for a middle term of negative 9x and i'm looking for a final term of negative 2. so when we think about negative 2 we know something's got to be positive and something's got to be negative okay so we also know that the factors there could be you know 1 and negative 2 or it could be negative 1 and positive 2. okay so let's say i put my negative 2 here and my positive 1 here would that work so in this case 5x times negative 2 would give me negative 10x and then 1 times x would be plus x so this does work itself out negative 10x plus x would be negative 9x and then if you think about 1 times negative 2 of course that's negative 2. so this is our correct factorization here and now we can just solve this pretty easily we're going to take each factor and set it equal to 0. so 5x plus one equals zero and then we're also going to have that x minus two is equal to zero let me kind of scroll down real quick get a little room going and so to solve this one i'm gonna subtract one away from each side of the equation i'll have five x is equal to negative one divide both sides by five and i get x is equal to negative one fifth okay then over here let me add two to both sides of the equation and i get that x is equal to two okay now these are just proposed solutions we actually need to check them in the scenario so let me erase all this and i'm just going to copy these so i'm just going to put this over here we have x is equal to 2 and then comma negative 1 5 we'll check them in a minute we're going to find that some of these won't work so let's move this over here and let's work on this now we'll get the solutions for this and then we'll check everything at once okay so for this guy we can also solve this with factoring so let's go ahead and set this up so we can again do 5x and x again i'm looking for the outer and the inner such that i get a positive 9x as my middle term and a negative 2 as my final term so instead of doing a negative 2 here i'm going to do a positive 2 and instead of doing a positive 1 i'm going to do a negative 1 right because my outer would be 10x and my inner would be minus x so 10x minus x would give me my positive 9x and then negative 1 times 2 would be negative 2. okay so we're good to go there all right so now i'm just going to set each of these equal to 0. so 5x minus 1 if that equals 0 we're going to find that we add 1 to each side of the equation we'll get that 5x is equal to 1 divide both sides by 5 and we're going to get that x is equal to 1 5th okay so let me just write that in here so we have one fifth and i'll just erase this we don't need this information anymore and then we also have x plus 2 we're going to set that equal to 0. of course that gives me a solution if i subtract 2 away from each side that x is equal to negative 2. okay let's put negative 2 in there all right so let's erase everything and now we just want to check our solutions and see what works and what doesn't okay so we have 5 times for x i'm going to start by plugging in a 2. so you'd have 2 that guy squared then minus 9 times the absolute value of 2 right plugging in for x and minus 2 and this equals 0. okay so 2 squared is 4 so you'd have 5 times 4 then minus 9 times the absolute value of 2 the absolute value of 2 is 2 so 9 times 2 then minus 2 and this equals 0. 5 4 is 20 then minus 9 times 2 is 18 then minus 2 and this should equal 0. and this is going to work out right 20 minus 18 is 2 2 minus 2 is 0. so you get 0 equals 0 so this guy checks itself out now let's work on negative 1 5 now so let's erase this and we'll do negative one-fifth so i'm gonna erase that and that's i'm gonna put negative one-fifth in there all right so essentially if i square negative one-fifth i would get one-twenty-fifth so i have five times that guy then minus nine times the absolute value of negative one-fifth is one-fifth and then minus two and this should equal zero so this guy is going to cancel with this guy and give me a five so you've got one-fifth minus nine-fifths you can go ahead and do that real quick because they have a common denominator one minus nine is negative eight so that would be negative eight-fifths so you have negative eight-fifths and then we're subtracting away two so i can go ahead and write that with the common denominator and say minus 10 over 5 right 10 over 5 is 2. and this should equal 0 and obviously it doesn't right if we do this it would be negative 18 fifths okay negative 18 fifths and that is not equal to zero this is false so we're going to reject that solution now i don't actually need to check one-fifth because i'm going to get the same kind of result here if i squared one-fifth is the same as squaring negative one-fifth in either case i'm gonna get 125th if i take the absolute value of one-fifth it's the same as taking the absolute value of negative one-fifth so this would be the same and this would be the same so you can go ahead and reject that as well okay so get rid of that then the other guy that you're gonna have was negative two so let's go ahead and check that but again you know that's going to work because 2 worked and you're going to end up with the same scenario right because negative 2 squared or 2 squared each give you 4. so in each case you're going to have 20 there because 5 times 4 is 20. then you're subtracting away 9 times 2 in each case because whether i put a negative 2 in there or positive 2 when i take the absolute value of that i get 2 right either way so 9 times 2 is 18 and then you've got minus 2 and this equals 0. again 20 minus 18 is 2 2 minus 2 is 0 so you get 0 equals 0 so this does work out as well so this gave us two solutions x equals 2 or x equals negative 2. so i can just erase those just kind of tighten this down and say negative 2 there and now let's think about doing this in alternative way so for this method we're going to focus on using a substitution technique we use substitution to solve things earlier in the course so it shouldn't be something new for you basically what we're thinking about here is the fact that x squared is always non-negative and so is the absolute value of x squared right so if i think about x squared and i think about the absolute value of x being squared this would produce the same thing no matter what i plug in for x right if i plugged in let's say negative 3. let's say i plugged in a negative 3 in each case well here negative 3 squared is 9 and here negative 3 if i take the absolute value first i get 3. when i square that i get 9 as well right so what i can do here is i can just rewrite this and i can say that we're going to have 5 times the absolute value of x that's squared then minus 9 times the absolute value of x then minus 2 and this equals 0. okay so now i have the absolute value of x here and i have the absolute value of x here okay so what i'm going to do is just make a little substitution so i'm going to let u be equal to the absolute value of x okay so now i'm going to replace the absolute value of x in each case with u so i'm going to have that 5 u squared minus 9 u minus 2 is equal to 0. so this is something we can again solve using factoring let me kind of scroll down for a minute so i'm going to put a 5u here and a u here this equals zero and of course i need my outer and my inner to sum to negative nine u and my final to be negative two okay so we know we're going to do that with a negative two here and a positive one here we saw that earlier okay so we know that 5u times negative 2 would be negative 10u and that 1 times u would be plus u so this gives me my negative 9u and of course 1 times negative 2 is negative 2. okay so let's erase this and let's kind of solve this real fast so i would set five u plus one equal to zero subtract one away from each side of the equation you get five u is equal to negative one we're going to divide both sides by five and we're gonna get that u is equal to negative one fifth okay so let me write that over here u is equal to negative one-fifth okay so let me erase this i don't need that anymore the other guy is really easy you can solve that mentally you have u minus two equals zero so u minus two equals 0 add 2 to both sides of the equation and we get u is equal to 2 okay so let's put a 2 there so you might be thinking you didn't get all the same solutions that you got using the previous method well we did we just have to go back and substitute again so let me erase all of this we don't really need it anymore okay we're just going to scroll up and we're going to think about what u means so i told you that u is equal to the absolute value of x so if u equals negative one-fifth i can say the absolute value of x because that's u is equal to negative one-fifth okay then i can also say that the absolute value of x because again that's u is equal to two now here's where we save some time we can throw this guy out right away right throw this out because the absolute value of x cannot be equal to a negative value okay if the absolute value operation is isolated on one side you can't have a negative on the other okay the absolute value operation will always give you a non-negative answer so that doesn't work for this one the absolute value of x equals 2 gives me two solutions right it could be true that x is 2 or x is negative 2 which is exactly what we found with our other method this one i think is just a little bit quicker because at least are you rejecting kind of the extraneous solutions right away and you're just left with the good ones in this lesson we want to discuss how to solve more advanced absolute value inequalities so over the course of the last few lessons we've discussed some methods that can be used to solve some more advanced absolute value equations now again we're just going to go one step further and think about how to apply these same strategies when inequalities are involved so before we kind of jump into the examples i just want to take a minute to refresh your memory on absolute value inequalities just the basic processes that we use when we're solving these guys because it's been a while since we looked at them so i think it's important just to get a little quick refresher in so we should know that if a is greater than zero so a just represents some positive real number if we have the absolute value of x is greater than a okay greater than this positive real number this gives us an or right a compound inequality with or so x can be greater than a or x can be less than the negative of a so as a quick example let's say we have the absolute value of x and this is greater than seven we should know that again this sets up as x is greater than seven or x is less than negative seven graphically it's really easy to figure this out again i'm thinking about what values here for x have a distance from 0 that is greater than 7. well if i start at 0 and i go 7 units to the right it can't be 7 itself but it can be anything larger right so i just put a parenthesis at 7 to say it's not included and i shade everything to the right same thing goes in the other direction if i start at 0 and i go 7 units to the left i put a parenthesis at negative 7 and i just shade everything to the left like that so this would be my solution graphically and we also have this solution here kind of in our standard notation and you can even use interval notation if you want you can say from negative infinity up to but not including negative 7 and then the union with 7 is not included again out to positive infinity okay so let's take a look at kind of the other scenario that happens so if a is a positive real number again if a is greater than 0 if the absolute value of x is less than this positive number a well then x is going to be between negative a and a right x is greater than negative a and it's less than a so here we have a compound inequality with and so let's look at this quick example here suppose we have the absolute value of x and now it's less than seven okay so we're going to have a compound inequality with and so basically x is going to be greater than the negative of seven and it's going to be less than seven itself again basically to think about this graphically i'm thinking about all the values whose absolute value or whose distance from 0 is less than 7. so starting at 0 i could basically travel up 2 but not including 7. so again at 7 i just put a parenthesis there and again i could go the opposite direction so starting at 0 i could go down 2 okay down 2 but not include negative 7 so i put a parenthesis there as well so that would be my graphical solution again from negative 7 to 7 with neither of those included and then in interval notation i'd have negative 7 comma 7 again with a parenthesis next to each all right so two special cases that come up that we need to be aware of again if the absolute value operation in this case is the absolute value of x if we say this is less than zero there's no solution here okay there's nothing i can plug in for x and take the absolute value of and say that it's going to be less than zero because the result of this operation here is going to be zero which is obviously not less than zero or it's going to be some positive number which is also never going to be less than zero so in this situation there is no solution in our last special case we're going to come across if a now is a negative number if it's less than 0 so some negative real number then if we have the absolute value of x is greater than this negative real number well then we're going to have a solution that's all real numbers right if i have the absolute value of x is greater than negative 5 well that's always true right no matter what i plug in for x the result of this will either be 0 which is greater than negative 5 or it'll be positive which is also greater than negative five so this is always true or you could say true for all real numbers okay so now let's get to kind of the first example and hopefully you saw the lesson where we solved more advanced absolute value equations if you did that lesson already this is going to be a breeze for you we're basically using the same concepts okay so suppose you had something like the absolute value of x plus 3 is greater than the absolute value of 2x minus 1. again when you look at these what you want to do is you want to think about where the expressions inside of the absolute value bars are going to be 0. okay you're thinking about kind of the turning point so where could this guy go from negative to positive or from positive to negative well that can only happen when that expression is equal to zero okay so if i say x plus three is equal to zero and i solve that i get that x is equal to negative three okay and then alternatively if i say that 2x minus 1 is equal to 0 i add 1 to each side of the equation i get 2x is equal to 1 divide both sides by 2 and i find that x equals 1 half okay so we're going to use these numbers negative three and one-half to kind of split the number line up into three intervals okay and i'm not going to do it on this number line here i'm going to use a table because i find it to be a little bit easier so the numbers we were working with again are negative 3 and positive 1 half so we're using that to split the number line up into three intervals i just prefer to do this on a table because it's just easier for me to organize things so the first interval will be from negative infinity up to but not including negative 3 okay so that's my first interval then my second interval negative 3 i'm going to include that okay and this is going to be out 2 but not including 1 half then my third interval will be anything that includes one half or larger okay so from one half out to positive infinity now what we're going to do here let's go back up so we have the absolute value of x plus 3 is greater than the absolute value of 2x minus 1. let's go ahead and write that real quick so the absolute value of x plus 3 is greater than the absolute value of 2x minus 1. okay so what we're doing here is we're looking at each interval and we're thinking about is the expression inside the absolute value bars negative is it zero or is it positive okay that's what i want to consider so for this guy right here think about the fact that in this interval it's going to be negative right i know that x plus 3 is equal to 0 when x equals negative 3. so anything less than negative 3 means it's going to be negative right so in this interval i will replace this expression here this absolute value of x plus 3 with the negative of x plus 3 okay so i just wrap it in parenthesis and put a negative out in front to express that okay for this guy right here this 2x minus 1 it's also going to be negative right because 2x minus 1 is going to be 0 when x is one half so if we're less than a half which we are in this case it's going to be negative so again i just wrap it in parenthesis so 2x minus 1 put a negative out in front okay and we'll deal with this scenario in a second let's just kind of move on and write everything so in this interval here we have from negative three up to but not including one-half so at negative three we know that this guy would be zero but anything larger than that means it's going to be positive okay so whether it's zero or some positive value i can just take the expression itself so i can just say it's x plus three here and again this is still in a format of where it's larger than negative three so in this interval it'll also be just x plus three okay then for two x minus one again it's made zero when x is one half in this interval between negative three including negative three up 2 but not including 1 half it's still going to be negative right because that's less than one half so i would say the negative of 2x minus 1 and then here it could either be 0 if x is a half or some positive value so i just drop the absolute value bars and write the expression so 2x minus 1. okay so a very simple process again if it's zero or positive in that interval you replace the absolute value expression with just what's inside the absolute value bars okay it's all you need if it's a negative okay then what you want to do is drop those absolute value bars wrap it in parentheses put a negative out in front you're basically changing it into its opposite all right so now we've got a lot of work to do we've got to go through the different possibilities so the first possibility is that both of these are negative so let's kind of scroll down and crank that out so i'm going to have the negative of x plus 3 is greater than the negative of 2x minus one now if we want we can clear the negatives by just dividing both sides by negative one but again because we're working with an equality and we divided by a negative we've got to flip that guy okay so this is going to cancel and this is going to cancel you have x plus 3 now is less than 2x minus 1. let's go ahead and subtract 2x away from each side of the inequality and let's also subtract 3 away from each side of the inequality so this is going to cancel this is going to cancel we'll have x minus 2x which is negative x and this is less than you've got negative 1 minus 3 which is negative 4. as a final step let's divide both sides by negative 1 and again we divided by a negative so flip this guy you're going to get that x is greater than 4 okay so x is greater than 4. now we want to compare x is greater than 4 to the interval that we're working with the interval that we're working with specifically says that x is basically coming from negative infinity up to but not including negative 3. x is greater than 4 is not in that interval so this solution is going to be rejected okay let me erase all this and i'll kind of write it in the column so i'm just going to write x is greater than 4 here and again this is not within our interval so we reject that solution okay now let's think about this interval here again where the x plus 3 is positive and we're going to use the negative of 2x minus 1 okay so let's scroll down and we're going to say that we have x plus 3 is positive and then this guy is going to be negative so the negative of 2x minus 1. okay so let's go ahead and solve this guy i'm going to distribute this this will be negative 2x plus 1. so if i add 2x to both sides of the inequality and i subtract 3 away from each side of the inequality this will cancel this will cancel x plus 2x is 3x this is greater than 1 minus 3 is negative 2. divide both sides by 3 and we get that x is greater than negative two-thirds okay so let me erase this real fast x is greater than negative two-thirds let me write that up here so x is greater than negative two-thirds okay is that in our interval the interval is from negative three including negative three up to but not including one half well yeah it is so this is a partial solution okay so for right now i'm just gonna write this guy as again not including negative two-thirds because it's strictly greater than up to but not including one-half okay and then we'll think about this solution and if we have a solution there we'll have two partial solutions we can put together as a full solution okay so in this case everything's positive so we can just drop the absolute value bars and go with what we have so you'd have x plus 3 is greater than 2x minus 1. let's subtract 3 away from each side of the inequality so that would cancel let me subtract 2x away from each side of the inequality so that would cancel x minus 2x is going to give me negative x this is greater than negative 1 minus 3 is negative 4. let's divide both sides by negative 1. so we know that this is going to flip right so this is going to end up being a less than so you have x is less than positive 4. okay so again let me erase all this so we say x is less than 4 does that fit in our interval yeah it does right because we have from one half including one half out to positive infinity x is less than four fits in there you could basically say it's another partial solution you would include one half and you would go up two but not include four okay so if i take these two partial solutions and i just take the union of the two so again from negative two thirds not including that up two but not including one half and the union width you have one half which is included up to four okay so this would be your solution set and basically when you combine the two you can get rid of this part right here and just kind of slide this down and that's what you're going to get right you're going to get from negative two-thirds not including that up two but not including four okay so let me kind of drag this up to our original page and we'll paste our solution here again from negative two-thirds not including that up to and not including four so negative two-thirds let's just go ahead and say that's about right here and i'm just gonna mark that and say this is negative two-thirds so i'd put a parenthesis there because it's not included four is not included just shade everything in between okay and then we can also write this saying that x is greater than negative two-thirds and less than four okay so this is your interval notation this is your standard notation and then this is graphically so those are the solutions for again the absolute value of x plus 3 is greater than the absolute value of 2x minus 1. all right let's go ahead and take a look at another example so this guy here is going to be a little bit more complicated but we're going to solve it using the same strategy so we have the absolute value of x plus 5 and then plus we have the absolute value of x minus 3 and then this is greater than 14. so the first thing we want to do is just take what's inside of the absolute value bars in each case so this expression here x plus 5 and this expression here x minus 3 we just want to set those guys equal to 0 and see what the result is so if we have x plus 5 equals 0 very simple subtract 5 away from each side of the equation we get x is equal to negative 5. then for the other guy we have x minus 3 so x minus 3 equals 0 add 3 to both sides of the equation we get x is equal to 3. so again these two numbers are going to help me split the number line up into kind of three intervals we're going to have from negative infinity to negative 5 with negative 5 not being included then we'll have negative 5 2 3 with negative 5 included and 3 not and then 3 included in anything larger okay so let's set this up using a table again i don't like to use the number line itself i like a little table it helps me organize things so again our numbers i'm just going to write them here we have negative 5 and positive 3. so again from negative infinity up to but not including negative 5 and then negative 5 will be included here so negative 5 is included and then up 2 but not including 3 and then 3 will be included and then out to positive infinity okay so let's rewrite our inequality again if we go back up we have the absolute value of x plus 5 then plus the absolute value of x minus 3 and this is greater than 14. so let's write that so we have the absolute value of x plus 5. and then plus we have the absolute value of x minus three and this is greater than fourteen okay so we know in this first interval the one all the way to the left where we're coming from negative infinity going up two but not including negative both of these guys are going to be negative right this guy right here is going to be negative because it's 0 or x equals negative 5 right so we're less than negative 5 so this guy's negative this guy is also negative because it's 0 when x equals three and we're definitely less than three in this case so we're going to say that this guy is the negative of x plus five and this guy is going to be the negative of x minus three okay and then in this case right here now we're from negative 5 including that up to 3 but not including that well now this guy right here this x plus 5 that expression is going to be either 0 or positive so i can replace it with the expression itself so we'll put just x plus five and then this guy right here this x minus three is still going to be negative right because again it's zero when x is three we're less than three so it's still going to be negative so we put the negative of x minus three and then this last case where we're 3 or anything larger well then this x plus 5 is going to be positive and then for x minus 3 that could be 0 or positive so again we just put x minus 3 there okay so again if it's 0 or positive you just take what's inside the absolute value bars and you can write that down basically think about just drop the absolute value bars if it's negative well then you drop the absolute value bars wrap it in parentheses and then put a negative out in front okay so that's all we're doing all right so let's go through the possibilities now again this is the tedious part so in the first case both are negative so i have the negative of x plus 5 then plus the negative of x minus 3 and you can just put the minus there instead of plus negative doesn't really matter then greater than 14. okay so i'm going to distribute this negative to each term i'm going to have negative x minus 5 and then minus x plus 3 and this is greater than 14. let me scroll down get a little room going so i know that minus x minus another x is negative 2x i know that minus 5 plus 3 is negative 2. this is greater than 14. let me go ahead and add 2 to both sides of the inequality we know this will cancel negative 2x is greater than 16 so then when i divide both sides by negative 2 remember we got to flip this so this becomes a less than now you'll have x is less than negative eight okay so x is less than negative eight is that in the interval that we're working in right now well yeah we said that our current interval is anything less than negative five negative eight is definitely less than negative 5 so we're good to go right this is going to be a valid solution for us but again it's just part of our solution because we might have more to come okay so we just consider it as a partial solution for now all right so let me erase all this so again for this part i said that x was less than negative eight and i'm gonna go ahead and just put a check mark there so we know that we accepted that solution all right for the next interval again from negative five and including that up two but not including three this guy's positive this guy's negative so let's go ahead and do that so x plus 5 again that's positive and then plus this x minus 3 that expression is going to be negative so i'm going to put minus the quantity x minus 3. and before we continue let's just go ahead and distribute this so this would be minus x and plus 3. make that a little quicker then this is greater than 14. okay so x minus x that's going to drop out so this is gone right so you basically have that 8 is greater than 14 which is false okay that's never going to be true so nothing's going to work in that interval so you don't even have to put anything you just say it doesn't work all right so the next one we're going to look at we have from 3 and including 3 out to positive infinity and again both expressions are positive here so we can just drop the absolute value bars in this case so we just say x plus 5 plus x minus 3 is greater than 14. so x plus x is 2x and then 5 minus 3 is 2 so plus 2 and this is greater than 14. all right so let's subtract 2 away from each side of the inequality and then we can cancel this we'll have 2x is greater than 12. let's divide both sides of the inequality by 2. we'll find that x is greater than 6 okay x is greater than 6. is that in our interval well yeah this interval is from 3 out to positive infinity x is greater than 6 that is in the interval so we're basically good to go with that so let's erase this we'll say x is greater than 6. let me put a check mark there so basically we would just combine these two as one statement we would say that x is less than negative 8 or x is greater than 6. so let me just copy this real quick we go back up and let me just paste this in so again x is less than negative 8. so find negative 8 i'm going to put a parenthesis there i'm just going to shade everything to the left or also x could be greater than 6. so find 6 put a parenthesis there shade everything to the right okay in interval notation again i could just say from negative infinity up to but not including negative eight and the union with anything larger than six right so in both cases i'm using a parenthesis to show that it's not included all right so let's kind of change things up and look at another common type of problem that you'll see in this section so let's suppose we saw something like the absolute value of 3 over x minus 1 and this is greater than 5. so how can we solve this type of problem well one approach that we can use we realize that this guy right here this absolute value operation first and foremost there's one of them and it's already isolated for us on one side of the inequality so you have this absolute value operation and this is greater than 5 okay so for right now don't even worry about what's inside of there just think about the fact that earlier in the lesson we said if we had the absolute value of x and this is greater than some positive real number a this led to x being greater than a or x is less than negative a right we saw this with our example we said if the absolute value of x was greater than 7 we said x was greater than 7 or x was less than negative seven okay so this should be pretty crystal clear at this point now the thing is you can do this if you have a more complex expression inside of the absolute value bars okay so i can replace this x with something generic like u we'll just say u is some algebraic expression so it could be a rational expression like we have here it could be a quadratic expression it could be something more complicated than the typical linear expression that you're working with when you solve basic absolute value inequalities so what i can do is erase this and erase the x and just put a u there okay so it's the same kind of concept it just takes more time to solve it it's just going to be way more tedious okay so let's use this to set this up so basically i'm going to take my rational expression inside the absolute value bars and i'm going to say 3 over x minus 1. since i have a greater than i'm going to say it's greater than this positive number 5 then or i'm going to say 3 over x minus 1 i'm going to say is less than the negative of 5. okay i just used this rule here that's all i did okay so let me erase this we don't need this anymore so basically at this point all we need to do is solve two rational inequalities and then kind of combine our two solution sets that we have so i'm going to start out with this one right here so i'm going to copy this we're going to go down to a fresh sheet all right so let me just paste this in here and i want you to recall that the very first thing you want to do when you're solving a rational inequality is get in the format where you have a single rational expression on one side and zero on the other okay so the way we're going to do that here is we're just going to subtract 5 away from each side of the inequality and so this would cancel and become 0. so essentially i would have 1 i would have 3 over x minus 1 and then we'd have minus 5 and this is greater than 0. okay so now i want to get a common denominator going let me just kind of slide this down a little bit and what i'm going to do is just multiply this by the quantity x minus 1 and then over the quantity x minus 1. so now i'll have a common denominator negative 5 times x is negative 5x and then negative 5 times negative 1 is going to be plus 5. okay so kind of continuing now i have a common denominator so i can just combine the numerators so we have 3 and then let me make this more clear this is negative 5x and i'll just put a plus here so we have 3 plus 5 which is 8 okay you have 8 and you have your negative 5x so let me write negative 5x plus 8 as the numerator this is over the common denominator which is x minus one and this is greater than zero okay so now we've accomplished kind of the first task we have a single rational expression on the left side in this case and we have zero on the other okay so the next thing we want to do is find out where the critical values are or some people call them endpoints or boundary whatever you want to call them okay i'll just say critical values so that is going to occur where the numerator is going to be 0 and where the denominator is going to be 0. i would take this numerator which in this case is negative 5x plus 8 and i would set it equal to 0. so that's one of them then i would take the denominator which is x minus 1 set that equal to 0. so both of those guys when we solve that the two solutions there are the critical values or again endpoints so let me add one to both sides of this equation that gives me x equals one so that's one of those guys let me subtract eight away from each side of this equation this is going to cancel you'll have negative 5x is equal to negative 8. divide both sides by negative 5 and you're going to get that x is equal to eight fifths so these are our two critical values so we're going to use that to split the number line up into three intervals so basically you'll have from negative infinity up to but not including one and then between one and eight fifths right neither is included and then anything greater than eight fifths okay so those are the three intervals you're gonna have you're gonna pick a number in each interval okay you're going to test that number in the original inequality you're going to see if it satisfies it if it does that interval works it's part of your solution set if it doesn't it gets rejected it's not part of the solution set so you can use a table for this if you want or you could do a number line for these i prefer to actually use a number line so let's do that all right so the critical values again we have one and we have eight fifths so one is right here so let me just draw a nice little vertical line and it won't be perfect but we'll put that in there and then eight fifths is 1.6 now i've kind of divvied the number line up in a different way you'll see that the notches here are separated by point five or a half so from one we go to 1.5 or you could say 3 halves so 8 fifths again is 1.6 if this is 1.5 1.6 let's just say it's about right there let's say this is eight fifths okay so let's go ahead and put another kind of vertical line there we're going to divvy up the number line and just basically say that let me kind of make that a little better we'll say that this is interval a basically anything less than one anything between one and eight fifths we'll call that interval b and anything larger than eight fifths we'll call that interval c so we'll have a b and c so the original inequality was three over x minus one is greater than five so just take values in each interval check them see if they work again if they do it's part of your solution if they don't it's not part of your solution very very easy just a very tedious process so from interval a i'm going to pick 0 plug that in there you would have three over zero minus one is negative one so three over negative one is that greater than five no this would end up being negative three negative three is certainly not greater than five so this is false right in interval a nothing's going to work so for interval b again i'm between 1 and 8 fifths or between 1 and 1.6 i think the easiest number to use there would be 1.5 so let's go ahead and plug in a 1.5 there 1.5 minus 1 would be 0.5 so you would have 3 divided by 0.5 is greater than 5. you know if you divide 3 by 0.5 it's the same thing as multiplying 3 by 2 right if you divide 3 by a half you could say when you crank that out you get 3 times 2 which is going to be 6. is 6 larger than 5 yes it is so this is true all right for the last one let's take a look at something in interval c i'm going to pick 2 so 2 minus 1 is 1 3 over 1 is 3. so you basically have that 3 is greater than 5 which is false okay let me write that a little better so false okay so only numbers or only values in interval b are going to satisfy this inequality it's a strict inequality so the end points we don't even need to consider okay they won't work they will be excluded so the solution for this part is going to be anything larger than 1 up 2 but not including 8 fifths so let's just copy this we'll go back up and again this is just going to be a partial solution so this is for this one and we could really write that x is greater than one and it's less than eight fifths okay so this is just a partial solution we're going to work on this guy now and see what else we can find so let's copy this and let's go down alright so let's paste this other scenario here we have 3 over x minus 1 and this is less than negative 5. so again i want to get one rational expression on one side or one fraction and 0 on the other so same process as last time so i'm going to add 5 to both sides of the inequality i'm going to have that 3 over x minus 1. then plus 5 is less than 0. i'm going to get a common denominator so let me again move this down so i'm going to multiply this by x minus 1 over x minus 1 and this is going to lead to 3 over x minus 1 plus you have 5 times x which is 5x and then minus 5 times 1 is 5 and this is over x minus 1. okay then this is less than zero so let me erase this and we don't need this for right now we'll write it again when we need it so let me just drag this up and we make that 3 a little bit better okay so now we're just going to kind of combine the numerators we have a common denominator so we'd have 5x and then 3 minus 5 is negative 2 so minus 2. this is over x minus 1 and this is less than 0. okay so at this point again i want to take my numerator set it equal to 0 take my denominator set it equal to 0. that will give me the two endpoints or critical values or boundaries okay so we would have 5x minus 2 is equal to 0 add 2 to both sides of the equation i'm going to get that 5x is equal to 2. we would divide both sides by 5 and find that x is equal to 2 fifths okay so that's one critical value and then x minus 1 we know if we set that equal to zero we would get that x equals one right we solved that earlier okay so let's go up and again the two critical values we have one and then we have two-fifths okay so one is right here let's just put a vertical line there two-fifths in decimal form is point four so i have a notch for one-half point four would be just a little bit less so let's just say it's right there so this is going to be my two-fifths let me just put a vertical line there okay so we have interval a interval b and interval c again interval a in this case is going to be anything that's less than two fifths interval b will be between two fifths and one and interval c will be anything larger than 1. so the inequality we're working with is 3 over x minus 1 and that's less than negative 5. so let's choose a number from interval a again i'm going to choose 0 because it's easy to work with so 0 minus 1 is negative 1 you would have that 3 over negative 1 is less than negative 5. that's going to be false right this would be negative 3 is less than negative 5. negative 3 is a smaller negative value than negative five so it's actually a larger number right so this is going to be false okay what about for interval b well it's kind of a tight window but between two fifths and one you can use one half so if i do 0.5 there i'm going to have 0.5 minus 1 which is negative 0.5 so i would have 3 over a negative 0.5 and this is less than negative 5. of course if i divide 3 by negative 0.5 it's like multiplying 3 times negative 2 so this would basically be negative 6 here negative 6 is a larger negative than negative 5 so it is less than negative 5. so this is true then for interval c again i'm just going to choose 2 nice and easy to work with 2 minus 1 gives me 1. so i would have that 3 is less than negative 5 which is obviously false okay so my solution here is going to be for interval b anything between two fifths and one so between two fifths and one okay so for this guy my solution is again between two fifths and one so now it's going to be the union of these two solution sets so let me erase this i don't need any of this anymore because i'm thinking about the problem as a whole now so i'm just going to start with this one so i'm going to say between two fifths and one again neither is included and the union with this guy which is between one and eight fifths neither is included okay now you could also write this with our kind of standard way using that or we can say that x is going to be greater than two-fifths and less than one and then or you have this solution so you have that x is going to be greater than one and less than eight fifths okay so let's copy this and go to the number line we'll make a nice little graph to finish this off all right so the first thing i'm going to do is just make a little notch for two-fifths so again that's 0.4 so i'm going to put it right there and just say this is two-fifths again this value here for this notch is one-half so two-fifths would be less than that then for eight-fifths i'm gonna say again that's 1.6 as a decimal so 3 halves is 1.5 so it's going to be a little bit more than that so let's say it's right there so this will be my eight-fifths okay so basically two-fifths is not included so i'll put a parenthesis there we're gonna go up 2 but not including 1. so i'll put a parenthesis there but then again at 1 i'm not including it so i'm going to put another parenthesis facing the opposite direction then i'm going to shade up 2 but not including 8 fifths and i'll put a parenthesis there so that's a nice little graph you can make and in this case it might be a little easier to use open circles so what we might want to do to make the graph a little cleaner is put an open circle here at two-fifths put an open circle here at one and put an open circle here at eight fifths okay so you could do it this way it's a little cleaner or the other way using the kind of parenthesis for each it doesn't matter either way it's the same solution so we see that x is greater than two-fifths and less than one or x is greater than 1 and less than 8 fifths in this lesson we want to review the coordinate plane so in this section of our course we will review graphing equations with two variables involved and we're also going to discuss functions so before we can kind of get into those topics we need to make sure that everyone understands the basics so today we're just going to start out the section by reviewing the coordinate plane and again how to plot an ordered pair so before we jump in and start talking about the coordinate plane let's first review the concept of an ordered pair so pretty much up to this point in our course we have exclusively worked with equations that have only one variable involved in this new section we're going to look at a lot of equations with two variables involved normally x and y okay so suppose we saw the following equation suppose we had something like 2x and then plus 9y is equal to 20 okay so this type of equation is known as a linear equation in two variables most of you have seen this if you've taken an algebra one or an intermediate algebra course you've seen this type of equation and you know when you graph this type of equation you get a line right henceforth the term linear equation okay so when we work with these types of equations we find that there's an infinite number of solutions okay so there's an infinite number of kind of x and y combinations that satisfy the equation or again you could say make the equation true so in order to get kind of a visual representation of the solution set it's normal for us to graph this type of equation on our coordinate plane to kind of see what's going on so we're going to graph linear equations in a few lessons but most of you already know this is a pretty easy process for now we just need to understand how we can write solutions for our equation using ordered pairs okay so basically an ordered pair consists of two components which are written inside of parentheses and separated by a comma so when we write an ordered pair the order of the components are important now normally in our studies with algebra you're going to see them written like this you're going to see again inside of parentheses your x value comma your y value so it's easy to remember because it's in alphabetical order x is before y in the alphabet and x is before y in the ordered pair okay so you can just remember it like that so as an example one solution for this linear equation in two variables would be that x equals 10 so i'm just going to write this over here i'm going to say x equals 10 and y is equal to zero so you've got to have two values there one for x and one for y we're going to plug both of those in and show that this is a solution so if i plug in a 10 here and i plug a 0 in there let me kind of scooch this over and out of our way you'd see that 2 times 10 is 20 plus 9 times 0 is 0. so you can basically just say this is 20 is equal to 20. so we know that this guy is correct so to write this solution as an ordered pair again i just follow this format all i'm going to do is put my x value first so in this case the x value is 10 then comma my y value comes second so that's going to be a 0 there so you've got 10 comma 0 right again enclosed inside of parentheses your x values first your y value is second okay so that's how you write your ordered pair as another example we could show another solution so let's say that x is equal to 3 and y is equal to 14 9. now don't worry about where i'm getting these solutions from we'll talk about these types of equations in a few lessons right now we're just getting the basics down we're learning how to write some ordered pairs so again if you wanted to check this you could plug in a 3 for x you could plug in a 3 for x and you could plug in a 14 9 for y let me erase this so we can just check this real quick 2 times 3 is 6 plus 9 times 14 9 we know that the 9's would cancel so we'll go ahead and just cancel this with this i'm left with a 14. and essentially you have 6 plus 14 which is 20. so you would have that 20 is equal to 20. so this guy is going to check out as well okay so let me erase this we don't need this information anymore so how would i write x equals 3 y equals 14 9's as an ordered pair again just follow this form the x value comes first so i'm going to put a 3 then comma the y value is second so it's 14 9. again they're separated by a comma and they're inside of a set of parentheses all right so now that we understand what an ordered pair is and kind of how to write an ordered pair let's move on and talk about the coordinate plane and then we'll get into how we can plot an ordered pair on our coordinate plane so we've all seen this basic number line we pretty much haven't done much in the way of graphing up to this point in the course we pretty much just use our number line for working with inequalities right so if we had something simple like let's say 2x minus 3 equals 7 we could actually graph our solution for something simple like this using a horizontal number line if we just add 3 to both sides of the equation we would find that 2x is equal to 10 we divide both sides by 2 and we would find that x is equal to 5 right so this is our solution very easy to get everyone at this point should understand that so graphically on a horizontal number line i can just say that at five i could just put in a filled in circle or filled in dot however you want to think about that that represents my solution right this value here of five but when we move in and start talking about an equation with two variables we're now going to need two number lots so we're going to need to look at something like this right so this is our coordinate plane you have two number lines involved and when i say coordinate plane i don't want you to get confused because there's a lot of different names floating around for this thing you might hear the cartesian coordinate plane okay that's named after its founder you might hear the rectangular coordinate plane there's a lot of different names floating around for this i'm just going to refer to this as the coordinate plane okay just to make it real simple now what you'll notice here is that you have these two intersecting number lines so we've got our horizontal number line that we're used to okay this one's going kind of left and right and then we have this new addition this vertical number line okay we're not used to that but what you'll notice is that they intersect at the point where each number line is zero so this spot right here is the point of intersection it has a special name it's known as the origin let me write this off to the side because i'm going to end up erasing this so this is the origin and this occurs at the point 0 comma 0. the x value is 0 the y value is 0 okay so that's why i wrote 0 comma 0 as an ordered pair so this is the origin x is 0 y 0. all right so let's talk a little bit more about this coordinate plane so the horizontal number line the number line that we're used to working with up to this point is known as the x-axis okay you see this little x here i can put x-axis for the sake of completeness and basically this number line represents our x values okay so what we can see here just like we're used to numbers increase as we move to the right and they decrease as we move to the left so if a number on this x-axis is to the right of another then it's a bigger number if a number on the x-axis is to the left of another then it's a smaller number okay so anything to the right of this kind of origin or to the right of zero is going to be a positive number okay we need to note that so anything in this direction is positive then anything to the left of the origin on the x-axis or to the left of zero is going to be negative okay so it's very important to understand that for the purposes of what we're going to do in a minute when we start talking about quadrants so to the right of the origin or to the right of zero on the x-axis is positive to the left of 0 or to the left of the origin on the x-axis is going to be negative okay now we also have this vertical number line basically we just took this number line and rotated it okay so this is known as the y-axis and it represents y-values so i can just say this is the y-axis and essentially what we see is that numbers increase as we move up okay so they increase as we go up so if a number is above another number then it's a bigger number they decrease as they go down okay so if a number is below another number then it's going to be a smaller number okay now anything above 0 or anything above your origin is going to be positive so this is going to be your positive values here kind of going up and then anything below the origin or anything below zero is going to be negative so now let's move in and talk about the quadrants okay so we can split our coordinate plane and i'm just going to erase this up into four quadrants and i'm gonna erase this two we don't need this anymore so essentially what i have is a movement that goes counterclockwise starting with the top right so this guy right here would be quadrant one okay that's the top right we move counterclockwise so we're going this way this guy right here will be quadrant two we're moving this way this guy right here will be quadrant three we're moving this way this guy right here will be quadrant four okay when you first start working with your coordinate plane you're going to be asked to plot ordered pairs and then also determine what quadrant it's located in so the main thing is to understand the kind of signs of your numbers involved so when we look at quadrant one we can see that x values are positive and y values are positive how do we know that well if we look at our x axis or to the right of zero if we look at our y axis we're above zero okay so any x value here would be positive any y value would be positive as well when we move into quadrant two when we look at our x axis we're to the left of zero now so our x values are going to be negative our y values are going to be positive though because we're still above 0 on the y axis so we're good to go there then for quadrant 3 we're still to the left of 0 on the x axis so that means x values would be negative we're below zero on the y axis so that means y values will also be negative then when we get into quadrant four we can see that our x values are going to be positive right because we're to the right of zero on the x axis and then our y values are going to be negative because we're below zero on the y axis okay so just remember this you might want to write this down in your notes for quadrant one it's positive posit right x value is positive y value's positive for quadrant two it's going to be negative positive okay so you have x values that are negative y values that are positive for quadrant three we have negative negative so x values are negative y values are negative and then for quadrant four kind of wrap this up we see that our x values are positive and our y values are negative okay so something you might want to write down because something you just want to remember for kind of reference sake if you don't want to write it down just remember that the top right is the first quadrant and then it goes counterclockwise so you could just derive this you can say okay this is 2 this is 3 this is 4 and then you can just look and see okay are my values positive or my values negative for each quadrant now it's very important to note this because this comes up as kind of a trap question if you were located on the x-axis or the y-axis right so let's say i got a point that was something like 6 comma 0. so that's going to be right here we'll talk about plotting ordered pairs in a minute but this is 6 comma zero so this is on the x-axis this is not located in any quadrant okay so if you can ask that question it's not located in any of them a lot of students will make the mistake and say okay well it's in quadrant one right because they'll say okay well the ordered pair 6 comma 0 they'll incorrectly say well 6 is positive that's true but they'll incorrectly say well 0 is positive 0 is not a positive number 0 is not positive it's not negative it's neither okay so don't make the mistake of trying to match that up and say well in quadrant one things are both positive and then six comma zero each thing is positive that's incorrect right because zero is not a positive number all right so now let's move on and look at a little exercise that's going to help us understand how to plot an ordered pair most of you know how to do this and you understand the process is very very simple so an ordered pair when we work with them it's usually referred to as a point okay so we could say we're going to plot the point that's just meaning we're going to plot an ordered pair okay so we're just finding the meeting point of the x location and the y location we're going to draw a little closed circle at that point and then we're sometimes going to write the ordered pair next to it in some cases you'll also have like a capital letter associated with it you might want to write that there just depending on what they give you we'll see the capital letters and the exercises that are associated with this lesson okay so what we want to do here again is just plot each ordered pair and then we're going to determine its quadrant okay so we're going to start with negative 5 comma 5. so this ordered pair here so i'm going to come down to my coordinate plane so negative 5 comma 5 and then for my quadrant i'm just going to put q and i'm going to question mark for right now we'll figure that out in a minute so how do we plot this or how do we graph this ordered pair well essentially i've got to think about the fact that again let me kind of slide this down this guy right here is an x value okay this is an x value this guy right here is my y value think about how this is constructed the x values are on the horizontal axis or the horizontal number line the y values are on the vertical axis or the vertical number line so the x value tells me how much i'm moving horizontally right left and right the y value tells me how much i'm moving vertically up and down if you have a negative you're going to the left if you have a positive you're going to the right in terms of the x values with the y values a positive is going to go up and a negative is going to go down okay and it's a lot of information at first but after you do a few of these it's very intuitive okay so if i start at the origin and i just kind of use my information here to kind of move my x value is negative 5. so just go 5 units to the left or just go to negative 5 on the x axis so that's right here this is negative 5. then my y value is 5. so that tells me vertically i want to move 5 units up it's a positive 5. so from negative 5 i'm just going to go 1 2 3 4 5 units up and what you see is that this is basically the meeting point of an x value of negative 5 and a y value of positive 5. okay if i drew kind of a little line extending this and a little line extending this you can see that's the meeting point right so this is negative five comma five so that's one way you can do it there's obviously quicker ways what i basically do when i do this now and i kind of erase my points let me draw that back in so this is going to be right here what i do when i do this now is i just go to the x location first and then i move up or down based on the y location i'm given okay so in this case i would just go to negative five first i don't need to start the origin and then my y location is five so i just go up five units okay that's all i need to do now what quadrant does this lie in remember the top right quadrant one and we go counterclockwise so this is going to be in quadrant two right this would be quadrant three this would be quadrant four okay so the quadrant that it's in would be quadrant two all right let's grab the next one so the next one is 3 comma negative 7. so 3 comma negative 7. so let me erase this and i will put 3 comma negative 7 there i'll erase this because we don't know what quadrant it's in yet so again you could start at an x value of 3 so that's right here and then the y value is negative 7. so a negative in terms of a vertical number line just tells me i'm going down right i need to go down so i'm going to go down by 7 units so start at 3 on the x axis and go down by seven units so basically right here right you can see the y value here is going to be negative seven so if you extend from here you see that's the meeting point between those two so this is going to be three comma negative seven and if you wanted to do this differently you could you could start out at the y location so i could have started out at negative 7 so right there and then my x location or my x value is 3 so it just tells me i want to move to the right by 3 units again that will get me to 3 comma negative seven so that's going to be my location now for the quadrant this guy is located in quadrant four so the quadrant is quadrant four all right so now we want to look at negative four comma negative four so let me erase this and i'll put a negative 4 here and a negative 4 here so what we want to do again is go 4 units to the left i want to start out at negative 4 on my x axis this is negative 4 and then i have a y value of negative four so that just means i'm going down by four units so i'm going to find basically negative four on the y axis it's the meeting point between the two so four units to the left and four units down that's going to be right there so this is negative four comma negative four and this is located in quadrant 3. okay right now we have negative 2 comma 8 so negative 2 comma 8. so erase this and this so negative 2 8. so we would again i'll kind of mix this up just because you can do it either way i'm going to start out by going to 8 on the y-axis right because i'm going 8 units up so this is 8 i'll just kind of circle that now my x value is negative 2. so horizontally i just want to move 2 units to the left again if this is negative it's telling me to go to the left so i go up eight units and then i go left two units and this is kind of in the way so let me kind of move this so again i would just go two is to the left so one two so this right here would be negative 2 comma 8. so negative 2 comma 8. it's kind of squished in there so let me erase that and i'll put it up here so negative 2 comma 8 that's a little better so let me write that the quadrant here is going to be quadrant two okay and again you could have done this a different way you could have started at the origin and move two units to the left and eight units up that'll get you there also or you could start at negative two and just go eight units up or again like we did you could start at 8 and just go two inches to the left a lot of different ways to do these again for me i always just start the x location so i would have started at negative two my y location is eight so i would just go up eight units all right the last one is six comma zero and if you remember we talked about this earlier in the lesson so let me erase this and put 6 comma 0. so 6 here and 0 here so what are we going to see so again my x value is 6. so from the origin i can move 6 units to the right or i could just go to 6 so that's right here now what am i going to do when i have a y value of 0 well essentially if i look at just the vertical number line if i highlight that where is it 0 it's 0 on the x-axis right so on the x-axis where i don't move it all up or down so i'm just going to stay at that point i'm just going to stay right here i move horizontally but i don't move vertically right in terms of the origin i can go six units to the right or again i could just start at six and i don't move it all vertically okay so this guy right here is on the x-axis and when it's on the x-axis or on the y-axis you are not in any quadrant so we can say not in any quadrant okay and that's a common trap question on a test to give you a point on an axis either the y-axis or the x-axis and say what quadrants it in it's not in any quadrant in this lesson we want to review finding the distance between two points using the distance formula so in our last lesson we reviewed the coordinate plane and how to plot points or ordered pairs on the coordinate plane so now what we're going to do is just discuss how we can find the distance between two points or again two ordered pairs on our coordinate plane so to accomplish this task we have a very simple formula it's known as the distance formula now this is not to be confused with the distance formula the distance equals rate of speed times time traveled that we used with motion word problems although it does have the same name it accomplishes a very different task so this formula we're going to get today is something we could just plug into we're going to evaluate and we're basically going to have our answer right away okay but the purpose of this lesson is not to just give you this easy formula we want you to understand where it comes from and essentially it's going to come from the pythagorean theorem or you could say the pythagorean formula so we talked about the pythagorean formula already in our lesson on applications of quadratic equations this is also something you would have learned in an elementary algebra course or an intermediate algebra course it's something that comes up a lot so basically the pythagorean formula relates the lengths of the sides of a right triangle or a triangle with a 90 degree angle so on our screen what we have here is a right triangle and we know it's a right triangle because of this symbol right here this guy is telling us that we have a 90 degree angle so you'll also notice that we've labeled each of the three sides of the right triangle a right triangle is going to consist of two legs and we've labeled each leg here so we have leg a which is basically from here to here let me just highlight that real quick so this is leg a and then we have leg b which is basically from here to here okay let me highlight that as well so this is our leg b and then we have a hypotenuse okay so the hypotenuse is the side that's opposite of the 90 degree angle so in this particular case it's going to be from here to here okay so that's going to be my hypotenuse so the right triangle is going to have two legs and a hypotenuse and again the hypotenuse is always opposite of the 90 degree angle the hypotenuse is always going to be the longest side okay so the two shorter sides are known as legs the longest side is known as the hypotenuse now our pythagorean formula tells us that if we sum the squares of the two legs okay the two shorter sides it will be equal to the square of the hypotenuse or again the longest side so for the pythagorean formula we get that this distance here for leg a that guy squared plus this distance here for leg b that amount squared if we sum these again two shorter sides those amounts squared we get the hypotenuse or c squared okay so this is the relationship between the sides and a right triangle okay so let's take a look at a quick example so one common application of this formula is that if you have a right triangle and you know two of the three sides you can solve for the third unknown okay so this is basically what we're going to be doing when we derive our distance formula so as a quick example suppose that we know that leg b is going to be eight okay so that's the distance there from here to here that's my leg b that's eight we don't know what leg a is so we don't know what this to this is but we do know what the hypotenuse or c is let me kind of write this in as c there so this guy's going to be 17. so again we know that a squared plus b squared equals c squared so what i can do for this example is plug in 4b we know that's 8. so i'm going to plug that in there and i could plug in for c we know the hypotenuse is 17. so we would end up with a squared plus 8 squared is 64. and this is equal to 17 squared which is 289 to solve for a what i'd want to do is get this guy by itself first so i would subtract 64 away from each side of the equation we would have a squared is equal to 289 minus 64 is 225. now here's where you got to kind of pay attention to what's going on we know how to solve this type of equation so let me write this over here we have a squared is equal to again 225 up to this point we've pretty much just taken the square root of each side but when i take the square root of this side on the right i have to go plus or minus to kind of account for all the possible solutions now there's an issue with that in this particular case we'd end up with a is equal to plus or minus 15. now a here is again a length it's a distance so it can't be negative 15 so you'd want to throw this solution out and just say that a here is going to be 15. okay so let me erase this we'll kind of come back up here and i'll just write here that this guy is 15 okay so this triangle here we have leg b that's eight we have leg a that's 15 and we have c our hypotenuse that's going to be 17. all right so now let's move on and talk about how to find the distance between two points on a coordinate plane using our pythagorean formula we're going to start with this example here so we want to find the distance between six comma four and zero comma negative four i'll show you how to do this the long way and then we'll move into kind of deriving the distance formula and then from then on once we use that formula it'll be very very easy so let's go down to the coordinate plane and let's first write our ordered pairs we have six comma four and then we have zero comma negative four okay so for six comma four if we plot that we start at the origin we go six units to the right and four units up so that's right there so this is six comma four for zero comma negative four starting at the origin i just dropped down four units so this would be zero comma negative four okay i'll put that there and essentially what we're going to end up doing is finding the distance between these two points so let me just draw a little line that connects the two so this line right here is going to end up being our hypotenuse or longest side of the right triangle so how can we end up drawing our right triangle well what we're going to need is one additional endpoint right for the right triangle or one additional corner it could say one additional vertex now how do we get that some of you can see that it's going to end up being right here okay but the way you can draw this is you can take an x value from one of the points so let's just say i take 6 then i'm going to take a y value from the opposite point okay so now i'm going to take a y value of negative 4 so that's how i get this point here which is 6 comma negative 4. okay so i can use this to kind of complete my right triangle now i can also make a right triangle by using alternate kind of coordinates so i can take the x value from here which is 0 and the y value from here which is 4. so if i erase this and i go to zero comma four we could also draw a right triangle that way this way is going to be a little bit better for me to kind of show things so i'm just gonna stick with the original one that i came up with so an x value of six and then a y value of negative four okay so that'll be kind of our third point on this right triangle okay so let me kind of draw this in okay so not perfect but we get the idea here so let me kind of draw my symbol here for the 90 degree angle so this is our right triangle and essentially we know that this side right here that's opposite of the 90 degree angle is the hypotenuse so i'm going to label that as c and then the two shorter sides i can label that as a and b so i'm just going to stay consistent with what i've done previously and i'm going to say that this horizontal leg this kind of side that's horizontal or parallel to the x-axis i'm going to call that leg b this vertical leg or this kind of side that is parallel to the y-axis i'm going to call that leg a okay but you can interchange these two it's not a big deal right as long as the hypotenuse is labeled as c the formula i gave you will work so it's pretty easy to find the length of the legs for b the horizontal leg we just find the distance by subtracting the x values okay and the reason we do that is because the x axis or the horizontal axis are movements left and right right you think about this line as being a horizontal line so all we need to really consider is the x value here of 0 which is kind of the x value for that point there and the x value here of 6 which is the x value for this point there so if you want to think about this you can think about this as being on a horizontal number line like this let's say that this point right here is 6 and this point right here is 0. so what i'd end up doing here is just subtracting and i want to do this inside of an absolute value operation so that i end up getting a positive result right distance is never going to be negative so that's why you do that so i could do the absolute value of 6 minus 0 which is the absolute value of 6 which is just 6 or i could do the absolute value of 0 minus 6 which is the absolute value of negative 6 which is also 6. so i know that my b here let me kind of write this here my b is going to be 6. okay so what about a well with a this is our vertical leg and we want to find the distance by subtracting y values okay because again when we think about y it's our vertical axis it's our vertical number line the y values are movements up and down so if i think about kind of this point here and this point here the y values are going to be 4 and negative 4 okay so i want to think about the fact that to go from here to here i would have to go from four to zero which is four units and then another four units down would give me another four so four plus four would give me eight but again you find that through subtraction so you could do 4 minus a negative 4 so 4 minus a negative 4 is the same thing as 4 plus 4. you do that inside of an absolute value operation so this ends up being the absolute value of 8 which is 8. or you could do it the other way you could say that you have negative 4 minus 4 again inside of absolute value bars so this ends up being the absolute value of negative 8 which is also 8. okay so you see that distance from here to here is going to be 8. so i'm going to put that a is 8. now once we have this information we're basically good to go right because we could just plug into the formula we know that a squared so a squared plus b squared equals c squared so i know what a is it's 8. so i'm going to plug that in there i know what b is that's 6. so i'm going to plug that in there i don't know what c is but again i can find it pretty easily here so i'll have 8 squared plus 6 squared equals c squared so 8 squared is 64. this is 64. then plus 6 squared is 36 this equals c squared 64 plus 36 is 100 so you get 100 is equal to c squared i'm just going to take the principal square root of each side i don't need the plus or minus over here on the left so i'm just going to have that 10 is equal to c okay so let me erase all this and i'll write that c here equals 10 so that's this value right here so this guy is going to be 10. and i can write the other values in if i want we know that a is 8 and we know that b is 6. okay instead of having to graph things each time we have this very easy formula that we can use we can just basically plug into this guy this is called the distance formula and what it is is we've taken the pythagorean formula and we've taken our points that we're working with and we've kind of set everything up so essentially if i had these two points again we were working with 6 comma 4 and we're working with 0 comma negative 4. i'm going to label one of the points as x sub 1 y sub 1 and the other as x sub 2 y sub 2. it doesn't matter which i label as which so i'm just going to say this is x sub 1 y sub 1 and i'm going to say this is x sub 2 y sub 2 okay so if i plug in here for x sub 2 i've got a 0 and then minus for x sub 1 i've got a 6. so you'll notice that what this gives us is the kind of horizontal distance if i go back up we see that from here to here we have a horizontal distance of six right we can find that by doing the absolute value of six minus zero or we can find that by doing the absolute value of 0 minus 6. either way we get a distance of 6 units now in this particular case you don't see absolute value signs and that's because we end up squaring this guy so if we get a negative it doesn't matter right it ends up making it positive if i did 6 minus 0 i would get 6 if i square it i get 36 if i do 0 minus 6 i get negative 6 if i square it i still get 36 okay so that's why we don't have absolute value signs involved then plus here you have the difference in y values so we have y sub 2 which is going to be negative 4 and then minus y sub 1 which is going to be 4. again we're squaring this result if i go back we see that's the difference in y values right or the vertical distance so we see that we have what negative 4 down here and positive 4 here so i do the absolute value of negative 4 minus 4 to get that distance okay that gives me the absolute value of negative 8 which is 8. again we don't need to use absolute value bars here because we're squaring the result so let me just kind of do this off to the side we have d is equal to the square root of 0 minus 6 is negative 6 negative 6 squared is 36 then plus negative 4 minus 4 is negative 8 negative 8 squared is 64. so you remember this is what we ended up getting when we squared a and b okay so if we were at this point right here let me label this as c so right now we would have c squared is equal to what you'd have a squared plus b squared so to solve for c or the distance between those two points all i did was i renamed this c as d right to count for distance and i just took the square root of each side so this becomes just d by itself this becomes the square root over here okay so that's where this comes from so now we just quickly say 36 plus 64 is 100 we take the square root of 100 and we get that d is equal to 10. okay so you can see i found that in just a few seconds or you could say pretty much instantly with this formula versus having to go to a coordinate plane and kind of you know draw a right triangle and go through it each time okay so this is definitely we're going to be using for application purposes but we wanted to make sure you understand where it came from all right so let's run through a few examples now so we want to find the distance between each pair of points and we're just going to use our distance formula so we have 2 comma 1 and we have negative 3 comma negative 11. okay so the distance formula we say that d is equal to the square root of you've got this x sub 2 minus x sub 1 this quantity squared again this is the difference in x values or your horizontal leg right that distance there then plus you have y sub 2 minus y sub 1 this quantity squared again this is the difference in kind of y values that's going to give you your vertical leg okay let me kind of make this better okay so i'm going to label one of these points as x sub 1 y sub 1 and the other as x sub 2 y sub 2. it doesn't matter which i label is which so we'll say this is x sub 1 y sub 1 we'll say this is x sub 2 y sub 2 okay just to keep it nice and simple so for x sub 2 i have negative 3 for x sub 1 i have 2 so let me erase these and just plug in so this is negative 3 minus 2 there and then i can erase this for y sub 2 i have negative 11 so i have negative 11. for y sub 1 i have 1. so let me plug that in so negative 3 minus 2 is negative 5 negative 5 squared is 25 so this is 25 negative 11 minus 1 is going to be negative 12 negative 12 squared is 144. so if i sum 25 and 144 i get 169 and if i take the square of 169 i get 13. okay so the distance between these two points is 13. let's take a look at another so we have 19 comma negative 4 and we have 3 comma negative 34. so the distance between these two points again the d for distance is equal to the square root of you've got x sub 2 minus x sub 1 this quantity squared then plus you've got y sub 2 minus y sub 1 again this quantity squared for this guy right here i'll just switch it up and say this is x sub 2 y sub 2 and i'll say this is x sub 1 y sub 1. so let's just plug in so we have that x sub 2 is 19. so this is 19. we have that x sub 1 is 3. then we have that y sub 2 is going to be negative 4 and we have that y sub 1 is going to be negative 34. so minus the negative 34 is plus 34. let's write that in like that 19 minus 3 is 16 so you'd have 16 squared which is going to be 256. so let me just write this as 256 here and then negative 4 plus 34 is 30. if you square 30 you get 900 okay so this is 900 here so what is 256 plus 900 that's 1156 and if you take the square root of that you end up with 34 okay so we'll put that the distance here is 34. all right let's take a look at one more example again when you use the formula it's extremely easy so the distance between these two points you have negative 10 comma 15 and you have 12 comma 17. so my distance my d is equal to the square root of you've got your x sub 2 minus your x sub 1 squared then plus your y sub 2 minus your y sub 1 squared okay so i'm just going to label this guy as x sub 1 y sub 1. i'll label this guy as x sub 2 y sub 2 okay so i'm just going to plug it so we have that x sub 2 is going to be 12 we have that x sub 1 is going to be negative 10 so minus a negative 10 is plus 10. we have that y sub 2 is going to be 17 okay 17. we have that y sub 1 is going to be 15. okay so pretty easy here so 12 plus 10 is 22 22 squared is 484 and then 17 minus 15 is 2 2 squared is 4. so if i add 484 and 4 i get 488 so 488 now 488 is not a perfect square so to kind of simplify this 488 divided by 4 is 122. so i'm going to write this as the square root of 4 which is 2 times the square root of 122. now i can't do anything with the square root of 122 i'm just going to leave that as it is so this is my kind of simplified answer we have the distance is equal to 2 times the square root of 122. in this lesson we want to look at determining whether three points are the vertices of a right triangle so in our last lesson we learned about the distance formula this formula allows us to calculate the distance between any two points on the coordinate plane and basically we derive this formula from our pythagorean formula okay this relates the sides of a right triangle again if you have a right triangle like this guy right here again you know it's a right triangle because of this 90 degree angle here then essentially the pythagorean formula relates the sides of a right triangle so you have these two shorter sides which are called legs and you have the longest side which is opposite of the 90 degree angle which is your hypotenuse and essentially it tells us that a squared plus b squared is equal to c squared now we can basically use this guy here to come up with a distance formula to find the distance between this point and this point when the two legs are known right so if you know the two legs which are easy to find kind of on a coordinate plane then you can easily find the distance between two points which would be your hypotenuse okay so to kind of summarize here we found that the distance between two points we label it as d it's equal to the square root of you have x sub 2 minus x sub 1 that quantity squared then plus you have your y sub 2 minus y sub 1 that quantity squared okay what we can do here is if we're given three points okay so let's say there's a point here and let me kind of do this in a different color so let's say there's a point here here and here if it's true that we have a right triangle then the distance for this guy right here for leg a if we square that plus the distance for this guy right here for leg b if we square that should be equal to the distance of the hypotenuse squared okay so we know how to get the distance between any two points just using this distance formula okay so let's look at an example so we want the three points here we have them labeled as a b and c so we have 2 comma 4 negative 1 comma 0 and 2 comma 0. we want to know if these are the vertices or the endpoints or the corners of a right triangle so what i need to do is look at the distance between each pair of points involved okay so basically the longest distance that i get if it's a right triangle would be my hypotenuse so i would square the two kind of shorter distances i would sum those amounts and it should be equal to the longest distance that i have or the largest distance squared okay if it's a right triangle so what i'm going to do is i'm going to find three distances i'm going to say distance 1 i'm going to say distance 2 and i'm going to say distance 3 okay so first i'm going to do the distance between a and b so i'm going to say this is going to be the distance between a and b and what i'm going to do is i'm going to use again these two points here and i'm going to use my distance formula so the distance formula is what it's the quantity x sub 2 minus x sub 1 squared plus the quantity y sub 2 minus y sub 1 squared and again we're taking the square root of this and this would give me my distance between those two points so i'm just going to plug in here and i'm going to label this as x sub 1 y sub 1. i'm going to label this x sub 2 y sub 2 okay so i'm just going to erase and just kind of plug in for x sub 2 i have negative 1. for x sub 1 i have 2. for y sub 2 i'm going to have 0 for y sub 1 i'm going to have 4 okay so if i run through this real quick negative 1 minus 2 is negative 3. square that and you get 9. and then 0 minus 4 is negative 4 square that you get 16. if i sum 9 and 16 i get 25 the principal square root of 25 is 5. okay so the distance between these two guys here a and b is going to be 5. now let's do a distance 2 we're going to say this is the distance between let's say a and c so a and c so again i use my distance formula so the square root of you've got x sub 2 minus x sub 1 that quantity squared plus you've got y sub 2 minus y sub 1 that quantity squared okay so for x sub 2 let me kind of label this guy over here i'm going to say this is x sub 2 y sub 2. so for x sub 2 i'm going to say that's 2. for x sub 1 i'm going to say that's 2 so 2 minus 2 is 0 0 squared is 0. so you just get rid of this then for y sub 2 i have 0 for y sub 1 i have 4. so basically i have 0 minus 4 which is negative 4 negative 4 squared is 16. so the square root of 16 is 4. so this guy's going to be 4 right here now for the distance 3 i'm going to say the distance between y well i need to do b and c now so b and c okay so what's that going to be let me kind of slide this over and say this is now x sub 1 y sub 1. so again it's the square root of you've got x sub 2 minus x sub 1 that quantity squared so 2 minus a negative 1 2 minus a negative 1 that amount squared 2 minus negative 1 is 2 plus 1 2 plus 1 is 3 3 squared is 9. so this is going to be 9 and then plus you've got your y sub 2 which is 0 minus your y sub 1 which is 0 that quantity squared 0 minus 0 is obviously 0 0 squared is going to be 0. so you get rid of this square root of 9 is just 3. now the largest number is 5. so if this is a right triangle this guy would represent my hypotenuse and these two guys right here would be the shorter sides the a and b so it would be true that 3 squared plus 4 squared is equal to 5 squared if these are the endpoints of a right triangle and you see that works itself out so this is going to be true right we're going to end up with a 3 4 5 right triangle 3 squared is 9 plus 4 squared is 16. if you sum these you do get 5 squared which is 25 right you get 25 equals 25. so these are the endpoints for a right triangle all right let's take a look at a few more of these so we have these points here a b and c so you've got negative 2 comma 5 12 comma 3 and 10 comma negative 11. again we want to find out if these are the end points or the corners or the vertices of a right triangle so again all i want to do is find the distance between the points so my first distance i'm going to say this is the distance between a and b and then my second distance i'll say that's the distance between a and c and then my third distance i'll say that's the distance between b and c okay so let's kind of crank this out really quickly i'll just kind of set this up in each case real fast and then we'll go through and do it so for the first kind of situation we have we're looking at the distance between a and b i'll go ahead and say this is x sub 1 y sub 1 and i'll say that this is x sub 2 y sub 2. so if i just plug in x sub 2 is going to be 12 and x sub 1 is going to be negative 2 so this would be plus 2 right minus a negative is plus a positive for y sub 2 i'm going to have 3. for y sub 1 i'm going to have 5 okay so if i crank this out 12 plus 2 is 14 14 squared is 196. so you've got 196 here and then 3 minus 5 is negative 2 negative 2 squared is going to be positive 4. so 196 plus 4 is 200 the square root of 200 so we know that the square root of 200 is not a rational number but i could simplify this 200 is what it's 2 times 100 and i know that the square root of 100 is 10 so i can say this is 10 times the square root of 2. so i'll put that this is 10 times the square root of 2. okay let's move on to the distance between a and c so let's kind of slide this over and say that this is x sub 2 y sub 2. so for x sub 2 i'm going to plug in a 10. for x sub 1 i'm going to plug in a negative 2 so this would be plus 2. for y sub 2 i'm going to plug in a negative 11. for y sub 1 i'm going to plug in a 5. so 10 plus 2 is 12 12 squared is 144 so this is 144 and then negative 11 minus 5 is going to be negative 16 and negative 16 squared is 256. so this is plus 256 if i sum 144 and 256 i get 400 so the square root of 400 is 20. so this would be 20 here okay so now let's talk about the distance between b and c so i'm just going to kind of slide this over so for x sub 2 i'm going to have 10 for x sub 1 i'm going to have 12 and 10 minus 12 is negative 2 negative 2 squared is 4. let's go ahead and do that y sub 2 is going to be negative 11. so this is negative 11. y sub 1 is going to be 3. so negative 11 minus 3 is negative 14. if i square negative 14 i get 196. so 4 plus 196 again is 200 and we already know that the square root of 200 is going to simplify to 10 times the square root of 2. now let's go ahead and show that if we have a right triangle again we know that 20 is the biggest number here right so this would be the hypotenuse if this was a right triangle so it would have to be true that this guy squared so the 10 times the square root of 2 squared plus again your 10 times the square root of 2 squared would be equal to 20 squared so is that true let's kind of scroll down and get some room going so if i square 10 i get 100 if i square the square root of 2 i get 2. so this would be 100 times 2 which is 200. so you basically have 200 plus 200 which is 400 is that equal to 20 squared yes it is 20 squared is 400. so we can say that these three points we had negative 2 comma 5 12 comma 3 and 10 comma negative 11 are the vertices of a right triangle all right let's take a look at one more problem so we have a b and c here for our points a is four comma one b is two comma four and c is zero comma two so again what i want to do is find the distance between each kind of pair of points and to do that i've already kind of written it so let me just paste this in i'll put this over here just save us a little bit of time so our first distance will be the distance between a and b the second will be the distance between a and c and the third will be the distance between b and c so all i got to do is just kind of label things and then plug in so i'm going to label this one as x sub 1 y sub 1 to start i'll label this as x sub 2 y sub 2. so i'm going to start with the distance between a and b so for x sub 2 i can erase this and put in a2 for x sub 1 i can erase this and put in a 4. over here for y sub 2 i can put in a 4 for y sub 1 i can put in a 1 ok so 2 minus 4 is negative 2 negative 2 squared is going to be positive 4 and then 4 minus 1 is 3 3 squared is 9. so 4 plus 9 is 13. so this would end up being the square root of 13. so square root of 13. okay so now we're looking at the distance between a and c so let me erase this and i will put this as x sub 2 y sub 2. so for x sub 2 i've got 0 for x sub 1 i've got 4 and then for y sub 2 i've got 2 and for y sub 1 i've got 1. so 0 minus 4 is negative 4 negative 4 squared is 16. so this is 16 here and then 2 minus 1 is 1 1 squared is obviously 1. so 16 plus 1 is 17. so this would be the square root of 17 so square root of 17 and lastly we want the distance between b and c let me kind of erase this i'll do this as x sub 1 y sub 1. so for x sub 2 i've got 0 for x sub 1 i've got 2 for y sub 2 i've got 2 and for y sub 1 i've got 4 okay so we just crank this out 0 minus 2 is negative 2 negative 2 squared is 4. so this is 4 here 2 minus 4 is negative 2 negative 2 squared is again 4. so you've got 4 plus 4 which is 8 here so this would be the square root of 8. okay now i can simplify the square root of 8 because it's what it's the square root of 4 times square root of 2. i could write this as the square root of 4 times 2 square root of 4 is 2. so you could write this as 2 times square root of 2 okay just to make it simple all right so what's the largest kind of value that we have here that would be the hypotenuse if we had a right triangle well the square root of 13 is approximately 3.61 square root of 17 is approximately 4.12 and 2 times the square root of 2 is approximately 2.83 okay these are all irrational numbers so you can't get an exact value but if we approximate it it's it gives us a good idea of what would the largest one be obviously the square root of 17 would be the clear winner so this guy right here should be the hypotenuse if we have a right triangle okay so what i'd want to do because this guy right here and this guy right here those would be the legs right if we were working with the right triangle i would take the square root of 13 squared then i would add that to 2 times the square root of 2 squared and i would say does this equal the square root of 17 squared okay so let's scroll down and get a little bit of room going square root of 13 squared is obviously 13. then plus 2 times square root of 2 squared 2 squared is 4 square root of 2 squared is 2 4 times 2 is 8. so this would be 13 plus 8 which we know is 21. so this is 21. does that equal the square root of 17 squared no it doesn't all right square root of 17 squared is 17. and i forgot to put a 1 there so this is 21 and this is not going to equal 17 right we can put this as false so this tells us that basically these are not the vertices of a right triangle okay so these points that we had 4 comma 1 2 comma 4 and 0 comma 2 they're not the vertices or the end points or the corners of a right triangle in this lesson we want to discuss how to determine whether three points are collinear all right so in our last lesson we used our distance formula to determine if three points were the vertices or you could say the corners of a right triangle now we're going to look at another application of our distance formula and basically we're going to use our distance formula to determine if three points are collinear which just means they lie on the same line now for most of you that again took lower level algebra courses you know that there are many different methods we can use for this task this is probably one of the more tedious methods generally at this kind of level where we're talking about the coordinate plane you would use slopes for this process right it would be a little bit faster but again for the sake of completeness we'll use this method and then as we move throughout the course we'll look at collinearity with slopes and then a few other methods that are much more efficient so we're going to start off with this first example where we have point a which is 0 comma negative 4 point b which is 3 comma negative 2 and point c which is 6 comma 0. so we want to know if these three points are collinear meaning they lie on the same line so i'll just tell you for the first example they will be collinear we're going to look at a graph in a minute we'll see that i've kind of already sketched the line that goes through the three points but the idea here is that to find the distance between each pair of points we use our distance formula so with three points you're going to have three distances let me kind of paste this in here so you're going to have the first distance which is the distance between point a and point b the second distance which is the distance between point a and point c and the third distance which is the distance between point b and point c okay so the idea here is that the sum of the two smaller distances is going to be equal to the largest distance if these three points are collinear okay so visually we can see that that would be true okay so i've already sketched the line that goes through these three points again for reference sake 0 comma negative 4 this is going to be point a so this is point a 3 comma negative 2 this is going to be point b and then 6 comma 0 this is going to be point c okay all we're saying is that the largest distance here which we can see is from a to c okay that's the largest distance that should be made up of these two smaller distances sum together okay in this case that's going to be from a to b and then from b to c so if i sum those two amounts together i should get the largest one which is from a to c if these three points are collinear or lie on the same lock so let's go back and plug into our formula and we'll prove that this is the case okay so i'm just going to label point a as x sub 1 y sub 1. let me do this in a different color so we don't get confused so this is going to be x sub 1 y sub 1. i'm going to label point b as x sub 2 y sub 2. okay let me erase this we're just going to plug into the formula and we'll get our first distance so the distance between a and b for x sub 2 i'm going to plug in 1. i'm going to plug in a 3. for x sub 1 i'm going to plug in a 0. so 3 minus 0 is 3 3 squared is 9. so let's put a 9 there then plus for y sub 2 i'm going to have negative 2 and then for y sub 1 i'm going to have a negative 4. so minus a negative 4 is plus 4. so this is plus 4. negative 2 plus 4 is 2 2 squared is 4. so you have 9 plus 4 which is 13. so this is going to be the square root of 13. okay so that's the distance between point a and point b so let me go back to this right here let me kind of erase this and again this is a this is b and this is c so from point a to point b so basically from here to here this guy right here we've said that's square root of 13. well now we're looking at the distance between a and c so let me erase this and i'll kind of put it over here so this is my x sub 2 y sub 2 now so for x sub 2 i'm going to plug in what i'm going to plug in a 6. for x sub 1 i'm going to plug in a 0 and then for y sub 2 i'm going to plug in what i'm going to plug in a 0 and for y sub 1 i'm going to plug in a negative 4 minus a negative 4 is plus 4. so 6 minus 0 is 6 6 squared is 36 so this is 36 and then 0 plus 4 is 4 4 squared is 16. so if you have 36 plus 16 that's going to give you 52. now the square root of 52 can be simplified right 52 is 4 times 13. so this is 4 times 13 underneath my square root symbol square root of 4 is 2. so i can say this is 2 times the square root of 13. now let me go back we know that from a to c so this guy right here from a to c this is 2 times square root of 13. okay from a to b is square of 13 and from a to c is 2 times square root of 13. so these points will be collinear if this guy right here from b to c is square root of 13. let's go ahead and prove that's the case so this guy right here for the distance between b and c let me kind of erase this and let me write it in here we'll say this is x sub 1 y sub 1. so x sub 2 now if i'm plugging in is going to be 6. x sub 1 is going to be 3 and we'll do this in a second y sub 2 is going to be 0 and y sub 1 is going to be negative 2 minus the negative 2 is plus 2. so 6 minus 3 over here is 3 3 squared is 9. so this is 9 and then 0 plus 2 is 2 2 squared is 4. so again 9 plus 4 is 13 so we do get square root of 13 here okay so you can see that the sum of the two smaller distances from a to b and from b to c square to 13 plus square root of 13 does give me the largest kind of distance right so 2 times square root of 13 which is the distance between a and c right so you can show this square root of 13 plus square root of 13 is equal to 2 times square root of 13. right this over here on the left would become 2 times square root of 13. so this is a true statement so we can say these three points are collinear or rely on the same line right so if we go back again we're looking at the distance between b and c so from here to here and we found that this was square root of 13 as well so again just to kind of wrap this up we have the distance from a to b so let me just kind of write this out so the distance from a to b so between those two points is again the square root of 13. then we also have the distance from b to c so let me write that out so the distance from b to c that's also the square root of 13. and again what we're saying here is that because these three points are collinear when i sum these two smaller distances together what i get is the largest distance okay which is from a right here to c right here which again is 2 times square root of 13. so the distance between again a and c is going to be 2 times the square root of 13. so this is always going to be true when we have three points that lie on the same line where you could say when they are collinear let's go ahead and take a look at another one very easy concept just something that's very very tedious so we have our first point a which is negative two comma zero where our second point b which is two comma seven and a third point c which is negative one comma four so let me kind of paste this in and to start i'm just gonna label this one as x sub one y sub one and this one x sub two y sub two and let's just plug into the formula so this guy right here x sub 2 is going to be 2 and x sub 1 is going to be negative 2 minus negative 2 is plus 2. and then for y sub 2 i'm going to have 7. for y sub 1 i'm going to have 0. so 2 plus 2 is 4 4 squared is 16. 7 minus 0 is 7 7 squared is 49. so what is 16 plus 49 that's going to be 65 so we'd have square root of 65. all right then for the distance between a and c let me erase this and i'll put this one over here let me kind of make this a little bit better so i'll say that this guy over here is going to be x sub 2 y sub 2 now so for x sub 2 we're going to plug in a negative 1. for x sub 1 we're going to plug in a negative 2 minus a negative 2 is plus 2 and then for y sub 2 we're going to plug in a 4. for y sub 1 we're going to plug in a 0. okay so negative 1 plus 2 is 1 so that's 1 squared which would be just 1. 4 minus 0 is 4 4 squared is 16. okay so 1 plus 16 is going to give me 17 so this is the square root of 17 here so square root of 17. so for my third distance the distance between b and c let me kind of slide this down real quick so b will now be x sub 1 y sub 1. so let me erase this x sub 2 will be negative 1 then x sub 1 will be 2 negative 1 minus 2 would be negative 3 negative 3 squared is 9. so let's just put that there y sub 2 is 4 and then y sub 1 is going to be 7 okay so 4 minus 7 is going to be negative 3 negative 3 squared is not okay so you have 9 plus 9 which is 18. now i can simplify them 18 is what it's 9 times 2 and 9 is a perfect square 9 is 3 times 3. so basically square root of 9 is 3 so i can write this as 3 times square root of 2. so 3 times square root of 2. so which one of these guys is the largest well square root of 65 is going to be the largest again this is an irrational number but if you approximate it it's about 8.06 for square root of 17 if you approximate that it's about 4.12 and then for 3 times square to 2 if you approximate that it's about 4.24 so this guy right here should be the largest distance right from a to b so basically we would see that the square root of 17 plus 3 times the square root of 2 would be equal to the square root of 65 and that's going to be false right this is not going to work itself out if you want to use an approximation for each we could say that again the square root of 65 is about okay about eight point okay and then the square root of 17 is about 4.12 and then 3 times the square root of 2 is about 4.24 okay just generally speaking approximately speaking are these equal well 4.12 plus 4.24 is 8.36 right so these aren't going to be equal okay so we can just erase this we don't need this okay even though we used approximations we can clearly see that these aren't going to be equal this is false so these three points are not going to be collinear all right let's just take a look at one more of these so we have point a which is negative four comma three point b which is two comma five and point c which is negative one comma 4. again are they co-linear so let's paste in our distance formulas and i'm just going to start out with the first distance between a and b again i'm going to label this guy as x sub 1 y sub 1. i'll label this as x sub 2 y sub 2. so x sub 2 is 2. let me kind of write that in and then x sub 1 is negative 4 minus a negative 4 is plus 4. y sub 2 is 5 and then y sub 1 is 3 okay so this is 3. all right so 2 plus 4 is 6 6 squared is 36 so this is 36 and then 5 minus 3 is 2 2 squared is 4. so 36 plus 4 is going to give me 40. so this is the square root of 40. now so we can simplify this because 40 is 4 times 10 we know the square root of 4 is 2. so i can write the answer here as 2 times square root of 10. all right let's kind of slide this down here and look at the distance between a and c and let me kind of move this over so it's not in the way so 2 times square root of 10. okay so for x sub 2 we have negative 1. for x sub 1 we have negative 4 so minus a negative 4 plus 4. for y sub 2 is 4. then for y sub 1 it's 3 okay so we have negative 1 plus 4 which is 3 3 squared is 9. then we have 4 minus 3 which is 1 1 squared is 1. so you have 9 plus 1 which is 10 so this is square root of 10. so this is square root of 10. okay so now let's look at b and c so let's move this down so for x sub 2 i'm going to plug in a negative 1. for x sub 1 i'm going to plug in a 2. for y sub 2 i'm going to plug in a 4 and then for y sub 1 i'm going to plug in a 5 okay so negative 1 minus 2 is negative 3 negative 3 squared is 9. okay so this is 9. 4 minus 5 is negative 1 negative 1 squared is 1. so this is 9 plus 1 or 10. so this is going to be the square root of 10. so let me make that a little cleaner and just say this is the square root of 10. so you can already see this is going to work out right because 2 times square root of 10 is the largest distance here and the two smaller distances are going to sum to that right so you can kind of say that the square root of 10 which is the distance between a and c plus the square root of 10 which is the distance between b and c is going to be equal to kind of the largest distance which is the distance between a and b which is if i can fit this on the screen here let me kind of move this down again which is 2 times square root of 10. okay so this does work itself out you end up saying that the left side here is 2 times square root of 10 and the right side here is 2 times square root of 10. so this is going to be true right these three points a b and c a being negative 4 comma 3 b being 2 comma 5 and c being negative 1 comma 4 are collinear or they lie on the same line in this lesson we want to discuss the midpoint formula all right so over the course of the last few lessons we've been talking about the coordinate plane and various related topics so we already talked about how to find the distance between two points we talked about finding if three points of the vertices of a right triangle and we talked about how to determine if three points were collinear again meaning they lie on the same line so now we want to talk about how to find the midpoint or the middle of a line segment so first and foremost let's clarify what we mean by a line segment so a line on our coordinate plane extends indefinitely in both directions so you'll see on the coordinate plane here i've graphed a sample lot and most of you work with lines in your algebra 1 or intermediate algebra course but in case you haven't we'll get to lines in a few lessons for now we just need to know that again lines continue in each direction forever that's why we have an arrow at each end so this guy right here and this guy right here again that's just telling me that the line continues indefinitely in each direction so because of that fact we're not going to be able to find the middle or the midpoint of a line but what we can do is we can find the middle or the midpoint of a line segment okay so a line segment is just a piece of a line it's going to have two endpoints and so because it has a defined length we'll be able to find the middle of it or again the midpoint okay so the midpoint is just the point that is equidistant which means it has the same distance okay from the endpoints of our line segment so again just to be clear what this means is that the midpoint will cut the line segment completely in half so from the midpoint to one end point is going to be the same distance as from the midpoint to the other end point so if we look at our graph here you're going to see that we have three plotted points so let me erase this and you'll see that we have a plotted point here and here and here okay so for reference sake let's label these using a b and c so i'm going to take this point right here i'm going to call this point a and this is going to be at negative 5 comma negative 6. so point a is at negative 5 comma negative 6. and then i'm going to call this guy right here this is going to be b let me write this down here just so it's a little cleaner so we'll put b down there so for point b this is going to be at 0 comma negative 3. and then this guy right here i'm going to call this c so c is going to be at 5 comma 0. so what we're going to see here is that b is going to be the midpoint for this line segment that's made up of from a to c so from right here to right here this line segment if i kind of draw that in what's going to happen is b is going to be the midpoint so right here if i just kind of cut this down like that you'll see that the distance from a to b and the distance from b to c that's going to be the exact same distance okay so how can we find these coordinates in general right the coordinates of the midpoint well let's go down and we'll come back up all right so i just want to show you this on a clean graph with no numbers involved basically we're going to derive our formula and you're going to see once you have the formula this is very very easy so suppose we have two known points so we have two known points and this is usually the scenario where you're trying to find the midpoint you know two of the points so let's call them x sub 1 y sub 1 and we're going to say x sub 2 y sub 2. okay so again these points are known then what you don't know is the midpoint so the coordinates for that for the midpoint we're just going to say this is x comma y okay so you see i have this labeled on the graph so this is your x sub 2 y sub 2 this is your x sub 1 y sub 1. so these are the endpoints of our line segments and this guy right here this x comma y this is your midpoint okay so again think about what this means if i draw like a little vertical line here you could say that the distance from here to here is the same as the distance from here to here right so it cuts it in half or it's the middle now how can we mathematically determine what the midpoint is well if i think about how to find the x value here for the midpoint just think about this mathematical fact if i kind of come up here and say this x value here to this x value here so the x value for x sub 1 and the x value for x the distance there would be the same as the distance from this x to kind of this x value here for x sub 2 okay so those distances would be equal so mathematically we can set up an equation that says that x sub 2 minus x is the same as or is equal to x minus x sub 1 because again those distances are equal from here to here is going to be equal to or the same as from here to here okay so let's just go ahead and solve this equation and you'll see that we derive the x value for the midpoint okay so x sub 2 minus x equals x minus x sub 1. so what's unknown here is x so we want to solve for that so the first thing we want to do is get all the x's to one side and i know it's confusing here because you have a lot of different notation but basically the x sub 2 and the x sub 1 that needs to be a one side and all the kind of x's need to be on the other side so i'm going to subtract x sub 2 away from each side of the equation and what's going to happen is this would cancel and i'm also going to subtract x away from each side of the equation so that's going to cancel all right so what we're going to have here is that you have negative x minus x which is negative 2x this will be equal to you're going to have negative x sub 1 minus x sub 2. okay so how can we solve this well the first thing i'm going to do is i'm going to factor out a negative from the right hand side so i'm going to write this as negative 1 times the quantity x sub 1 plus x sub 2 okay so all i did was i factored out a negative 1. if i went back through and said negative 1 times x sub 1 i'd have my negative x sub 1 and if i did negative 1 times x sub 2 and i have my negative x sub 2. okay so i didn't do anything illegal then over here i have my negative 2x and now i just want to solve for what's unknown so i want to solve for this x right here and the way i'm going to do that since i'm multiplying by negative 2 i'm just going to divide by negative 2. okay i'm going to do that to both sides we kind of make that better so what's going to happen is this is going to cancel and i'm going to just have x on the left hand side and this will be equal to on the right hand side you can cancel this negative 1 with this negative here and you're going to have that x sub 1 plus x sub 2 over 2. so this is how you find the x value or your x coordinate for the midpoint so we're just basically finding the average of the two x values if we go back and we look at these two endpoints so you have a and you have c so a has an x value of negative 5 c has an x value of 5. so if i add those two together and i divide by 2 negative 5 plus 5 is 0 0 over 2 is 0. so look at the x coordinate for the midpoint it's 0 right we can see that right there so that's how that was found so let me erase this let's talk about the y value so for the y value it's going to be the same concept so what's going to happen is if i kind of draw a line going over here so this guy right here this y value this is my y sub 2. this y value right here this is my y and this y value right here this is my y sub 1. so this guy is known and this guy is known and again we know that the distance between here and here and the distance between here and here are going to be the same okay so i can do basically the same thing and i can say that y sub 2 minus y is equal to y minus y sub 1 okay and when we solve this we're going to see the same thing right we basically want to take the average of the y values so let's subtract y sub 2 away from each side of the equation and let's subtract y away from each side of the equation so we know that this is going to cancel we know that this is going to cancel negative y minus y is negative 2y this equals you're going to have negative y sub 1 minus y sub 2. again i'm going to do the same thing just going to factor a negative 1 out from the right side so this would be negative 1 times the quantity y sub 1 plus y sub 2. over here i'll have my negative 2y and when i solve this i just divide both sides of the equation by negative 2. and so this is going to cancel with this i'll have y is equal to the negative 1 cancels with the negative here you have your y sub 1 plus your y sub 2 over 2. all right so again this is just the average of the y values so if we go back up so if i wanted to find the y coordinate this negative 3 here in the midpoint i would take my negative 6 the y coordinate of a one of the endpoints then add to that my zero the y coordinate from c another endpoint and i would just divide by two okay so negative six plus zero is negative 6 negative 6 divided by 2 is negative 3 and so that's how you get this negative 3 right there okay so a very very easy formula to kind of come up with we would say that capital m so capital letter m not lowercase m that's used for slope is equal to we'll have x sub 1 plus x sub 2 divided by 2 then comma you have y sub 1 plus y sub 2 divided by 2. okay so this is how you can get your kind of coordinates for the midpoint very very easy you're just taking the average of the x values that's how you get your x coordinate and then you're taking the average of the y values that's how you get your y coordinate okay so very very simple now if we really wanted to prove this we could calculate the distance between a and b and we could calculate the distance between b and c and again if we've correctly found this midpoint as b then the distances will be the same okay so i'm not going to do that in the interest of time i'm just going to tell you that the distance between a and b is the square root of 34 and the distance between b and c is also the square root of 34. okay so this is a way you can check this the distances are equal so we know we've found the correct midpoint all right let's run through a few examples real quick so suppose you had 2 comma negative 6 and 8 comma negative 1 and you wanted to find the midpoint well again all you need to do is label one of the points as x sub 1 y sub 1 and the other is x sub 2 y sub 2 and all we're going to do is plug into our formula so capital m okay capital m is equal to you've got x sub 1 plus x sub 2 over 2 and then comma you've got y sub 1 plus y sub 2 over 2 okay so if i plug into this guy x sub 1 is 2 so m will be equal to you're going to have a 2 plus your x sub 2 is going to be 8 okay and we're going to divide by 2. and then for the y you're going to have y sub 1 which is going to be negative 6 and then plus your y sub 2 is going to be negative 1 so you might as well do just negative 6 minus 1. okay and this is going to be over 2. so 2 plus 8 is 10 10 divided by 2 is 5. so this would be 5 and then negative 6 minus 1 is going to be negative 7 negative 7 divided by 2. you can just write negative 7 halves if you want so you can say this is negative 7 halves like this or you can say negative 3.5 if you want a decimal okay so this is going to be the midpoint 5 comma negative 7 halves of these two kind of points this line segment that they form so from 2 comma negative 6 to 8 comma negative 1. all right let's take a look at another one so we have negative 4 comma negative 13 and 5 comma 9. so again we want to find the midpoint so the midpoint formula capital m is equal to you've got x sub 1 plus x sub 2 over 2. let me make that better and then comma you've got y sub 1 plus y sub 2 over 2. okay so if i label this one just kind of change things up as x sub 2 y sub 2 and this one as x sub 1 y sub 1 and i just plug into my formula so m capital m or midpoint will be equal to you're going to have your x sub 1 which is 5 plus your x sub 2 which is going to be negative 4 so 5 plus negative 4 would be 5 minus 4 which is basically just 1 okay then over 2 so that's one half then for the y coordinate you've got y sub 1 which is 9 plus y sub 2 which is negative 13. so you might as well say you have 9 minus 13 so 9 minus 13 which is negative 4. so you're going to have negative 4 then over 2 so over 2 and negative 4 over 2 is going to be negative 2. so my midpoint is going to be at one half comma negative 2. all right let's take a look at one more of these and then i'll give you a different type of problem so we have 11 comma 7 and 9 comma 2. so again our midpoint i'm going to put capital letter m is equal to you've got your x sub 1 plus your x sub 2 over 2 comma you've got your y sub 1 plus your y sub 2 over 2. okay so let's label this as x sub 1 y sub 1. let's label this as x sub 2 y sub 2. again just plug into the formula very very easy you've got capital m for midpoint is equal to for your x you've got x sub 1 which is 11 and x sub 2 which is 9. so 11 plus 9 is 20. so you'd have 20 over 2 which is just 10. okay so this is 10 then comma for the y value you've got y sub 1 which is 7 and y sub 2 which is 2 so you've got 7 plus 2 which is 9 over 2. so the midpoint here is going to occur at 10 comma 9 halves and if you wanted a decimal form for 9 halves you could just do 4.5 all right for the last problem we're going to look at we're going to know the midpoint and we're going to know one end point but one of the endpoints is going to be unknown and that's what we're going to try to figure out so our midpoint is going to occur at 5 comma 1 and then the known endpoint i'm just going to label this as x sub 1 y sub 1. this guy's going to occur at negative 6 comma negative 3. and the unknown endpoint we just have x sub 2 y sub 2 as kind of a stand in okay we don't know what it is yet so again if i plug into the midpoint formula capital m is equal to you have your x sub 1 plus your x sub 2 over 2. let me make that 2 a little better and then comma you have your y sub 1 plus your y sub 2 over 2. now let's think about this for a second we know that the x value here for the midpoint is 5. so i can take this guy right here this x sub 1 plus x sub 2 over 2 and i can set it equal to 5 right because if i knew what x sub 2 was i know what x sub 1 is but if i knew what x sub 2 is i could plug in here and here and i could evaluate and i should get a 5. okay so let's use this to kind of get our unknown value so i know that x sub 1 is negative 6 so i know this is negative 6. so let's just solve this equation let's multiply both sides by 2. let me kind of scroll down and get some room going we'll come back up so i'll multiply this by 2 let me kind of scooch this over just a little bit i'm going to multiply this side by 2. we know that this would cancel with this so i would have negative 6 plus x sub 2 is equal to 10 and to solve for x sub 2 i just got to add 6 to both sides of the equation and so what i'm going to have is that x sub 2 is equal to 16. okay so let me erase all this so again my x sub 2 i'm just going to put a 16 right there okay and you can verify that's the case because 16 plus negative 6 is going to give me 10 10 divided by 2 would give me this value here of 5 okay so that does work itself out now if i want y sub 2 i'm going to use a similar procedure okay so i know y sub 1 so i'm going to say that negative 3 that's y sub 1 plus my unknown which is y sub 2 divided by 2 would be equal to what it would be equal to 1. so if i multiply both sides by 2 let me kind of scooch this over just a little bit again if i multiply both sides by 2 so multiply this side by 2 and this side by 2 well this is going to cancel with this i'm going to have negative 3 plus y sub 2 is equal to 2. let me kind of scroll down just a little bit let me add 3 to both sides of the equation so that i can isolate y sub 2 that will cancel you'll have y sub 2 is equal to 5. okay so very very easy let me erase this and we said that y sub 2 was 5 okay you can verify that negative 3 plus 5 is going to give you 2 2 divided by 2 is going to give you 1. so we know that this is correct right our unknown endpoint here our x sub 2 y sub 2 is going to be 16 comma 5. okay that's what we're trying to find in this lesson we want to talk about plotting complex numbers on the complex plane so at this point in our course we already know and understand how to use the imaginary unit i again is defined as the square root of negative 1 or you could say i squared is equal to negative 1. so we found that this came in handy when we're solving quadratic equations right in some situations you won't be able to get a real solution and so you need to turn to the imaginary unit i to gain a solution now we have also talked about complex numbers and complex numbers are of the form a plus bi so we've seen these before we've performed operations with them before and we should know that a this part right here is known as the real part of the complex number and b this part right here that's multiplying i the imaginary unit is known as the imaginary part of the complex number now some books will say bi is the imaginary part but for the purposes of this lesson we're just going to stick to saying that b is the imaginary part now how can we plot a complex number such as a plus bi on the complex plane well let's go down to our complex plane and suppose we wanted to plot something like 5 plus 3i well i'm just going to kind of match this up real quick i'm going to say this is a plus bi so my a here is 5 and my b here is 3 okay so keep that in mind now the complex plane if you're looking at it you might say well that looks just like the coordinate plane i'm used to working with when we see real numbers well it does it looks exactly the same the difference is is that instead of the horizontal axis being labeled with x now it says real instead of the vertical axis being labeled with y now it says imaginary so it's very easy to plot 5 plus 3i again the a or the real part is 5. so on the real axis i'm just looking for 5. so that's going to be right here so i'm just going to circle that then i'm looking for 3 on the imaginary axis because the b here or the imaginary part is 3. so on the imaginary axis i'm just going up 3 so that's right here and i'm just finding the meeting point between the two so this is just like if i had 5 comma 3 in the real number system it's the same process okay so i'm going to go 5 units to the right on the real axis and 3 units up on the imaginary axis so i'm going to be right here this is going to be my 5 plus 3 i okay it's just that simple all right let's go ahead and try another one so again i'm going to write this in the format i'm going to say this is a plus b i okay now in this case my a is negative 2 so that means on the real axis i want to go 2 units to the left so that's going to be right here and let me use a different color because that doesn't show up very well so let me use kind of a reddish or purple i'm not sure what you call that color maybe it's magenta and then for my b i'm thinking about the fact that this is negative so really if i want to stay consistent with this i'm going to write this as plus negative 4i okay so my b here is a negative 4 so that means i want to start on the imaginary axis at 0 and i want to drop down 4 units so that's going to be right there so this is just like if you had negative 2 comma negative 4 if you were working in the real number system so i'm just going to go two units to the left on the real axis and i'm going to go four units down on the imaginary axis and that's going to be right there so this is going to be my negative 2 and then i'll put minus 4i again you could put plus negative 4i doesn't match the same thing right what about 8 minus 6i so again let's just match this up this is my a and again i'm going to put plus negative here this is my b okay so i want a real location of 8 so i'm going to go 8 units to the right on my real axis so that's here and then i want a negative 6 for my imaginary axis so i'm just going to go down 6 units so that's going to be right there so i go 8 units to the right on the real axis and 6 units down on the imaginary axis so that's right there and this is going to be 8 minus 6i or again 8 plus negative 6i all right so let's look at another one so we have negative 5 minus 7i so this guy right here is my a and again i'm going to put plus negative here so that's going to be my b so again on the real axis i'm going to negative 5 so that's going to be right here on my imaginary axis i'm going to negative 7 so i'm going to drop 7 units that's here so again 5 units to the left on the real axis 7 units down on the imaginary axis that's going to be right there and let me label this below it so i'm going to put negative 5 and i'm going to do minus 7 i so that's that point right there okay so what if they throw a curve ball at you this is one that kind of trips people up a lot let's say you're working on the complex plane and someone says to plot negative eight what do you do well some of you will immediately say okay well i know that negative eight is a real number so i'm probably just going to go to negative 8 on the real axis and i'm just going to put a point there and you'd be correct but another way you can think about this is you can still follow the format of a plus bi okay because a here is negative 8 but i can say plus zero i okay and so my b here is just zero so if i go eight units to the left on the real axis and zero units up on the imaginary axis i am at this negative eight on the real axis okay so this right here would just be negative 8 or you put negative 8 plus 0i if you wanted to write it in the complex number form what about something just like 9i well again we can use the same technique we can write this as a plus bi by just saying that this is my b here so i can put a plus here and i can just put a 0 here okay so my a here is just 0. so that means on the real axis i'm not going to move at all i'm not going left or right at all okay all i'm going to do is i'm going to travel 9 units up on my imaginary axis so i'm just going to go up 9 units so i'm just going to place my circle there and this will be 0 plus 9 i in this lesson we want to talk about finding the absolute value of a complex number so at this point in our course we should fully understand the concept of taking the absolute value of some real number okay we should know that the absolute value of a number is just a measure of the distance from zero to that number that we're taking the absolute value of on our number line okay so we know that if we take the absolute value of zero it's just zero if we take the absolute value of some positive number it's just the number itself if we take the absolute value of a negative number we just take the opposite of the number you can say make it positive and so we can see that very clearly with these two examples right so something like the absolute value of positive five is just equal to five right if i look at my kind of sample number line here i have zero highlighted and five highlighted so basically five is five units away from zero the number line this is true whenever you take the absolute value of any positive number it could be a million it could be a trillion whatever it is it's always going to be that number of units away from zero on the number line okay when we think about the absolute value of a negative number something like the absolute value of negative five we said that we just take the opposite of the number or again we just make it positive okay so in this case you see negative five is highlighted zero is highlighted so negative five is also five units away from zero on the number line again this is true for any negative number that you work with so the absolute value of negative one million would be one million right the absolute value of any negative number is just going to be the opposite of that number units away from zero on the number line so again that's how we find our answer we just take the opposite or again make it positive so now when we talk about taking the absolute value of a complex number the concept is the same okay the absolute value of a complex number is also a measure of its distance from zero the only difference is now we're going to be measuring the distance on the complex plane okay so this is something we talked about in the last lesson we talked about how to plot a complex number on the complex plane so hopefully you saw that lesson in case you didn't i could just kind of walk you through this really quickly basically with a complex plane it looks the same as the kind of coordinate plane that you've worked with throughout all of algebra right instead of having a y-axis the vertical axis that you're used to now this guy is labeled as the imaginary axis okay then instead of having the kind of horizontal axis or the x-axis that you're used to the horizontal axis is now labeled as the real axis okay so you've just got to know that the horizontal axis is the real axis and the vertical axis is the imaginary axis so once you understand those two things once you commit that to memory you should recall that a complex number flows in the format of a plus bi when it's in standard form okay so this number here this a is the real part in this case this is 3 okay this number b the number that's multiplying i the imaginary unit that's the imaginary part so that in this case is going to be 4 okay that's what's multiplying i let me make that a little bit better so to plot 3 plus 4i on the complex plane it's just like plotting 3 comma 4 on the coordinate plane we're used to working with i basically want to think about a real location of 3 so that's here and i want to think about an imaginary location of 4 so go up 4 units that's here what's the meeting point between those two so 3 inches to the right on the real axis and 4 units up on the imaginary axis so this is 3 plus 4 i okay so that's the point 3 plus 4i now when we think about getting the absolute value of 3 plus 4i so the absolute value of 3 plus 4i what is this well it's the distance from this 3 plus four i this point here to this origin here which has a real location of zero and an imaginary location of zero so you can say really it's zero plus zero i for the origin so we already know how to find the distance between two points on a coordinate plane we know that we can use our pythagorean formula to do this and we know we've already derived a distance formula right to find this but i want to go through this process again and we're going to derive a formula for finding the absolute value of any complex number let me just kind of draw a line connecting these and again we're just thinking about the measure of this line here the line that's going to connect those two points okay so we could find this by again dropping a point here where we're at three on the real axis or we could also drop a point at four on the imaginary axis either one of those could be used to complete our right triangle so we went ahead and just drew the right triangle already it's just a little bit cleaner when it's pre-drawn so you'll notice that we dropped a point in here at 3 on the real axis again you could have also dropped a point in at 4 on the imaginary axis you could have completed a right triangle in that manner as well so once we have this drawn we can kind of think about the concept using the pythagorean formula so we remember this is a squared plus b squared equals c squared where a and b are the lengths of the two kind of shorter sides we call them legs and then the c is the hypotenuse the longest side okay so this is the hypotenuse and that's what we want to figure out this is c and again this is going to represent what the distance from 3 plus 4i to 0 plus 0i or the origin so it's the absolute value of 3 plus 4i that's what we want to find so i'm going to go ahead and just say this is the absolute value of 3 plus 4i and i'm going to square that so what we need to figure out is a and b so we know that we could label this as a and this is b okay so a is just the measure from here to here now we can do this inside of absolute value bars but again because we're squaring the result it's really not necessary you're really thinking about what it's the difference between the imaginary values so at the top you're at 4 right this is a 4 here and at the bottom you're at 0. so 4 minus 0 is going to be 4 or again you could do 0 minus 4 which is negative 4 either way when you square you're going to get 16. right so i'm just going to go ahead and say this is 4 squared then plus you want to also think about b which is going to be this horizontal measure here so the difference between 3 and 0 right you have a real location here of 3 and a real location here of 0. so you could do 3 minus 0 which is 3 or you could do 0 minus 3 which is negative 3. again you don't need absolute value because you're going to be squaring it 3 squared is 9 negative 3 squared is 9 is the same either way right so i'm just going to put 3 squared here okay so we find that the absolute value of 3 plus 4i which is what we're trying to find squared is equal to 4 squared plus 3 squared okay so what i want to do is just take the square root of each side so take the square root of that side that's going to cancel this and take square root of this side and so 4 squared is 16 3 squared is 9 16 plus 9 is 25 here the principal square root of 25 is equal to the absolute value of 3 plus 4i and then we end up with 5 is equal to the absolute value of 3 plus 4i okay so very very easy to go through and figure that out but it was kind of a lengthy process so we don't want to have to again pull out a complex plane each time and go through this to figure out the absolute value of a complex number you've got to be thinking like there must be an easier way and in fact there is so just like we derived a distance formula okay when we worked with finding the distance between two points on a coordinate plane we can do the same thing for finding the absolute value of any complex number so let me erase everything and first i just want to rewrite kind of the distance formula that we know so d the distance between two points is equal to the square root of you have the difference in x values so x sub 2 minus x sub 1 squared then plus you have the difference in y values so y sub 2 minus y sub 1 squared okay so this is the distance formula that we know so d in this case would just take the place of the absolute value of some complex number okay so this would be a plus bi okay in this case it was just 3 plus 4i but it could be any complex number that you're trying to find the absolute value of so this is equal to one the square root of so when we think about the difference in x values well now we don't have x values we have real values okay so because i'm always working with this guy as a point zero plus zero i one of these guys is going to be zero okay it's easier to kind of see what's going on if i just put this as zero so if you have x sub two minus zero or in this case you could say your real value that you're given which is one of them is going to be a and one of them is going to be 0. so really i could just say it's a minus 0 squared okay you can easily see that in this case i can put 3 and say 3 minus 0 squared or again i can reverse that and say 0 minus 3 squared either way i'm going to get the same result because i'm squaring it so in either case this is going to simplify to just a squared okay so a squared so this value here gets squared then plus for y sub 2 minus y sub 1 again i don't have y values anymore i have imaginary values so now i'm thinking about okay well the difference here because i have zero as a location of one of the imaginary values i can erase this and just put zero okay i could just put 0. so again if i think about that my y sub 2 one of those guys when i think about this in terms of the imaginary values is just going to be b right i would have b as one of them then minus 0 as the other and this is squared this again simplifies to just b squared so it's very easy to find the absolute value of a complex number the absolute value of a plus bi is just going to be the square root of a squared plus b squared and we found that already we found that the absolute value of 3 plus 4i was just the square root of 3 squared plus 4 squared okay let me make that better okay so 3 squared is again 9 4 squared is 16. it was the absolute value of 25 which is 16 plus 9 which is 5. okay very very easy once you get this formula but i just wanted you to see where it came from a lot of times you know you're working through things you just get the formula you don't understand where it came from so it's kind of worth it to spend the you know 10 or 15 minutes to understand where this comes from that way if you're getting a pinch and you're asked to kind of you know explain where it comes from you're able to do it okay so i have a cleaner version where i've represented this again the absolute value of a plus bi is just equal to the square root of a squared so the real part squared plus b squared okay the imaginary part squared all right let's just go through a few examples real quick very easy concept overall so again let me just write the formula here the absolute value of a plus bi is equal to the square root of a squared plus b squared okay so this guy right here would be equal to one this is my a this is my b so it's the square root of a is negative 5 negative 5 squared is 25 plus 7 is my b 7 squared is 49. so what's 25 plus 49 well that's going to be 74. so this is equal to the square root of 74. now 74 is divisible by 2 but it's 37 times 2 and 37 is a prime number so we can't really simplify this any further so we just write the square root of 74. all right let's take a look at a number again let me write the formula the absolute value of a plus bi is equal to the square root of a squared plus b squared okay so in this case i have the absolute value of negative 16 plus 10i this is equal to the square root of a is negative 16. so if i square that i would get positive 256 then plus b here is 10 10 squared is 100 so this would be the square root of 356. so 356 is 4 times 89 so 4 times 89 so we know 4 is a perfect square so i could say this is 2 times the square root of 89 right 2 times square root of 89 and 89 is going to be a prime number so we can't do anything more with that all right so what about the absolute value of negative 3 minus 29 i so again let me write my formula it's the absolute value of a plus bi is equal to the square root of a squared plus b squared so again this is equal to a here is negative 3 negative 3 squared is 9. so the square root of 9 then plus b here is negative 29. you could write this as plus negative so negative 29 squared is 841. so if i sum those two i'm going to get 850 right under the square root sign so 850 is 25 times 34. it's 25 times 34 we know 25 is a perfect square square root of 25 is 5. so i can say this is 5 5 times square root of 34 and we know 34 is just 17 times 2 so you can't really do anything else with that all right for the last one i gave you one that students kind of struggle with what if i ask you for the absolute value of 9i well just so happens that you can still use the formula if you got the absolute value of 9i again the absolute value of a plus bi is equal to the square root of again a squared plus b squared so in this case a is just 0. right you can say the absolute value of 0 plus 9i this is the square root of 0 squared is just 0. so forget about it you would just take your b which is 9 and square that so that's 81 square root of 81 is just 9. okay why does that make sense well if you think about this graphically if i look at 9i on my complex plane that's right here okay so how far away is it from here to here well it's just nine units right if i made a line here kind of connecting these two points you would say that's 9 units right if i rotated that and you saw it as kind of a horizontal number line right if i was measuring from here to here you could easily see that that distance is 9 units it's the same thing if we're going vertical right so that should make sense to you that the absolute value of 9i is going to be positive 9. in this lesson we want to talk about the distance and midpoint formulas in the complex plane so over the course of the last two lessons we've been working with the complex plane we've already seen how to plot complex numbers and also how to find the absolute value of a complex number so now we're going to go to the kind of next topic and we're going to talk about how to find the distance between two complex numbers on the complex plane and then later on we'll talk about the midpoint formula on our complex plane so i want you to recall that we have a distance formula that we've been using throughout you could say lower level algebra or whatever algebra course you're in you should be familiar with this kind of distance formula again we talked about in this course it's derived from the pythagorean formula right so the d stands for distance between two points this is equal to the square root of you have your difference in x value so x sub 2 minus x sub 1 and then this is squared then plus you have your difference in y values so y sub 2 minus y sub 1 and this is squared okay so if we had something like 8 comma 3 and 4 comma 1 we could just label one of these as x sub 1 y sub 1 it doesn't matter which one gets labeled as which and the other as x sub 2 y sub 2. okay just plug into the formula so for x sub 2 you're going to have a 4 for x sub 1 you're going to have an 8. for y sub 2 you're going to have a 1 and for y sub 1 you're going to have a 3. okay so if we just kind of crank this out real quick the distance between these two points you would basically have what the square root of you would have 4 minus 8 which is negative 4 negative 4 squared is 16 then plus you'd have 1 minus 3 that's going to be negative 2. if i square negative 2 i would get 4 right so 16 plus 4 is going to give me 20. so this would be that the distance between these two points is the square root of 20. now we know that 20 is 4 times 5 and 4 is a perfect square so i could say this is 2 times the square root of 5 right if i wanted to simplify that all right so let's take a look at an example using complex numbers now you're going to see that the concept is exactly the same okay so we're going to do it using this formula but we're going to have some modifications we just want to remember a few things when we work with our complex plane we don't have x values anymore the horizontal axis is now labeled as the real axis and then similarly we're not looking at y values anymore the vertical axis is now labeled as the imaginary axis so we have imaginary values that we're working with okay so when we think about a complex number we have to think about it as a the real part plus b the imaginary part times i the imaginary unit okay so this right here is the real part this is the real part and this right here is the imaginary part so this is the imaginary okay the imaginary part all right so when we look at this kind of formula that's been modified the distance between two complex numbers is equal to the square root of you see a sub 2 minus a sub 1 squared then plus you see b sub 2 minus b sub 1 squared so this guy right here would be the difference in a values or the difference in real values okay and we have the a sub 2 and the a sub 1 that's just notating two different a values similarly we have the difference in imaginary values and again we're using b sub 2 and b sub 1 to kind of notate the two different ones so if i look at these two complex numbers i have z equals 3 plus 9 i and i have w equals 7 plus 5i so this is my a and this is my b okay this is my a and this is my b okay so i can label one of them as a sub 1 b sub 1 and the other as a sub 2 b sub 2. okay it doesn't matter which one gets labeled as which because we're squaring this guy so whether it turns out being positive or negative when you square it it's going to be non-negative okay so let's just say this is a sub 1 and this is b sub 1 and this is a sub 2 and this is b sub 2. so now we can just plug into the formula okay just like we did before and so for a sub 2 i'm going to have 7. for a sub 1 i'm going to have 3. for b sub 2 i'm going to have 5 and for b sub 1 i'm going to have 9 okay so the distance between these two points would be what the square root of you have 7 minus 3 which is 4 4 squared is 16. then plus you have 5 minus 9 which is negative 4 negative 4 squared is also 16. so 16 plus 16 is 32 okay 32 but we know that 32 factors into 16 times 2 because we just added 16 and 16 to get to that and we know that 16 is a perfect square so we could say the distance is equal to the square root of we could say this is 2 times 16 16 if i take the square root of that is 4 so i could say the distance is equal to 4 times the square root of 2 okay so very very simple very very easy one of the things you want to watch out for a lot of students mistakenly try to work with i in these formulas that we get you see that when you're taking the absolute value of a complex number you see that when you're trying to work with the distance between two complex numbers i is not involved a is the real part of a complex number b is the imaginary part of the complex number the imaginary unit i is not involved in any of these calculations all right so to see this visually let's look at our complex plane and you'll notice that i've already kind of gotten things started i've already pre-drawn the right triangle for us i plotted 3 plus 9i so that's a real location of 3 and an imaginary location of 9. so go 3 is to the right and 9 units up that's how i got that and then i've also plotted 7 plus 5i so that's a real location of 7 and an imaginary location of 5 so 7 years to the right 5 units up so that's how i got that and then i threw in a third point to form the right triangle so to do that i used a real location of three and i used an imaginary location of five so this right here would be three plus five i you could have also done seven plus nine i if you wanted to okay so that could be another way you could draw a right triangle i did this one because it's a little bit easier to see okay so now that we have the right triangle we should remember that we have these two shorter sides which are known as legs typically we'll say this vertical leg is a we'll say this horizontal leg is b and then the hypotenuse what we're trying to figure out here is c that's the distance between these two complex numbers 3 plus 9i and 7 plus 5i so we should know at this point the pythagorean formula is what it's a squared so this guy here the measure from here to here squared plus b squared again from here to here squared is equal to c the hypotenuse squared now typically when we're working with kind of trying to find a distance or replace c with d to stand for distance and we can solve for this we say d squared is equal to a squared plus b squared we take the square root of each side and we say d is equal to the square root of a squared plus b squared okay so let's work with it in this format let me kind of scooch this up a little bit and let's just think about why this makes sense so d the distance between the two is equal to the square root of what is a a is the measure from here to here so it's the difference in imaginary values okay so we see when we looked at this formula here we had b sub 2 minus b sub 1. each b here represents one of the imaginary values we had an imaginary value of 5 and an imaginary value of 9. so if i go back you see that's the case here you have 9 and you have 5 it doesn't matter the order you do this in because you're squaring it the result is going to be 9 negative so 9 minus 5 is 4 or 5 minus 9 is negative 4 either way when you square it you get 16. so let me just put this in for a i'm going to say this is 9 minus 5 and this is going to be squared okay and again all this is is my kind of b sub 2 minus my b sub 1 it's the difference in imaginary values that we saw in our formula okay then plus what are we going to have for b well that's going to be the difference in real values so in this case we have 3 and we have 7. again if we go back to the formula we had a sub 2 minus a sub 1 you have 7 minus 3. you could have also done 3 minus 7 because again when you square it you get the same result so let's just go ahead and say we have 7 minus 3 this quantity squared and you can see from where we have this here it's the same thing as the formula we had again it's just an application of the pythagorean formula very easy to kind of come up with this so the distance between these two complex numbers again is the square root of we already know this this is 4 and this is 4 4 square root of 16 4 squared is 16 so you get 16 plus 16 is 32 and we know that the square root of 32 is going to simplify to 4 times the square root of 2. okay so the distance is equal to 4 times the square root of 2. so if you wanted to you can kind of label this and say this is 4 times square root of 2. you can label this b and say this is 4 you can also label this and say this is 4. right so you get 4 4 and then 4 times square root of 2. all right so let's take a look at a few more examples so again if i have two complex numbers one is z and one is w so we're just going to go through and label one of these as a sub 1 and then b sub 1 and the other is going to be a sub 2 and then b sub 2. again it doesn't matter what you label as which i just want to make sure that you don't involve i okay just make sure you don't do that that's the common mistake so the distance is going to be equal to what the square root of if i plug in for a sub 2 i'm going to have a 13. if i plug in for a sub 1 i'm going to have an 11. 13 minus 11 is 2 so you'd have 2 squared and then plus for b sub 2 we have 2 for b sub 1 we have 4 2 minus 4 is negative 2 so you'd have negative 2 squared okay negative 2 squared so the distance is equal to the square root of 2 squared is 4 negative 2 squared is 4. so you basically have 4 plus 4 which is 8 okay 8 and we know that turns into what so the distance is equal to the square root of you have 4 times 2 under here square root of 4 is 2 so i can simplify this and say the distance is equal to 2 times the square root of 2. all right let's take a look at one more of these again a very simple concept overall so we have z which is just equal to negative 5 and w which equals 12 plus 7 on so don't freak out if you get this you might say okay well this is a real number this is a complex number every real number that you work with is also a complex number right the real numbers are a subset of the complex numbers so i can write this as a complex number by putting plus 0i okay very very easy so again i can just label now i can say this is a sub 1 and this is b sub 1 and i could say this is a sub 2 and this is b sub 2 and then i can just plug into the formula so the distance is equal to the square root of for a sub 2 we would say this is 12 for a sub 1 we would say this is negative 5. for b sub 2 we're going to say that's 7 and for b sub 1 we're going to say that's 0. so 12 minus a negative 5 is the same as 12 plus 5 that's going to be 17. so you would have 17 squared then plus 7 minus 0 is obviously 7 so you'd have 7 squared so the distance is equal to the square root of 17 squared is 289 plus 7 squared is 49. so if i sum these two amounts together i get 338 so the distance is equal to the square root of 338 and 338 is 2 times 169. so i can simplify this further i can say that the distance is equal to the square root of 169 times two the square root of 169 is 13 so i can finish this up and say the distance between these two kind of complex numbers z and w would be equal to 13 times the square root of 2. all right so now let's talk about using the midpoint formula with two complex numbers as kind of endpoints of a line segment on the complex plane so we previously saw that the midpoint formula can basically be used to find the coordinates of the midpoint of a line segment again this is just on the regular coordinate plane we're used to working with in the real number system if you have two points that are given one x to the one y sub one and the other x sub two y sub two you could just plug into this formula m or capital m you would say is equal to what it's the average of the x values and then the average of the y values those are your two coordinates right that should make sense because if i'm trying to find the kind of halfway point i would want to find the average of the x values and then the halfway point the average of the y values right so that's what i'm doing so for this guy this is x sub 1 y sub 1 this is x sub 2 y sub 2. again it doesn't matter the order that you label things so if i plug this in the midpoint here would be found at you would have 3 plus 7 which is 10 10 divided by 2 is 5 and then you would have 5 plus 8 which is 13 13 divided by 2 you could say it's 13 halves or you could say it's 6.5 either way so the midpoint of a line segment with these two points as endpoints 3 comma 5 and 7 comma 8 would be at 5 comma 13 halves or again you could say 5 comma 6.5 so again when we work with complex numbers it's basically going to be the same formula with some minor notational adjustments so a line segment with two endpoints let's say a sub 1 b sub 1 and a sub 2 b sub 2 in the complex plane we just adjust the formula right so this is one of the a values this is the other right you sum them and you divide by two you're just getting the average this is one of the imaginary values this is the other you sum those you divide by two you get the average okay so it's the same process so let's say we have a line segment and it's got endpoints z and w so z is 3 minus 2i w is 1 plus 6i okay so all i'm going to do is label one of these as a sub 1 and then b sub 1 and the other will label as a sub 2 b sub 2. okay so just like we did with the distance form so i'm just going to plug in so the midpoint is equal to one you're going to have one plus three which is four four divided by two is two and then you're going to have six minus two or you could say six plus negative two that's also four four divided by 2 is 2. okay so this guy has a complex number if i have a real location of 2 and i have an imaginary location of 2 it's 2 plus 2i so it's 2 plus 2i so to visually see this we can come to the complex plane we can show this line segment here with endpoints of 1 plus 6i so that's one endpoint and another would be 3 minus 2i again that's another endpoint so we're saying that the midpoint or the point that kind of cuts this line segment in half or you can say splits it into two equal halves is here this 2 plus 2i okay so we're saying this distance from here to here is equal to this distance from here to here okay so basically when we think about this if i think about where should the real location be well again if i think about the total movement along the real axis well i'm going from here right from 1 all the way to 3 so that's a total of 2 units in terms of how far i'm moving so half of 2 would just be 1. so if i start here i'm just going 1 unit 2 kind of the right if i start here i just go one unit to the left so that gives me a real location of two and again another way to think about that is if i just average the real values i have a real location of three and again a real location of one add those together three plus one is gonna give me four divide by 2 i get to 2. ok so either way you want to look at that the other thing to think about would be the imaginary location so here i have a value of 6 and here i have a value of negative 2 so that's a total distance of 8 units right i would go from negative two up to zero and then from zero up to six so that's a total value of eight units half of eight would be four so if i started here and went up four units i'd end up here well if i started here went down four units i didn't appear so that should make sense to you as to why this kind of imaginary location would be here at 2 okay and again you can also find this by taking 6 plus negative 2 write those 2 imaginary values you sum those together 6 plus negative 2 is 4 and then just divide by 2. 4 divided by 2 is 2. so again that's how you get this spot here as the midpoint 2 for the real location plus 2 for the imaginary location times i so 2 plus 2 all right so let's just look at one more example of this it's a very very easy concept once you know the formula you just kind of plug in so again the midpoint i'll put m capital m is equal to you're going to have the average of the real parts so a sub 1 plus a sub 2 over 2. let me kind of slide this down so it fits on my screen and then comma you're going to have the average of the imaginary parts so you're going to have b sub 1 plus b sub 2 again over 2. so that would be your location so what we're going to do here is find the midpoint of a line segment with these two endpoints so z and w so for z let's just label this 7 as a sub 1 and let's label this 3 as b sub 1. again i want to be clear we're not touching this i at all that's the most common mistake in this section where people try to do stuff with i we're just working with 7 as a and 3 the number that's multiplying i as b okay so this is a sub 1 this is b sub 1. then down here for w i'm going to label this 11 as a sub 2 and then we have i by itself so understand that i could write this as 1i right because 1 times i is still just i so then this part right here will be my b sub 2 okay that number 1. so if i just plug into the formula m is going to be equal to what well a sub 1 is 7 and a sub 2 is 11. so 7 plus 11 is 18 18 divided by 2 would be not so this part would be 9 and then for this part the imaginary part we have 3 for b sub 1 and we have 1 for b sub 2 3 plus 1 is 4 4 divided by 2 is 2 okay so this would correspond to a complex number that is 9 for the real part plus 2 for the imaginary part times i so this would be the complex number that's the midpoint of a line segment with endpoints z which is 7 plus 3i and w which is 11 plus i in this lesson we want to review finding the equation of a circle and we also want to review how to graph a circle on the coordinate plane so i know for most of you you've already taken an intermediate algebra course and so when you studied conic sections you talked about circles you talked about finding the equation of a circle you talked about graphing circles on the coordinate plane so this is just a review for you for others you might not have ever seen this topic before and i can assure you that this is a very very easy topic to learn and you'll be able to pick it up from just this lesson so let's just start off by talking about the definition of a circle so in your textbook you'll probably read that a circle is just the set of all points in a plane that lie a fixed distance from its center okay a fixed distance from its center so the fixed distance from the center to any point on the circle is a special name it's known as a radius okay so let's write radius here we're going to hear about this throughout our lesson and generally we denote the radius with the lowercase r okay so when you see the lowercase r we're talking about a radius and this is what we're referring to if you look at the center here which i've put as h comma k and we'll talk more about that in a second and you look at this point here which i've labeled x comma y we'll talk more about that in a second this distance from the center of the circle to this given point is going to be the radius r okay and it would be the same if i put a point here on the circle and i drew a line from here to here that distance would be the same as this one that's already drawn or if i put a point here and i drew a line from here to here all of these would be the same distance okay so let me erase this and let's talk about some other things you'll see with a circle so generally we denote our center with h comma k the h is the x value the k is the y value so h comma k all right in this particular case we have actual coordinates the center occurs at negative four comma four okay but generically we would say that this is the center when we're working with a circle okay and then a given point could be labeled as x comma y right if we don't know the coordinates in advance in this particular case what i've labeled as x comma y we already know this would be one comma 4 okay but we can refer to a point on the circle as just x comma y so the reason i talk about this is because we're going to use kind of the concept of the distance from the center to a given point to derive an equation for a circle and this comes from using our distance formula so let me show you this equation real quick and then we'll show you how it gets derived so if we look at this center radius form of a circle this is also called standard form you have this x minus h this quantity squared then plus this y minus k this quantity squared and this equals r the radius squared so if you have an equation in this format it's very easy to graph the circle because you're going to know the center of the circle which would occur at h comma k so you have your h and you have your k so immediately i can say okay this is my center and you're going to know your radius right it's the r not the r squared it's the r so i would need to take the principal square root of r squared to find the radius r now why don't i want the negative square root well because the radius is a distance and i don't want to end up with a negative distance so where does this equation come from a lot of times we see these equations and they're given to us and it's really cool to just be able to kind of plug in and do your homework but i want you to understand where it came from so let's talk a little bit about that let's revisit our distance formula that we work with so much in this course so we know the distance between any two points is equal to the square root of we have the x sub 2 minus x sub 1 squared plus the y sub 2 minus y sub 1 squared okay so the difference in x value squared plus the difference in y value squared this is under a square root symbol okay so in this particular case we know what let's say we have a given point of x comma y and we have a center that is h comma k so these are two points that we know okay and we know the distance between a given point on the circle and the center is the radius so i know that the radius would be my d my distance between these two points okay so i can say r is equal to the square root of and then i could just kind of plug in for these kind of points normally i work with what i have x sub 1 y sub 1 and i have x sub 2 y sub 2 that i'm dealing with well in this case i can just let x comma y take the place of x sub 2 y sub 2 and i can let h comma k take the place of x sub 1 y sub 1. so let's just set this up and say that we have x minus h okay i just took this guy minus this guy so it's still the difference in x values and then i'm squaring that and then plus i would take the difference in y value so i'm going to go y minus k so y minus k and then this quantity squared so let me erase all this just so we have some room to work okay so let me scooch this back up and so all we need to do to get into our center radius form is just square both sides if i square this side and i square this side we know that this guy right here is going to cancel with this index here and so i can just basically remove this and then i have my center radius form right i can flip this r squared around because normally it's on the right side of the equation so i can just say it's x minus h that quantity squared plus my y minus k that quantity squared and this equals r squared okay so exactly what we saw on the previous page right x minus h that quantity squared plus y minus k that quantity squared and this equals the radius squared so that's where this equation comes from so one of the first kind of tasks that you encounter in this section you're going to be given a center and you'll be given a radius and they're going to say write the equation of the circle and center radius form so all you have to do is just kind of match things up so again we have the quantity x minus h squared then plus we have the quantity y minus k squared this is equal to the radius squared okay so i'm just going to plug in i know my center occurs at h comma k so this guy right here this 4 is going to be my h and this guy right here my 5 is going to be my k okay so i'm just going to plug in i'm going to plug in a 4 here and i'm going to plug in a 5 there and then for the radius it's the square root of 7 so i'm going to be plugging that in there okay so what do we have we'd have the quantity x minus 4 squared then plus the quantity y minus 5 squared and this equals if i take the square root of 7 and square it i just get 7. so this is my center radius form for a circle with a center four comma five and a radius of square root of seven all right let's try one more of these and then we'll just kind of move on to something that's a little bit more complicated so we have a center that occurs at negative 13 comma negative two and a radius that occurs at the square root of 11. so again let me rewrite our central radius form this is x minus h this quantity squared then plus you have your y minus k this quantity squared and this equals the radius squared okay so again this is going to be my h and this is going to be my k now you have to be careful here because you have a negative 13 and a negative 2 okay so you have to match this up perfectly if i'm plugging in for h and h is negative 13 what i have here is minus a negative 13. okay so minus a negative 13 is plus 13. so i've got to be careful so i'm going to say x plus 13 this quantity squared okay a lot of students make the mistake when they have a negative involved they just plug in kind of in this case you'd say i just plug in x minus 13. okay that would be wrong so then plus i'm doing the same thing here i'm plugging in a negative two so y minus a negative two is the same thing as y plus 2 and this is squared this equals my radius i'm going to plug in the square root of 11 there so the square root of 11 squared is just 11. okay so x plus 13 this quantity squared plus y plus 2 this quantity squared and this equals 11 that's the center radius form for this circle that has a center of negative 13 comma negative 2 and a radius of square root of 11. all right so now let's move on and talk about something that's a little bit more complex so we also have a general form of a circle okay so when it's in general form generally you have to do a little bit of work you're going to use the method known as completing the square that we talked about with quadratic equations and we're going to put it back into our standard form or center radius form so general form of a circle looks like this you have x squared plus y squared plus c x plus d y plus e equals zero and basically this guy can be obtained if we start with our kind of standard form and we expand those two kind of binomials that are squared okay so for example let's say i start with this guy so i'm going to start with this i'm going to put it in general form and then i'm going to go backwards and put it back in kind of standard form so the first thing i'm going to do is just expand these two right here so if i do that then x minus 3 that quantity squared we know that's x squared we know it's minus 2 times this guy times this guy so 2 times 3 is 6 6 times x is 6x and then we know it's plus this guy squared then we'd have plus i'm going to have y plus 2 that quantity squared so this guy squared then plus 2 times this guy times this guy 2 times 2 is 4 4 times y is 4y and then plus we have this last guy 2 squared which is 4. this equals 9. now i can combine some things here i can write this as x squared and then minus 6x and then i'm going to put plus y squared and then plus 4y and then 9 plus 4 is going to be 13. but before i do that realize you have a 9 here and a 9 here if i subtract 9 away from each side of the equation it would cancel right so i could just put plus 4 equals 0 and that would be my general form okay so from this i can go back to this so let's say i'm presented with this and i'm asked to kind of put it in standard form well essentially what i want to do is i want to complete the square right what is completing the square it's where you create a perfect square trinomial right a perfect square trinomial can be factored into a binomial squared right that's exactly what we have here we have a binomial squared and a binomial squared so we're going to need to kind of add a number to this guy to get a binomial squared we're going to add a number to this guy to get a binomial squared but there are some things we need to do to make sure that's legal so let me scroll down we'll come back up to this when we're done so the very first thing in case you forgot how to complete the square is to move the constant to the right side so i'm just going to subtract 4 away from each side of the equation and that's going to give me x squared minus 6x plus y squared plus 4y is equal to negative 4 okay that's the first step the next thing you want to do is make sure the coefficient of all the squared terms so we have x squared and we have y squared you want to make sure this is a 1. now in this case it's an easier example so it's already done force so we don't have the additional work but if it wasn't you would need to divide right both sides of the equation by whatever was necessary to kind of clear that all right so now what we want to do is group the x terms together and group the y terms together that's kind of already done for us but what i'm going to do is i'm going to just wrap them in parentheses so i'm going to say this is x squared minus 6x and then i'm just going to put a blank space here because we're going to add something to that and then plus you're going to have y squared plus 4y i'm going to put a blank space because we're going to add some of that this equals negative 4. all right so what we want to do now is kind of the step that everyone forgets we want to take one half of the coefficient of the first degree term squared and we want to add that to each side of the equation okay so in this particular case again this variable is raised to the first power this variable is raised to the first power so i look at the coefficient this is negative 6. what is one half of negative 6 a lot of times i say cut it in half just divide it by 2 that gives you negative 3. once you get that you want to square it so negative 3 squared would be 9. so i'm going to add 9 here but to make it legal i've got to add 9 on the other side of the equation so i'm going to add 9 over here as well then i'm going to do the same thing over here if i take 4 this coefficient for y to the first power and i cut it in half 4 divided by 2 is 2. if i square 2 i get 4. so plus 4 but again to make it legal i've got to add 4 on this side as well let me erase this let me kind of scroll down and so what i end up with is exactly what i started with right if i factor this now i would have x minus 3 that quantity squared then plus if i factor this i would have y plus 2 that quantity squared and then negative 4 plus 4 is 0 then i just have positive 9. so this equals positive 9. so if i look at this guy right here i can see it's exactly what i started with if i go back up so let me just kind of paste this in so we can kind of compare what we have so we started with the standard form x minus 3 that quantity squared plus we have y plus 2 that quantity squared and this equals 9. then we expanded our binomials so we got the general form right x squared minus 6x plus y squared plus 4y plus 4 equals 0 that's general form and then we went through the process and we completed the square for each and we ended up with the standard form again right so we went from standard form to general form back to standard form so it can be tedious in some situations but not really a hard process overall as long as you understand how to complete the square it's pretty easy all right let's look at another example of this and then we'll move on and talk about graphing okay so let's suppose you're given this equation here which is a circle in general form so let's say you're asked to put it in standard form again you're just going to go through the process so the first thing i want to do is move this guy this constant to the other side so let's say we have 2x squared plus 2y squared minus 4x minus 48y and this is going to be equal to if i subtract 218 away from each side of the equation i'm just going to say this is equal to negative 218. okay then what you want to notice is that the coefficient on x squared and y squared is a 2 okay you want that to be a 1. so what we're going to do is divide each side of the equation by 2 so i'm going to divide every part here by 2 and what are we going to get well this is going to cancel you'll have x squared then plus this is going to cancel you're going to have y squared then minus 4 over 2 is 2 so 2x then minus 48 over 2 is 24. so this would be 24y and this equals negative 218 divided by 2 is going to be negative 109. okay let me scroll down to get some room going so what do we want to do now now we want to complete the square so let me group all these x terms together so i'm going to put this as x squared minus 2x i'll put a little blank here and then plus i'm going to group my y terms together so y squared minus 24y put my blank and this equals negative 109. let me kind of slide that down just a little bit so we have enough room okay so what i want to do is i want to take the coefficient for each variable raised to the first power so in this case this guy's raised to the first power this guy's raised to the first power so take the coefficient so i'm going to have negative 2 let me make that better so i'm going to have negative 2 and i'm going to have negative 24. i want to cut it in half and then square it i'm going to take that value and add it to both sides of the equation so negative 2 if i cut that in half so times a half the 2's cancel and i'm left with negative 1. if i square negative 1 i get 1. so i want to put plus 1 there okay but again to make that legal i want to add 1 over there now for negative 24 if i cut that in half if i divide it by 2 or multiply by half whatever you want to do that's going to end up giving me a negative 12. if i square negative 12 i get plus 144 and again i've got to do that over here i've got to add 144 to make it legal all right so let's see what we got now so this guy right here is going to factor into x minus 1 that quantity squared and plus this guy right here is going to factor into y minus 12 that quantity squared and this will be equal to if i take 1 and i add it to 144 i get 145 if i have 145 minus 109 i'm going to get 36 right 36 and essentially the square root of 36 is 6 so you know the radius is 6 in this case and you know your center occurs at 1 comma 12. right so my center is 1 comma 12 and my radius is going to be six okay so looking at this piece of information here i can get the center and i can get the radius if i looked at the original equation you can't really get information out of it because it's it's got too much going on right so when you put in this format it makes it easier to grab the center and the radius and then you can graph stuff pretty quickly all right so now let's talk a little bit about graphing circles this is probably the easiest thing you're going to do in this section basically to graph the circle if it's in standard form okay which we already know how to get to from general form or if we're just given standard form if we're lucky basically i know the center and i know the radius so i can just kind of use the definition of a circle and say if i plot my center okay i can use the radius to obtain additional points right because from the center to the additional points it's a distance of the radius so starting at the center i can go up by the radius starting at the center i can go to the right by the radius starting at the center i can go to the left by the radius and starting at the center and go down by the radius so i can use that to kind of plot four points and then i can just draw a smooth curve connecting those four points and i'll have my circle drawn so to make this a little bit more tedious and to match something you're probably going to get in your course i'm going to give you this in general form we're going to derive the standard form and then we're going to graph it okay so we have x squared plus y squared plus 2x plus 2y minus 23 equals 0. let's just kind of blow through this we already know how to do it let's add 23 to both sides of the equation so i'll have x squared plus y squared plus 2x plus 2y is equal to 23. okay so at this point i would check to make sure that the coefficient of the squared variables so x squared and y squared are one and in this case each one is a one right so we don't have to worry about that step so now we just group the x's together and group the y's together so i'll say we have x squared plus two x plus i'll put a blank here because i'm going to be adding something in there so then plus over here i'm going to have y squared plus 2y plus again i'll put a blank there let me kind of slide this down because i'm going to run out of room and i'll say this is equal to 23. okay so now to complete the square i'm looking at the variables that are raised to the first power so this one and this one and i want to look at the coefficients so it's a 2 and a 2. so in each case it's actually the same number so that makes it nice and easy so i want to cut the number in half so divide it by 2 or multiply by half whatever you want to do and then square that so 2 divided by 2 is 1 1 squared is 1. so i'm going to add 1 here and here to complete the square but to make that legal i've added a total of 2 to the left side so i can add 1 individually like this or i just add a total of 2. right either way it's the same thing so i know that now i can just factor this so x squared plus 2x plus 1 that factors into x plus 1 quantity squared this guy is going to factor into y plus 1 quantity squared right because it's basically the same thing with just a different variable and then this is equal to 23 plus 2 is 25 okay so now we can go through and we can figure out what is the center what is the center and what is the radius okay so i know the radius is the principal square root of this number so that's going to be 5. so that's pretty easy for the center again it occurs at h comma k so if i just kind of match this up it's x minus h quantity squared plus y minus k quantity squared equals the radius squared we already know the radius is 5 but for this guy if you try to match this up remember you have a plus here and a plus here it is not in the format of minus some number okay so what you got to do is make a little adjustment and i can write this as minus a negative 1 minus a negative 1 and i kind of can't fit that in there so let me make that a little better so minus a negative 1 and then i'll put minus a negative 1 over here so again you've got to match the format that you're given okay so now i have a minus and a minus a minus and a minus so what i can do is i can just grab the number negative one and negative one okay so my x value for the center will be negative one that's my h my y value for the center would be negative one that's my k so the center here occurs at negative 1 comma negative 1 the radius is 5. all right so looking at the graph of this guy it's already pre-drawn for us to make it a little bit cleaner also i'm a terrible drawer so i want you to notice that you have the center which is plotted here we already know that this occurs at negative one comma negative one okay so this is my center and we already know that the radius is five right we know that already so this is five so from the center if i go up by five so up one two three four five i can make a point okay so that's one point on the circle and this guy is going to occur at what it's going to be negative 1 for the x location and it's going to be 4 for the y location then from the center i can go five units to the left so one two three four five so that's another point this would be at negative six comma negative one then from the center i could drop by five units so one two three four five so right there that's going to be at negative one comma negative six and then lastly i can go five units to the right from the center so one two three four five so that's a point there that's going to be at four comma negative 1. okay so with those four points i can just kind of draw my circle and i'm just going to trace over what i already have hopefully i don't mess it up too badly and you can see that i am terrible at tracing and drawing but you get the idea of what's going on that's how you can draw a circle again if you're doing this for homework or for a test the person grading this just wants to know that you know how to find the center and the radius that you can display those on the graph and then you can make a somewhat accurate kind of portrait of what's going on all right let's take a look at one more of these so we have 2x squared plus 2y squared plus 12x plus 4y plus 2 equals 0. again i can just subtract 2 away from each side of the equation and i'll have 2x squared plus 2 y squared plus 12x plus 4y is equal to negative 2. notice here that the coefficient for x squared and y squared is a 2. we need that to be a 1 okay in each case so just divide everything by 2 and basically i'll have what x squared plus y squared plus 6 x plus 2 y is equal to negative 1. okay so from here now i can just go into the step where i'm going to complete the square and again i want to group my x terms together group my y terms together so we'll have x squared plus 6x and then plus something and then plus you're going to have y squared plus you have your 2y and then plus something and this equals negative 1. okay so i'm looking at the coefficient for the variables that are raised to the first power so this x this variable is raised to the first power this y this variable is raised to the first power so the coefficient is a six and a two okay so that's what we're looking at so cut it in half and square it six divided by two is three three squared is cut in half and square 2 divided by 2 is 1 1 squared is 1. so this would factor into x plus 3 quantity squared this would factor into y plus 1 quantity squared this equals again when i added 9 here and i added 1 here i've got to add a total of 10 to this side of the equation to make that legal so i'm going to put plus 10 here and so negative 1 plus 10 is 9. and so now we have this guy in center radius form okay so what we want to do again is match it up because you have this plus sign here and plus sign here it's going to throw you off okay i know already because i've solved so many of these problems the center occurs at negative 3 comma negative 1 and the radius is 3. but again if you're just starting out write your formula and match it up so x minus h quantity squared plus you have y minus k quantity squared and this equals you're going to have your radius squared okay so the radius is easy just take the principal square root of 9 that's going to be 3. so the radius is going to be 3. and then this guy right here again i'm just going to write it as minus a negative so x minus a negative 3 so that it matches up with what we have there so my center is going to occur at negative 3 comma again this guy i'm going to write it as y minus a negative 1 okay so this matches up with that so negative 3 comma negative 1 is going to be the center okay so this is my center all right so let's plot negative 3 comma negative 1 and then use the radius to get additional points so here's negative 3 comma negative one again i'll just label that it's negative three comma negative one that's our center and then my radius is three so i go up by three one two three plot this point here which is negative three comma two and then you go to the left by three so one two three so this is gonna be negative six comma negative one and then you're gonna go down by three so this guy right here is going to be negative three comma negative four and then you're gonna go to the right by three so this guy right here is going to be zero comma negative one okay so those are your four points and again i can just connect them with a smooth curve and i'll just have my circle drawn all right so let's wrap up our lesson and talk about just one more commonly occurring scenario that you're going to get when you're working in the section on circles so you might be asked to find the center radius form of the equation of a circle given that you have these kind of endpoints that represent the diameter of the circle so let's suppose that we have a circle that has a diameter with endpoints negative one comma three and then five comma negative nine so again before i kind of jump in and show how to solve this let's look at a little graph real quick so i just took the first graph that we looked at with the center labeled as h comma k and a given point labeled as x comma y again i know there are real coordinates here but i'm just labeling this generically so we can get an idea that would work for kind of every scenario okay so we know that the radius is the distance from the center h comma k to a given point we said x comma y so that's this distance from here to here so that's r that's the radius if i doubled that amount okay so if i had a radius over here as well if i drew kind of in a different color from here to here two of those guys would be known as the diameter okay so the diameter is twice the radius the diameter is the length of a line that passes through the center and touches two points on the edge of the circle so if i just think about this as one line i'm going to do a final color here so if i think about these two end points and i do one line all the way across that distance there would be the diameter okay so that's what we're thinking about here now if i'm asking you to find the center and let's say i know this and i know this well i need the point exactly in the middle of this line segment how do we find that we use the midpoint formula so i want to go back here and i want to use my midpoint formula so i can find the center right i need to find the center radius form so i got to find the center and i got to find the radius so the center again i use my midpoint formula so the midpoint formula it's x sub 1 plus x sub 2 over 2 right the average of the x values and then the average of the y values so y sub 1 plus y sub 2 over 2. okay so i can label either as whichever you want right we can say this is x sub 1 y sub 1. we'll say this is x sub 2 y sub 2. and let's just plug in so you'd have negative 1 plus 5 that's going to give me 4. 4 divided by 2 is 2. so this would be 2 and then for y i would have y sub 1 which is 3 plus y sub 2 which is negative 9. 3 plus negative 9 is negative 6 negative 6 divided by 2 is going to give me negative 3 okay so this is negative 3. so this guy is going to be the location of my center so this is my center okay so i want you to remember let me just kind of erase this and i'll put this over here so i'll say my midpoint and actually i'm just going to label it as the center so my center is going to occur at 2 comma negative 3. and we know our center radius form it's going to be x minus my h in this case h is going to be 2 right so it's going to be 2 that quantity squared then plus i'm going to have my y minus k in this case k is negative 3 minus a negative 3 is plus 3 and this is squared this equals my radius squared so now what i need to do is figure out the radius i need to figure out the radius now by definition i know that the distance from the center to any point on that circle is going to be the radius so i can just use my distance formula with one point i'm just going to choose this one and the center and that's going to give me my radius i can plug it in and i'll be done so we've got 5 comma negative 9 5 comma negative 9 and we've got 2 comma negative 3 okay and again we can switch up the labeling we can say this is x sub 1 y sub 1 and this is x sub 2 y sub 2 and just so don't get confused i'm going to erase this right this has nothing to do with what we're doing now so let me scroll down all right so the distance between these two points again from a point on the circle and the center is going to be my radius so instead of saying d for distance i'm going to say r for radius this is equal to the square root of the difference in x value so you have x sub 2 minus x sub 1. so 2 minus 5. so 2 minus 5 that quantity squared plus the difference in y values so y sub 2 which is negative 3 minus y sub 1 which is negative 9 minus a negative 9 is plus 9 this quantity squared okay so 2 minus 5 is going to be negative 3 negative 3 squared is positive 9. so this is 9 negative 3 plus 9 is 6 6 squared is 36. so if i add 9 and 36 i get 45 so i get the square root of 45 here let me just kind of clean this up a little bit so i know that 45 is what it's 9 times 5 so it's 9 times 5. square root of 9 is 3 so i can say this is 3 times square root of 5. and so this is my radius so i can plug that in here so i would have 1 i would have 3 times square root of 5 but then this guy is squared okay so let me erase all this so to finish this up if i square this guy remember to square each part 3 is going to be squared which gives me 9 and the square root of 5 is going to be squared which gives me 5. so 9 times 5 is 45 so this becomes my equation of this circle and center radius form you have x minus 2 that quantity squared plus you have y plus 3 that quantity squared and this equals 45. in this lesson we want to review relations and functions so i know most of you took algebra 1 and algebra 2 before joining this course and essentially you already know what a relation is and what a function is right you have those concepts down so this is basically a review for you but it never hurts because functions are something that are so important and used throughout the study of algebra and all your higher level math courses so a review is not something that's going to really kill you right you want to make sure you really understand this topic so we'll talk about functions here and then over the course of the next few lessons we'll talk about function notation we'll talk about domain and range and we'll talk about using the vertical line test all right so let's start out with the basic definition of a relation and then we'll kind of move into functions so a relation is any set of ordered pairs now before i kind of move on from this let's think about what an ordered pair is used for normally we see ordered pairs when we work with equations with two variables involved normally x and y so something simple like x plus y equals 3 i can display the solutions for this equation using ordered pairs right let's say x was 0 i would have y equals 3 right so an x value of 0 and a y value of 3 those guys are related to each other because they represent a solution to this equation so i could write this as an ordered pair 0 comma three right an x value of zero is related to a y value of three that again solves this equation i could also do an x value of one and a y value of two so i could have an x value of one and a y value of two right so on and so forth i could do 2 comma 1 right if i wanted to an x value of 2 and a y value of 1. now there's an infinite number of ordered pairs as solutions for this equation so i'm not going to continue further beyond this point but i want you to understand that ordered pairs are really just a nice way to write two pieces of associated information or two pieces of related information here the x and the y values are associated with each other because they represent two values that when plugged in for this equation are going to give us a true statement right they're solutions for the equation but we could also have some kind of random information let's say we had set a here and set a just contain these kind of four random ordered pairs okay so i have 3 comma 5 9 comma 6 1 comma 3 and 8 comma 7. so we don't know the association between these numbers we're not given any type of equation or any type of like back information on this but this is something you'll see in this section pretty often just so you can get the concept down of what a relation is again it's just a set of ordered pairs something like set b again is equal to we have three ordered pairs here one comma negative four six comma negative seven and negative four comma four now ordered pairs don't have to just relate kind of numerical information it doesn't have to be solutions for equations or it can really be any two pieces of related information kind of put together nicely okay so for c here what we have is numerical grades and then associated letter grades so a grade of 38 if you get that on a test you get an f right so 38 is associated with a letter grade of f or it's related to a letter grade of f 73 with a d 81 with a c 95 with an a then something like set d let's say we related kind of people in a group and their pets so let's say john has a cat mary has a dog james has a bird lacey has a hamster so in each case the ordered pair is just giving us two pieces of related information it's wrapped inside of parentheses it's separated by a comma so whenever we have a set of ordered pairs we have a relation all right so let's look at a kind of real world situation we'll set up a little equation we'll go through some of the kind of vocabulary that's going to be involved in this section so we have here that heather purchases gas from the local station for five dollars per gallon which includes all taxes and fees so five dollars per gallon okay so let's set up a little equation and we're going to use x and y and i've already kind of pre-written this stuff for us so we're going to let x equal the amount of gas in gallons that she's going to choose to purchase and then y is going to equal the total dollar amount of her purchase so the equation we end up setting up is y again this is the total dollar amount for purchase is equal to or the same as five the cost per gallon of gas times x the amount of gas in gallons she's going to buy okay and you can think about this if i plugged in a one here well it's five dollars per gallon she buys one gallon this guy is going to be five right if i plug in a two here she buys two gallons of gas and multiply it by five the cost per gallon i get a y value of ten so on and so forth now there's a couple of things that i have related to this first i've said that y is the dependent variable and i've said that x is the independent variable what do those two terms mean y is the dependent variable because it depends on her choice of x right the independent variable okay so the independent variable is something she chooses the dependent variable is something that comes as a result of her choice she chooses how much gas to buy right she rolls up to the station and she chooses to buy what she chooses to buy no gas she gets upset at the price and says i can't afford it or i don't like it or whatever she leaves she doesn't buy anything okay so she chooses in that case to buy zero gallons she pays zero dollars okay if she rolls up to the station and buys one gallon she pays five dollars right i multiply one times five i get five again i showed you that already so again the independent variable here is x is something she chooses and y is the dependent variable something that is a result of her choice okay now we also have other names for the independent and dependent variables for the independent variable we call it an input it's something we're going to plug in so in this case again let's just say she buys three gallons of gas that's my input okay in this case it's three and my output is another name for the dependent variable so i plug in a three that's my input i multiply it by five and then i get 15 as an output so we would say this is 15 or this is my out point okay so you'll hear that a lot in this section now in this particular case i gave you something known as a domain restriction we're going to talk about domain and range very shortly i know most of you already know what a domain is but again we'll cover in a second i said that x here the independent variable has to be greater than or equal to zero so i put that in there in this particular case because she can't go on and buy a negative amount of gas okay when you describe a real world situation you have to think about these things if she rolls up to the pump and she decides not to buy anything that's fine she could buy zero gallons she can buy some positive amount up to the point to where her car is full or the station has no more gas right those are the kind of constraints in the real world but she can't roll up to the station and pump a negative amount of gas right she can't siphon gas out or something silly like that so we put this constraint in here to say that x the amount of gas in gallons she's going to buy is going to be greater than or equal to 0. in this example here i'm going to further constrain x and i'm going to say that x is going to be greater than or equal to 0 and less than or equal to 5 and x is an integer okay so that means you can roll up to the pump and buy no gas a gallon 2 3 4 5 gallons of gas those are her only choices okay just to make it nice and easy so we can end up with a set of these ordered pairs so we have zero comma zero one comma five two comma ten three comma fifteen four comma twenty and five comma twenty five right each of these represent solutions to this equation again given these two constraints these are the only possibilities okay and you can see that these values are related to each other buying zero gallons of gas is related to spending zero dollars choosing to buy one gallon of gas is related to spending five dollars right so on and so forth now a cool way to look at this is with a little table a lot of times you'll kind of set this up with real world information so if we have gallons of gas here as x amount paid is y and again our ordered pair x comma y so we have zero gallons of gas means she pays zero zero comma zero right going all the way through this again if she bought five gallons of gas she pays 25 total it's five comma 25 as an ordered pair now we're going to revisit that example in just a second so i can give you a definition of a function but before we do that i want you to understand what the domain and range represent when we're talking about a relation or function okay so the domain is the set of allowable values for x and the range is the set of allowable or possible values for y so previously when we talked about domain in this course it's been things that have been restricted right dividing by zero if you had x in the denominator you couldn't put an x value of 0 there because division by 0 was undefined so as an example here we have y equals 2x and i think about what's possible for x is there anything that would restrict me from plugging in something for x the answer to that is no right i can plug in anything for x and multiply by 2. so the domain here would be all real numbers and because i can plug in anything i want for x i can make y as small or as large as i want so for the domain i could say that it's all real numbers and i could write it out like this okay but that's a little bit lengthy and i don't really like to do that so i'm just going to use interval notation and i'm going to say this is from negative infinity to positive infinity because it just includes all real numbers and then for the range is going to be the same thing for this guy so from negative infinity to positive infinity all right let's think about this guy right here so we have y equals the square root of x minus 1. so what would my restriction be here think about the fact that you cannot in the real number system take the square root of a negative number okay that's not allowed you can do it with the complex number system but if we're working in the real number system which we're doing right now we're not going to be allowed to take the square root of a negative number so what i would have to do is i would have to take this radicand here this x minus 1 and i would have to say this needs to be greater than or equal to 0 okay i would add 1 to both sides of this inequality and so for my domain my domain i would basically say that it needs to be greater than or equal to one okay so it could be one that could be included and then anything out to positive infinity right so if i plugged in a 1 there i'm okay 1 minus 1 is 0 i can take square root of 0 that's fine it's just 0. and then i can take the square root of any positive number so if i plugged in a 2 right 2 minus 1 is 1 square root of 1 is 1 i'm fine but if i plugged in something like let's say zero zero minus one is negative one square to negative one again in the real number system that's not possible okay so let's erase this and this so my domain basically is from one including 1 out to positive infinity real quick as a side note you might see this in your course you might see them write it this way so they might say something like the set of all x such that x is greater than or equal to 1. now this is set builder notation okay and all this means is that x is a real number and this vertical bar means such that the x that we said was a real number has to be greater than or equal to the number one okay so it's just giving you the conditions the set of all x such that x is greater than or equal to one okay so that's set builder notation i'm not going to use it here but you might see it in your book so now let's think about the range so what is the range so if we think about this carefully again the smallest value i can plug in for x is going to be a 1. if i plug that in 1 minus 1 is 0 square root of 0 is 0 so the range would start at 0. that's the smallest it can be and then it can be any positive number out to positive infinity so from 0 including zero out to positive infinity okay so we'll go more into depth on domain and range in the next lesson we'll go way more into depth on it actually but for right now this is just kind of the basic idea of domain and range the domain is the set of allowable values for x or basically all your x values the range is the allowable values for y or all the y values for that relation or that function all right so kind of revisiting these ordered pairs that came from our y equals 5x if i asked you for the domain and range for this example all you would do for the domain is find the x values for the ordered pairs you have so the domain is just basically one it's each first component so each first entry so the leftmost entry of each each x value so i'm highlighting all those now so 0 1 2 3 4 5. so that's your domain right this set would be 0 1 2 3 4 and then 5. okay so that's the domain and then the range is very simple also it's just the kind of y values that are involved here so 0 5 10 15 20 and 25 so for the range we have 0 5 10 15 20 and 25 okay so that's your range all right so now that we understand the concept of domain and range let's talk about the definition of a function so this is something you really want to pay attention to and there's several different definitions that kind of go with this so i'll read a few of them and you just take the one that you like so essentially a function is a special type of relation where each x value or member of the domain is associated with only one y value or member of the range okay so all functions are relations but not all relations are functions now let me read you a few alternative definitions you may hear so a function is a relation in which for each value of the first component of the ordered pairs there's exactly one value of the second component another definition a function is a set of ordered pairs in which no first component is repeated so you'll see this in these examples here if you have a duplicate first component or a duplicate x value you're not going to have a function then you might also hear a function is a rule or a correspondence okay that assigns exactly one range value that's a y value to each domain value again that's an x value the definition that i like the most the definition that always stuck with me is for each x there's one y okay that's as simple as it can get for each x there's one y so for each x or remember the domain there's one y or member of the range that's going to be associated with it so what i did was i took the kind of ordered pairs we're working with from the y equals 5x and basically i set up a mapping diagram so we're showing the association between the members of the domain and the members of the range and this is very common to see this in this section of your textbook so an x value of 0 is corresponding to or linked up with or associated with a y value of 0. an x value of 1 is associated with a y value of 5. an x value of 2 is associated with y value of 10 x value of 3 is associated with y value of 15 and x value of 4 is associated with y value of 20 and then an x value of 5 is associated with the y value of 25. so this relation this set of ordered pairs represents a function because for each x value or member of the domain there's exactly one y value remember the range that it's associated with so this guy is a function all right so let's go through and just look at a few examples and then again as we move on throughout the next few lessons we'll start working with this in the format of an equation it'll be a little bit more challenging but we'll get through it so here we have f and f is again just a set of ordered pairs so we have 2 comma 6 9 comma 1 2 comma 13 and 8 comma 7. is this a function remember one of the definitions i read to you you basically find out if you have a duplicate x value you don't have a function right here you have a 2 and here you have a 2. well what's the problem 2 is associated with 6 and then 2 is associated with 13. you can see that very clearly in the mapping diagram an x value of 2 is associated with or linked up with a y value of 6 and also a y value of 13. so there is no clear association here if i said hey if x is 2 what's y your answer could be 6 or it could be 13 you don't know what to answer for a function essentially if i give you a value for x if i give you an input you have to be able to give me a very specific output right or a value for y or a value from the range so for 2 again it's associated with 6 and also 13 so this is not a function right the fact that nine is associated with one and a is associated with seven those are fine it doesn't violate the definition but because this fails the whole thing fails so we would say this is not a function let's look at another one so we have g here and it equals again just a set of ordered pairs one comma four negative seven comma three two comma five and eight comma six so any duplicate x values no right so you can say that you have a function here an x value of one is associated with a y value of 4. an x value of negative 7 is associated with y value 3. an x value of 2 is associated with y value of 5 and an x value of 8 is associated with y value of 6. so for each x there's one y value that's associated with it so this is a function let's take a look at another one this one i'm going to tell you in advance is a little bit tricky so we have h is equal to again we have a set of ordered pairs so negative 2 comma negative 1 3 comma 4 6 comma negative 1 and 9 comma 8. okay so some of you would immediately say okay you have a duplicate y value here of negative 1 not a function b would be wrong if you said that we say that for each x value it's associated with or linked up with or corresponds to one y value if i said hey what is y when x is negative two there's a clear answer it's negative one what is y when x equals six there's a clear answer it's negative one it doesn't matter that these two x values are linked up to one single y value that's all right it doesn't violate the definition of a function right the problem comes when one x is linked up to more than one y but two different x values or you could say two or more x values could be linked up to the same y right that's fine so for each x here there is one y value that's associated with it negative two is associated with negative one six is associated with negative one three is associated with four nine is associated with eight so this is a function let's look at two examples with some non-numerical information you'll get these also so what this set is representing this guy f it's going to represent take off cities so you're kind of flying somewhere and destination cities okay so the first guy is where you're taking off from the second guy is where you're kind of landing or your destination so you have new york common miami you have boise comma san diego baton rouge combat orlando and denver common dallas so again i'm just looking for duplicate first components here right duplicate input value so the idea here is that i input a value for the takeoff city and i should get an output as the destination city so i have new york boise baton rouge and denver none of those are repeated right so this guy's gonna be a function we can see it more clearly with this mapping diagram again the acts of the domain those are the takeoff cities the y of the range that's going to be the destination cities a flight from new york is associated with the destination of miami a flight from boise is associated with a destination of san diego a flight from baton rouge is associated with a destination of orlando and a flight from denver is associated with a destination of doubt so each acts or member of the domain each takeoff city is associated with one and only one y or member of the range or destination set so we can say this is a function let's look at one more problem and then we'll kind of move on again in the next lesson we'll start looking at things that are much more complicated so for g here again we're going to have the same thing take off cities and destination cities so it's equal to we have this set with portland common chicago seattle common houston los angeles common pensacola and los angeles comma little rock so immediately you see the first component here and here those are the same so this violates the definition of a function right i'm taking off from los angeles and there's no clear association between my destination city am i going to pensacola am i going to little rock you can see this more clearly in the diagram where we show the mapping again from x the domain the takeoff cities to y the range the destination cities so a flight from portland lands in chicago a flight from seattle lands in houston a flight from los angeles lands in pensacola and little rock okay so that's where you get the problem this x value this member of the domain los angeles is associated with pensacola and also little rock so if i take off from los angeles i don't know where i'm going to land am i landing in pensacola am i landing in little rock that might be a problem if i'm trying to book a hotel or have somebody pick me up so this guy is not a function in this lesson we want to review domain and range so in the last lesson we talked about relations and functions and we also touched on domain and range but we didn't quite go as far as we needed to essentially in our course we're going to come across a lot of different functions and we need to understand the thought process involved in determining what the domain is and what the range is so before we kind of get into this let's start out with an easy example and just recap what we know so we have f here and f is a relation and it's also a function so we talked about again in the last lesson that a relation is any set of ordered pairs so f here again it's equal to we have four ordered pairs three comma seven two comma six eight comma one four comma five then we also talked about the fact that a function is a special type of relation where for each x or member of the domain there's one y or member of the range that's associated with it or you could say linked up with it or paired with it whatever kind of word or phrasing you want to use for that essentially with f here we know that it's a function because an x value of 3 is linked up with just a y value of 7 then an x value of 2 is linked up with just a y value of 6 an x value of 8 is linked up with a y value of 1 and then an x value of 4 is linked up with a y value of 5. so for each x value here or remember the domain it's linked up with exactly one y value or member of the range now this is a very easy example we're going to look at some equations here and that's what we're going to see kind of moving forward these examples are just so that we can get the concept of you know relations and functions and then domain and range under our belt so real quick what's the domain and what's the range here again the domain of a relation is just the set of allowable x values okay and then the range for a relation is the set of kind of y values that you're going to have so for this guy f if i ask for the domain and i say okay the x values here are 3 2 8 and 4. that's all the possible values because i only have 4 of them right so the domain here if i just want to put it inside of a set of curly braces i would say it's 3 two eight and four okay so it's a set with four elements then the range okay the range is going to be very similar so it's just the y values so it's seven six one and five okay so seven six one and five so everything's constrained here because we have four ordered pairs involved so i just have four x values and four y values now as we move forward in our course obviously things get more complex we're going to get equations that are relations we're going to get equations that are functions and we need to know how to find the domain and the range so for something like this we have y equals 3x minus 4. so most of you know this is a linear equation in two variables and it's also considered a linear function because for each x value here we're going to have one y value but we'll talk more about how to determine if a relation is a function in kind of an equation format in the next lesson when we look at the vertical line test here we're specifically going to just focus on domain and range so for this guy here if you want to find the domain you want to start by asking yourself a very simple question since the domain is the set of allowable values for x is there anything that i can't plug in there for x is there some type of restriction that's what you want to ask yourself is there something that's not allowed okay something illegal well i'm multiplying x by 3 and then the result of that i'm just subtracting 4 away from it and that's what's going to give me my output or y so the answer to that is no i can make x negative i can make x positive i can plug in square root of 2. i can do whatever i want so for x there really it could be any real number to be plugged in right so the domain is a set of all real numbers so the domain is the set of all real numbers so there's a lot of ways you could write that i'm going to choose to write this in interval notation because i feel like this is the most convenient way to do it you might see a variety of different ways to write this in your textbook you might see set builder notation which i'll show you in a little while and you might also see them just put all real numbers inside of some kind of curly braces as long as you notate that it's all real numbers it really doesn't matter how you do it then for the range for the range let's think about the possible outcomes for y so i'm plugging in something for x it's the independent variable it's the input it's what i'm choosing i choose something for x i multiply by 3 i subtract a 4 i get a result or an output for y right that's the dependent variable so the range is going to be all real numbers as well because again if i can make x as large as i want or as small as i want then y can be as large as i want or as small as i want right if i chose negative 1 trillion and i plugged it in for x i could make y really small right and i can make y even smaller than that by choosing even smaller numbers for x i can also make y very big by choosing 1 trillion right positive 1 trillion for x and again i can go above that if i want to make y even larger so for the range it's all real numbers as well so from negative infinity to positive infinity all right now graphically we see this this is the equation y equals 3x minus 4 and this is graph force already we'll review how to graph linear functions or linear equations and two variables in a few lessons for right now just take the graph as given and essentially you can see that the arrows at each end of the line indicate that the line continues in each direction forever so we can see visually that the domain or the set of all possible x values is going to be all real numbers and the same for the range or the set of all possible y values that's going to be all real numbers as well let's look at one that's a little bit more challenging so now we have y equals 1 over x minus 4. so again if you want to find the domain start with x and just say is there something i can't do well if you have x in the denominator we already know from our section on rational expressions we have a restricted value here because we can't divide by zero okay so when a variable's in the denominator you've got to think about what would make this denominator zero so i would set x minus four equal to zero and i would solve that very simple equation so add four to each side this cancels you get x is equal to 4. so essentially the domain here is any real number okay because i can do anything else i want except for 4 right i can't plug in a 4 there because i'm dividing by 0 but anything else is fine so for the domain for the domain i could put in interval notation from negative infinity up to but not including 4 and the union width from 4 again not including that out to positive infinity okay so that's one way to write it this guy right here would probably be faster to write in set builder notation you might see this in your textbook so it looks like this so inside of some set braces you have x and you say the set of all x it just means that we have real numbers and then you see this vertical bar that means such that and then you'll have your condition okay so the condition here is that x does not equal four okay so basically the set of all x just means that you have real numbers involved such that that real number okay is not equal to four that's all this is saying it's just a fancy way to say it in this case it's a little bit shorter but i'm going to stick with my method okay so for the range it's a lot more complex to think about when you look at this graphically which we'll see in a minute it's very easy to see but when you think about this just by looking at the equation it can be a little bit tough at first so i want you to think about the fact that you're going to get an output for y based on what you plug in for x okay i have 1 which is a fixed value divided by x minus 4 which is not fixed right because x can be changed so if i divide by larger and larger values here i'm taking one i'm dividing it by larger and larger values y is going to get smaller and smaller but it's never going to actually get to zero okay and the reason for that is is that if i want to divide and get a result of zero i've got to do zero divided by some non-zero number this is fixed it will never be zero okay so the result of this or you could say y will never be zero so automatically i know that y will not be zero okay because i'd have to have a zero in the numerator and it's fixed at one okay so that's not possible so now you can start thinking about can i make y really really big can i make y really really negative the answer to that question is yes i can basically do anything except make it 0. so i could plug in something like 5 for x 5 minus 4 is 1 1 over 1 is 1. so i know that y would be equal to 1 if x was equal to 5 okay so if x equals 5 y is 1. as i increase x what happens is again i'm dividing 1 by a larger and larger number this y gets closer and closer to 0. so let's say i said x was 14. okay well in this case y is going to be 1 10 right because 14 minus 4 is 10 you'd have 1 over 10. so that's 0.1 again as i increase x i'll get closer and closer to 0. now if i wanted to make y really large on the positive scale then what i do is i choose a number between 5 and 4 okay so let's say i chose 4.5 just to give you an example 4.5 minus 4 would be 0.5 1 divided by 0.5 is equal to 1 times 2 okay and that equals 2. so i'm increasing now i'm starting with 1 and i'm going to 2. so as i get closer and closer to 4 without touching 4 what happens is i make y infinitely large so let's say that i chose a value like let's say 4.0001 okay just to make it interesting well i'm subtracting away 4 so i'm going to end up with 1 divided by .0001 and what's that equal to well that's going to give me 10 000 right as a result so you can see how large i made this and again i can make this infinitely large if i just keep getting closer and closer to 4 without touching it i could erase this 1 here and make a lot of zeros i can keep going out forever and ever and ever and then put a 1 at the end again when i subtract away 4 i'm left with kind of this part of the number in my denominator 1 divided by that's going to be really really large okay so i know that y can't be 0 i know that y can be infinitely close to zero and i know y can be as large as i want so now let's think about what happens on the negative side of things can i get close to zero from the negative side meaning can i get something like negative point zero zero zero zero one or something like that and then can i go to negative infinity well the answer to that is yes right you're just coming from the other side so you think about plugging in a three there three minus four is negative one one over negative 1 is negative 1 okay so we know we can do that as i increase 3 going towards 4 right getting very close to 4 but not touching it what happens is i'm going to make y infinitely negatively big okay so i'm going to have a really really small number let's say i did something like 3.999 as an example well if i subtract that what i'm going to get is 1 over negative .001 and if i do this i'm going to end up with negative 1000 okay but i can go out further i can do a 9 a 9 a 9 a 9 i can just keep going and going and going so i can make y as big negatively or you could say a really really small number i can do that as far out to negative infinity okay so that's not an issue and again i can also get close to zero coming from the other direction if i plug in smaller and smaller numbers for x let's say i plug in a negative 6 here well negative 6 minus 4 is negative 10 y would be equal to what it would be 1 over negative 10 which is negative 1 10 or negative 0.1 and you can keep decreasing x as much as you want you'll get as close to zero as you want but you'll never actually touch it so from this kind of experimentation that we've done here we can see that y just can't be zero it can be as small as we want it can be as large as we want it can be as close to zero as we want but it just can't ever be zero so we'll use the same kind of notation we'll say from negative infinity up to but not including zero in the union with not including zero out to positive infinity again you could see this in set builder notation the set of all y again this is just meaning that y is some real number such that in your condition y is not equal to zero okay so that's your set builder notation all right so graphically this is going to be a lot easier to see and i would say that in most cases you're just going to graph these things and just kind of observe what the domain is what the range is if we look at the x values again that's the horizontal axis notice that the graph from either direction does not touch an x value of 4. this is an x value of 4 right there you have this vertical line that's kind of drawn here it's a dashed line at x equals 4 and you see that coming from the left and going to the right we approach the line but don't touch it coming from the right and going to the left we approach the line but don't touch it this vertical line is known as an asymptote and we'll talk more about these when we start graphing polynomial functions and rational functions this is going to be in the next section of our course but for now you can see that the graph of this equation does not ever go where x is 4 okay it does not go where x is for again this is y equals 1 over x minus 4 and the reason for that is if we plug in a 4 for x we have division by 0 which is undefined okay now additionally you have the same phenomenon that's horizontally occurring at a y value of zero okay see where y is zero that's right there okay you have a horizontal dashed line again that's an asymptote you'll see that coming from the left and going to the right our graph approaches that line but won't ever touch it and it's the same thing if we say coming from the right and going to the left it's going to approach the line but not touch it so again graphically we can see that the domain consists of all values except for an x value of four okay you can see horizontally there's no x value of four on either part of that graph then again vertically if you look at the y values the range here it's going to just exclude 0. so all real numbers except for 0. all right let's look at another example so here we have y is equal to the square root of x plus 1. so what do we know about square roots what can't we do we can't in the real number system take the square root of a negative so that means when i'm considering here my x or my domain values what can i do i can't take the square root of a negative so that means this radicand here this x plus 1 has to be 0 or positive so it has to be non-negative so i would say that x plus 1 must be greater than or equal to 0 i would solve this inequality i'd have find that x has to be greater than or equal to negative 1. so my domain here my domain is from what it starts at negative 1 and includes negative 1 and it's out to positive infinity now let's think about the range this one isn't too difficult we think about the fact that if i plugged in a negative 1 for x negative 1 plus 1 would give me 0. so i know 0 is possible because the square root of 0 is 0. so for the range i know that i could get a value of zero okay that would be included can i get something larger well yeah i can just keep making x larger and larger i can make it infinitely large if i want and so the y values would increase infinitely right it would go out to positive infinity i can't ever get below 0 because if i decrease x below negative 1 i get something that has to involve a complex number system and we're only dealing with real numbers now so i don't want to take the square root of a negative so my range is kind of constrained to be zero at the smallest and zero's included out to positive infinity okay so the domain again from negative one including that out to positive infinity and the range from zero out to positive infinity all right so this guy we can see it graphically so again we have y is equal to the square root of x plus 1. so you can see that the smallest value again when you look at this horizontally when you look at the x values this guy is negative one that's the smallest value this graph takes on and then it's anything going out to kind of positive infinity i'll put my arrow there let me use this color here it's a little bit closer and for the range although it doesn't seem like it it's going to keep increasing right it's going to keep increasing so it would also go out to positive infinity the smallest value on the kind of y-axis is here right that's 0. so it starts at 0 and goes out to positive infinity for the x values it starts at negative 1 and goes out to positive infinity so again the domain is anything from negative 1 out to positive infinity the range is from 0 to positive infinity all right let's look at one more of these and then we'll get into some more complicated examples so now we have y equals x squared so y equals x squared so let's think about this again is there anything that i'm restricted from when plugging in for x well no i can plug in anything for x and square it i can plug in a negative i can plug in a positive i can plug in 0 i can plug in square root of 2 square root of 10 whatever i want right it doesn't matter so my domain would be all real numbers my domain would be all real numbers so from negative infinity to positive infinity there's no issues but what about the range okay what about the range well when i square something i know that the result is what it's either 0 or it's positive it can't be negative if i take a negative and i square it i get a positive if i square 0 i get 0 square positive i get a positive so the range the smallest this guy can be is a zero and that only happens when i square zero so the range would be from zero and including zero out to positive infinity so we can see this again if i think about the x values and i'm just gonna put these arrows in here so the x values i think about the horizontal axis this guy's going to continue going this way and this way forever okay so this is going to keep expanding left and right so my domain values it's just all real numbers anything i want to choose i can plug in it for x and i can square it for the range values again what happens is when i square my x it makes it non-negative so the smallest value that y can take on again if i look at this guy let me use a different color because it's not showing up why the smallest it is is zero and then it can take on any value larger than zero so again the domain consists of all real numbers and the range consists of zero and then anything positive out to positive infinity all right so let's look at a typical example you'll see in this section so in this example i'm only going to ask you for the domain so y equals we have 2x squared plus 9x minus 1 over we have the square root of 5x minus 3. now if we think about just the domain you have an x here here and here so are there any restrictions in the numerator if i just gave you y equals 2x squared plus 9x minus 1 is there anything i can't plug in for x there the answer to that is no right i can plug in whatever i want when i think about the denominator there's a problem because i have the square root operation and also in a denominator i can't divide by zero so there's two things you really have to consider here the first thing is that this radicand 5x minus 3 has to be greater than or equal to 0 and then also we have to have that 5x minus 3 is not going to be equal to 0 okay so those are two different restrictions we have to consider so for the first one let's add 3 to both sides of the inequality we'll find that 5x is greater than or equal to 3. let's divide both sides of the inequality by 5 and we're going to get that x is greater than or equal to three fifths so at this point we know that x needs to be greater than or equal to three-fifths but also we have to consider this value here that's going to be restricted so let's solve this equation let's add 3 to both sides and we'll get that 5x is equal to 3. let's divide both sides by 5 and we're going to get that x is equal to three-fifths so what we found is that what we found that x is greater than or equal to three-fifths allows us to plug in a three-fifths here five times three-fifths would give me a three three minus three is zero square root of zero is 0. so that's okay we can take the square root of 0 right that's fine but i can't divide by zero so what i have to do is take this guy into consideration and remove this kind of non-strict inequality and make it a strict inequality right because three-fifths has to be restricted because it makes the denominator zero so here x needs to be strictly greater than three-fifths so for this one i'll just put that the domain is going to be anything larger than three-fifths so three-fifths is not included out to positive infinity all right let's wrap up the lesson with one that's a little bit more tedious so again i'm just looking for the domain so y equals we have the square root of 2x minus 3. this is over 7x squared minus 36x plus 5. so again when i think about my x values involved we're considering two things here for one i have a square root operation involved so 2x minus 3 2x minus 3 which is my radicand has to be greater than or equal to 0. then additionally since this is in the denominator the 7x squared minus 36x plus 5 it can't be equal to 0. so 7x squared minus 36x plus 5 cannot be zero okay so let's think about this one first and then we'll consider this one and then we will kind of merge the two so for this guy right here let me add 3 to both sides that's going to cancel i get 2x is greater than or equal to 3 i'll divide both sides by 2 and find that x needs to be greater than or equal to 3 halves so x is greater than or equal to 3 halves so let me erase this let me just write that up there so again for this first part x is greater than or equal to three halves now let us consider this guy right here this is a quadratic equation that we need to solve again the values here that i find that satisfy the equation are going to make the denominator 0 so they've got to be restricted so let's solve this with factoring this one's pretty easy to use reverse 404 because 7 is a prime number so i'll do 7x and x and what do i want well i know that this last guy is five here right this is five and five can only come from one it can come from negative one times negative five or it can come from positive one times positive five now this middle term here is negative so i know i need a negative 1 and a negative 5. now what's the order going to be well we know that you're going to need a big number here that's negative so i would put a negative 5 here and a negative 1 there the outer would be negative 35x the inner would be negative x negative 35x plus negative x would give me the negative 36x and negative 1 times negative 5 will give me positive 5. so this is the correct factorization so all i need to do now is just set each factor equal to 0. so 7x minus 1 equals 0 or x minus 5 equals 0. for this one let me add 5 to both sides i'll get that x is equal to 5. so that's one restricted value so x can't be 5. and then for the other one let me add 1 to both sides of the equation i'll get 7x is equal to 1. let me divide both sides by 7 i'll get x is equal to 1 7. so x can't be equal to 1 7. now let's think about all these restrictions together and we'll come up with something that's conclusive now we said specifically that x needs to be greater than or equal to three halves in decimal form this is 1.5 x can't be 1 7 but 1 7 is already less than 3 halves right it's less than 1.5 if i divide 1 by 7 i'm going to get about 0.14 okay not exactly but about 0.14 so this guy i don't even need to list it because it's going to be less than this and i've already restricted the domain to say it's got to be greater than or equal to 3 halves so what i can say since 5 is larger than that is that the domain is going to consist of numbers starting with three halves that would be included up to but not including five and then the union with anything larger than five out to positive infinity in this lesson we want to review the vertical line test so over the course of the last few lessons we've been reviewing relations and functions along with related topics such as domain and range so at this point we should fully understand the definition of both a relation and a function and also how to find the domain and range of a relation so now we're just going to go a step further and review the vertical line test so again i know most of you took algebra 1 and algebra 2 prior to joining this course and essentially you already know what the vertical line test is but in case you've never seen this topic before the vertical line test is just a visual way to determine if a relation is a function now before we go too far into this i want to make sure that you understand the concept of a vertical line itself so with a vertical line no matter what value you choose for y x always just equals some number so the way this equation is set up is it's x equals some real number let's just call it a so a could be square root of two a could be negative a trillion a could be positive one billion it could be any real number you want it to be essentially no matter what you choose for y x just equals a and a lot of people just say oh well where's y there's no y involved in this equation we can use a little trick and say that this is x plus i can put a 0 as the coefficient of y and then put y and this equals a so now i've written it as a linear equation in two variables and i can put it on my coordinate plane so this is a way that we can kind of graph this guy and see what it looks like visually so again no matter what i choose for y i'm multiplying it by zero so it goes away and x just equals this real number a so to see an example of this suppose we're given x equals negative four the quickest way to graph this is just to go to negative 4 on the x-axis and draw a vertical line this is what you're going to do throughout most of the time that you're graphing these guys but when you first start out you might see it in this format just to make it crystal clear what's going on so again you might expand it and say x plus zero y equals negative four you might choose some values for y just to see what's going on so you might say let y be negative four let y be zero let y be positive four well okay if i plug in a negative four for y what's x well 0 times negative 4 is 0 x is negative 4. same thing goes if i plug in a 0 for y 0 times 0 is 0 x is negative 4 and with 4 as well 0 times 4 is 0 x is negative 4. so the idea here is that x is always negative 4 no matter what you choose for y so this is going to be important for us to understand when we look at a vertical line it's always going to have the same x value okay the x value will not change and we'll talk about why that's important in just a minute but let's just write these ordered pairs out so this one's negative four comma negative four this one's negative four comma zero and this one's negative four comma four okay so let's go down to the coordinate plane so here's our graph of x equals negative four and i've already pre-drawn this stuff just to make it a little quicker we found some ordered pairs we said we had negative four comma negative four which would be right here right four units to the left four units down we had negative four comma zero which would be here right just 4 inches to the left and we had negative 4 comma 4 so 4 units left and 4 units up so again no matter what i choose for y x is always just negative 4 okay and the quick way to do this is just to go to negative 4 on the x axis and sketch your vertical line as another example of this suppose you've got something like x equals seven again just find seven on the x-axis draw a vertical line it's really just that simple now the main takeaway from this is that i want you to understand that again with a vertical line this one single x value is always the same and you can see that it corresponds to an infinite number of y values so we don't have a function here but the basis of how this vertical line test is going to work is that if a vertical line intersects the graph of a relation at more than one point then the relation is not a function this is because the single x value again a vertical line has a single x value it's going to correspond to more than one y value because you're impacting the graph in more than one location now if this doesn't make sense right away that's okay when you see an example it will so let's look at y equals 2x minus 3. so this guy is a linear function i'm just going to tell you in advance this is a function and whenever you work with a linear equation in two variables as long as it's not a vertical line like we just saw it's going to be a function okay so for each x there's one y now when we look at this kind of graphically i can see that no matter where i draw a vertical line it's only going to impact the graph once and so for each x there's one y and you can go through and just make some vertical lines and that should be enough and i'll just put some arrows at each end just for the sake of completeness but again if you observe these look they only hit the graph once in each case so here here here and here right so when you don't have a function it's going to hit the graph more than once so this guy is a function we'll just label this as a function all right let's look at one that's not a function so we have x squared plus y squared equals 36 we already talked about circles in our course we know this is the graph of a circle that has a center zero comma zero and has a radius that is six okay but essentially what we want to understand here is that visually we can determine this is not a function by graphing this guy and then we can draw some vertical lines we'll see that the vertical lines hit this circle in more than one location and i'll just show you with three lines again you can just do one as long as one hits the graph in more than one location you know you don't have a function right so this guy hits here and here this guy hits here and here this guy hits here and here so let's think about an x value of 0 corresponding to a y value of 6 and negative 6. so for that one x value you've got two associated y values again 6 and negative 6. so think about why this happens let's take this equation and solve it for y so we have x squared plus y squared equals 36. how do we solve it for y well if i want to solve for y i want to subtract x squared away from each side of the equation i'll have that y squared is equal to negative x squared plus 36. to get y by itself i'm going to take the square root of each side but remember i've got to take plus or minus the square root of the right side and this is kind of associated here so i've got to move this down manually so negative x squared plus 36 i'm going to go plus or minus the square root of this and i'm going to take square root of this and so i'll have y is equal to plus or minus the square root of negative x squared plus 36. so where does my problem come in at well it's from this plus or minus right it's from this plus or minus because let's say i plug in my 0 here for x we already said it was associated with a y value of 6 and then also negative 6. so if i plug in a 0 there i plug in a 0 there 0 squared is 0 the negative of 0 is 0 so i just have 36 so i'd have the positive square root of 36 which is 6 and the negative square root of 36 which is negative 6 so you can see that this guy right here is what creates the problem okay for that one x value of 0 you get 2 associated y values again 6 and negative 6 so this guy is not a function so we'll come back up here and say this is not a function but again you can clearly see that from the graph if a vertical line impacts it in more than one place not a function all right let's look at another example so this guy is going to be a parabola so y equals we have the quantity x minus 3 squared plus 2. we're going to work a lot with parabolas in the next section of the course this is going to be the graph of a quadratic equation okay so let me put some arrows here and here and what do we notice here we notice that this is a function right if i go through and i draw vertical lines no vertical line is going to impact this guy in more than one spot okay so i'll just do four you can see that this guy hits it here let me put my arrows at each end just to make this complete okay so this guy hits it here this guy hits it here and this guy hits it here okay so any vertical line you draw is only going to hit the graph in one location so this guy is going to be a function okay well there's something i want you to observe here and this confuses a lot of students remember we had these problems that we looked at where a single y value was sometimes associated with more than one x value and we said that that was still a function right that was okay so in other words i could have a set of ordered pairs where i had two comma five and then let's say i had three comma five this is a function right because 2 is associated with 5 so for this x value of 2 i know that y is 5 and 3 is associated with 5. so for this x value of 3 i know that the y value is 5. so this is ok the same y value can be linked up to more than one x value it's just that an x value can't be linked up to more than one y value okay so in this particular case what you'll notice is that with the exception of a y value of 2 okay this would be a y value of 2 you'd have an x value of 3 there every other y value is linked up with more than one x value that's because of this squaring operation here okay that's what's creating that if i plugged in something like let's say 2 for x 2 minus 3 is negative 1 negative 1 squared is 1 1 plus 2 is 3. so i would get a y value of 3 is if x was 2 right but also if x was 4 i would also get a y value of 3. so i plugged in a 4 there 4 minus 3 is 1 1 squared is also 1 1 plus 2 is going to give me 3 as well so that's ok it's ok that an x value of 2 is associated with a y value of 3 and also an x value of 4 is associated with a y value of 3. that's fine it doesn't violate the definition of a function so i don't want you to get confused it's a very common mistake this guy is going to be a function all right let's look at another example where we don't have a function so this is a sideways parabola we will talk about these later on also so in this case you have the quantity y minus 3 squared is equal to x plus 2. so clearly if i draw vertical lines they will impact this graph in more than one location so i'll just draw those two to show you and we'll talk a little bit more about this we'll solve this equation for y and see what's going on but essentially you can see that an x value of 7 is associated with a y value of 0 and also a y value of positive 6 and then an x value of 2 is associated with a y value of 1 and also of 5. okay so this violates the definition of a function so this guy is not a function all right so to see why this is the case let's go down to the next page and i'll rewrite this equation i want to solve it for y i want to think about this for a minute with you so y minus 3 quantity squared equals x plus 2. how can we solve this for y well what's going on with y y is inside of these parentheses here and 3 is being subtracted away from it and it's being squared right so the first thing i would do is i want to undo the squaring operation so let me take the square to this side but again i've got to go plus or minus over here before i take the square root of this side which again was x plus 2 okay so this is going to cancel with this so now what i have is y minus 3 is equal to plus or minus the square root of x plus 2 and if i add 3 to both sides i can get rid of this and i can essentially say that i have y equals plus or minus the square root of x plus 2 plus 3. now this is what's going to give me the problem again if i think about what i saw up here an x value of 7 gave me a y value of 0 and then also a y value of 6. so if i plugged in a 7 there i can see why 7 plus 2 is 9 the principal square root of 9 is 3 3 plus 3 is 6. then if i do 7 plus 2 again i get 9 the negative square root of that is negative 3 negative 3 plus 3 is 0. okay so that's how the x value of 7 is corresponding to or linked up with or paired with two different y values right in this case it's zero and then also six so again this guy is not a function for the last one it's a little bit tricky we have y minus 1 this quantity cubed equals x so far every time we've seen y wrapped in some parentheses we haven't had a function right so this is one that kind of throws students off so let me just put some arrows in here real quick and let's think about this for a second if i went through and drew vertical lines you can see that none of them would intersect the graph in more than one location with the exception of you might think that it would hit more than once if you drew it right here right where an x value is 0. it's because of the way the graph gets drawn but if you really look at if you had a computer model and you know you're online and you use one of those kind of graphing calculators you can zoom way in you can clearly see that this guy right here is doing this okay it's not ever stagnant okay so this guy a vertical line is not going to intersect that graph in more than one location if i drew a vertical line at x equals zero it would only impact the graph in one location it's just the way it's drawn okay so i can go through and make some vertical lines and i'll draw one at x equals zero i'll just draw one more over here and again in each case this is going to impact the graph in one place only now to show you this completely let's solve this for y and again we can think about an x value of zero so let's go down here so we have y minus 1 cubed equals x how am i going to solve this for y well what i'm going to do is i'm going to take the cube root of each side so this is going to what this will cancel with this and what i'm going to have is just y minus 1 and this will be equal to the cube root of x and i can add 1 to both sides of the equation so what i'm going to end up with is just y is equal to i'll have the cube root of x and then plus 1. okay so let's think about an x value of 0 now if i plugged in a 0 there am i going to get two different outputs for y the answer to that is no the cube root of 0 is just 0 y would just be 1 and 1 only so for the given input of zero the y value is only going to be one and that's the one that we really call in a question if i kind of erase all this and we go back a lot of times when you look at this graph again if you look at an x value of zero that's where it looks like a vertical line would kind of hit the graph in more than one location but once you solve it for y and you plug in for x you can see that that's not the case right an x value of 0 just gives you a y value of positive 1 and positive 1 only so this guy is going to be a function and i erased my arrow so let me draw that back in so once again this is a function in this lesson we want to review function notation so up to this point in our course we've worked with functions where we used y as the dependent variable and x as the independent variable so now we want to go a step further and think about another way to kind of notate a function this notation is known as function notation so let's begin by looking at a few basic functions so we have a linear function y equals five x minus seven we have a quadratic function y equals x squared minus three x plus four and we have the square root function y equals the square root of two x minus one so at this point we should know these three relations or functions if i input something in for the independent variable x i'm going to get a unique output for the dependent variable y right so for each x there's going to be one y now when we work with functions you're going to typically see the name of the function as f g or h although other letters could be used and then what we're going to do is we're going to replace y with a notation such as f of x g of x or h of x so i'm just going to use a different one in each example so i'm just going to change my y into f of x i'm just going to say f of x is equal to 5 x minus 7. okay so these are the same i'm just replacing y with f of x and i want to note here that these parentheses do not mean multiplication this is read f of x okay it's not f times x a lot of students get confused on that for this one i'm going to replace it with g of x so instead of y i'll just say g of x is equal to we have x squared minus 3x plus 4. and then for this one i'm going to use h of x so i'm going to say that h of x is equal to the square root of 2x minus 1. so i don't want this to confuse you all we did was we replaced the dependent variable in each cache with either f of x g of x or h of x so in other words this function up here on the top its name is f and it's a function of x so we just say f of x right instead of y here we need our function as g so we say g of x here we name the function as h so we say h of x right we're just saying that the function depends on x that's what this notation is for so you might be asking why would this notation be used what can we do with it well one immediate use for this is the ability to quickly indicate what we would like to evaluate a function for given a value of x so you've seen this kind of exercise before where we say evaluate you know some equation in this case it's y equals 5 x minus 7 for a given value of the variable in this case we have x equals 2. so we know at this point we just plug in a 2. it's very easy so y equals 5 times 2 is 10 10 minus 7 is 3 right so y equals 3 when x equals 2. but using function notation we can kind of shorten this process what we can do is first say that this is f of x is equal to 5x minus 7. so this is my function here and if i want to ask for the function's value when x is 2 all i'm going to do is put a 2 in for x in these parentheses here so i would say f of 2. this is the function's value when x is 2 this equals 5 times i'm going to plug in a 2 for x then minus 7 and this equals what 5 times 2 is 10 10 minus 7 is 3. so f of 2 equals 3 tells me this is the function's value when the independent variable x is going to be 2. okay so f of 2 here is 3. so suppose i wanted to change this up and i wanted to plug in a 9 here what am i asking for so f of 9 equals what well i'm asking for the function's value when the independent variable x is 9. so again i'm just plugging in it's just like if i said for x equals 9. it's the same thing okay so i'm going to have 5 times 9 minus 7 5 times 9 is 45 then if i subtract away 7 i get 38 so this equals 38. so again f of 9 equals 38 is telling me that the function's value when the independent variable x is 9 is 38. all right let's look at an example with our quadratic function so instead of y we have g of x now and this equals x squared minus 3x plus 4. so again all i did was replace y with g of x so if i want to find g of negative 2 g of 5 g of a and g of a plus 1. all i'm doing when you get this type of example you're just plugging in for the independent variable what you're given inside of those parentheses so for example g of negative 2 all i would do is plug in a negative 2 everywhere i see an x so here and here okay so i would have negative 2 that quantity is squared make sure you wrap it in parenthesis because i'm plugging in for x a negative 2. okay so the whole thing's got to be squared then minus you've got 3 times negative 2 then plus 4 okay so negative 2 squared is 4. let me just erase this and put 4. negative 3 times negative 2 we know that's 6. so this is plus 6. and if we go through here we know 4 plus 4 is 8 8 plus 6 is 14. so g of negative 2 is going to be 14. all right so let me just write this off to the side g of negative 2 is going to be 14. okay so again that's the function's value when the independent variable x is negative 2. let me erase this and use some highlighting instead that might be a little cleaner all right so let's go to the next one so now we have g of 5. again all i'm doing is if i have g of 5 i'm saying what is the function's value when the independent variable x is 5 okay so just plug in you'd have 5 squared so 5 squared minus 3 times 5 then plus 4. again all i did was i plugged in a 5 there and there 5 square is 25 minus 3 times 5 which is 15 and then plus 4 okay so 25 minus 15 is 10 and then 10 plus 4 is also 14. so we can say g of 5 is 14 as well all right so now let's look at one that might confuse you so what if you get g of a okay g of a what do i do for that so g of a let me write the whole thing we have what we have x squared minus 3x plus 4. so what is g of a again what did i do when i had g of 5 i plugged in a 5 for each occurrence of x my independent variable well if i have a i'm just going to do the same thing okay don't let this confuse you whatever they give you inside the parentheses plug it in for the independent variable okay so i'm just going to replace this with a and i'm going to replace this with a and g of a is just equal to a squared minus 3a plus 4. that's all it is all right so let me kind of slide this up and let's look at the last one so now we have g of we have the quantity a plus 1 this equals what again just plug in so i have x squared i'm going to kind of give myself a little room then minus 3 times x give myself a little room then plus 4. the reason i gave myself some room in each case is because i'm plugging in this a plus 1 for each occurrence of x so remember this has got to be squared so it's going to be the quantity a plus 1 that's getting plugged in for x and that whole thing is squared then minus 3 times x so x again this is the quantity a plus 1 and then nothing else to do there so let's scroll down and let's kind of simplify this so a plus 1 that quantity squared we know that would be a squared plus 2 times a times 1 2 times 1 is 2 times a is just 2a and then plus 1 squared is 1 and then you'd have this basically negative 3 times each term so you'd have minus 3a and then minus 3 and then plus 4. okay so this is my g of a plus 1. so don't let this get you confused okay very easy whatever they're giving you inside the parentheses just plug it in for the independent variable in this case it's x but it could be any situation that you're given right whatever the independent variable is of that function plug it in so now i'm going to simplify so g of a plus 1 i have a squared nothing to combine with that i have 2a minus 3a that's negative a so minus a then i have 1 minus 3 which is negative 2 then plus 4 which is positive 2. so g of a plus 1 is equal to a squared minus a plus 2. all right let's take a look at another example so we have h of x is equal to x minus 1 over x plus 1. so we want to find h of one h of negative one h of z and h of z minus three so again in each case i'm just plugging in right so if i want h of one i can do this right here there's enough room this is equal to one plug in a one here and here okay let me use a highlighter because again that's a little cleaner so what am i going to have i'm going to have 1 minus 1 which is 0 over 1 plus 1 which is 2 0 over any non-zero number is 0. so h of 1 is going to be 0. okay and i can leave that up here all right what about h of negative 1 well most of you see that you're going to have a problem there right negative 1 is restricted from the domain because if you plug in a negative 1 for x there you get negative 1 plus 1 that's zero if you have zero in the denominator you have a problem right it's undefined so this guy is restricted from the domain we could say h of negative one is undefined okay now let's talk about h of z and again don't get confused when they give you another kind of letter to plug in because it's the same thing if i had h of 1 i plugged in a 1. if i had h of negative 1 i plugged in a negative 1. if i have h of z i just plug in a z so h of z is equal to what it's z minus one instead of x minus one over z plus one okay that's all it is and then h of z minus three let's kind of scroll down and get some room so h of z minus 3 is going to be equal to y again so rewrite this i know it's cut off from the screen it's x minus 1 over x plus 1. so just plug in for x so i'm going to plug in a z minus 3. so i would have z minus 3 and then we would have minus 1. and then again i'm going to have z minus 3 and then plus 1. so z minus 3 and then plus 1. okay so z minus 3 minus one again you can just drop the parentheses i just did that to show you that i'm plugging in negative three minus one is negative four so you just put z minus four in the numerator in the denominator negative three plus one is negative two so let's put z minus two and let me make the z a little clearer because it looks like a 2. so let me make that crystal clear so we end up with h of z let me make the z a little cleaner again it looks like a 2. so h of z minus 3 is equal to z minus 4 over z minus 2. all right so let's wrap up the lesson and talk about something that you might see in your course so sometimes you're going to see functions in what is known as implicit form okay this means the function is not solved for the dependent variable y so when this occurs we're going to first solve the equation for y and then we can just replace y with f of x or g of x or h of x or whatever you're using okay so it's just an extra step so let's say we had something like x squared plus 2y equals 12. the first thing i would do before i tackle any of these kind of examples i would need to solve it for y so i want this guy by itself on one side of the equation so we know the first thing would be to isolate the 2y okay the variable term with y involved so let me kind of do this down here i would have x squared plus 2y equals 12. let me subtract x squared away from each side of the equation and we know that this part right here is just going to cancel so we're going to have that 2y is equal to negative x squared and then plus 12. let me scroll down a little bit and we'll come back up in a minute all right so again we're trying to solve for y so what we're doing here is we're multiplying y by 2. so to get y by itself i can just divide both sides of the equation by 2 and we know that this would cancel with this i would have y is equal to you'd have negative x squared over 2 and then plus 12 over 2 is 6. so this is my function solved for y okay y equals negative x squared over 2 plus 6. let's erase everything and copy this so let me just paste this in right here and i'm just going to put a visible border here so we started with x squared plus 2y equals 12. i can erase all this now we solved it for y we get y equals negative x squared over 2 plus 6. and all i'm going to do is i'm going to change y into f of x okay so i'm just swapping this out and saying this is f of x is going to be equal to negative x squared over 2 plus 6. so that's all you do when this guy is in implicit form okay just solve it for y and then you can replace y with f of x g of x h of x whatever you're working with so now we can do the same process as we used in the previous examples if we want to find f of negative 1 well then i just plug in a negative 1 for x and i evaluate so f of negative 1 is equal to i have the negative of you're plugging in a negative 1 for x now be careful here because whatever x is it's being squared so i want to wrap my negative 1 inside of parentheses and i want to square that this is over 2 and then plus 6. so negative 1 that quantity squared the negative and the 1 would be positive 1. so i can erase this and just put one pay close attention because you still have the negative out in front so really what i have here is negative one-half plus six so let me get a common denominator going i'm just going to multiply this by two over two this would be twelve halves okay this is twelve halves so 12 minus 1 is 11 so basically this is going to be 11 halves okay so we can say f of negative 1 or the value of this function when the independent variable x is negative 1 is going to be 11 halves okay so let's erase this and we'll move on to the next one so now we have f of 8. okay so let me write my function f of and i'm just going to put 8 in this place this equals negative x squared over 2 plus 6. and again whatever i have for x i'm just going to put parentheses around there just leave a blank and just plug this in so i want an 8 plugged in for where x was and so i have 8 squared right and the negatives out in front okay so i'm not squaring that negative if it's inside the parentheses i'm going to square it if it's outside i'm not going to square it you can really think about it like this i have the negative of 8 squared so 8 squared is 64. so basically you have negative 64 divided by 2 which is negative 32 so we get negative 32 plus 6 which would give me a final value of negative 26 okay so again f of 8 is negative 26 or you can say the function's value when the independent variable x is 8 is going to be negative 26. all right let's erase this now we have f of q so f of q is equal to what again wherever x is just plug in the q it doesn't matter that it's a different letter okay you're just replacing x with whatever's inside of parentheses so you're going to have negative of q squared over 2 plus 6 so that one's really really simple all right let's look at the last one all right so we're going to have f of q minus one so again let me start by writing f of and i'm just gonna put a blank here instead of x because i have to fill it in with this this is equal to we have the negative of x squared over two and then plus 6. okay so inside the parentheses again i'm going to use q minus 1. and you have to be careful here because you've got to plug the q minus 1 in for x and that whole quantity is being squared so let me scroll down we don't need any of this information anymore so what i'm going to have is f of q minus 1 is equal to i have the negative of inside of parentheses i'm going to put q minus 1 this quantity and this guy is being squared okay so remember you have to expand this you can't just say it's q squared minus 1 squared that's a big mistake okay then this is over 2 and then we have plus 6 over here all right so let's scroll down a little bit and the first thing i'm going to do is just kind of expand this so we already know how to do the special products formula on that i would have q squared right the first one squared then minus right because we have a minus involved 2 times the first term times the second term 2 times 1 is two two times q is just two cubed okay and then we would have plus the last term which is one squared one squared is obviously just one okay so you have to make sure here that you pay close attention to what you have so let me kind of slide this out of the way for a second we have f of q let me make that better again f of q minus 1 is equal to the negative of this guy okay this is what's going to be replacing this right here but you have to make sure you take the negative of it it's very important that you don't make a sign mistake so the negative is outside of the parentheses and then you have q squared minus 2q plus 1 inside the parentheses this is all over 2 and then we have plus 6 over here all right so let's scroll down and get a little bit more room going so what we want to do now is drop our parentheses and distribute the negative to each term inside the parentheses so i'm going to have f of q minus 1 is equal to if i make each term here into its opposite you'd have negative q squared then you'd have plus 2q and then you'd have minus 1 right so the negative of q squared is negative 2 squared the negative of negative 2q is plus 2q and the negative of 1 is just negative 1. okay then this is over 2 and then we have plus 6 out here so i'm not done here because i could get a common denominator going so i can multiply this by 2 over 2 okay and what i'd have is f of q minus 1 is equal to you'd have your negative q squared you're not going to be able to do anything with that then plus your 2q can't do anything with that then you have minus 1 but look over here now you're going to have 12 over 2 here so i have a common denominator so i can just do plus 12 over here on the numerator and then let me kind of scroll down this is going to be over the common denominator of 2. okay so what i can do now is just kind of get a simplified answer where negative 1 plus 12 is going to give me positive 11 and i can officially say that f of q minus 1 where again the function's value when the independent variable x is q minus 1 is going to be negative q squared plus 2q plus 11 and again this is all over 2. in this lesson we want to review graphing linear functions so most of you already know how to graph a linear equation in two variables this will pretty much be a review for the majority of you but in case you've never seen this topic before it's something that's very easy and you can just pick it up from the lesson today so let's start out with just a basic definition of a linear equation in two variables we basically have two variables involved typically we use x and y so we have ax plus b y equals c where a b and c are going to be real numbers and we can say that a and b the coefficients for x and y can't both be 0 at the same time now when we work with a linear equation in two variables as long as we don't have a vertical line which we'll talk about later on in today's lesson you have a function so for each x you're going to have one unique y okay so we can call them linear functions we can say we're graphing linear functions or we're graphing linear equations in two variables it's the same thing and i could replace this f of x with just y right i could say y is equal to in this case a times x and then plus b okay so if this guy has solved for y we can replace y with f of x or g of x or h of x or whatever you want to use if you're writing this in function notation all right so how do we sketch the graph of a linear function well right now we have f of x equals 2x minus 5 and essentially the quickest way to do this is to solve the equation for y which in this case it actually is right i could replace this f of x with y equals 2x minus 5. so you solve it for y and then you're going to have it in something known as slope-intercept form you can plot the y-intercept and you can get additional points using the slope of the line now because it's not typically taught like that in the beginning i'm going to go through kind of the earlier steps kind of the more tedious methods that you use to do this so let's start off by just making a table of values and then towards the end of the lesson i'll show you this kind of quicker technique it's something that you pretty much use after you get out of this section because it's the fastest way to kind of accomplish this task so when you first start out you might be told to make a table of values so in other words i would choose a value for x and solve for y or choose something for y and solve for x the problem with using this method is you have to be able to kind of choose these smaller integers something that's going to fit on the coordinate plane that you can draw yourself right so you want smaller integers you want to make sure they're integers you don't want to get fractions or decimals things that are really hard to kind of put into a graph so what i'm going to do is i've already kind of pre-chosen some points that are going to work well for us and additionally if it's solved for y i can just plug things in for x and see what i'm going to get for y so if i choose let's say values for x of negative 2 let's say 0 and 5 i'm just going to plug in for x and see what i get so let's start out with an x value of negative 2 so i can notate this by saying f of negative 2 that's just the function's value when x is negative two this equals what it's two times negative two and then minus five two times negative two is negative four negative four minus five is negative nine so f of negative two is equal to negative nine so basically we have an x value of negative two and an f of x value or a y value of negative 9. so x comma y would just say this is negative 2 comma negative 9. so this is going to be a point on the line then we have an x value of 0 so i want f of 0 so what's that going to give me so let's plug in a 0 here and what are we going to get well 2 times 0 is just 0. so i can just kind of line that out and say okay we just have negative 5. so if x is 0 y is negative 5 so we can say 0 comma negative 5 that's my x comma y and the last one i want to check is an x value of 5 so i want f of 5. so let's see what that gives me so 2 times 5 is 10 10 minus 5 is 5. so if x is 5 y is going to be 5 as well okay so this gives us three ordered pairs that we can kind of plot on the coordinate plane and then once we've plotted those you draw a straight line through them and you put arrows at each end to indicate the line is going to continue forever in each direction the main idea here when you use this kind of table of values method you only need two points to make a line right but a third one is usually something you want to do just to get a check just to guard against any errors that you might make all right so i've kind of pre-drawn everything just to make it a little quicker so again we had negative 2 comma negative 9 and on my coordinate plane if i go 2 years to the left 9 units down that's here i had 0 comma negative 5 which is starting at the origin i just dropped 5 units so that's right there and then i also had 5 comma 5 okay so at the origin i go 5 units to the right and 5 units up so that's right there you can label your points if you want so we can say this is 5 comma 5 we can say this is 0 comma negative 5 and over here i'll label this one to the left i'll say this is negative 2 comma negative 9. okay now what you would do is you would make sure that those three points line up right the reason you do a third point as a check is let's say you get a point somewhere over here well you know that you wouldn't have a line that does that right that doesn't make any sense so that third point is guarding against errors for you and believe me everybody makes errors when they're doing math right there's a lot of silly mistakes that just happen because you might not be paying attention or you might have not be completely focused at the time lots of things can happen so it's good to have three points again to guard against errors so provided these things line up which in this case they do you just sketch a line through the points and you put arrows at each end and you're done right it's a very simple process all right so let's just trace over this and i'm not a good drawer or a good tracer but i do the best i can so you can label this we'll say this is f of x is equal to 2x minus 5. okay let me make that a little bit better i'll just kind of label it over here again f of x or you could say y if you wanted to equals 2x minus 5 okay so that's the graph of that line now let's talk about intercepts x-intercepts y-intercepts these are concepts that again you probably know from your elementary or intermediate algebra courses essentially the intercepts are the kind of spots on the graph where you intersect the axis right it could be the x-axis that's going to be the x-intercept or the y-axis that's going to be the y-intercept so on the next page i've kind of pre-drawn these in for you so essentially what you'll see is that the x-intercept again if i look at the x-axis where do i hit the x-axis well i hit right here okay right there so the coordinate is 2.5 comma zero okay and i'll talk more about this in a second for the y-intercept again i'm thinking about the y-axis so for this guy where am i going to hit well i'm going to impact i'm going to impact right there okay now i want you to notice something and this is very important something you definitely need to know when you look at the x-intercept where is y in terms of its coordinates when you're on the x axis well it's always going to be a zero right if i think about my y values if i'm on the x axis basically i'm here or i'm here or i'm here my y value is always zero so to find the x-intercept all i need to do is plug in a 0 for y and solve for x so when the equation we had was y or you could say f of x so y equals 2x minus 5 plug in a 0 for y plug in a 0 for y and solve for x so add 5 to both sides of the equation you get that 5 is equal to 2x divide both sides by 2 and you find that 5 halves is equal to x and again that's just my x coordinate when y is 0 or where it's going to impact the x axis so that's my x intercept again that's right here 2.5 is the same thing as 5 halves in decimal form and then comma 0. okay so 2.5 comma 0 or 5 halves comma 0 that's the x intercept now similarly let me just erase this if i want to find the y intercept think about what happens when you're on the y axis the x value or the x coordinate is always going to be 0 right because i'm here or i'm here or i'm here or i'm here right i'm not moving at all from the origin horizontally i'm just moving vertically right somewhere on the y-axis is where i am so what happens is i can solve my equation for y once i plug in a 0 for x that'll give me my y intercept so let me erase this and we'll say again we had y was equal to what it was 2x minus 5 and what i would do is plug a 0 in there 2 times 0 is 0 i'm just left with negative 5 right so if x is 0 y is negative 5 so that's my y intercept now the good thing about kind of the equations we work with it makes it quicker if you have it solve for y if you have y equals 2x minus 5 you'll notice that this part right here gave me the y-coordinate of the y-intercept that is not a coincidence okay that's always going to be the case because when it's solved for y when y is by itself on one side what happens is you have something times x and then plus some value here in this case it's plus negative 5. well plugging in something for x here of 0 okay makes this go away and i'm just left with this right so that gives me my y intercept every time later on we're going to talk about y equals mx plus b which is the slope-intercept form of the line this is the fastest way to graph a line because we'll immediately know the y-intercept which is one point on the line and we can generate additional points on that line using our slope okay we'll get to that later all right let me show you something known as the intercept method this is something you would see in your textbook typically in algebra 1 or algebra 2 as you get further along in algebra in a college algebra you get into something like pre-calculus they won't spend a whole lot of time on this method because again it's much faster just to use the slope-intercept form of the line to sketch the graph but the intercept method just basically relies on you finding the x and the y-intercept as points a third one to guard against errors if you want to check but you really only need two points to make a lot so in other words i would plug in a 0 for x and i would plug in a 0 for y so i would find the x and the y intercept that gives me two points again if i don't want a third point as a check i'm pretty much good to go right i can graph my line so let's go ahead and do this real quick we have y equals x minus 4 again i just replace f of x with y and if i want to find out what is the function's value when x is 0 again what i'm going to do here is just plug in a 0 for x so it's just like if i had f of 0 and this equals what 0 minus 4 which is negative 4. so if x is 0 y is negative 4 so we have 0 comma negative 4 okay let me erase that and then what about if y is 0. so basically this guy right here is 0. let me just erase this and i'll just kind of do this over here let's replace f of x with y to make it a little a little bit easier for you so y equals x minus 4. replace y or again f of x with 0 what do we get we'll just add 4 to both sides of the equation you're going to find that x is equal to 4 okay so this guy's 4 so we would have 4 comma 0. so you get a third point as a check again i always recommend this because you might make some mistake so let's go ahead and say that we want to find the value of y when x is just 2 okay so f of 2 is what we're trying to find so we'd have 2 minus 4 which equals negative 2. so we would basically have negative 2 here for y so 2 comma negative 2 is our ordered pair i'm just going to copy those let's go to the coordinate plane all right so again i've pre-drawn this to make it a little faster so those are our three ordered pairs so 0 comma negative 4 again the x value is 0 here so what does that tell me that's my y intercept right that's where we're going to cross the y axis so from the origin just drop 4 that's right there that's going to be 0 comma negative 4 and let me make that a little better and again notice how that's where we're crossing the y-axis okay that's the y-intercept then you have four comma zero so that's going to be up here right go four units to the right you don't move it all vertically so this is four comma zero let me make that better and again this is my x intercept because y is 0 right so this guy is where we cross the x axis all right so the last point is just for a check so we have 2 comma negative 2 so 2 units to the right two units down so that's right there this is my two comma negative two okay so i've already sketched the line that goes through these three points again just make sure it lines up and it's straight if you get a point somewhere over here and you have these other two points there you know it doesn't make sense it's not going to be a line so then you know you made an error you've got to go back and kind of recalibrate some things all right so let's go ahead and just sketch the line okay and again you want to put arrows at each end and then a lot of times people like to label their line so let me just kind of draw an arrow here and say this is the graph of f of x is equal to again this was x minus four all right so as i alluded to using a table of values is fine when you get started but it's very slow and it's very inefficient you've got to go through and generate all these points it's kind of a waste of time to be honest with you once you learn kind of how to put an equation in slope intercept form it's the preferred way to graph a line okay so basically slope intercept form is like this we have y is equal to m the slope times x plus b the y intercept so this guy's the slope and the y intercept will occur at 0 comma b okay again if i plug in a 0 for x that's how i find my y intercept this is gone i'm just left with y equals b okay so that's why that works now the reason we have m as slope we'll talk about over the next two lessons we're going to talk about slope in general in the next lesson in the lesson after that we're going to talk about equations of lines so we'll talk about the standard form of the line we'll talk about slope intercept form we'll talk about point slope form i'll show you where all this stuff comes from but for right now we just need to know the basics and that is that the coefficient of x is the slope and this b right here is our y-intercept all right so what is the slope of a lock well again we're going to go way more into detail on this in the next lesson but slope is basically the ratio of the vertical change which we call the rise to the horizontal change which we call the run so let's say we had f of x equals three fifths x minus two or again we can write y equals three fifths x minus two it makes no difference if you use this function notation or you use y okay it doesn't matter it's the same thing so the coefficient of x is my slope and my y-intercept again is this guy over here so it's this guy over here so y equals mx plus b this guy is my m three-fifths my b my y-coordinate for the y-intercept is negative two so all you need to do is plot the y-intercept first so the y-intercept occurs at zero comma negative two so let's plot that guy so from the origin i'm just going to drop down two so it's going to be right there that's my 0 comma negative 2 and i'm going to use my slope to get additional points so slope is rise it's rise over run okay so if this is three fifths what this means is that my rise is telling me how much i move vertically to get to the next point my run is telling me how much i move horizontally to get to the next point so a rise of 3 tells me to move vertically 3 units so go up 1 2 3. a run of 5 tells me to go to the right by 5. so 1 2 3 4 5. so that would be the next point on the line so you would have 5 comma 1 okay so that's an ordered pair and then you can do this backwards if you want so i can also say this is negative 3 over negative 5 because negative over negative is positive now i want you to think about this when you think about rise or vertical movement a positive means you're increasing so you're going up right on the y axis we increase as we go up a negative is going to mean we're going to go down right we decrease on the y axis as we go down so if i have a negative 3 a rise of negative 3 is going to tell me to decrease by 3 units or go down 3. so 1 two three then a run of negative five again with the x-axis you go to the right you're increasing you go to the left you're decreasing so what i want to do now from here is go to the left five units because i have a negative five so i wanna go one two three four five and so this would be a point on that line so this would be negative five comma negative five so negative five comma negative five and you can check these points to make sure that they work we know zero comma negative 2 would work because if you plug in a 0 for x you get negative 2 for y but let's check these other ones so 5 comma 1. does that work well if i plugged in let me kind of do this down here if i had y equals three fifths times i plug in a 5 for x would i get a 1 for y well these cancel 3 minus 2 is 1 so that one's good to go let's check the other one so again 3 5 times x minus 2 the other one was negative 5 comma negative 5. so if i plugged in a negative 5 here what would happen so this guy would cancel with this guy but it would be negative right so i'd end up with a negative 3. so you'd have negative 3 minus 2 which would give you negative 5. so that one's good to go as well so look how quickly we found kind of three ordered pairs we're basically good to go now and so we can draw a line connecting these points okay let me just put arrows at each end so that would be my line right and again i can label it if i want i can say this is f of x is equal to three fifths x minus two okay look how much quicker that is versus generating a table of values and a lot of times it's not solved for y so it's kind of hard to figure out what's going to be an integer answer and what's not with this you have your y-intercept and you can plot additional points using the slope again this kind of concept of rise over run okay let's look at another one of these so we have f of x equals negative four thirds x plus one again if you don't like the f of x you can say y equals negative four thirds x plus one and again when it's in this format we know y equals m x plus b m is the slope is the coefficient of x so it's negative four thirds b is the y coordinate for the y intercept so it's going to occur at zero comma one okay so let's plot the y intercept first zero comma one is going to be right there let me use a different color because that doesn't show up very well so i can label that if you want it's zero comma one and then you get additional points using your slope so m the slope is equal to negative four thirds and again this is my rise over my run okay so my rise is negative 4. again if i'm talking about rising or moving vertically if it's a negative i'm going to go down if it's a positive i'm going to go up so go down by 4 units so 1 2 3 4. my run is positive 3 so that means i'm going to the right by 3. 1 2 3. so that's a point so this is 3 negative 3. so 3 negative 3 that's a point and then you can do this again if you want so you can go down 4 1 2 3 4 to the right 1 2 3. so you can put a point there that's going to be at 6 comma negative 7. and you can also reverse this remember if i have a negative over a positive it's the same as a positive over negative so i could start here and i could go up 4 i could rise 4 1 2 3 4 and i could go to the left 3 right a run of negative 3 just tells me to go to left 3. so 1 2 3 so that's a point there it would be at negative 3 comma 5. okay and you can check all these points if you want i've already checked them i know they work so it's up to you if you want to pause the video again you'd plug in a negative 3 for x verify you get 5 for y plug in a 0 for x verify you get 1 for y plug in a 3 for x verify you get negative 3 for y and plug in a 6 for x verify you get negative 7 for y okay so now all we want to do is just draw our line through these points okay and again you want to put arrows at each end to indicate the line continues forever in each direction and i'm just going to label it i'll say this is f of x is equal to we've got negative four-thirds x plus one all right let's take a look at an example where it's not solved for y in other words we don't have y equals something or we don't have f of x equals something right so in this particular case as a linear function it's not solved for the dependent variable y so we're saying it's in implicit form right so if we want it in the format of y equals m the slope times x plus b the y intercept we just solve it for y okay it's pretty simple so i would just what isolate this term with y involved first so the negative 2y just subtract 3x away from each side of the equation to do that this part would cancel you'd have negative 2y is equal to negative 3x and then plus 6. to completely isolate y it's being multiplied by negative 2 so let me divide both sides of the equation by negative 2. and let me slide this down so we have enough room to kind of complete this so we know this is going to cancel you have y is equal to negative or negative is positive so 3 halves times x 6 over negative 2 is negative 3 or minus 3 okay so our slope here m is three halves our y intercept will occur at zero comma negative three so now this guy is solved for y or it's in explicit form so in this form we can replace y with f of x or g of x or whatever we want right if we want to use function notation so f of x equals 3 halves x minus 3. okay so let me erase everything and let me just write my slope up here m the slope is equal to three halves and again this is the rise over the run okay so just one additional step when you get this format not very difficult still faster to solve for y and just kind of use the slope and y intercept to get your kind of line graphed so the y intercept is at zero comma negative three so i would go down three units from the origin so that's right there and then my slope is three halves so again the rise is three the run is two so i'm gonna go up three one two three to the right one two so there's a point rise three one two three to the right one two so there's a point so this guy is going to be 0 comma negative 3. again that's the y-intercept that's on the y-axis this guy is going to be 2 comma 0. so this is 2 comma 0. that's the x-intercept that's on the x-axis and this guy is going to be 4 comma 3. so 4 comma 3. all right so let's sketch our line real quick right let me just put an arrow at each end and then we're basically done if you want to label it you can use the 3x minus 2y equals 6 or you can use the y equals 3 halves x minus 3. or you can use the f of x equals 3 has x minus 3. it doesn't really matter you're just labeling it and saying hey this equation or this function goes with this line so i'll just say f of x is equal to three halves x and then minus three and we'll be done all right so now that we know how to work with most of the cases the special cases that come up you have horizontal and vertical lines these are very easy to draw okay so a horizontal line occurs when y or you could say f of x is set equal to some constant in your book you'll probably see it called k or it could be called something else it doesn't matter so basically you have f of x is equal to some number let's just say it was 5 okay or you can see this says y equals 5. these are the same i just replaced y with f of x so basically what you can do here to think about this as a linear equation in two variables if you had y equals 5 you could say you have 0x plus y equals 5 okay no matter what you choose for x when you multiply it by zero it goes away so you're still just left with y equals five so what happens here is that no matter what value you choose for x if you're thinking about ordered pairs y will always equal five so i can choose a million for x y equals five i can choose three for x y equals five i can choose negative one trillion for x y equals five so what happens is you end up with a horizontal line that basically crosses the y y-axis at this value here in this case it's five it could be you know whatever you have if it was seven it would cross the axis at y equals seven you know so on and so forth so to see an example of this let's say we had f of x equals three okay or again you could write y equals 3 if that makes you more comfortable so all you need to do to graph this line is go up to 3 on the y axis so let's just say right there and we draw a horizontal line that crosses through that point 0 comma 3. let me make that a little better just going to use my eraser and just kind of knock that part off and i'll just put an arrow at this end an arrow at this end it continues indefinitely in each direction so if you wanted to kind of prove this to yourself again you can write this as 0x plus y equals 3 and you can pick values for x and kind of generate some ordered pairs so think about this if i picked let's say negative 3. so negative 3 times 0 is 0 so that's gone so y equals 3. so you go to negative 3 3 so that's a point on the line you could say if x was 0 y equals 3 so 0 3 would be this you could say if x equals 3 okay plus 3 0 times 3 is 0 you'd have y equals 3. so no matter what you choose for x it's not going to matter because y is always going to be 3. so that's what creates this horizontal line again that impacts the y-axis at 3. all right let's look at another example so suppose we have f of x equals negative four again all i've got to do is go down to negative four on the y axis so that's going to be right there this is my negative four and just draw a horizontal line that impacts that y-axis at negative four so let's do that real quick okay so this would be f of x equals negative 4. all right let's wrap up the lesson and talk about the last special case scenario this would be when we encounter a vertical line so this is going to occur when x is set equal to some constant value in this case we're not going to have a function so this is the only case we have a linear equation in two variables where you don't have a function and you might say you don't have two variables but again you could write it as x plus 0 y equals a if you wanted to so you have two variables involved this is the only case where you're not going to have a function this one single x value in this case it's a but it could be whatever right you replace that with 3 or 10 or a million or trillion whatever it is basically that one single x value is going to be associated with or paired up with an infinite number of y values okay so it's definitely not a function so let's say we had x equals negative 6 and again you could write this as x plus 0 y equals negative 6 if it makes you feel more comfortable so for anything that i plug in for y here it just goes away right so i'm left with x equals negative 6. so essentially for this guy you go to negative 6 on the x-axis now and i can kind of just highlight that basically i'm just going to draw a vertical line there okay and i went a little bit over so let me just kind of erase some of this and i'll put my arrow at each end so this is x equals negative 6. again just find out where this number here whatever it's equal to is on the x-axis draw a vertical line right because no matter what value i choose for y x will always be negative 6. if i choose 9 for y x is negative 6. if i choose 6 for y x is negative 6. if i choose 0 for y x is negative 6. right if i choose negative 1 trillion for y x is negative 6. all right so for x equals 8 how would we graph this again look at 8 on the x axis eight is going to be here right it's one notch to the left of nine so again i'm just going to draw a vertical line okay and that one's pretty good so put an arrow at each end and this is my x equals eight so again for a vertical line x equals some number find the number in this case it's 8 on the x axis draw a vertical line that crosses through that on the x axis that's all you need to do in this lesson we want to review finding the slope of a line in our last lesson we reviewed how to graph a linear function or you could say we reviewed how to graph a linear equation in two variables here we're going to go one step further and we're going to review how to find the slope of a given line which is just a measure of the line's steepness now officially when we talk about the slope of a line it's nothing more than just the ratio of the vertical change which is known as the rise to the horizontal change which is known as the run as we move along the line from one point to another so generally speaking we're going to use this lowercase m to denote slope okay so this is denoting slope so we have here that our m our slope is equal to the rise over the run or we could say it's equal to delta y over delta x so before we go any further let's just kind of explain this for a minute when we say the rise we are talking about the vertical movement on our coordinate plane so from one point to the other how far did we move vertically we know that the vertical axis is the y-axis so that's why we have this delta y here which just basically tells us the change in y values now when we talk about run we're talking about a horizontal movement how much did we move left and right we know the horizontal axis is the x-axis and so that's why we're saying the change in x okay this symbol right here in case you're unfamiliar with it is the greek capital letter delta and in math we generally use it to say the change in so don't let this scare you it just says the change in y over the change in x or again you can remember this as the rise over the run so using this concept of rise over run or again delta y over delta x we can come up with this little slope formula which allows us to find the slope of any line given two points on the line so m the slope is going to be equal to the change in y values y sub 2 minus y sub 1 over the change in x values x to 2 minus x sub 1 and this is given the fact that we have this restriction x sub 2 minus x sub 1 cannot be 0. why can't this be 0 because you'd be dividing by 0 and that's undefined and this scenario here where you have x sub 2 being equal to x sub 1 is only going to happen when you have a vertical line so you can just remember when you have a vertical line the slope is undefined because you'd be dividing by zero in your slope formula all right so let's use this slope formula real quick it's a very easy formula to kind of work with the first thing that you might struggle with if you haven't worked with kind of notation where you say you know x sub 1 y sub 1 and x sub 2 y sub 2. this might be foreign to you you might be asking what does this mean well essentially we know that an ordered pair is what it's x comma y okay and this one is x comma y as well so how do we keep track of kind of which x i'm talking about which y i'm talking about if i'm working with multiple ordered pairs well i can just use the kind of sub 1 sub 2 notation okay so i can just call this first point x sub 1 y sub 1. so the x value i'm talking about here if i say x sub 1 is 2 the y value i'm talking about here if i say y sub 1 is negative 7. and then this guy i can just say it's x sub 2 y sub 2. okay and then for the purposes of this formula it doesn't matter which point gets labeled as which you get the same answer either way and i'll prove that to you in a second so the slope formula let me just copy it down over here m is equal to again the change in y values y sub 2 minus y sub 1 over the change in x values x sub 2 minus x sub 1. i'm not going to write the restriction that x sub 2 can't equal x sub 1. you know that that's basically going to give you division by zero so that's something easy to remember okay so let's go ahead and find out what the slope is so i'm just going to plug into the formula so m is equal to for my y sub 2 i have what i have a 3. so just plug that in then minus for my y sub 1 i have a negative 7. here's where you've got to be careful it's a very common mistake to just put 7 in because you already had a minus but that's wrong i have a minus and then i have a negative seven so i want minus a negative seven here okay then this is over x sub two minus x sub one x sub two is negative three then minus my x sub one is going to be two so three minus a negative seven is the same as 3 plus 7 and that's going to give me 10 then over negative 3 minus 2 is going to be negative 5. 10 divided by negative 5 is negative 2. okay so the slope of the line that passes through these two points is going to be negative 2 and i'll show you that in a moment graphically but really quickly i just want to show you this works the same if i do it in the opposite way so i'm just going to change the labeling up i'm going to say this is x sub 2 y sub 2 i'll say this is x sub 1 y sub 1 and basically we're going to get an answer of negative 2. okay so if i do y sub 2 minus y sub 1 now y sub 2 has changed it's now negative 7. y sub 1 has changed it's 3. x sub 2 minus x sub 1 x sub 2 has changed to 2. x sub 1 has changed to negative 3 again minus a negative 3 is plus 3. so be careful there okay let me make that better so negative 7 minus 3 is negative 10 this is over 2 minus a negative 3 is 2 plus 3 that's 5. negative 10 divided by 5 is negative 2. you see we get the same answer either way all right so to think a little bit more deeply about this concept of slope let's go ahead and look at the graph of the line that passes through those given points so again the points we were working with we had negative three comma three and we also had two common negative seven so let me just plot those real quick negative three comma three starting at the origin if i went three units to the left and three units up that would be here so that's your negative 3 comma 3. and then 2 comma negative 7 starting at the origin i could go 2 to the right and 7 units down so that would be your 2 comma negative 7. so this line that we've already graphed that passes through those two points and basically what we can see from this graph is that our slope m when we say it's equal to negative 2 again when we think about this in terms of rise over run so m is equal to the rise over the run i'm going to say this is negative 2 over positive 1. so let me explain when we think about the rise it's the vertical movement or the vertical change as we move from one point to the other so let's say i'm going from this point to this point well how far do i move vertically well from 3 2 this guy is going to be 1 i'm moving down by 2 units okay i'm moving down by 2 units so on the y axis if i'm going down that's a negative so it's really negative 2. then how much do i move kind of horizontally let me erase this because it's kind of tight this here is negative 3 and this guy here is negative 2 so i'm only moving one unit to the right and again if i'm moving to the right on the horizontal axis it's a positive okay so basically i can say that it's negative two over positive one that's my rise over my run okay so every time i move from one point to another i can just drop by two units so i can drop one two and then i can just go to the right one and i'm back on the line drop two one two go to the right one drop two one two go to the right one okay so that's what a slope of negative two is telling me if i had a slope of negative four i could drop four units and go to the right one if i had a slope of negative 10 i could drop 10 units go to the right one now another thing you can do is you can change this up we know that negative over positive in terms of a fraction is negative so i can also say this is positive over negative because this is also going to be negative so you might be confused by that and say how is that possible well if i start at a given point on the line let's say i start right here let me use a different color if i rise to so a positive 2 means i'm going up so 1 2 and then i run negative 1 it means i'm going to the left one okay so a negative 1 run is to the left so that's right here so i'm back on the line up 2 1 2 to the left one up two one two to the left one right i'm kind of going backwards so you can always use that trick to kind of get additional points if you have a positive over negative or negative over a positive the net result is it's going to be negative now one last thing i want to draw your attention to as we move to the right okay as we move to the right you'll notice that this line is falling so as we go this way the line is going down this is known as having a negatively sloped line and again you see that you have a negative 2 there okay if this was a positively sloped line as we move to the right the line is going to rise we'll see an example of that in a minute all right let's take a look at another example and i'll show you this graphically as well so we have these two points zero comma negative two and five comma negative six again let's plug into the slope formula so let's just say this is x sub one y sub one let's say this is x sub two y sub two so again my m my slope is what it's the change in y values over the change in x values so y sub 2 minus y sub 1 is what y sub 2 is negative 6 y sub 1 is negative 2. so negative 6 minus a negative 2 is going to be negative 6 plus 2 and then x sub 2 minus x sub 1 x sub 2 is 5. x sub 1 is going to be 0. 5 minus 0 is just 5. so negative 6 plus 2 is going to be negative 4 and this is over five okay so m here our slope is equal to negative four-fifths so what this means is my rise is negative four so from whatever point i'm at i can drop four units my run is five so i would run to the right five units so again to look at this graphically the points where we're looking at we had 0 comma negative 2 and we had 5 comma negative 6. so 0 comma negative 2 is right here and 5 comma negative 6 if i go 5 units to the right and six units down that's right here you can label this we'll say this is five comma negative six and we'll say this is zero comma negative two and again to get from this point to this point i can fall four units one two three four and then run to the right 1 2 3 4 5. so again my m my slope is negative 4 5. another thing you can do as we talked about we could say this is 4 over negative 5. so if i started at this point i could go up 4 one two three four right a positive rise and i could go to the left five i could go one two three four five units to the left and that gets me to another point so remember if you're in a situation where you have a positive rise that means you're going up if you have a negative rise you're going down if you have a positive run you're going to the right if you have a negative run you're going to the left and if that gets confusing just think about how numbers increase and decrease on the x-axis right they increase going to the right decrease going to the left and on the y-axis the increase going up and decrease going down and then once again you'll notice as we move to the right this line is falling so this is another example of a negatively sloped line and of course you can see that because the slope itself is negative all right let's look at one more of these and this guy is going to have a positive slope so as we move to the right the line is going to rise so m is equal to y sub 2 minus y sub 1 again change in y values over x sub 2 minus x sub 1 again change in x values so let's just label this as x sub 1 y sub 1 and this as x sub 2 y sub 2 and what are we going to have m my slope is equal to we have y sub 2 which in this case is 4 minus y sub 1 which in this case is negative 6 minus a negative 6 is plus 6. then we have x sub 2 which in this case is 2 and then minus x sub 1 which in this case is negative 2 minus a negative 2 is plus 2. so we get 4 plus 6 which is 10 over 2 plus 2 which is 4. 10 divided by 4 we can basically say is 5 halves right each is divisible by 2. 10 divided by 2 is 5. 4 divided by 2 is 2. so my slope m is positive 5 halves so essentially my rise is 5 my run is 2. so from any point in that line i can go up by 5 units and to the right by 2 and get to another point on the line all right so again the points we're working with we had negative two comma negative six and we had two comma four okay so negative two comma negative six starting at the origin two units to the left and six units down that's going to be right there again negative two comma negative six and then two comma four again starting at the origin two units to the right four units up is right there so that's two comma four okay so again if you think about this the slope we found m is equal to the rise over the run and we said this was five halves so starting at any point on this line i can go up five units so let's say we start here one two three four five so that's my rise and i can run two so that means i'm going to go to the right 2 1 2. so i'm back on the line i can go up 5 1 2 3 4 5 to the right 1 2 back on the line and again you can do this in reverse so this might be something you haven't thought of you can say m is equal to the rise over the run and negative over negative is positive so i can really say negative 5 over negative 2 this is the same thing as positive 5 halves right negative or negative is positive so that means if i start at this point on the line i can fall five units one two three four five and go to the left two units one two back on the line fall five one two three four five go to the left two one two back on the line okay so it's just as simple to kind of use that trick to get extra points as well now you'll notice that as we move to the right this line is increasing right it's going up so again this is an example of a positively sloped line and again you see that you have a positive slope right it's positive 5 halves all right so let's look at these kind of special case scenarios and then we'll move on and talk about slope intercept form so you're going to have horizontal lines that you're going to encounter and you're going to have vertical lines that you're going to encounter i'll tell you right off the bat a horizontal line has a slope of 0 and a vertical line has an undefined slope but we want to think a little bit more deeply about why that's the case so horizontal lines we know they're of the form either in function notation f of x equals k or kind of normally we just see y equals k right but we could write this as a linear equation two variables if we wanted to we could say zero times x plus y equals some value k right so an example i'm going to give you is y equals four so i could say this is 0 times x plus y equals 4. and the reason you can write it like this is this allows you to kind of think about some ordered pairs you could get going and basically no matter what you choose for x you're multiplying it by zero so it's gone and you're just left with y equals 4. so in other words i could say let's say i plug in a negative 1 for x negative 1 times 0 is 0. so you're just left with y equals 4. so negative 1 comma 4 that would be an ordered pair right so another one would be let's say i choose i don't know positive 4. positive 4 gets plugged in for x you multiply by 0. it's gone you're left with y equals 4. so 4 comma 4. so any value you choose for x doesn't matter right the y value will be 4. so let's go look at this on a coordinate plane so this is the graph of y equals 4 and again it just crosses through the y axis at 4 and again it's just a horizontal line right so any x value i choose let's say i choose an x value of 6 the y value is 4. let's say i choose an x value of negative five the y value is four let's say i choose an x value of a million the y value is four right so what happens is the vertical change is always going to be zero right in our slope formula we think about again the rise or the change in y values over the run or the change in x values okay so delta y here the change in y values is zero and delta x can be whatever you want it to be okay it's not going to matter because zero over any non-zero number is always zero so in this case for a horizontal line the slope is 0. and i'll prove this to you mathematically let's just pick two points on the line and plug into the slope formula so y sub 2 minus y sub 1 over x sub 2 minus x sub 1. and again two points let's say we use the ones we have here so this guy is going to be six comma four so you have a point six comma four and then you also have a point over here i have negative five comma four so negative five comma four okay so those two points let me kind of slide this down and let me slide this down okay so let's go ahead and say this is x sub 1 y sub 1 and let's say this is x sub 2 y sub 2 and again when you plug in here your y sub 2 is 4 your y sub 1 is 4 4 minus 4 is 0 right so you automatically have 0 in the numerator and again as long as you don't have 0 in the denominator you know the result there is 0. for this one x sub 2 is negative 5 so it's negative 5 and then minus x sub 1 which is 6. if i had a negative 5 minus a 6 i'd have negative 11. but again it doesn't matter as long as this isn't 0 down here 0 divided by any nonzero number is 0. so again the slope for any horizontal line will just be zero all right so for a vertical line basically you're going to have an undefined slope so this is a line where we have x equals some value right so x equals a could be x equals 10 it could be x equals a million it could be x equals a billion it could be x equals negative two trillion right whatever it is it's not going to have a slope as we normally think about it because you end up dividing by zero which is undefined so as an example suppose we had x equals negative two again you think about the fact that the x values here are always the same it's always going to be negative 2. so let's say you pick two points on the line so again my slope formula m equals y sub 2 minus y sub 1 over x sub 2 minus x sub 1. again if you pick two points on the line let's say you pick negative 2 comma 3 and you pick negative 2 comma negative 3 just as an example go ahead and plug that in so negative 2 comma 3 and you have negative 2 comma negative 3. so let's just say this is your x sub 1 y sub 1 this is your x sub 2 y sub 2. let me make that a little bit better and if i plug in here y sub 2 minus y sub 1 you have negative 3 minus 3. so negative 3 minus 3 would be negative 6 so that's negative 6 and here's where you have the problem for x of 2 minus x sub 1 you're going to have negative 2 minus a negative 2 which is the same thing as negative 2 plus 2 which is 0. so this part right here tells us it's undefined right this is going to be undefined so anytime you have a vertical line you end up with division by zero in your slope formula so the slope for a vertical line is undefined all right so something we talked about in the last lesson when we were graphing linear equations two variables or you could say graphing linear functions we talked about how to get the slope and the y-intercept directly from kind of the equation of the lot okay so this is pretty popular something you need to know how to do this is known as slope intercept form when you solve the equation for y and it comes directly from the slope formula right so basically if you have m the slope is equal to y sub 2 minus y sub 1 over x sub 2 minus x sub 1. we'll go into more detail on this in the next lesson but essentially let's say i know a point on this line let's say i know the y intercept occurs at 0 comma b we know that the x value for the y intercept is always 0. so one point is going to be 0 comma b and i'm just going to let another point just be x comma y okay so my normal notation is what it's x of 1 y sub 1 and x sub 2 y sub 2. i'm going to let this guy take the place of x sub 2 y sub 2 i'm going to let this take the place of x sub 1 y sub 1 and i'm just going to plug in so what we would have is what for y sub 2 i'm just plugging in a y for y sub 1 i'm plugging in a b for x sub 2 i'm plugging in an x and for x sub 1 i'm plugging in a 0 minus 0 is just nothing you can get rid of it right so essentially this turns into m the slope equals y minus b over x and all i'm going to do from this point is just solve for y and what's going to happen if i multiply both sides by x i get that mx is equal to y minus b this guy right here has cancelled right that's gone so to solve for y i just add b to both sides of the equation and we get our slope intercept form which is that y is equal to mx plus b so what do we know we know that m is the slope because we got this directly from the slope formula so the coefficient of the variable x is my slope when it's solved for y right y equals m the slope times x plus b b is going to be the y coordinate for the y intercept so the y intercept when you have it in this form is going to occur at 0 comma b and again you can see this because if you plugged in a 0 for x here what would happen 0 times whatever m is it doesn't matter is 0. so you're just left with y equals b right so that's my y intercept so this is my y intercept okay so let's try two of these i'm going to show you that it's the same either way we'll solve it for y we'll get the slope and then i'll show you kind of generating two points and using the slope formula it's the same thing so let's erase this and let's kind of try this out so the first thing i want to do is put it in slope intercept form so that's y equals m the slope times x plus b let me just label this again this is my slope so how do i do this i want to solve the equation for y and to do that i want to first isolate the variable term with y involved so start by isolating the negative 2y so i have this 3x that's out here let me just subtract that away from each side of the equation i'll have negative 2y is equal to negative 3x plus 12. and then i get y by itself i just need to divide both sides by negative 2 and what i'm going to get is that y is equal to negative 3 over negative 2 is 3 halves times x and then 12 over negative 2 is minus 6. so what this tells me is that the slope is 3 halves and the y intercept occurs at 0 comma negative 6. so m my slope is three halves now that was pretty quick let's suppose we were given two points on the line and we had to use the slope formula so i'm just gonna give you two points i'm gonna give you a point that's zero comma negative six and i'm gonna give you a point that's two comma negative 3. you can easily verify those two points work if i plugged in a 0 here you'd have negative 2 times negative 6 which would give you positive 12. if you plugged in a 2 for x and a negative 3 for y you would have 6 here and you would have 6 here right so 6 plus 6 would give you 12. so those are two valid points so again if i use my slope formula i would have x sub 1 y sub 1 and i would have x sub 2 y sub 2 and again i'm just plugging in m the slope equals what y sub 2 which is negative 3 minus y sub 1 which is negative 6 minus a negative 6 is plus 6. this is over x sub 2 which is 2 minus x sub 1 which is 0 so you can just erase this and essentially we have what negative 3 plus 6 which is positive 3 over 2 which is exactly what we found right there by just solving for y so if you have an equation okay and you have the option of generating points and plugging into the slope formula it's a lot quicker to just solve it for y and observe the slope as the coefficient of x all right let's wrap up the lesson and just look at one more of these again i can find my slope by just solving for y y equals m the slope times x plus b again this is called slope intercept form and we'll talk more about this in the next lesson so to solve for y i want to first isolate the variable term with y involved so i'm just going to subtract 2x away from each side of the equation i'll have negative 5y is equal to negative 2x minus 60. let me divide both sides of the equation by negative 5 and what we're going to have is y is equal to negative 2 over negative 5 is 2 fifths and this is times x and then negative 60 over negative 5 is plus 12 okay so the slope here is two-fifths let me make that a little cleaner okay that's my m and my y-intercept occurs at zero comma 12. okay and again i'll just prove this to you real fast i'll give you two points on the line we can use our slope formula so one point we could use would be 0 comma 12. again if you plug in a 0 for x you'd have negative 5 times 12 negative 5 times 12 would give you negative 60. so that's a valid point and another point is 5 comma 14 okay 5 comma 14. if i plug in a 5 for x 2 times 5 is 10 then minus you have 5 times 14 which is 70 and this does equal negative 60 right 10 minus 70 is negative 60 so that's a valid point as well okay so let's say i label this as x sub 1 y sub 1 and this is x sub 2 y sub 2. again m the slope equals y sub 2 which is 14 minus y sub 1 which is 12 over x sub 2 which is 5 minus x sub 1 which is 0. 5 minus 0 is just 5. 14 minus 12 is going to be 2 so you get 2 fifths again the same thing as you got here so once again putting it in slope intercept form is a really quick and efficient way to find the slope of a line again it's just going to be the coefficient of the variable x in this lesson we want to review equations of lines so a pretty common task in algebra is being able to write the equation of a line given certain information so additionally you may have to change between different forms of align to perform a specific task so as we kind of progress through the course you'll see that in some cases you need one form of the line it's more advantageous in other cases you will need a different form of the line so what we're going to do today is discuss the slope-intercept form the point-slope form in the standard form and we'll show how to kind of go back and forth between those different forms so we're going to begin with slope intercept form which we previously talked about so basically slope intercept form allows us to immediately know the slope and the y-intercept by kind of observing the equation okay so we have y equals m which is our slope this is our slope times x plus b which is the y-intercept right you can see that if you plugged in a 0 for x you would get 0 comma b as the y-intercept right so where does slope intercept form come from well essentially if you take your slope formula which we already talked about m the slope is equal to the difference in y values so you have y sub 2 minus y sub 1 over the difference in x values x sub 2 minus x sub 1 again where you say x sub 2 minus x sub 1 is not equal to 0 because this if this is 0 you have a vertical line and the slope is undefined but following this slope formula if i just say i know that the y intercept would occur at zero comma b so let's just say i have that and let's say my other point is just x comma y okay so let's say i replace zero comma b with my x sub one y sub one so let's say this is x sub 1 y sub 1 let's say this is x sub 2 y sub 2. we're just going to plug in okay we're just going to plug in so for y sub 2 i'm going to have just y right so that's what i'm plugging in so m equals y minus for y sub 1 i have b this is over for x sub 2 i have x and then for x sub 1 i have 0. x minus 0 is just x right so we can go ahead and just write that like that and then what i can do is i can multiply both sides of the equation by x to clear that denominator so this will be gone and what i have is i have m times x is equal to y minus b so how can i get y by itself i can just add b to both sides of the equation let's just kind of finish this up we would have what we would have mx plus b is equal to y which i could flip around and put y out in front and say y equals m the slope times x plus b my y intercept so pretty simple to derive that and we know our slope is m since again this came from the slope formula and we know that b is the y coordinate for the y since we plugged in a 0 comma b the y-intercept that got plugged in as a point all right so a pretty common task in this section is to be given a slope and a y-intercept and to be asked to write the slope-intercept form of that line so suppose we're given a slope m of negative 3 4 and a y intercept of 0 comma 7. so again this right here this is my b we already know what m is so just match up the format y equals m times x plus b again just plug in it's very simple y equals m which is negative three fourths times x and then plus b b is seven right it's the y coordinate for the y intercept so plus set all right what about m equals four fifths and a y intercept that occurs at zero comma negative four so again this guy right here the y coordinate for the y intercept that's your b so you get y equals m the slope times x plus b so y equals m m is four-fifths times x and then you could do plus negative four you could just do minus four so y equals four-fifths x minus four all right so in some cases we're not given enough information to immediately write the equation of a line and sometimes we're just given the slope and a point on the line or we might be given two points on the line when this occurs we can use point slope form and then from there we can go into slope intercept form okay so point-slope form this is something you should remember a point on the line and the slope okay that's the normal situation to be given when you want to use this form so this kind of x sub 1 and y sub 1 that's going to be our known point and m our slope is going to be known or sometimes we'll get two points on the line and from that we can use one of those points and we can calculate the slope using slope formula all right so where does this guy come from well if we start with our slope formula and let's say we say m is equal to again y sub 2 minus y sub 1 over x sub 2 minus x sub 1. and let's just say that x sub 1 y sub 1 is the known point and x comma y is the unknown point so i'm just going to change this notation up just a little bit but it's the same thing right you've got the difference in y values over the difference in x values all i'm going to do let me just kind of slide this down to get this i'm just going to multiply both sides by the quantity x minus x sub 1. so multiply this by x minus x sub 1. and again i can put the m out in front so put that out here and this cancels with this what i'm going to have is i'm going to have m times the quantity x minus x sub 1 is equal to y minus y sub 1 okay it's exactly what we have here it's just in a different order i can erase this and kind of flip this around it makes no difference i can say this is y minus y sub 1 is equal to m times the quantity x minus x sub 1 okay so this came straight from the slope formula all right so let's suppose we wanted to write the equation of a line given the following information we're given the slope m which is three and we're given one point on the line which is five comma seven so all we need to do is just plug into the point slope form so again that's y minus y sub one is equal to m the slope times the quantity x minus x sub 1. again m is 3 so i would just plug in a 3 there let me put y minus y sub 1 equals i'm going to put a 3 there times the quantity x minus x sub 1. now this is my given point this is what you would label as x sub 1 y sub 1. so what i'm going to do is i'm going to erase this i'm going to plug in my y sub 1 which is 7 i'm going to erase this and plug in my x sub 1 which is 5 and then from here i can just solve for y right so i'd have it in slope intercept form so y minus 7 is equal to 3 times x is 3x then minus 3 times 5 is 15. and let me scroll down and get a little room going so let me add 7 to both sides of the equation and i'll be done right so this is going to cancel so my slope intercept form is y is equal to 3 times x and then negative 15 plus 7 is going to be negative 8 so you just put minus 8. so y equals 3x minus 8. so again if you get confused by this just remember for point-slope form just follow the name i know a point and the slope or it could be that you get two points right you can calculate the slope from that so you would still know a point in the slope if you're given a y-intercept and a slope directly then you can immediately put it in slope-intercept form okay so that's the difference between these two and where you're going to use it in which situation so again i've given another situation where i have a point which is 2 comma negative 4 and the slope m equals one-fifth so again i have a point in the slope so i want to use point-slope form so y minus y sub 1 equals m times the quantity x minus x sub 1. okay so just plug in y minus for y sub 1 again just label this as x sub 1 y sub 1. so y sub 1 is negative 4 minus a negative 4 is plus 4. and then m is 1 5 and then times the quantity we have x minus x sub 1 which is and then i can just solve this for y and i'll have my slope intercept form so y plus 4 is equal to we'd have 1 5 x and then minus you'd have 1 5 times 2 which is two-fifths okay so let's go ahead and solve this guy for y so y is going to be equal to what if i subtract 4 away from each side of the equation let me do this over here as getting a common denominator if i multiply 4 by 5 over 5 it would be minus 20 over 5 okay 20 over 5 again is the same as 4 so i just subtracted 4 away from each side so i'm going to have 1 5 times x and then you'd have basically negative 2 minus 20 which would be negative 22 so i'll put minus 22 then over the common denominator of 5. all right so a more tedious scenario occurs when you're just given two points on the line and you're asked to write the equation of a line so this is one of those scenarios that can get kind of confusing you still want to use point-slope form because you know a point right you have two of them you can use either one and you know the slope because you can calculate the slope using a slope formula so m is equal to what it's the change in y values so y sub 2 minus y sub 1 over the change in x values x sub 2 minus x sub 1 right so the change or difference in y values over the change or difference in x values so i can label this as x sub 1 y sub 1 and this is x sub 2 y sub 2. again you can change that up it doesn't matter so for y sub 2 i have 7. so let's plug that in for y sub 1 i have a negative 5. then for x sub 2 i have 7 and for x sub 1 i have 3. so what i'm going to have here is what 7 minus a negative 5 is 7 plus 5 that's 12. this is over 7 minus 3 is 4. so 12 over 4 is 3. so now i know my slope is 3. so let me erase this and just put m equals 3. so here's where it can get confusing a lot of students will at this point be confused as to which point they need to plug in and the answer is you can plug in either one right they're both points on the line so it doesn't matter which one you choose but this labeling will get a little confusing so let me just show you this i'm going to use each one i'm going to say this is x1 y1 to start and i'm going to plug into my point slope form and then i'll do the other point i'll show you it's the same so y minus y sub 1 is equal to m times the quantity x minus x sub 1. okay so we're just going to plug in so let me just do this above y minus y sub 1 is going to be negative 5 so that's plus 5. and this equals m is my slope it's 3 times the quantity we're going to have x minus again for x sub 1 i have 3. so let me erase this real quick let's solve this for y so y is going to be equal to what if i subtract 5 away from each side of the equation over here on the right 3 times x is 3x and then minus 3 times 3 is 9 and then minus 5. negative 9 minus 5 is going to be negative 14. okay so you get y equals 3x minus 14. so let me erase this so we have let me just kind of put this over here we have y equals 3x minus 14. okay so let me use this guy as a point now i'll show you get the same thing so x sub 1 y sub 1 okay so when we plug in we have y minus y sub 1 equals m times the quantity x minus x sub 1. okay so y sub 1 is 7. so this is 7 and x sub 1 is going to be 7 as well and m my slope is 3. so at this point you might say hey that doesn't look the same but when you solve it for y you're going to get the exact same equation as you got using this point over here okay so y minus 7 equals you'll have 3x minus 21 add 7 to both sides of the equation and again you're going to get that guy you would have what this cancels you'd have y is equal to you'd have your 3x and then negative 21 plus 7 is negative 14. so again either way it's the same right so don't get confused about which point to use no matter what point you use you end up with the same equation when it's put in slope intercept form all right let's try another one so now we're given two points 2 comma 14 and 6 comma 4. again just go ahead and calculate your slope first so m is equal to i'm going to go as this one with x sub 2 y sub 2 and i'll say this one is x sub 1 y sub 1. again it doesn't matter so we'd have y sub 2 minus y sub 1. so 14 minus 4 is going to be 10 right so i'll just write 10 here just save a little time and then x sub 2 is 2 x sub 1 is 6. so 2 minus 6 x of 2 minus x of 1 so 2 minus 6 is negative 4. so my slope here is going to be what if i divide each by 2 i get 5 over 2 and then it's negative so negative 5 halves okay so now that i know my slope and a point on the line so a point on the line and pick either one again it doesn't matter i just showed you that let's just pick this one we'll say this is x sub 1 y sub 1. again m my slope is negative 5 halves plug into the point slope form so we have y minus y sub 1 is equal to m the slope times the quantity x minus x sub 1. so for y sub 1 i'm plugging in a 14. for x sub 1 i'm plugging in a 2 for m i'm plugging in a negative five halves okay so let's just solve this for y and we'll be done so we'll have y minus 14 is equal to negative five halves times x is negative five halves x and then negative five halves times negative two let's just do this off to kind of the side let's put plus negative five halves times negative two we can clearly see the twos would cancel negative times negative is positive so you basically have positive five there okay so we'll put plus five let me scroll down and get some room going so at this point i just want to add 14 to both sides of the equation so we can see that this would cancel over here and now you have y is equal to negative 5 halves times x and then plus 5 plus 14 is 19. okay so that would be your equation in slope intercept form y equals negative 5 halves x plus 19. all right let's talk about something that's a little bit confusing because the definition is going to vary from textbook to textbook and also based on which classroom okay so something known as the standard form of a line this is something very useful it looks like this you have a times x plus b times y is equal to c where a is the coefficient of x b is the coefficient of y and c is some constant okay so an example would be something like 5x plus 2y equals 3. okay my a here is 5. my b here is 2 and my 3 here is c so we need to understand what a b and c represent because they're going to be used a lot in our formulas so for most textbooks when we talk about a b and c we say that they need to be integers okay this is kind of a stricter definition so we would say a b and c are integers okay or integers but as you move higher in math this is kind of relaxed okay and we say a b and c are just any real numbers okay it just needs to be in this format now we also say that again in the stricter definition this is something you generally see in high school that a is greater than or equal to 0. so a the coefficient of x b the coefficient of y and c the constant are integers and a the coefficient of x is going to be non-negative right so it's 0 or some positive number now from this form we can quickly get a lot of information we can see that if i plugged in a 0 for x okay a 0 for x this would go away i would have b y is equal to c i would divide both sides by b and i would find out that my y intercept occurs at what so there y would be equal to c over b so c over b would be the y coordinate for my y intercept now i can also get the slope pretty quickly from this as well if i start with ax plus b y equals c i know if i solve this for y the coefficient of my variable x is my slope so if i subtract ax away from each side of the equation i would have what i would have b y is equal to negative ax and then plus c so if i just divide both sides of the equation by b what do i get what i'm going to have is what this cancels i'll have y is equal to the negative of a over b times x and then plus c over b again if you match this up with slope intercept form we're using different kind of letters but it's the same thing y equals m the slope times x plus b the y intercept m here is negative a over b and then b here my y intercept occurs at zero comma c over b which we just talked about a minute ago and i'll prove this to you later on with some examples that we can get the slope either as the coefficient of x when this guy solved for y or if it's in standard form we could just have the negative of a the coefficient of x over b the coefficient of y all right so let's take a look at an example so suppose you're given y equals 5 3 x minus 3 and you're told to write this in standard form again with the stricter version where a b and c are integers and a is not negative so ax plus b y equals c just want to put in this format so here's my something times x this is my ax right so let's subtract this away from each side of the equation and that's going to give me what i'll do this over here you'll have negative 5 3 times x and then plus y is equal to negative 3. now at this point okay at this point if you're using kind of a looser definition where a b and c are real numbers then you're basically done right this is in standard form and i can show you really quickly okay i can show you really quickly that the slope is the same again my slope when it's in standard form m is equal to the negative of a over b in this case a is negative five thirds and b is going to be positive one right this guy has an implied coefficient of one so negative of negative five thirds so the negative of negative five-thirds is positive five-thirds and this would be divided by one anything divided by one is just itself so you see you get a slope of five-thirds which is exactly what you get here right the coefficient of x so it's a quick way to calculate the slope if it's in standard form it's the negative of a the coefficient of x over b the coefficient of y okay but again for our purposes if we want a b and c to be integers and a to be non-negative we need to do a little bit of work okay so the first thing is to consider how can i make this into an integer well i can multiply both sides of the equation by 3 to clear the denominator but at the same time i want it to be non-negative so what i could do is i could just multiply by negative 3 and that would fix both problems right so if i multiply this side by negative 3 and this side by negative 3 to make it legal negative 3 times negative 5 thirds the negative times negative would be positive and then the denominators would cancel right 3 over 3 is 1 so i would just have 5 times x and then you'd have negative 3 times y which would be negative 3y and this equals negative 3 times negative 3 is positive 9. okay and again even though i transformed it to look different the slope is still the same it's again m is equal to negative a over b what is a a in this case is five now what is b b is negative three what is negative five over negative three again that's five thirds so either way our slope is the same right we have five thirds here we have five thirds here all right so let's put everything together in one kind of last example so suppose you're given two points on a line and you're told to write it in standard form and again you're going to use the stricter definition so you're given points 10 comma 6 and negative 5 comma negative 12. well you're not going to immediately put this in standard form and you can't immediately put it in slope-intercept form you've got to use your point-slope form again if you know a point and the slope you use point-slope form in this case i can generate a slope and i know a point so my slope formula again m equals y sub 2 minus y sub 1 over x sub 2 minus x sub 1. so let's label this as x sub 1 y sub 1. let's label this as x sub 2 y sub 2. so let's go ahead and plug in so for y sub 2 i have a negative 12. for y sub 1 i have a 6. for x sub 2 i have a negative 5 and for x sub 1 i have a 10. okay so if i think about negative 12 minus 6 that's negative 18. so this is negative 18. negative 5 minus 10 is negative 15. so i know immediately this is going to be positive right because negative over negative is positive each here is divisible by 3. if i divide 18 by 3 i get 6. if i divide fifteen by three i get five so it looks like my slope is going to be six fifths okay so my slope is six fifths okay so from here i can just use a point it doesn't matter again which one i use let's just go ahead and take this one we'll say this is x sub one y sub one and i'm going to plug into my point slope form y minus y sub 1 is equal to m times the quantity x minus x sub 1. so for y sub 1 it's 6 for m it's six fifths and for x sub 1 it's going to be 10 okay just scroll down just a little bit all right so we'll have y minus 6 is equal to 6 fifths times x is 6 fifths x and then minus 6 fifths times 10 we can cancel this 5 with this 10 that'll be a 2 up here 2 times 6 is 12. so we'll have minus 12 there and again to finish this up let me add 6 to both sides of the equation this will cancel and we'll have y is equal to six fifths x and then negative 12 plus six is minus six okay so let me just scroll down so now we want it in standard form and to do that again i want ax plus b y is equal to c so let's move this over here on the left so we're going to subtract six fifths x away from each side of the equation and of course on the right side it cancels you'll have negative six-fifths x and then plus y is equal to negative 6. so again at this point it matches this format if you were using a definition where a the coefficient of x b the coefficient of y and c the constant were real numbers but again we want to use the stricter definition so what i'm going to do is a little bit of additional work so if i want to clear this denominator here right because i don't want this to be a fraction i want it to be an integer and i want to get rid of this negative here because again i want it to be non-negative i can just multiply both sides of the equation by negative 5 right it would change the negative to positive and it would clear the denominator so let's multiply this by negative 5 to make it legal so negative 5 times negative 6 fifths would just be positive 6 then times x negative 5 times y would be minus 5y this equals you have negative 6 times negative 5 which is 30. again you can use this 6x minus 5y equals 30. i can show you that everything is the same right we found that y was equal to 6 fifths x minus six okay again my m my slope is what it's the negative of a the coefficient of x over b the coefficient of y okay so make sure you take the negative and the five what do we get we get six fifths right which is exactly what we got here okay so that's pretty quick and then again the y-intercept occurs at c over b so the y-intercept is c over b so then here we know it's negative 6. c is 30. c is 30. b is negative 5 and again you get the same thing 30 over negative 5 is negative 6 so i know the y intercept would occur at 0 comma negative 6 just as it tells me here in slope intercept form in this lesson we want to review parallel and perpendicular lines so usually when we study linear functions we come across the topic of how to determine if two lines are parallel perpendicular or neither so for the lesson today we're going to start by thinking about parallel lines so most of you already know that parallel lines are any two lines on a plane that will never intersect we can determine if two lines are parallel by examining the slope of each and essentially the rule is any two non-vertical parallel lines are going to have slopes that are equal and again the reason i say non-vertical is because when we have a vertical line we know the slope is undefined okay so let's start out with an example here again with parallel lines so i'll go through and show you how to determine if you have parallel lines and then i'll show you this graphically so let's suppose you're given this and you have 25x minus 5y it equals negative 20 and then you also have 5x minus y equals 3. so in each case we have a linear equation in two variables and what we want to do is solve each equation for y and put it in what is known as slope-intercept form so we've been talking about this form for a while now and in case you don't know slope-intercept form occurs when we solve for y so we have y equals m the slope times x plus b the y intercept so this m right here the coefficient of x is known as the slope and again this b this is going to be the y coordinate for the y-intercept so we'll say this is the y-intercept all right so let's start with the first one we have 25x minus 5y equals negative 20. so we have 25x minus 5y is equal to negative 20. i'm going to scroll down and get some room and i'll come back up okay so what i'm going to do is isolate this y here right on one side of the equation so to do that i want to start by isolating the variable term with y involved so i'm going to isolate this guy so i'm just going to subtract 25x away from each side of the equation and so what's going to happen is this will cancel we'll have negative 5y is equal to negative 25x and then minus 20. all right so to complete this i want y by itself so think about what's multiplying y that's going to be negative 5. so i'm just going to divide both sides of the equation by negative 5 and let me get a little bit more room going so this is going to cancel and we'll have y is equal to you have negative 25 over negative 5 that's positive 5 then times x and then negative 20 over negative 5 is plus 4. so again this is in the format of y equals m the slope times x plus b the y-intercept so my slope here is 5 right that's the coefficient of x and my y intercept will occur at 0 comma 4 right so b again the y intercept is 4. all right so let's erase everything and bring this equation up to the top all right so let me just drag a line over here and say this is y equals 5 x plus 4 again in slope intercept form and i'm just going to label 5 as m my slope and you can label 4 as the y intercept if you want to okay let's do the same thing for this one it's going to be less tedious so we have 5x minus y equals 3. so 5x minus y is equal to 3. okay so with this guy again if i want to solve it for y the first thing i want to do is just get this guy by itself so let me subtract 5x away from each side of the equation that's going to give me what this will cancel you'll have negative y is equal to negative 5x and then plus 3. let me go ahead and divide both sides of the equation by negative 1 right so i can get rid of this guy and let me scroll down a little bit all right so this will cancel with this i'll have y is equal to if you have a negative 5 over negative 1 that's positive 5 and then times x and then 3 over negative 1 is minus 3. so we have y equals 5x minus 3. so again this coefficient of x this 5 here that's going to be my m or my slope and the y-intercept would occur at what 0 comma negative 3. so negative 3 there would be my y intercept right and you could make that more clear by just putting plus negative 3 if you wanted to or you could leave it in this form like we have okay either way all right let me erase everything all right let me drag a line over here and i'll just say that we have y equals we have 5x and then minus 3. okay i can just kind of match this up and say my m here my slope is 5 my y-intercept would have correct 0 comma negative 3. let me make this kind of line going to it a little bit more accurate and so we have our two equations there one is y equals 5x plus 4. the slope is 5. one is y equals 5x minus 3. the slope is 5. so again i said that two parallel lines if they're non-vertical would have the same slope and obviously they would have different y-intercepts right if you have the same slope and the same y-intercept then you have the same line okay but with parallel lines again they're never going to intersect so it's going to have different y intercepts and the same slope now let's look at this graphically real quick and then i'm going to come back up and show you a little trick that's going to make this a little faster so looking at the graph of this guy we can clearly see what parallel lines look like right you can see they have the same slope or steepness so that's what causes them to never intersect or never touch each other so again i can put down that we have y equals 5 x plus four and that's going to be this guy right here right the one on the left i can go through and plot the y intercept which occurs again at zero comma four this is my y-intercept this is my slope my m so 0 comma 4 is right here right that's my y intercept the slope is 5 so i can either go up 1 2 3 4 5 and to the right 1 to get another point or i could start here and go down 5 1 2 3 4 5 into the left one to get a point so you can see this is again the graph of you have y equals 5x plus 4 okay then this guy on the right here you can see this is the graph of y equals 5x minus 3 okay so my m my slope is 5 which is the same and then my y intercept occurs at 0 comma negative 3. so 0 comma negative 3 is here again slope is 5 so i go up 1 two three four five to the right one go up one two three four five to the right one okay so this guy right here is y equals my five x minus three so that gives you a little insight into what two parallel lines look like again they don't intersect because they have the same slope or steepness alright so i want to show you a little shortcut generally speaking when you work with these problems the problems are given to you where the lines are in standard form okay so what is standard form again we talked about the different forms of a line in the last lesson but in case you don't know basically standard form is where we have a times x plus b times y is equal to c our constant okay so usually we'll define a b and c to be integers and we'll say a is greater than or equal to zero but for what i'm going to show you here a b and c could be any real number that you want it to be okay so generally speaking let's say that we solve this equation for y and we put it in slope intercept form we're going to get a little insight here so let's subtract ax away from each side of the equation i'll have that b times y is equal to negative ax plus c let me go ahead and divide both sides by b we're going to find here that y is going to be equal to the negative of a over b times x plus you have c over b okay so what we can say is that if the line is in standard form the slope is going to occur at what this coefficient on x is negative a over b so it's the negative of the coefficient of x over the coefficient of y right the negative of a over b and the y intercept would occur at c the constant over b the coefficient of y so let's try this out real quick on this example and i'll show you how easy this is so for this first one the coefficient of x again that's considered a that's 25 in this case so a is my 25. the coefficient of y is b so b is negative 5. let me erase this and then c is my constant so that's negative 20. okay so that's the first case so how can we find the slope again it's negative a over b so let me write that rule down m is negative a over b so the negative of a would be negative 25 over b b is negative five so this is my m and what does it give me positive five so look how quickly we found the slope versus having to solve it for y now you can still solve it for y it's still valid this is actually a little bit quicker now let me erase this and we'll do the other one so we would have what an a value here that's going to be 5 a b value here that's negative 1 and a c value here that's 3. again a is the coefficient of x b is the coefficient of y and c is your constant so in this case m is what it's the negative of a a is 5. so negative 5 over b b is negative one again negative five over negative one is positive five so that's my slope so again for your y intercept you can find that quickly as well so for the y intercept for the first one it would be what c over b c is negative 20 negative 20 over b which is negative five so that gives me positive four and that's what we found right in the first case it was y equals five x plus four so the y intercept occurred at zero comma four then for the second case what did we find it was y equals five x minus three so again if i have c over b c is three b is negative one again we get negative three there it occurs at zero comma negative three just as we found so this is a very quick way to kind of work with lines when they're in standard form if you're trying to perform this task let's talk a little bit about perpendicular lines now so with perpendicular lines they're going to intersect at a 90 degree angle so if we have two non-vertical perpendicular lines they're going to have slopes whose product is negative 1. so let's take a look at this example so we can do this the long way we could solve for y or we could just really quickly get our slopes again if it's in the format of ax plus b y is equal to c then the slope is going to occur where m is going to be equal to the negative of a over b okay so that's how quick we can find our slope so a in each case is the coefficient of x so this is a and this is a b in each case is going to be the coefficient of y so this is b and this is b okay so what is the negative of a over b for the first case so the negative of 4 over negative 7. so my slope here is what negative over negative 7 is 4 7. so that's for the first one and then m is equal to again negative a over b so for the second one you'll have negative 7 is my a over 4 which is my b so this is negative 7 4. so much quicker than solving for y so let's erase this and let's think about this we know the slopes are not the same 4 7 is not the same as negative 7 4 so immediately you can rule out that you have parallel lines but what we notice is that these guys are negative reciprocals of each other right if i multiply them together i end up with negative 1 as my product so 4 7 times negative 7 4 we know that this 7 would cancel with this 7 and leave me with a negative 1 and this 4 would cancel this 4 and leave me with a 1 right so you basically have 1 times negative 1 which is negative 1 as your answer so that tells me these lines are perpendicular now if i want i can get my y intercept and we'll look at this graphically so let's get the y-intercept for each one again that occurs at c over b so c is your constant right so in this case it would be 14 over b is the coefficient for y so negative 7. so 14 over negative 7 is negative 2. right so the y intercepts would occur at 0 comma negative 2 for the first one using that information i can go ahead and write this in slope intercept form i can say this guy is going to be y equals m the slope which is 4 7 times x and then minus 2 right my y intercept part then let me just erase this we have the information now for this one the y intercept occurs at again c over b so c is 28 over b which is 4 28 over 4 is going to be 7 so this guy would be y equals m the slope which is negative 7 4 times x and then plus 7 your y intercept and you can pause the video and solve each one of these for y and you will completely verify that i got the correct answer in each case okay so let's look at this graphically again one equation was y equals 4 7 times x and then minus 2. the other was y equals negative 7 4 times x plus 7. so if we look at 0 comma negative 2 which is the y-intercept for this first one so 0 comma negative 2 is going to be right there and then my slope is 4 7. so i would go up 1 2 3 4 and to the right 7 so that would put me right there and again i could start at this 0 comma negative 2 and go down 1 2 three four into the left seven okay so that's right there so that's this line here again this is going to be y equals four sevenths x and then minus two for this other line here again we have a y intercept that occurs at zero comma seven so that's right here and my slope is negative seven fourths so i can drop seven so drop seven and go to the right four so that's there i can drop seven go to the right four so that's there so that's this line here which is y equals negative 7 4 x plus 7. so what you notice is that where they intersect you'll have a 90 degree angle so that's that guy right there again that's a 90 degree angle so that's how you know you have perpendicular lines all right let's look at a few examples real quick we have 2x minus 3y equals 3 and we have 6x plus 4y equals 44. so again you have your choice you can solve each one for y or you can use my little shortcut for lines that are in standard form either way you want to do it i prefer to use the shortcut so we'll say m our slope is equal to the negative of a over b again where a is the coefficient of x and b is the coefficient of y so for this first one my a is equal to 2 and my b is equal to negative 3. for the second one my a is equal to 6 and my b is equal to 4. so for m again it's equal to the negative of a over b so a is two and b is negative three so negative two over negative three is just two-thirds okay so that's for the first case so let me just kind of erase this and put right here that m is equal to two thirds okay for the second case again my m is equal to negative a over b a is six so let's put six here and b is 4 so let's put 4 here so what we have here is going to be negative 3 halves divide 6 by 2 when you get 3 divide 4 by 2 and you get 2 right so m here my slope is equal to negative three halves so these guys are negative reciprocals of each other so these are perpendicular lines so we can say that two-thirds times negative three-halves is equal to negative one right because this cancels with this and gives me negative one this cancels with this and gives me one so you'd have one right this guy's completely cancelled times negative one that will give you negative one okay so once we know this is perpendicular we can just state that we can just say these are perpendicular locks all right let's take a look at another one so we have 5x minus 12y equals 20. we have 12x minus 5y equals 20. okay so let's determine again if they're parallel perpendicular or neither so what i want to do again if it's in standard form which generally it will be you can solve it for y or you can use the shortcut so the slope m is equal to the negative of a over b so this is a my coefficient of x that's 5 over this is b it's going to be negative 12 my coefficient of y so the slope here is 5 12. for the second guy what's my slope again negative of a which is 12 right coefficient of x over b which is negative 5 coefficient of y so this is going to equal 12 fifths now you see these guys are reciprocals of each other but they're not negative reciprocals of each other so basically what you have is 5 12. if i multiply this by 12 fifths it's going to give you positive 1 not negative 1 okay so these are not perpendicular lines and obviously the slopes are not equal so they're not parallel lines so we can go ahead and say these are neither all right let's take a look at another one so we have 9x minus 4y equals 48. we have 18x minus 8y equals negative 120. so again for my slope m it's equal to the negative of a a is the coefficient of x so that's 9. over b b is negative 4 that's the coefficient of y so the slope is just 9 4. then in the second case again the negative of a a is 18 the coefficient of x over b b is the coefficient of y that's negative eight so this would give me what so each would be divisible by negative two right so if i divide negative eighteen by negative two i get positive nine if i divide negative eight by negative two i get positive four you can see the slopes are equal right and the y intercepts are going to be different so we're going to have parallel lines right these are parallel ones okay all right so another scenario you might see in this section you might be asked to write the equation of a line given that it is parallel or perpendicular to some other line so let's write the standard form of the line that passes through this point negative 1 comma negative 4 and it's parallel to 3x minus y is equal to 5. all right so to tackle this task basically what i want to do is i want to figure out the slope of this line here because once i have the slope of this line here i'll know the slope of the line that passes through this point negative 1 comma negative 4 because they're parallel lines right so they're going to have the same slope so i can solve this for y or again i can use my shortcuts and the slope occurs at the negative of a over b a in this case is 3 right the coefficient of x over b in this case that's going to be negative 1 the coefficient of y so the slope here will occur at 3 so again the slope for this line that passes through this point negative 1 comma negative 4 will be 3 as well okay so now that i know my slope and a point i can get rid of this okay this is what confuses people so just cross this out forget that it exists so with a slope and a point i can use point slope form hopefully you saw the last lesson when we talked about equations of a line but if not when we look at point slope form it looks like this y minus y sub 1 is equal to m the slope times the quantity x minus x sub 1. so your given point is x sub 1 y sub 1. so this is x sub 1 y sub 1. so i'm just going to plug in so we have y minus for y sub 1 i have negative 4 minus the negative 4 is plus 4. this equals m my slope which i found was 3 times the quantity x minus x sub 1 which in this case is negative 1 minus a negative 1 is plus 1. all right so i want this in standard form so let me erase this and let me drag this up here so again standard form we want kind of ax plus b y equals c and again typically a b and c would be integers and a is greater than or equal to 0. so to match this format let me just distribute this on the right side first 3 times x is 3x and plus 3 times 1 is 3. we have y plus 4 over here on the left let me subtract 3x away from each side of the equation and let me just kind of move down a little bit so what i'm going to have is this cancels i'll have negative 3x plus y plus 4 is equal to 3. let me subtract 4 away from each side of the equation so this is going to cancel and let me just kind of write this up here we'll have what we'll have negative 3x and then plus y and this is equal to negative 1. now typically we want a the coefficient of x to be a non-negative integer so in this case it's a negative integer right so we have negative 3 here but we can fix that by just dividing both sides of the equation by negative 1. that's going to give me what positive 3x minus y is going to be equal to positive 1 and you can erase this you don't need that anymore so it follows that format of ax plus b y equals c and again if you want to check the slope real quick again the slope should occur at what m should be equal to the negative of a which is the coefficient of x so negative 3 over b which is negative 1 right the coefficient of y so this is negative 3 over negative 1 which is 3 which is exactly what we found there all right let's try another one so we have a given point that's negative 5 comma negative 3 and it's perpendicular to 3x plus 9y equals 36. so again perpendicular lines have slopes whose product is going to be negative 1 right so when we think about this if i have 3x plus 9 equals 36 again the slope will occur where it's going to be at the negative of a which is 3 in this case right the coefficient of x over b which in this case is not so this would be what if i divide each part here by three three divided by three is one nine divided by three is three so my slope is negative one third okay so my slope is negative one third now negative 1 3 times what gives me negative 1. well most of you can understand that it would be positive 3 right but if you didn't know that you can use a variable and to not confuse ourselves let's just use z so to solve this i would multiply both sides of this equation by the reciprocal of this which is negative three over one and so this would cancel and z would be equal to what positive three okay so that tells me that my slope is going to need to be positive 3 because again 3 times negative 1 3 is going to give me negative 1. okay so m is going to be positive 3. so now that i have my slope and a given point you can line this out don't think about it anymore don't consider it it's just going to confuse you so plug into the point slope form so we have y minus y sub 1 is equal to m the slope times the quantity x minus x sub 1. okay so for m we have 3. for x sub 1 again that's this guy right here we have negative 5 minus a negative 5 is plus 5. for y sub 1 again that's this guy right here minus a negative 3 would be plus 3. okay so let me distribute so we have y plus 3 is equal to on the right side 3x plus 15 and again if you want this in standard form i want ax i want ax plus b y equals c again a b and c we want those to be integers and we want a to be greater than or equal to 0. so what i can do to kind of make this a little quicker let me subtract 15 away from each side of the equation and that's going to cancel over here and 3 minus 15 is going to be negative 12. and let me subtract y away from each side of the equation so that's going to cancel over here i know this is a little messy but let's scroll down and get some room going so what's going to happen is on the left side i just have my negative 12. on the right side i would have 3x minus y and you can flip this around and say this is 3x minus y is equal to negative 12. so a my coefficient for x is 3 that's positive b my coefficient for y is negative 1 and c my constant is negative 12 so i basically fit my definition right so let me erase this okay so you can check the slope again make sure that it is three again it's going to be m occurs at the negative of a which is three over b which is negative one negative three over negative one is 3 so we're good to go there and we can say this line is going to be 3x minus y equals negative 12 again in standard form in this lesson we want to review the graphs of basic functions so at some point in a college level algebra course or pre-calculus course you're going to come across a section where they talk about how to graph the most commonly occurring functions that you're going to kind of come across now there's more of these than what we're going to show you today but essentially the functions here are known as elementary functions or some books will say basic functions or parent functions it's important to kind of look at these functions because they just keep coming up over and over and over again so if you know the basic shape and you know the domain and range it's just going to help you a ton when you have to keep considering problems like this you know with these functions now as we progress through the course we'll kind of add some of these to our list we're not going to talk about exponential functions and logarithmic functions and kind of sine and cosine these are things we're going to talk about later in the course once we have the foundation okay but for right now we're going to look at some ones that you should be comfortable with things that we've already kind of seen up to this point so let's start out today by looking at something known as the identity function so this is just where y equals x it's a very simple linear equation in two variables and again i can write this in function notation and say this is f of x is equal to x so whatever i plug in for x i get for y so that's why it's called the identity function and it's very easy to graph this remember if we have a line in slope intercept form it's solved for y it's y equals m the slope times x plus b the y intercept in this case we don't have a visible coefficient for x so we can really just say it's y equals 1 right i can write anything times 1. so 1 times x plus i don't have anything visible here so i just put plus 0. so lining things up i can see that my m my slope is 1 my b is 0. so this tells me that the line has a slope of one and a y-intercept that occurs at zero comma zero well in fact the x-intercept is also going to occur at zero comma zero so this is essentially a line that passes through the origin you can see that right there and it has a slope of one right so if we wanted to graph this we could graph that zero comma zero the origin we could go up one to the right one we'd have two points and we could draw our line it's already drawn four so i'm just going to skip that part all right so let's talk a little bit about the domain and range and then we'll kind of move on we know that when we work with a linear equation in two variables that's not a special case scenario so that's not a horizontal line and not a vertical line we're going to have a domain that's all real numbers as well as a range that's all real numbers so you can see this graphically and you can also see this by just kind of inspecting the equation if i look at x you think about that for your domain is there anything i can't do the answer is no so the domain the domain is all real numbers so from negative infinity to positive infinity as well as the range right because whatever i plug in for x i get for y so i can make x as small as i want i can make it zero i can make it as large as i want so the range is all real numbers as well now you can also see this graphically although we're limited to kind of the space that's available on this kind of graph that i drew you see that you have an arrow at each end so that's telling me i'm extending to the right forever to the left forever up forever and down forever okay so i'm going in all directions forever so the domain again i can have anything for x my range i can have anything for y all right now let's move on and talk about the squaring function so this is y equals x squared essentially i plug in something for x i square it i get y again i can write this in function notation so f of x equals x squared we haven't talked about how to graph quadratic equations yet in this course i know most of you saw it in algebra 2. essentially this is the shape of a quadratic equation this graph is known as a parabola we'll do a deep dive into parabolas in the next section of the course for now we just want to kind of observe what it looks like and think about the domain and range and i put two arrows there because that is going to extend this way and this way kind of forever okay so we want to think about those arrows to indicate that now if i think about kind of some points that would be on this graph we know that squaring something gives me a result that's non-negative so if i plug in a 0 for x y would be 0. so that's this point right here and notice how that's the lowest point that y goes and that's because if i plug something in for x that's negative i get a positive if i plug in a positive i get a positive so plugging in a 0 is going to be the smallest result i can get for y okay so that's why this 0 comma 0 is on the graph and it's the lowest point now for every other y value other than 0 it's going to be linked up to more than one x value it's going to be linked up to two x values so let's see what's going on and why it's going on so you see this point here you have negative 1 1 and you have this point here 1 1. so two different x values one is negative one and one is positive one they're both linked up to the same y values so this one is one and this one is also one okay remember this doesn't violate the definition of a function two different x values can link up to the same y value you just can't have the same x link up to multiple y's okay but we could really save some space here and just say plus or minus plus or minus one like this for x and then a y value is 1. okay and you can see the same effect as we kind of move up this graph if i had an x value of 2 i'd have a y value of 4. also if an x value of negative 2 i'd have a y value of 4 okay it's just because of the square right if i take 1 and i square it i get 1. if i take negative 1 and square it i get 1. if i take 2 n squared i get 4 if i take negative 2 and square i get 4. same thing goes for 3 comma 9 and negative 3 comma 9. same thing so i could put plus or minus 3 and then this is 9. i forgot to write 4 here but you know that's going to go there okay so that's how we get some ordered pairs that kind of gives you an idea of the shape and we can sketch our graph you already see what it looks like so i don't need to kind of draw that in let's think about the domain and range of this guy so the domain and the range now again you can do this graphically if you have it already drawn or you can just think about the kind of equation of what's going on so for the domain i always think about is is there something i can't do with x am i restricted is there something illegal well in this case i'm squaring x so there's there's nothing wrong with plugging in a zero nothing wrong with plugging in a negative or a positive so the domain is all real numbers so from negative infinity to positive infinity and the graph confirms that right we're going to the left forever we're going to the right forever but where we're limited is our range right because it's the result of squaring something so we know that kind of the lowest or the smallest it can be is 0. we see that on the graph if we think about the vertical axis the y values the lowest it is is right here and that's going to be a y value of 0. so from 0 and including that out to positive infinity okay so the domain all real numbers the range from 0 and including 0 out to positive infinity all right now let's talk about the cubing function let me put these arrows in here now so i don't forget so y equals x cubed so obviously i'm plugging in a number and i'm cubing it and i get my y or again my f of x if you want to write that notation so when i think about this guy okay it's kind of harder to get points because when you look at this your graph is usually constrained to what you can fit on a sheet okay so i always go up to 10 and down to 10 to the left to 10 and to the right of 10 so i can really only get about 5 points on this guy right so if i plug in a 0 for x i get a zero for y so one point that i didn't put on here would be zero comma zero let me just kind of list this out so zero comma zero another point if you plug in a one and you cube it you get one so one comma one and then another point if you plugged in a negative one you'd get a negative one okay and then you can kind of do two and eight and you can do negative two and negative eight okay so that's the points that we have again we have zero comma zero we have one comma one we have two comma eight we have negative one comma negative one and we have negative two comma negative eight so from those five points you get an idea of what the shape looks like and you could sketch it that way again it's already pre-drawn so we don't need to kind of draw that in and let's just think for a minute about the domain and range because this one isn't very complicated you can see that because we're cubing this guy and cubing something does not affect the sun right it's an odd exponent so it's not going to change a negative to a positive like when we square something so i can plug in anything for x and i can also get an output of anything for y because let's say i squared for example negative 2 well i'm getting negative 8 right i squared a negative i got a negative okay so for this guy the domain the domain is again all real numbers from negative infinity to positive infinity and the range is also going to be all real numbers because again this exponent is odd so it's not going to have an effect on the sine like it does when we square something so from negative infinity to positive infinity now again you can see this graphically this guy is obviously rising forever and it's falling forever and although it doesn't look like it's going to the right forever and to the left forever it is it's just doing it kind of slowly right so it is going to the left forever it is going to the right forever so the domain is all real numbers and the range again is all real numbers all right now let's talk about the square root function so this guy is y equals the square root of x let me put my arrow in and again you could say f of x is equal to the square root of x if that's better for you and again all i'm doing here is i'm plugging in something for x i take the square root i get y okay so you can see some points that we can make here if you plug in a zero you get zero so again if i kind of think about it this way a zero gives me a zero and then if i plug in a one i get a one so that's why this point is here one comma one to get more values you got to kind of think about what i can plug in here that's going to be a perfect square so you can plug in a 4 4 is a perfect square and you would get a 2. right so 4 gives me 2. plug in a 4 for x get a 2 for y and then the next perfect square not five not six not seven not eight you come to nine and so nine if i took the square root of that would be three okay so that's where this nine comma three comes in there okay so that gives you a general idea of the shape and you could sketch it from there and you could keep going if you wanted to the next perfect square is going to be 16 so you could have 16 and then four but obviously i don't have enough space here to kind of plot that point so that's our that's our graph that's our general shape and i erased that and i should have so let me just erase this and now let's think about the domain and range so let's think about the domain and our range okay so what do we got going on here again if i just look at the equation i'll look at the graph in a second the square root of x is there anything i can't do well yeah i'm restricted in the real number system from taking the square root of a negative we know that we worked problems when we're talking about the quadratic formula where we ended up with the square root of a negative we used our complex number system we involve the imaginary unit i but when we think about domain we're thinking about the real number system it's very important that you know that because a lot of students will kind of mess that up the domain is always the real number system unless your teacher specifically tells you otherwise okay so we can't take the square root of a negative so that means the domain is constrained to be non-negative so it's from zero out to positive infinity you can confirm that by looking at the graph if we think about horizontal values if we think about on that x-axis this guy is zero that's where it starts and it goes out to the right forever so from zero including zero out to positive infinity it's the same thing with the range right it's from 0 out to positive infinity again graphically here's the y value kind of think about things going up and down starts at 0 and increases forever now we know that just looking at the equation we can see the same thing because if i plugged in a 0 for x that's the smallest i can plug in for x y would be 0. and i can't make it any smaller right because i can't go negative and as i increase x y is going to increase so the domain is from 0 to positive infinity including 0 so is the range right from 0 including 0 out to positive infinity all right now let's talk about the cube root function so we have y equals the cube root of x or again you could write f of x equals the cube root of x let me make that 3 a little bit better now let me put these arrows in here so here and here okay so what you'll notice here is that when you make ordered pairs let's just make some real quick so x and y if you plug in a 0 you get 0 right so if you plug in a 0 for x you get 0 for y if you go through and plug in a 1 for x you get a 1 for y if you plug in a negative 1 for x you get a negative 1 for y so that's these points here i didn't put 0 comma 0 but here's 1 comma 1 and here's negative 1 comma negative 1. now for these points they're going to start to seem very familiar let's say you do 8 and then 2. so if i plug in an 8 for x i take the cube root i get 2. if i plug in a negative 8 for x i take the cube root i get a negative 2. why does this look familiar let's go back up notice how with y equals x cubed i plugged in a 2 for x i got an 8 for y right that's because i'm cubing 2 and i get 8. well if i take the cube root of 8 i go back to 2. same thing goes for this negative 2 comma negative 8. so these points kind of seem familiar and the graphs also look familiar so what you're going to find is that these guys are inverses okay and we're not going to talk about this now but later on in the course we're going to talk about inverses and we're going to see how we can quickly graph inverses and kind of all these things that are related to inverses but essentially the x and y values are just swapped so in y equals x cubed you plugged in an x value of 2 and got a y value of 8 here you plug in an x value of 8 you get a y value of 2 okay same thing goes here you plug in a negative 8 for x you get a y value of negative 2 and y equals x cubed it's reversed right you plug in a negative 2 for x you get a negative 8 for y so we get the general shape of this again it looks just like the other one it's just sideways now and essentially what we want to think about is the domain and range so the domain and the range but again i'm working with kind of an odd number here for the index so the domain is all real numbers right i can plug in anything for x and take the cube root there's no constraints there and obviously on the graph you can see this goes kind of to the left forever to the right forever so from negative infinity to positive infinity the range is the same thing and a lot of students don't kind of get that they think this is constrained on the graph but this is falling over here and this is rising it's doing it at a slow pace but it is doing that so it's from negative infinity to positive infinity and again you can think about the equation here can i make y as large as i want well yeah i can plug in as big of a number for x as i'd like so i can make y infinitely large same thing goes for making y infinitely small right i can plug in a negative there there's no problem and take the cube root so i can make x as small as i want and make y as small as i want by plugging in these gigantically large negative numbers so the domain is all real numbers and the range is all real numbers all right the last one we're going to look at is known as the reciprocal function so we have y equals 1 over x and again i could write this as f of x equals 1 over x so it's essentially the reciprocal of x right if i take the reciprocal of x i end up with 1 over x so that is why you call this the reciprocal function now obviously when i think about this guy i can't divide by zero okay so if i think about restrictions here i can't divide by zero and i'll also never get a result of 0. if you have 1 over x and you say this result how can i make that 0 i can't right because if i divide here i know x can't be 0 that's undefined but i have a fixed numerator here of 1. think about how you get 0 from a division problem you have to take 0 and divided by something that's not 0. this is fixed at 1 so this has no solution right so it's not possible for 1 over x to give you a y value of zero so i want you to notice two kind of things that are going on with this graph so let me put my arrow here and my arrow here and my arrow here my arrow here to make this simple i'm just going to think about the positive for right now okay when i get an x value that's really really really large let's say i erase this and put something like i don't know let's say 100 1 divided by 100 is .01 okay but we can make this number as large as we want and the larger we make it the closer we actually get to zero so that's what's going on kind of here where we go to the right you're noticing that the x values are getting bigger and bigger we're getting closer and closer to a y value okay vertical value here of 0 but we're never going to touch it so that's where this dashed red line comes into play this is a horizontal asymptote so we're approaching that value where y equals 0 but we're never going to touch it we'll talk more about this later on in the course in the next section we're going to talk again a lot about quadratic and rational functions and how to graph them for now you just need to know that this is an asymptote and that the graph is approaching that but will not touch it now the same thing is happening when we go in this direction in terms of trying to reach an x value of zero we know we can't divide by zero but as we get closer and closer to an x value of zero y becomes infinitely large okay but x never actually can get to zero so what happens is as i put in something like let's say point five for example one divided by .5 is going to be 1 times 2 which is 2. okay we know that but if i put in something smaller like 0.1 well now i'm multiplying by 10. so now i've got 10. if i put in .01 i'd be multiplying by 100 right and i can keep going keep getting closer and closer to zero without touching it and y gets infinitely large okay it gets as big as you want to make it just as you get closer and closer and closer to zero okay so that's what we're doing here we're approaching an x value of zero so that's this vertical asymptote here at x equals zero okay that vertical line we're approaching that but we're not going to touch it again we'll talk more about this in the next section of the course all right let's think about the domain and range it's pretty obvious what they are again from what i just said and then also from kind of looking at the graph we know we can't divide by zero so the domain is kind of the set of all x such that x can't be zero or we could also use our interval notation so from negative infinity to zero the union with zero to positive infinity again zero zero's not included in either okay so that's how we kind of restrict that now for the range it's going to be the same thing okay the range can't be zero because again one over x will never produce zero because one is fixed okay so this guy from negative infinity to 0 not including it the union width again 0 to positive infinity again you can see this from the graph that all the y values are covered going from not 0 but anything larger up to positive infinity again not zero but anything less right down to negative infinity all the x values are covered except for zero so starting kind of anything larger than zero out to positive infinity and then starting here on the other side so from anything less than 0 out to negative infinity so the domain and range just both exclude 0. in this lesson we want to talk about the piecewise defined functions all right so in the last lesson we talked about some of the most commonly occurring functions that we're going to come across we examined the graphs and then we talked about the domain and range here we're just going to go a step further and talk about the absolute value function the greatest integer function and piecewise defined functions in general so we're going to start out today with the absolute value function so we have y equals the absolute value of x or we could write this as f of x is equal to the absolute value of x now i've already graphed this guy for so we can see what it looks like and let me put an arrow here and an arrow here those two are going to indicate that they extend in each direction forever and essentially you can see that you have this v like shape and what's creating this is the absolute value operation itself if you think about what happens here we can say that y really equals x if x is a non-negative number so in other words if i plug in an x that's greater than or equal to 0 well then y just equals x that's this part of the graph right here okay we saw this in the last lesson where we talked about y equals x that's the identity function okay but what happens is if x is less than zero so if x is less than zero then y really equals negative x okay so what happens is instead of this guy continuing on this path down here it gets flipped up because of the absolute value operation okay so it takes the negative that you're plugging in here for x and it flips it and makes it positive so what happens let me kind of erase this let's say i plug in something like a negative 2 well it flips it and instead of it being a negative 2 down here like you would have had with y equals x well it flips it up here and then you get this point there okay so you get negative 2 comma positive 2 because it made it positive okay so let's think about some ordered pairs here this is already graphed but i want you to see something so let me make an x y table here so we see that the lowest point occurs at zero comma zero right here so a zero for x gives me a zero for y that should make perfect sense for you because the absolute value operation is going to make things non-negative so the smallest i can make y is by plugging in a 0 for x if i plug in a 0 for x i take the absolute value i get a 0 for y if i plug in something positive i'm going to get a positive and if i plug in something negative i'm going to get a positive so absolutely the lowest we can go with y is by plugging in a 0 for x now another thing you can consider with this v like shape every y value other than zero is going to be linked up with or associated with two different x values so in other words if i pick an x value of two it's associated with a y value of two so an x value of two is associated with the y value of 2 but then also the negative 2 that x value is associated with a y value of positive 2 as well so again i can say plus or minus 2 for x gives me a y value of 2. you could do the same thing as you kind of move up the graph so if you go to four you have four comma four so an x value of four gives you a y value of four and then an x value of negative four gives you a y value of positive four as well so again that's how you're ending up with this v like shape it's taking this line right here and as you go to the left of zero right on the x-axis it's kind of flipping this graph back up so instead of you having y equals x when x is less than zero you have y equals negative x all right so let's talk a little bit about domain and range i think it's fairly obvious from looking at the graph that for x values it can basically be anything right and thinking about this from the equation you always just think about what you can plug in for x that's your domain right what can i plug in what am i restricted from doing the answer here is nothing right i can do whatever i want so the domain is all real numbers so from negative infinity to positive infinity then for the range okay for the range what am i going to have well now i'm constrained because this absolute value operation is going to make things non-negative so the smallest it can be a 0 and then anything out to positive infinity and again you can see that clearly graphically if i look at my y-axis the lowest point here is going to be at 0 on the y-axis so my lowest y value would be 0 and then of course it extends out indefinitely going up so it's out to positive infinity all right now let's go through kind of the definition of a piecewise defined function so let's again stay with f of x equals the absolute value of x and essentially what happens with a piecewise defined function or some people will say just piecewise functions some people say split definition function essentially with this type of function it's defined by different rules over different intervals of the domain so i kind of already showed this to you earlier so essentially you could say y is equal to this absolute value of x which kind of splits into two different rules if x is greater than or equal to 0 then really y just equals x so i can say it's just x if x is greater than or equal to zero and then really it's the negative of x if x is less than zero because again what happens is if i have a positive number or zero that i plug in there i just get the same thing for y so if i plug in 6 i get 6 for y if i plug in 144 i get 144 for y right so on and so forth but if i plug in a negative there i take the negative of the negative that's why you have a negative out in front of that and you end up with a positive so if i want to plug in a negative 144 there again all i say is it's the negative of negative 144 and i get 144 as my output or my y value so this is an example of a piecewise defined function it's got different rules over kind of different intervals of the domain here we have one rule where we're kind of at zero or to the right of zero and another rule if we're to the left of 0. all right so let's look at another common function that you're going to come across this guy is called the greatest integer function and it's also known as the floor function essentially the greatest integer function is going to pair every real number x with the greatest integer that's less than or equal to x so when we think about this guy the floor function is actually a better name in my opinion because what you're doing is you're actually rounding it down to the nearest integer sometimes you hear greatest integer function you get a little bit confused about what it's doing so we have y equals the greatest integer of x here and there's a few different ways we can write this let me just start by saying it's f of x equals i can use the brackets here around x to indicate this you might also see these kind of lines drawn in okay that might be a way that you would see it another way you would see it this is pretty common as well you'll see something that kind of looks like an l essentially it's a bracket where the top part's cut off so all of this is doing is just telling you that you have the floor function okay this is for the floor function or again the greatest integer function for the purposes of what we're going to do today i'm just going to use the brackets so let's think a little bit about what happens here and then we'll look at the graph so if i plug something in for x again it's going to round it down to the nearest integer so let's say i plugged in 2.5 as an example well just think about which two integers this guy is between it's between two and three so it always rounds it down so it's going to round it down to two if i plugged in something like let's say 1.7 okay 1.7 a lot of people make the mistake and think that it's going to round normally it always just rounds down okay so it's going to round down to 1. so we can really write some rules here to kind of define what's going on and you're not going to write all of them because there's an infinite number but let's just write a few we can really say that if x is let's say greater than or equal to negative one and less than zero then what's going to happen if i'm in that range it's going to route it down to negative one right so we can say that our f of x here is going to be equal to negative 1. if x is now greater than or equal to 0 and less than 1 again if i'm in this range i'm going to go round down to 0. so here f of x will be equal to zero and we can continue this pattern so if x is greater than or equal to one and less than two again it's going to round it down to one so f of x will equal one so let's think about this graphically now so if we think about let's just start with values that are greater than or equal to zero and less than one so if we kind of come here and say okay if i choose a value for x that's zero i know y is going to be zero so let's just say we have that point zero comma zero that's filled in now as i choose values for x that are larger than zero but not one okay not one i'm just kind of filling this in because i'm staying at zero for my y value then when i get to an x value of actually 1 what happens is my y value is now going to be 1 so i've got to jump up here so when x is 1 y is 1. so that's why that's filled in and this is not filled in and then again it continues so y is going to be 1 as we increase x going up 2 but not including 2. so that's why that's not filled in we've got to jump up here if x is 2 now y is 2 and again it's always going to be 2 up to the point where we get to 3 then i've got to jump up here and again this process continues forever and ever and ever and also going down here as well so that's why you see these three dots here and these three dots here that just tells you that the pattern continues forever it's similar to seeing arrows when you're graphing a line or some other kind of shape now let's talk a little bit about the domain and range let's talk a little bit about the domain and range let me erase all of this so we can write that in let me put this back okay so what's the domain and what is the range well the domain is all real numbers because i can plug in anything for x i'm not restricted there so it's going to be from negative infinity to positive infinity but for the range i'm always going to get an integer value if i plug in an integer for the domain i get an integer but if i plug in something that's not an integer it gives me an integer right as an output so the range we could say is the set of all y such that y is an integer okay so that's one way to write that now let's talk about something else real quick and then we'll look at a few more of these we want to talk about the fact that there's something known as a discontinuity in this graph so this is a concept that you're going to need to know for calculus and in fact you're going to get a much more advanced definition when you get to calculus but essentially a function is continuous over an interval of its domain if we can sketch the graph of that interval without lifting the pencil from the paper okay or pen or whatever you're using when you get into calculus again you're going to get a more advanced definition that generally is going to involve limits there's some other ways to think about it too but for now this is going to give us a basic understanding so when a function is not continuous at a given point we can just say that it has a discontinuity there so notice how when the greatest integer function is graphed again we've got to keep picking up our pencil i start here i go here i got to pick up my pencil start here go here pick up my pencil start here go here so everywhere there's an integer value for x i have a discontinuity all right let's look at another example of a piecewise defined function and what i want you to do here and this is a common task we want to look at the graph and we want to determine the domain and range and we also want to come up with kind of the rules that are associated with this function so what we can see here is that there's a discontinuity at x equals 2. here's x equals 2. you see that if you were graphing this guy you could graph this going down you're not lifting your pencil and then all sudden you get here and you've got to pick up your pencil and come up here okay so there's a discontinuity at x equals 2. so what this shows me is that there's some kind of border here where essentially you have one rule on one side and another rule in the other so if i'm writing a rule i can say f of x is equal to and let me just kind of put a bracket in there so let me do that so on one side again the border is two so i know this guy is including two and this guy is not so we'll say one rule is where x is less than or equal to two and then another rule is where x is going to be greater than 2 okay so those are the two scenarios we're considering now they're only going to give you stuff at this point that you know how to find so you can see that this guy is basically a line and this guy is a line so we just need to figure this out again from slope intercept form this guy is going to cross at negative 2 on the y axis so if i have y equals m x plus b this guy crosses at negative 2 so i know my b is negative 2 and we have y equals what's m well to get from this point to this point i go up one two three and to the left one so it's negative three over one which is just negative three right so i have y equals negative three x minus two so essentially my f of x or my y is equal to negative 3x minus 2 if x is less than or equal to 2. so negative 3x let me kind of slide this down a little bit and let me move this down and i can just say minus 2. okay so what about this guy over here on the other side we don't have a visible y-intercept but that's okay we can still go through and figure this out we can see that our slope here if i start here i would go up one two and to the right one two so the slope is one right it's two over two so y equals one x plus what's the y-intercept going to be well i can continue with this slope i can just go down one to the left one down one to the left one down one to the left one i know that i would impact the graph here right you can see that that would be the case that would be that line so the y intercept would occur at y equals 4 okay so this would be a 4 here so essentially what i'd have is y or f of x equals x plus 4 so x plus 4 if x is greater than 2. okay now let's copy this real quick and we're going to look at another task that you'll come across and then we'll come back up and talk about the domain and range so you might be asked to kind of evaluate this guy and so you'll have f of 3 f of 0 and f of negative 1. essentially what you have to do is go through and think about where do i fall in the domain if i have f 3 does 3 fall in the category of less than or equal to 2 or greater than 2. well it falls in this category here so i follow this rule so f of 3 is equal to 3 plus 4 which equals 7. and that's something you can verify on the graph if i go back up and look at 3 this is an x value of 3 i have a y value of 7 okay so that's true then we look at the next one which is f of 0. so f of 0. so now i'm in the category of here which is less than or equal to 2 so i have negative 3 times 0 minus 2. negative 3 times 0 is 0. so i just have negative 2. and i can verify that when x is 0 i do have a y value that's negative 2 right that's negative 2 right there so let's go back for the last one we have f of negative 1. so f of negative 1 and again i'm in this category here so i'd have negative three times negative one minus two negative three times negative one is going to be three three minus two is going to be positive one and again we can check that so f of negative one that's here and that is positive one so we're good to go on that all right so let's think about the domain and range for a second that's something that does trip students up so the domain and the range we're going to look at the graph and then we can also verify that by looking at these kind of pieces here so for the domain a lot of students incorrectly say that 2 is not included that's wrong you go to the left forever you go to the right forever 2 is included it's included right there the kind of closed circle means it's included the open circle means it's not so every real number is going to be part of the domain including that 2. so from negative infinity to positive infinity now the range is going to be limited it continues up forever here and it continues up forever here but the lowest point is right here right where y is negative eight okay so i can see that negative eight is included and it's out to positive infinity and looking at the equations here this should make a little bit of sense so we know that if x is greater than 2 we use this guy so x plus 4. i can't plug in a 2 but if i did i would get 6 here and then anything larger than 2 would give me a larger result so this guy's going up now if i think about the lowest i can make this this negative 3x minus 2 what happens is the smallest i can make this is actually when x is 2. if i plug in a 2 here negative 3 times 2 is negative 6 negative 6 minus 2 is negative 8. if i keep decreasing x going this way you can see that y is actually increasing right and the reason for that is because i'm multiplying a negative times a negative so that gives me a positive and then i'm just subtracting away two from that so as this negative gets bigger and bigger the result of this is going to get larger and larger so that's why this guy is increasing as we move to the left on the x axis all right let's go ahead and look at one more of these and we'll just kind of wrap up the lesson so let me put an arrow here so this one's a little bit more tricky it involves a lot of different stuff so you'll notice that in this case we have a discontinuity here at x equals 0. again if you were graphing this you would have to pick up your pencil here and jump up here and then you have another one where x equals 3. if i was graphing this i'd have to stop here and jump up here okay so we know that there's kind of two different borders involved here so we'll have f of x is equal to let me kind of make my rule here so we know that one of these guys let me kind of slide this down because i'm going to need some room we know that one of these guys is going to be where x is less than or equal to 0. so if x is less than or equal to 0 another one is going to be if x is what it's going to be equal to 3 and then less than that up to but not including zero so we could say it's greater than zero and less than or equal to three so that's going to be another interval and then lastly we would say if x is greater than three okay let me use different colors here make this a little bit better so i'll say if and then i'll say if and then i'll say if right three different colors so three different intervals we're going to be working with let me make this a little bit better so what are the rules here well essentially if x is less than or equal to zero i'm dealing with this graph here okay and this shape here you might remember from the last lesson where we talked about a parabola the y equals x squared right so f of x is going to equal x squared you can see that because again if you start with 0 you get 0. if you plug in a negative 1 you get 1. right so if i took negative 1 and squared i get 1. if i plug in a negative 2 and i square that i get 4 if i plug in a negative 3 and i square that i get 9. so you just have to go through a few points there and it becomes obvious that they're taking the x value and they're squaring it to get a y value so this is f of x is going to equal x squared if x is less than or equal to zero now when x is larger than zero up to and including three so zero is not included here you see that's why there's an open circle three is included that's why that's filled in you have just a straight line here and it's a horizontal line so y is constant there it's just equal to 3. so f of x is just equal to 3 if x is greater than 0 and less than or equal to 3 that's all it is okay then what happens is if x is greater than 3 okay that's why this is again not part of this so if it's strictly greater than 3 we have this here again i don't know what the y intercept is but i can determine the slope pretty easily i can look at this and this and say i'm going up one two and to the right one so i would have y equals two over one or two times x plus what would my y intercept be well if i continue moving down i would go down one two to the left 1 down 1 2 to the left 1 down 1 2 to the left 1. okay so that would be my y intercept it would occur at negative 1 on the y axis so i can say 2x minus 1 would be my rule here okay and it's hard to kind of fit this in so let me just slide this down a little bit so i'll say 2x minus 1 if x is greater than 3. so these are our rules here again x squared if x is less than or equal to zero it's going to be just three if x is greater than zero and less than or equal to three and then it's going to be two x minus one if x is greater than three okay let's again take this and evaluate some stuff and then we'll look at the domain and range so f of negative two how could we find that again figure out where it comes into play so that falls into this category where x is less than or equal to zero so f of negative two i'd plug in a negative two for x and square it f of negative two is going to be negative two squared which is four again you can verify that here's negative two here's four this is negative 2 comma 4 on the graph now the next one is f of 1 so f of 1 and that falls in the category of being greater than 0 and less than or equal to 3 so here's just 3 right and you can verify that again by just looking at the graph here's 1 if i go up i'm at 1 comma 3 that's where i'm going to be all right let's look at the last one so f of 7 is going to be 1. well in this case if x is greater than 3 i fall in this category so it's 2 times 7 minus 1 2 times 7 is 14 14 minus 1 is 13 so i'd have a y value of 13. now we don't have that value on our graph unfortunately because it stops at a y value of 10 but if it did you could verify that and you could do a computer generated graph if you wanted to and verify that all right so let's think about the domain and range and just kind of wrap things up here so the domain and the range so again the domain is going to be all real numbers okay because this is going to the left forever this is going to the right forever again where you get confused let me kind of erase all this stuff i drew in where you get confused is these open circles here they throw you off notice how you have this guy is closed at x equals zero this guy's closed at x equals three so we do have zero and 3 represented on the graph it's just that we jump in each case right there's a discontinuity at each of those so the domain again is all real numbers so from negative infinity to positive infinity we're covered for the range we are limited right so what happens is the smallest value here is zero okay we can see that from the graph and then a lot of students get this messed up as well because they see this open section right here and they think it's not involved it's involved here in this graph there so it is covered and then it's going all the way up from there so you're covered for every y value from 0 out to positive infinity so that's going to be your range okay so you can say the domain here is all real numbers and the range is from 0 to positive infinity in this lesson we want to determine where a function is increasing decreasing or constant from its graph alright so when you start talking about functions in their graphs you're going to come across a section where you want to look at the graph and find the intervals where kind of the graph is increasing it's decreasing and then where it's constant now this is something you can do without looking at a graph when you get to calculus you're going to talk about derivatives and you're going to find methods where you can kind of use the derivative to determine the same information without looking at the graph but that's a bit beyond what we can do at this point so we're just going to stick to looking at the graph and observing the intervals where again we're increasing decreasing or constant now you won't get any linear functions in this section your book but i want to start by talking about linear functions because they're very easy to work with and they're going to demonstrate kind of something that's fundamental to what we're talking about here so for this guy right here this line let's just get the equation going we know the y-intercept occurs where let me make that better the y intercept occurs where it's going to be right here so it's 0 comma 3. so 0 comma 3 and the slope is what if i started at this point right here and i want to go to this point i would rise 1 2 3 4 and i would run to the right 1 2. so m my slope again it's rise over run and in this case it's going to be 4 over 2 which is 2 right and specifically here it's positive 2. so we can say the equation is y equals 2 my slope times x plus 3 my y-intercept or if you want to use function notation you could say f of x equals 2x plus 3. that's up to you but the main takeaway here is that with a positively sloped line essentially as x increases or as we move to the right y also increases okay and with a line because the slope is always the same it's always increasing it's always constant or it's always decreasing right it's one of those three things unless you have a vertical line where the slope is undefined but in this particular case again because we have a positively sloped line as x increases right as we move to the right on the x axis y is going to increase so we can say that this function this y equals 2x plus 3 or f of x equals 2x plus 3 is increasing kind of over its whole domain okay so the way we want to write this let me just erase all this when you get in this section they're going to ask you for the intervals where it's increasing decreasing or constant so we're going to say increasing and i'll say on the interval well for here it's going to be from negative infinity to positive infinity again across its entire domain right so if i go all the way to the left as far as i can go we have this arrow here that indicates it continues forever in the negative direction as i go to the right as far out as i can go it's always increasing okay so from negative infinity to positive infinity it's never decreasing and it's never going to be constant all right we also have negatively sloped lines and for this guy the reverse is going to be true so for its entire domain this function is going to be decreasing as x values increase or we move to the right y is going to decrease so let's kind of do this formally so this guy right here the y intercept is going to occur at 0 comma negative 1 and the slope m is going to be what if i want to go from this point to this point i've got to fall 1 2 3 run to the right 1 so that's negative three for the rise and it's a positive one for the run so this is just negative three so i would say y equals negative three x and then minus one or again you could say f of x equals negative three x minus one if you wanna use your function notation but the main takeaway here is that again as x increases as we move to the right on the x axis y is always decreasing okay and this is from negative infinity out to positive infinity so i'll say decreasing again on this interval from negative infinity to positive infinity again it's not increasing anywhere and it's not constant anywhere all right so just to recap when working with lines the process is really easy in fact you're never going to get a line in the section i'm just showing you this for the purposes of getting our thought processes going but it's easy because the slope is always the same okay so when the line has a positive slope it increases across this entire domain again from negative infinity to positive infinity and when a line has a negative slope it decreases across this entire domain and then with a horizontal line it's constant across its entire domain now in most cases and i would say pretty much in every case you're going to be dealing with something that's not as simple as a line so sometimes our graph is going to make all these weird twists and turns and so over the span of its domain you may have several changes between increasing decreasing and constant again when you get to calculus you'll be able to use the derivative to kind of find the increasing intervals the decreasing intervals and the constant intervals but for right now that's a bit beyond us so let's just observe this kind of general definition and then we'll move on and look at some kind of graphs so f our function is increasing if across this given interval that we're looking at for any two x values let's call them x sub 1 and x sub 2 we could say that x sub 1 is less than x sub 2 and f of x sub 1 is less than f x sub 2. so what this means is that this x value is smaller than this x value so this x value is to the left of this one okay then plugging in this x sub 1 this smaller x value for the function and kind of getting the output results in a smaller value than using the larger x value as an input right the output there is large so to see this graphically we see again that x sub 1 is to the left of x sub 2 so we know that means that x sub 1 is less than x sub 2. and we see that if i kind of go up here the f of x sub 1 or the y value associated with plugging in an x sub 1 for each occurrence of x in that function is smaller than f of x sub 2 right the y value associated with plugging in the larger x or the x sub 2 for each occurrence of x in that function so we can say that this function is increasing right as we go from left to right the function is increasing over that interval if the y values are increasing over that interval all right now also you'll see f is decreasing if kind of the reverse is true so we have x sub 1 is less than x sub 2 so this part's still the same x sub 1 is still smaller but now f of x sub 1 is greater than f of x sub 2. so plugging in the smaller x value results in a larger y value or a larger output than when we plug in the larger x value right now we're getting a smaller output here so when we look at this again x sub 1 is less than x sub 2 we see that f of x sub 1 is higher than f of x sub 2. so this graph is falling as we move to the right so it's decreasing over this interval now lastly we might see where it's constant so we could say f our function is constant if x sub 1 is less than x sub 2 and f of x sub 1 is equal to f of x sub 2. so we all know what a horizontal line looks like and essentially over this interval you have any given x sub 1 and x sub 2 x sub 1 again is less than x sub 2 and the y values here are the same so f of x sub 1 is the same as f x sub 2. this generates a horizontal line so here the function is just constant all right let's take a look at the first example here so i'm just going to put some arrows on this and i'm going to say that this is y equals x squared we've already seen this graph before in a previous lesson so with this guy again i want to determine the intervals where it's increasing decreasing or constant so what i'm going to do is i'm going to start from kind of out here all the way to the left from negative infinity to kind of going this way what happens well if i look at my y values if i kind of think about it i'm decreasing right as i'm going to the right i'm decreasing decreasing decreasing decreasing decreasing until i get to zero okay and then after zero then i'm increasing increasing increasing increasing increasing a lot of people will draw like a stick figure or car or something like that to kind of illustrate i'm not a good drawer but i'll try my best so my stick figure there would be walking downhill until he gets to zero and then after zero he'd kind of be walking uphill so let me do that so you can see that there it's not a good drawing but you get the idea another way you can think about this is if i kind of think about an x value of negative 3 corresponds to a y value of positive 9 if i increase the x value by 1 if i go to negative 2 well now my y value is positive 4. so increasing x resulted in me decreasing y so again that's how i know i'm decreasing again until i get to 0 and then i start increasing right if i look at an x value of 2 i get a y value of 4 if i look at an x value of 3 i now get a y value of 9 okay so that's another way you can kind of think about this so we can say that in terms of the intervals we are decreasing on the interval from a negative infinity up to but not including zero and then we're increasing we are increasing on the interval again not including zero out to positive infinity so you might be asking why didn't i include 0 and you might also be arguing saying that your book includes 0. so these are both valid points so the first thing is that i didn't include 0 because if you think about what happens to this function when you're at zero you're not increasing and you're not decreasing in fact the slope there if you took the derivative and you looked at it specifically where x was zero you'd find that the slope is zero okay so it's not positive it's not negative it's zero so that means it's not increasing and it's not decreasing so that's why i didn't include it i just used the open notation so that it's excluded the second point is your book might show this your book might do this now is this wrong the answer to that question is i would go with whatever your teacher tells you to do because ultimately they're grading your tests most books i would say the majority of them and most teachers are going to want you to use this notation here for the point that i just made right at 0 it's not increasing and it's not decreasing so it's not included in either of these intervals okay but if your book uses this kind of bracket notation follow your book follow your teacher go with what they tell you now the other thing that you want to consider here is that when we list the intervals it's in terms of x values we're not listing ordered pairs and we're not listing y values so don't make that mistake i am considering what is happening to the function what is happening to kind of the y values as i kind of look at these intervals on the x axis so from the interval negative infinity up to but not including zero the function is decreasing right so the y values are decreasing then when x is larger than zero we can say the y values are increasing again so the function is increasing over that interval from zero out to positive infinity so it's always in terms of the x values that's what you're writing all right let's take a look at another one so this is the graph of y equals we're going to have x cubed minus we're going to have 3x and then plus 2. for right now we just want to observe from the graph where it's increasing where it's decreasing and where it may be constant well again if i kind of think about things coming from negative infinity going to positive infinity so again i'm thinking about what's happening as i move across the x-axis as i go to the right so starting from negative infinity going up to this kind of negative one on the x-axis i'm increasing right so i can again draw my little stick figure this guy would be walking uphill the whole time when he gets to an x value of negative one he would kind of change and then right here he would be walking downhill so he'd be walking downhill until he got to an x value of positive one and then he would be walking uphill again from positive one that x value out to positive infinity okay so if you think about it this way for the increasing intervals so it would be increasing on okay the intervals one interval would be from negative infinity up to but not including negative one and then another interval would be from positive one and then out to positive infinity now let's talk about the decreasing so we'll say it's decreasing on what interval so that's going to be from negative 1 to positive 1. again i'm not going to include either one of those so at negative 1 and positive 1 you kind of have these turning points here and the slope here the derivative of the function when x is negative 1 and when x is positive 1 it's going to be 0. so i'm not going to say it's increasing and i'm not going to say it's decreasing okay before we move on to the next problem i want to talk about another common source of confusion so if your book specifically asks you for the intervals where this function's increasing decreasing or constant you don't want to put a union symbol in here okay you see that a lot so sometimes you have more than one interval involved in one of your answers so you'll see students go okay it's this guy and then union with this guy that's not correct so if i'm asking you for an interval what am i asking you for let's go to a number line real quick and think about this let's say i had an interval from 3 to 6 with 3 and 6 not included so that's this interval here essentially it's a set of numbers that's kind of between two numbers where the endpoints could be included or not included you can't have a gap right so you can't say well i have another interval from negative 5 to negative 1 and let's say i have this guy right here i can't just use a union symbol and say okay this is now one long kind of interval okay that doesn't make sense this is a set that contains the union of those two intervals okay but it's not one long interval that wouldn't make any sense okay you can't have a gap like that so if they're asking you for the intervals okay intervals plural then just put a comma between them or use the word and okay this is a common trap question because they might give you a multiple choice and they might put a union symbol in between the two and they might give you a choice like this and you might be scratching your head and say okay well i've used this a lot so you might check this one as correct but what happens is this one would be wrong okay because they asked you for the intervals okay so if we go back we want to list it like this or just put the word and kind of in between them if the question is asked differently okay this is the thing that gets confusing then you might be able to use your union symbol if we're just talking about where's the function increasing and there's no mention of give the intervals well then you could give it with that kind of union statement because you're just looking for the set that kind of satisfies that answer and that set would contain these two intervals and you could join those with a union symbol that would be fine okay so if you're given the question in that way it's okay but if you're specifically asked for kind of the intervals where the function is increasing decreasing or constant you want to make sure to just kind of use a comma if you have kind of two different intervals there or use the word and okay and that's all you want to do all right let's look at another so this guy is going to be y is equal to the square root of the absolute value of x so where is it increasing where is it decreasing and where is it constant well again as i move to the right again if i draw my stick man figure this guy is going to be walking at a slight decline it's not a massive decline but it is decreasing so as i move to the right it's decreasing from negative infinity up to the point where i get to zero and then he's going to be increasing out to positive infinity okay so we can say that it's decreasing on the interval from negative infinity to zero again with zero not being included and then increasing okay on the interval from again 0 not being included out to positive infinity what about this guy right here let me draw some arrows in here we should recognize this guy as being the function y equals 1 over x or again you could say f of x equals 1 over x if you want so essentially what's happening as i move to the right well from here remember we're infinitely close to zero and as we move to the right we start decreasing decreasing decreasing decreasing decreasing as we approach zero but we'll never touch it okay so i know to kind of the left of zero i'm always decreasing okay so from negative infinity to zero i'm decreasing so decreasing on and from this guy right here as we move to the right what's happening well my y values start out up here basically going to positive infinity and i'm coming down down down down down as i move to the right and i'm still decreasing right i'm going towards a y value of 0 but i'm never going to touch it i'm going to keep decreasing forever and ever and ever so i can also say comma again i'm going to have another kind of thing to list here we're going to say anything larger than 0 out to positive infinity okay so from negative infinity to 0 not included and then from 0 to infinity again 0 not included again don't put a union symbol in there that's not correct all right for the final example i'm going to give you one that's just kind of made up this is a drawing that i came up with you'll see a lot of these in your textbook just to give you a couple of examples where you have some kind of constant some decrease and some increasing a lot of stuff going on so if we look at this guy again if i use my little stick figure what's going on here is that we start out as we come from negative infinity we're going to be increasing up to the point where we get to an x value of negative four okay so let me just write this we're going to say increasing on so from negative infinity up to not including negative four and then let me put decreasing on so decreasing on so next the stick figure is going to be walking downhill right starting at negative 4 that x value going all the way up to an x value that is negative 1. okay this is negative 1 here so we'll say from negative 4 to negative 1. again neither of those is going to be included let me make this a little better so negative 4 negative 1. and then this guy right here again let me kind of change colors he's going to start walking uphill now starting at an x value of negative one he's walking uphill until he gets to an x value of two okay so it's going to be increasing the other interval is going to be from negative one up to positive two again neither of those is included okay and then once he gets here he's walking flat okay so from an x value of two to an x value of six it's flat or it's constant okay so we'll say constant on okay the interval is going to be from two to six again neither is included and then lastly as we get past an x value of six and we go out to positive infinity this guy would be walking downhill so we could say it's decreasing over another interval again anything larger than six so again six is not included out to positive infinity in this lesson we wanna talk about graphing techniques we're gonna focus on stretching or shrinking a graph so what we want to do today is we want to start to think about ways where we can graph functions that are defined by altering the equation of one of our elementary functions so over the course of the next few lessons we're going to develop techniques for graphing these types of equations collectively we refer to these graphing techniques as transformations so we're going to begin by looking at stretching and shrinking a graph then on kind of our following two lessons we'll talk about reflections we'll also talk about horizontal and vertical shifts so we're going to begin with f of x equals x squared so you'll remember this is our squaring function so we plug something in for x our input we square it and we get our y or f of x as our output so we already see that i've kind of pre-plotted some points and graphed this guy but let's just go ahead and fill in this kind of table of values let's choose an x value of negative 2 to start i know that if i squared negative 2 i would get 4 and if we did negative 1 we'd get 1 0 would give me 0 1 would give me 1 and 2 would give me 4. now i've plotted additional points here i have my negative 2 4 here i have my negative 1 1 here my 0 0 here my 1 1 here and my 2 4 here i also have my negative 3 9 right there and my 3 9 there so i can't fit all that here because i'm out of space and we only need a few of these values to kind of compare in our next section anyway but i just want you to get the idea here these are the kind of basic points that you should associate with y equals x square or f of x equals x squared so every time you want to graph this guy you know these points by heart and you can kind of sketch this graph very quickly now knowing that we've committed f of x equals x squared comma to memory we know the points associated with it we know the shape suppose we had a new function where we take this original function and we multiply the right side by 2. so in other words we have f of x equals x squared so focus on this x squared part now we have g of x equals 2 times this x squared so essentially i can write g of x by saying it's 2 this number right here times f of x what's f of x it's x squared so times f x you'll see this notation in your textbook and what's going to happen is this is going to give us a vertical stretch okay and i'll explain why in a minute but essentially for the same x value the y value is now going to be twice as large so we filled out this table of values before we said negative 2 4 we said negative 1 1 0 0 1 1 and 2 4. so if i keep the same x values so negative 2 negative 1 0 1 and 2 the y values here now for g of x will be twice as large as the y values for f of x again given the same x value so instead of a y value of 4 when x is negative 2 now it's going to be 8. and again think about why that's the case i plug something in for x i square it now instead of getting my output i still got to multiply it by 2 okay so the original output is multiplied by 2 and now i have my output okay so that's why i can take this y value of 4 multiply by 2 to get 8. negative 2 squared is 4 then times 2 would give me 8. so this guy 1 would now be times 2 so that's 2. 0 times 2 is still 0 then 1 times 2 is going to give me 2 and of course 4 times 2 is going to give me 8. so again for the same x value y is going to be twice as large so this is going to create a graph that is vertically stretched when compared to f of x equals x squared you're going to see that it's skinnier or narrower so let's take a look at this visually so the guy in orange this is my original f of x is equal to x squared and this guy in light blue this is my g of x is equal to 2 times x squared or really we could say g of x is 2 times f of x okay so this multiplication of 2 times this function f of x is what's creating this vertical stretch okay this vertical stretch which just means that we are pulling the graph away from the x-axis a lot of times we like to think about it by saying okay well if this point here originally on f of x equals x squared was 2 comma 4. so let's kind of highlight that real quick now that same x value of 2 is associated with a y value that is twice as high so it's up here so you see how that's pulling the graph away from the x axis okay so that's what we talk about when we say that we have a vertical stretch again it's being pulled away from the x-axis or the horizontal axis now let's look at the other scenario so we have f of x equals x squared and now we have g of x is set equal to one half times x squared so now what i have is g of x is really one half times f of x again f of x is x squared so one half times x squared is this g of x so what's going to happen is for a given x value now my y value is going to be half as large so instead of having a vertical stretch right where it's narrower or skinnier right it's being pulled away from the x axis now what we're going to have is we're going to have a vertical shrink or a vertical compression okay so if this guy again is negative 2 4 negative 1 1 if it's 0 0 1 1 and 2 4 these guys if i stay with the same x values so negative 2 negative 1 0 1 2 these guys are going to be half as large as these guys so i would have 2 instead of 4 i would have one half instead of 1 0 would be the same right because 0 times a half is still 0. and then one again that would be one half and then four that would be two right everything i did here was i just took these y values and i just multiplied them by half to get these y values okay that's all i did now if we look at this visually we can see that this guy right here is my f of x equals x squared and then this blue guy this is my g of x equals one half times x squared okay one half times x squared let me make that a little bit clearer and again let me show you this with a point so if we have 2 comma 4 that's here now the x value of 2 is associated with a y value that's only 2. so what happens is this creates a vertical shrink or vertical compression it's a movement towards the x-axis okay it's a movement towards the x-axis all right so let's talk about the general rule this is something you want to write down in your notes and then from here it's very very easy to kind of answer questions about whether there'll be a vertical stretch or vertical shrink and then also it's easy to kind of graph things if you already know what the parent function looks like so g of x equals a okay some number a times f of x now for the purposes of what we're going to do today a is going to be positive okay i'll talk about negative scenarios when we get to reflection in the next lesson so a if it's greater than 1 it's a vertical stretch we saw that if we had 2 times f of x in this graph here we had a vertical stretch again it was a movement away from the x axis then if a is greater than zero and less than one we're going to have a vertical compression right we saw this with one half times f of x we can see that here again we had a movement towards the x axis all right so suppose we look at f of x equals the square root of x and then g of x is equal to 2 times the square root of x so really again g of x can be defined as 2 times f of x okay that's all it is because f of x is the square root of x well 2 times the square root of x well it's 2 times f of x right that's what g of x is so what's going to happen is for a given x value the y value is now twice as large right i can just multiply it by two so this will give me a vertical stretch right or a movement away from the x axis so if we look at the two graphs here this guy is the graph of f of x is equal to the square root of x then this guy here is going to be g of x let me make that g a little bit better the g of x is equal to again 2 times the square root of x so the way to graph this kind of quickly if you don't want to use kind of like a t table or something like that if you already know f of x equals square root of x if you already know what that looks like then essentially what happens is you take each y value okay for the same x value you just multiply by 2. so 0 comma 0 is still there okay but if i go to 1 right i know that square root of 1 is 1. so that's this point here 1 comma 1 but now in the g of x function if i take the square root of 1 i get 1 but now i'll multiply by 2. so now i have an x value of 1 corresponding to a y value of 2. and as we move to the right this point here that's 4 comma 2 would now be 4 comma 4 right because i took that y value of 2 and multiplied it by 2 to get a y value of 4. and here where we have 9 comma 3 just multiply the y value of 3 by 2 to get 6. so now it would be 9 comma 6. so once you have these kind of 4 points you can sketch the graph of this g of x very very quickly right but the main thing here is to understand that you have a vertical stretch here or again a movement away okay movement away from the x axis all right so now let's look at f of x equals the absolute value of x and then g of x equals one-half times the absolute value of x so again g of x can be defined as what i've got this one-half that is multiplying f of x right f of x is the absolute value of x one-half is multiplying f of x so g of x is one half times f of x okay so we know at this point because this guy is between zero and one it's going to create a vertical compression or a vertical shrink so looking at this guy this is the original f of x equals absolute value of x then this guy right here is what it's g of x is equal to one-half times the absolute value of x so again thinking about this if i want to graph it quickly without a t table for a given x value the y value is now half as large so 0 0 is still going to be there right because if i plug in a 0 i get 0 0 times a half is still 0. but if i move out let's say i do 2 well 2 is associated with a y value of 2 here okay but in the new g of x i just cut it in half and so 2 is associated with a y value of 1. you can just keep doing this to get enough points to kind of graph this pretty quickly but one of the main things is to understand that you would have a vertical compression here because this number right here that's multiplying f of x is going to be between zero and one right it's greater than zero and less than one so that's what's creating this kind of vertical compression or movement towards the x axis all right so we're also going to come across horizontal stretching and compression so with a horizontal stretch now it's going to be a stretching of the graph away from the y-axis okay we saw with a vertical stretch we stretch the graph away from the x-axis with a horizontal stretch we're now stretching it away from the y-axis okay with a horizontal compression this is now going to be a squeezing or compression of the graph towards the y-axis again with a vertical compression we were compressing it towards the x-axis now with the horizontal compression we are compressing it towards the y-axis now what's going to happen is we have g of x that's defined as being equal to f of ax so let's think about this notation for a minute i know we've all seen this at this point if we see something like f of 2 we know we just plug in a 2 for each occurrence of x and f of x so this is no more challenging i just take ax plug it in for each occurrence of x and f of x so if i had something like let's say f of 2x let's say i have that wherever where i have an x in my function just plug in a 2x that's all it is okay nothing fancy now here comes the confusing part and i'm just going to tell you in advance it's a little bit more challenging to understand a horizontal stretch and a horizontal compression versus a vertical stretch in a vertical compression what's going to happen is to obtain this g of x based on our original function f of x we can essentially just multiply each x coordinate of that original function by 1 over a so let me write that down so 1 over a and this part's going to be confusing we'll see when we get to the example that it takes a minute to kind of wrap your head around where this comes from so if a is greater than 1 now we're going to have a horizontal compression and if a is greater than 0 and less than 1 now we're going to have a horizontal stretch so the first thing that makes it confusing it's kind of hard to keep track of kind of what's going on okay and the second thing is this 1 over a is really confusing where did this come from and why does this kind of happen essentially for a given y value now the x values are obtained by kind of taking those x values from f of x and multiplying them by this 1 over a so let's look at an example so this can be more clear so we have f of x equals x squared we already know this function we've committed the points to memory so we know negative 2 4 we know negative 1 1 0 0 1 1 and 2 4. but essentially what i just told you was that for this g of x here it's equal to this 2x and this is being squared well what i can write this as is what this guy right here is essentially just plugged in for x this guy is plugged in for x so what i can say is that i have g of x being defined as f of this 2x okay this 2x and notice that it's this 2 that's multiplying the x so this is my value for a okay that's my value for a if i go back it's a times x that's inside these kind of parentheses in this case that value is going to be 2. okay so let me write this off to the side to say a is 2. now i told you that for a given y value i can find kind of the new x values that's associated with this g of x by taking these x values and multiplying them by 1 over a so in this case a is 2 so 1 over a is going to be 1 over 2 or one half and this is the part that's going to be tricky it's not going to make sense at first so let's copy these y values we have 4 1 0 1 and 4. and then i'm going to just find my x values by kind of taking these and multiplying them by half so let's put times one half there that kind of looks like an x so let me just put a times like that so negative two times a half is gonna be negative one negative one times a half is going to be negative one half zero times a half is still zero one times a half is one half and then two times a half is one okay so the intuition here is that for a given y value the x value should have been doubled right because this guy is a 2 but in fact they were caught in half so what caused that well what's going on here is that let me kind of erase this for a second we have this 2 here that is multiplying x before it gets squared okay so now what's going to happen is this y value is associated with an x value that is half as large as it was before again this is really tricky to wrap your head around it first but if you think about it the one half is just what's needed to kind of undo this multiplication by 2 and get this kind of new x value that's associated with that y value so in this example if x is one y is one but here if y is one well x needs to be one half and the reason for that is i have one half that gets multiplied by two first and then it's squared so what's going to happen is because of this we have that same y value that's associated with an x value that's half as large this is compressing the graph horizontally or moving it towards the y axis so let's look at the graphs so this is going to be my original f of x equals x squared and then in blue we have our g of x which is equal to 2x and this amount is squared so let's just take a point right let's just take let's say this 2 comma 4. so this 2 comma 4 on the original f of x equals x squared well now what's going to happen is this y value of 4 is associated with an x value of 1 now instead of 2. and i know it's also associated over here on the left but let's just focus with these points right here on the right so what's happening is we're getting a horizontal compression or a movement towards this y-axis because again we have this 2 that is multiplying x before it gets squared okay so that y value is now associated with an x value that is half as large okay so that's what's going on there now let me kind of address a point of some confusion you might be scratching your head and telling me that this graph is stretched vertically well in fact you would be correct because in this particular case it does work out because we could say that really g of x is what if i squared 2 i would have 4 if i squared x i'd have x squared well i can also say that g of x is really just 4 times f of x right if f of x is x squared so if i look at let me kind of erase this if i look at this this way well really for a given point let's say i chose the point one comma one on the original well now that x value is associated with a y value that's four times as large right one comma four but i want you to kind of not get into this method of thinking okay you want to stay consistent with how the problem is presented so if they give it to you as g of x equals f of ax then you want to show this as a horizontal compression or a stretch okay so in this particular case the way we got this problem was in this format it was g of x is equal to f of 2x so we're going to show this as a compression or a movement towards the y axis okay and again this is something you've got to stay consistent with because if you don't you're going to get problems later on when you get more challenging functions so suppose we have f of x equals x cubed minus x and then g of x is one half x cubed minus one half x okay so this guy right here is just plugged in essentially everywhere i have an x so what i can say is that this guy g of x can be defined as f of one half x okay so if i plugged in a one half x here i would have one half x that amount cubed and then minus if i plugged in a one half x here i'd have just one half x which is what i have okay so what i'm going to do is just get some points going we don't work with this one very often so we have to kind of go through the operations so if i had a 0 for x i'd have a 0 for y right that's pretty easy to see if i had a negative 1 for x i'd have negative one cubed minus negative one so negative one cubed is negative one minus the negative one is plus one this would be zero then if i had let's say a positive one what would i have so if this was plus one and this was plus one i would also get a zero right because one cubed is still one and then i'm subtracting away one which is also zero so then let's just do negative two and two so let's do negative two and 2. if i had negative 2 and that was cubed that would give me negative 8 and then minus a negative 2 is plus 2 so this would be negative 6. then if i had positive 2 cubed that would be positive 8 and then i would subtract away this 2 here and that would give me a positive 6. all right so we have some kind of ordered pairs going we have five of them and essentially what i want to do now is use my rule we know that for a given y value so let's copy those we have this 0 0 0 again negative 6 and 6. for these given y values i can just take the x values that are associated and i can just multiply them by 1 over a in this case my a the guy that's multiplying the x is one half so one over one half is what this is division with fractions it's basically one times two over one which is just two okay so now what's going to happen is i'm going to multiply each of these guys by 2. so 0 times 2 is still 0. negative 1 times 2 is negative 2 1 times 2 is 2 negative 2 times 2 is negative 4 and 2 times 2 is 4. so we have all these points here that are going to be associated with this kind of g of x function and again for a given y value the x values have just been multiplied by 2 from that f of x function so what's going to happen is we are going to horizontally stretch okay we're going to horizontally stretch this function so let's take a look so we've already kind of plotted the points and sketched everything this guy is f of x equals x cubed minus x and this guy in blue is g of x is equal to let's just go ahead and say it's f of one half x to make it a little bit more simple and essentially what i see is that what let's say i take an x value of positive 2. well here it's associated with a y value of 6. well now that same y value of 6 is associated with an x value that's twice as large right so this guy over here this x value is 4. so you can see that this guy has been horizontally stretched right it's moving away from the y-axis so that's what's going on and it's by a factor of again one over this one over one half which is two so when you work with this horizontal stretch and horizontal compression you've gotta be able to keep this straight in your mind you've gotta remember that this guy right here this ends up being a okay and it stretches it horizontally by a factor of one over a so what happens is if a is going to be kind of larger than one then you're going to end up with a fraction here right you're going to end up with some amount that is kind of smaller than one so you are horizontally shrinking or compressing it if your a value is less than one and greater than zero well then what happens is you're dividing by a fractional amount let's just say this was one fourth as an example one divided by a fourth is the same thing as multiplying by four and so we're kind of horizontally stretching this guy by a factor of four so those are the things you have to keep in mind when you work with a horizontal stretch or horizontal compression in this lesson we want to talk about graphing techniques and here specifically we're going to focus on reflecting a graph across an axis all right so in our last lesson we discussed the concept of stretching or shrinking a graph now we're just going to go a step further and discuss the next type of transformation this is where we're going to reflect a graph across an axis so we're going to begin today by looking at our square root function so we have f of x equals the square root of x this is a function that at this point you've probably committed to memory if you haven't it's something you should commit to memory because you're going to see it a lot okay so let's just get some ordered pairs going or some points if you want to call it that going because we need to compare them to something on the next page so i know my input x i'm going to plug something in there i'm going to take the square root of that and i'm going to get my output y or f of x so if i plug something in for x i want it to be a perfect square so i get a nice clean answer so i'm going to start with 0. take square root of 0 i get 0 then 1 take square root of 1 and get 1. and then for 4 i'd get 2. for 9 i would get 3 and then for 16 i would get 4. okay so kind of looking at this i've already plotted the first four points i can't fit 16 comma 4 on the graph it's just too too small right it's just going to be cut off so if i look at 0 comma 0 that's going to be here and then 1 comma 1 is here and then 4 comma 2 is here and then of course 9 comma 3 is here so these are the points that you should kind of commit to memory so that you can quickly graph this kind of square root function okay and it's already done for so i don't need to trace over it you just need to understand the general shape of this guy now suppose we introduce something g of x that looks very similar to f of x so first let's study these two functions so we have f of x again it's just the square root function it's the square root of x and then g of x is now the negative of the square root of x so this part right here and this part right here those are going to be the same the difference is this negative out in front so really based on this i could define g of x i could define g of x as being this negative of f of x because f of x is the square root of x and you're having the negative of the square root of x so this is the negative of f of x now how is this going to kind of affect or change the graph when i have this negative out in front well let's make a table of values for each one and then we can compare the two so we know this was 0 0 it was 1 1 it was 4 2 it was 9 3 and 16 4. so what's going to happen here i plug something in for x i plug something in for x i take the square root okay so that was my original output here but now i need to take the negative of it so there's an extra step so essentially for a given x value the y value will be its opposite so let's copy these x values we've got 0 1 4 9 and 16 and then just take these corresponding y values over here and change them into their kind of opposite so the negative of zero is just zero the negative of one is negative one the negative of two is negative two the negative of 3 is negative 3 and then the negative of 4 is negative 4. all right so kind of looking at these two tables here we should get an idea of what's going on essentially for the same x value i now have two different y values okay so i have one that's positive and one that's negative so let's look at this graphically and what we see here is that this is the original guy right here this is my f of x equals the square root of x then this guy down here let me just change colors this is my g of x is equal to the negative of the square root of x okay or again you could say this is g of x is equal to the negative of f of x if you want to define it that way that's fine what we see here again for a given x value let's just say i have an x value of 4 i now have a y value of 2 and then also negative 2 okay negative 2. so what happens is you can think about this as if we flipped the graph or folded the graph over the x-axis okay and this is what is known as kind of reflecting the graph across the x-axis i took this original guy here in orange and i flipped it over the x-axis or i folded it over the x-axis and so it creates what's known as a reflection across the x-axis so let's look at another scenario now so suppose we have f of x and g of x and now g of x is defined just a little bit differently so f of x is the same as the square root of x it's the square root of x but now g of x is the square root of negative x so this is a bit different essentially this negative is what's making it different so really i could say if i plugged in a negative x for x in that function i'd have my g of x right so i could now define g of x by saying it's what it's f of negative x again if i take a negative x and i plug it in for x i'm going to have my g of x g of x is the square root of negative x okay so now that we understand that let's think about what's going to happen with this guy so again these guys we should know by memory 0 0 1 1 we have 4 2 we have 9 3 and 16 4. so what's going to happen is again i'm going to plug something in for x i'm going to take the opposite of it now before i take the square root well i want you to think about the fact that with this guy right here it's defined for x values that are greater than or equal to zero right i can't take the square root of something that is negative in the real number system of course we can do it with kind of complex numbers but we're not working in that kind of frame right now for this guy i've got to think about the fact that this one right here is going to change things i plug something in for x i take the opposite of it so really i'm going to be restricted to values that are going to be 0 or less so 0 or less so we can say that x needs to be less than or equal to 0 here okay so thinking about that what's going to happen is for a given y value now i'm going to have kind of the opposite x about so if i copy these y values down 0 1 2 3 and 4. what i can say is that this would still be 0 because the negative of 0 is just 0. this guy would now be negative 1 this would be negative 4 this would be negative 9 and this would be negative 16. and if you don't believe me try that out let's say i plugged in a negative 1. the negative of negative 1 is 1. so i would have the square root of again the negative of negative 1 so that would give me the square root of 1 which is 1. okay so that's how i get my answer there if i have the negative of negative 4 okay of negative 4 that becomes the square root of positive 4 which is 2. so that's how i get that answer there so on and so forth so this kind of negative that's plugged in here with this x what it does is for a given y value now i'm going to have the opposite x value okay so let's look at this graphically so now this graph is going to be reflected across this y-axis okay and again when i say reflected you can just think about folding the graph so if i start with this orange graph here if i fold it across the y axis i'm going to end up with this guy over here that's what we mean when we say we're reflecting it across the y-axis it's a folding or flipping a graph over the line of reflection in this case the line of reflection is the y-axis the main thing here is to understand that for a given y-value i have x and then negative x so let's say i take a y value of let's say three okay so here's three so i have an x value of nine right nine with this point here and then i also have an x value of negative 9 with this point here okay so that's what you need to understand when you reflect over the x axis you're going to have for each given x value a y and then a negative y okay when you reflect over the y axis for any given y value you're going to have an x and then a negative x okay let's break this down to a general rule so as a general rule if g of x is equal to the negative of f of x so notice how this is outside it's the same as the graph of f of x reflected across the x axis so again that just means folded across the x axis so if you have f of x and on that f of x you have x comma y well then on g of x you're going to have x comma negative y again here was that scenario right this was reflected across the x-axis for a point on f of x like let's take this point here where it is four comma two so this is four comma two on g of x the same x value which is four now corresponds to a y value that's the negative of this one so it's four comma negative two so that's what that means now alternatively if g of x is equal to f of negative x so notice how this is inside now this is the same as the graph of f x reflected across the y axis okay so if you have f of x and there's a point on there x comma y well then for g of x it's going to be the negative of x comma y and again we saw this right here right i already showed you this so for this given y value of 3 you have this point here that is going to be 9 comma 3 and then you also have this point here which is negative 9 comma 3. so for that given y value of 3 you have negative 9 and you have positive 9 as x values that are kind of associated with that y value of 3. all right so let's suppose we wanted to work with g of x is equal to the negative of x squared and for reference sake i included this f of x equals x squared our squaring function as you kind of progress you're just going to be given one function you have to kind of pair it up with this parent function this is one that you're going to know you're going to have it committed to memory again this is the squaring function so essentially i can define g of x as being the negative of f of x notice how this guy again is outside okay it's multiplying the outside so the negative of x squared so how can we graph this really quickly well essentially i know that by rule if i go back up if g of x is equal to the negative of f of x it's the same as the graph of f of x reflected or again folded across the x-axis so all i need to do is graph f of x equals x squared and then just flip it over the x-axis so i've already kind of done this and essentially you should know what this graph looks like at this point this is again my f of x equals x squared and this is going to be my g of x which you could define as negative x squared or you could say negative of f of x either way is fine but essentially you'll notice that this graph is just folded over the x axis so let's fold it over this guy right here or you can say reflected across the x axis and again all you have to do is if you know this graph by memory just take these kind of points 0 comma 0 is going to be on there you know that but let's say i take something like this 1 comma 1 here let me highlight that well i know that for a given x value now the y value is going to be its opposite so instead of 1 comma 1 now it's going to be 1 comma negative 1. so i have that so instead of negative 1 comma 1 it's negative 1 comma negative 1. right so on and so forth so this guy right here instead of 2 comma 4 it's 2 comma negative 4. instead of negative 2 comma 4 it's negative 2 comma negative 4 and you can do that get enough points going and you can quickly sketch your graph all right so let's look at one that's a little bit tricky and this is going to come up again in our next lesson but suppose we have f of x equals 1 over x and g of x equals 1 over negative x so how can we define g of x so g of x is equal to what is it the negative of f of x or is g of x equal to f of negative x let's think about this again f of x is 1 over x and g of x is 1 over negative x okay so if i plugged in a negative x there would that get me here well yeah i would g of x could be defined as f of negative x now you might see that and say okay i'm done but there's an extra thing here because this is a fraction i could also define it this way okay why does that work suppose i said that i had the negative of 1 over x well really this just means that i have what i could have negative 1 over x or i could have 1 over negative x both of those would really be negative 1 over x because a negative in the numerator with a positive in the denominator or kind of flipping that a positive in the numerator and the negative in the denominator doesn't matter as long as one sine is negative one sine is positive you're going to end up with a negative overall so you can actually define this function either way okay and we're going to talk about this scenario in the next lesson when we talk about even and odd functions for now you can just grab either method you want and kind of sketch the graph it will work out either way so let me show that to you so this is our graph kind of pre-drawn this guy again is f of x equals 1 over x so that's the orange so you have this and then you have this over here okay again because of the way this guy reacts because you have this x in the denominator you can't ever get an x value of zero so that's why this guy has this behavior as you get closer and closer to zero kind of going towards zero coming from the right going to the left you notice the y values get really really really really big right they approach infinity because as i'm dividing by something that's kind of getting closer to zero you think about dividing by something really really small like point zero zero zero zero let's just say one if i divide one by this number it's going to be a really big number okay and i can keep making that infinitely close to zero as far out as i want right so i can make this as high as i want or over here if i think about it in the negative direction i can make it as low as i want okay the other thing also you'll see that this guy approaches a y value of zero but doesn't touch it and from this direction it also approaches a y value of 0 and doesn't touch it and that's because 1 over x can never be equal to 0. okay if i'm going to get 0 as a result of kind of this fraction here well 1 is fixed right i need for 0 to be in the numerator and for the denominator to not be 0 to get this result okay that's never going to happen so getting 0 as a result for 1 divided by x is impossible ok so that's why this function kind of looks like this we've talked about this previously but now what we want to do is reflect it over either the y-axis or reflect it over the x-axis again it's the same either way so i'm going to say this is let me kind of use a different color i'm going to say this kind of pink color is my g of x is equal to and we'll start by just saying it's negative 1 over x so this is and let me define that a little bit better so the negative of f of x so this will cause us to reflect this guy across the x x-axis so this is going to be my line of reflection and so let's say i take this point right here this is going to be 1 comma 1. if i reflect it over here it would be here right this point would be here this point would be here this point would be here this point would be here and this point would be here so you can see that allows you to kind of sketch this graph now the other way you could do this because it can be defined either way i could also say that g of x is again f of negative x and now it's going to be reflected over the y axis okay so then for kind of this point it would go here this point would go here right this point would go here this point would go here you know so on and so forth so you can sketch your graph that way the main thing is to understand what's going on here again if it's g of x is equal to the negative of f of x then it's reflected across the x-axis if it's g of x is equal to f of negative x it's reflected across the y-axis all right let's look at something really quickly that you might see as kind of a practice question or you might get this on your test suppose that f of x is defined as negative 2x squared minus 5 and then you're given these kind of g of x h of x and l of x so you've got to figure out what they are so g of x is defined as the negative of f x and essentially all they want you to do here is just multiply a negative by this right here the right side of the function okay that's how you're going to find g of x so i'm going to say this is equal to if i multiplied a negative by this right side just wrap it in parentheses and perform the operation this is going to run over but i'll come back to it so i'm going to distribute a negative to each term i'm going to have 2x squared plus 5 okay that's all it's going to be so let me erase this and drag this up and let me do this one now and i'll do it in a different color so now we have f of negative x so what would happen if i plugged in a negative x where i have that x there well the answer to that is nothing right because the negative is going to be squared it's going to have no effect right so essentially i can just write this as negative 2x squared minus 5. so negative 2x squared minus 5. essentially f of negative x is equal to f of x and there's a special name for this again when we get to even and odd functions in the next section we'll talk about this specifically all right the other one is the negative of f of negative x okay and you might see this you might be told to graph something in this way essentially what happens here is that first you're going to plug in a negative x everywhere you see an x so we already know that that results in this so you have negative 2x squared minus 5. it's the same thing then if i take the negative of the whole thing while i wrap it in parenthesis and put a negative out in front that gets me back to this first one here so i already know that's going to be 2x squared plus five okay so that's what you would do if you get that scenario all right so before we kind of wrap up the lesson i want to revisit stretching and compressing a graph when we did this lesson we didn't talk about kind of the scenario where you had a negative number involved so if g of x was defined as a times f of x we said that if a was greater than 1 this would be vertically stretched and if a was greater than 0 and less than 1 it would be vertically shrunk or be compressed now what happens if a is negative when you follow the same rules you just use an absolute value sign here if a ends up being negative you have to reflect it across the x-axis right because let's say i put a negative a there let's just say it was a negative 2 as an example well what happens is it's vertically stretched by a factor of 2 but it's then reflected across the x axis another way you can think about this too is for a given x value the y values are going to be changing by a factor of negative two okay so you can take the old y values just multiply them by negative two those would correspond to those kind of original x values and f x so let's look at an example of this and then we'll go on to the kind of horizontal stretching and compression so suppose we had f of x equals the absolute value of x and g of x equals negative one half times the absolute value of x so again i can define i can define g of x as being this negative one half times f of x so i've got this negative right here again if i had a negative and no one half this is a reflection across the x-axis but now when i put this one-half in here i know that i am vertically compressing this guy okay and that just means you're moving towards the x-axis that doesn't necessarily mean that i have to move towards from the top towards the x-axis which is this way it could also mean i'm moving from the bottom coming up towards the x-axis right either way is going to be appropriate for that definition okay so let's erase this i want to compare kind of some tables here so i know that zero would be zero one would be one we could do negative one and one we could do negative 2 and 2 and we could do let's say 2 and 2 right i plug something in for x i take the absolute value i get y essentially anything i plug in here just becomes non-negative now with this guy here i can take these same x values so 0 1 negative 1 negative 2 and 2 and i can just multiply the corresponding y values by negative a half right so zero there's no change multiply one by negative a half it's negative one half multiply one by again negative a half it's negative one half multiply two by negative one half it's negative one again two times negative one half is negative one so let's look at this graphically real quick and you have multiple things going on because i want to show you that there's a lot of different ways to kind of do this the first way you could kind of do it is again for a given x value you could just multiply the y values by negative a half so 0 0 would still be there okay if i took an x value of let's say 2 right now the y value is 2 okay but now if an x value of 2 goes on this guy the y value is going to be cut in half and it's going to be made into its opposite so instead of it being two multiply it by negative one half you get negative one so you're down there and that color is the same so let me change the color of that let me use like an orange so it stands out and i can use a point over here so in this situation i have an x value of four and a y value of four now that y value would be multiplied by negative one half so it would be negative two right same x value let me change the color again so that it's clear so that's one way to do it you can get some points and sketch the graph the other way to do this we could compress the graph vertically first again that's a movement towards this x axis so in other words for a given x value my y value is now kind of shrunk or multiplied by half so for a given x value of 4 the y value of 4 would now go to 2 okay so once you go through and get these points you can then take this guy right here this green graph let me kind of sketch over it and you could reflect it across the x-axis okay reflect it across the x-axis and you would get this red graph here okay so two ways to kind of go about that there's no right or wrong way to do that here you're going to get the same answer either way you just go with what is more comfortable for you all right let's quickly go over the horizontal stretch and horizontal compression again this is one that was a little bit confusing so in general if you have g of x and this equals f of ax we said in the lesson if a was greater than one now you had a horizontal compression and it doesn't make sense because you have this number multiplying by x and you think that it's going to stretch or grow but it doesn't right because it's by a factor remember this this 1 over a okay this 1 over a so what happens is let's say a was 3 as an example so i'd have a 1 3 there what this means is that for a given y value the x values are now a third the size so that has kind of the effect of compressing the graph horizontally okay and then we said that if a was greater than 0 and less than 1 well now we're going to stretch the graph horizontally and the reason for that is you end up putting a fraction in here for a you have 1 divided by a fraction so you're going to get something larger than 1. if i plugged in something like let's say 1 8th well 1 divided by an 8 is the same thing as 1 times 8 or 8. so now let's think about this and i'll erase that we don't need this let's think about this when we have negatives involved well all i need to do is just kind of say the same thing i would put the absolute value bars around it's the same thing i just need to reflect across the y-axis now okay that's all i need to do so to see this let's take a look at an example so in our last lesson we have this f of x equals x cubed minus x this is a good example to see kind of a horizontal stretch or compression and then we have g of x now which is f of negative one half x so essentially what i'm doing is i'm plugging in a negative one-half x here and here okay so for the table of values we can kind of use zero that would give me zero if i plugged in a negative one negative one cubed is negative one and then you'd have minus a negative one which is plus one so that's zero we know that one would also give me zero right if this was plus one one cubed is one and i would subtract away one so it also get zero and then we worked with two and negative two two cubed is eight then minus two would be six so this would be six here and then negative two cubed would be negative eight and then minus the negative two would be plus two so this would be negative 6 here okay so how can we get the values over here without kind of going through and actually writing everything out well again for a given y value for a given y value so 0 0 0 6 and negative 6 i know the x values can just be multiplied by one over this number again don't multiply them by negative a half you need to do one over that number okay very important to remember that so one over negative one half is the same as 1 times negative 2 which is negative 2 okay so what i would do is i would take 0 multiplied by negative 2 get 0 i would take negative 1 multiply by negative 2 and i would get 2 1 times negative 2 is negative 2. 2 times negative 2 is negative 4 and negative 2 times negative 2 is positive 4. okay so now we have some points going and this is something we probably don't know the shape of so if i took these values i could go to my coordinate plane and i could very quickly graph this guy and let's take a look at that real quick all right so graphically you're going to see three different graphs here and i did that to show you the two different ways you could do this the first way you can do this you don't even need the original one this is the original one in orange this is f of x equals x cubed minus x again if you know that and you're told this is your new function you can just make your table of values plot those points and then sketch the graph that's all you really need to do but if you're specifically told that you kind of have to graph the original one and then reflect it well then you can use this process here what you can do is you can horizontally stretch it first so i can say let's go with h of x let's say h of x now h of x is going to be f of and i'm just going to do one half x i'm not going to include the negative yet okay so what that does is it stretches it by a factor of 2 horizontally again it's 1 over this number 1 over this number 1 divided by half is 2. okay so it's horizontally stretching it by a factor of 2. so you can see this because a y value of 6 used to correspond to a x value of 2. but over here the y value of 6 corresponds to an x value that is 4. so it's gone from two to four so it's horizontally stretched or pulled away from the y-axis by a factor of two now once you've done that now we can reflect it across the y-axis okay so that's where this graph comes in over here this guy that's kind of in i guess you would call this a teal i might call it a green i'm not sure what color it actually is but let's just call this this is our g of x this is what we're trying to find this is our f of negative one-half x okay and essentially i got this by taking this guy and reflecting it across the y-axis so notice how this point here goes over here okay so essentially i have what i have 4 comma 6 the y value of 6 stays the same but my x value is going to get kind of the opposite put on it so you put a negative on it so you have negative 4 comma 6 now so negative 4 comma six and over here it's four comma six okay and you can do that everywhere right and i'll focus on these ones down here because in the middle it just gets kind of crowded but if i focus on this one this point was negative four comma negative six so negative four comma negative six okay well what happened to this point is it got flipped or folded or reflected across this y-axis and it went right there so now this is positive four comma negative 6. so again the y coordinate stays the same it's negative 6 in each case but we took this x coordinate and just made it into its opposite so a lot of different ways to kind of do this either way you're going to get the same result some of these graphs do get kind of involved but for now in the section we're in it's pretty easy overall in this lesson we want to talk about symmetry and we also want to talk about even and odd functions so what we want to do is start off by talking about symmetry with respect to kind of the x-axis the y-axis and the origin this is going to lay the foundation for us to talk about even and odd functions so we're going to start off by talking about graphs that are symmetric with respect to the y-axis okay the y-axis so this will occur whenever replacement of x with negative x yields the same equation so let's look at an example so here we have f of x equals x squared and of course this is our squaring function we've worked with this a lot we can visually see that this is symmetric with respect to the y-axis right the way you can think about this is let me highlight the y-axis if i folded kind of the right side of this graph over the y-axis you would see that the right side and the left side would match up perfectly okay and the reason for this is for any given y value you're going to have two x values that are kind of associated with the one exception to that being zero okay so that's because of this kind of squaring operation here let's say i take this point here and i take this point here so this point on the right this is going to be 2 comma 4 this point on the left is going to be negative 2 comma 4. so where does this come from if i plug in a 2 and square it i get 4. if i plug in a negative 2 and square it i also get 4. so because of this squaring operation essentially i can have two different x values that produce the same y value okay a negative and then a positive but that doesn't work for zero so this guy down here this y value of zero is only kind of associated with an x value of zero because zero squared is just zero right so that's the only exception to that guy every other y value is going to be associated with two different x values but the rule still works with zero right if i put a negative in front of zero i still get zero now in a lot of cases you're not going to have a graph you're going to have something kind of complicated you might not have time to graph or you might not be able to so you want to kind of go through a process where you replace the x with the negative of x and you see if you get the same thing so let's do that on a fresh sheet so if i have f of x and this is equal to x squared and what i want to do is see if f of a negative x that just means i'm plugging in a negative x for x if that's equal to x squared okay is that the same so what i'd have is the negative of x being squared now one thing i want to call your attention to i wrapped this in parentheses whenever you substitute in for a variable you want to wrap whatever you're plugging in inside of parentheses so i'm wrapping my negative x inside of parentheses and then i'm squaring it okay that's the proper way to do that if you make a mistake on this and just say it's negative x squared like that well this is negative 1 times x squared that's going to give you the wrong answer okay so you want to wrap it in parentheses and then square that okay so really what i can do is i can break this up and i can say this is negative 1 times x that's being squared now hopefully you remember your rules of exponents if we have a times b and this is squared this is a squared times b squared same thing here i have negative one times x these two guys these factors are both being raised to the second power so what i can do is i can split this up and say i have negative 1 again that's that whole thing is being squared so i'm wrapping in parentheses and then times i have my x squared we know that negative 1 that amount squared is just 1 so that would be 1 times x squared which of course is just x squared so these two guys are the same right if i replace x with negative x i end up with the same guy so we can say that it's symmetric with respect to the y axis later on in the lesson we'll hear this called an even function okay when this kind of property holds it's called an even function but we'll get to that later on all right so another good example of kind of an even function or a graph that's symmetric with respect to the y-axis is this f of x equals the absolute value of x again if i kind of highlight the y axis you can see that if you folded the right side over the y axis it would perfectly line up with the left side so again we can prove this pretty quickly we can say that f of negative x is equal to what well if i plugged in a negative x there inside of the absolute value bars again i have another rule with absolute value if i have let's say the absolute value of a times b this is the absolute value of a times the absolute value of b okay by rule so i can split this up and say that it's the absolute value of negative 1 right because this is really just negative 1 times x then times the absolute value of x okay that's all that would be the absolute value of negative one is always one so this is just the absolute value of x so these two are the same so again we've proved that this guy is symmetric with respect to the y axis all right so now let's talk about symmetry with respect to the x-axis so this is going to be the same concept but now the x-axis is going to kind of cut the graph in half again where each half is just a mirror image or reflection of the other now if a graph is symmetric with respect to the x-axis okay with respect to the x-axis then replacement of y with negative y yields the same equation one thing i'm going to tell you in advance is that if a graph is symmetric with respect to the x axis for a given x value you're going to have this y and negative y that are kind of associated with it so we know that from our definition of functions this will not be a function right it will fail the vertical line test now let's look at an example of this so we have y squared equals x right we're used to seeing y or f of x is equal to x squared and we saw this earlier but now we're seeing something a little bit different so now i plug something in for y i square it and i get a result for x so each x value now is going to be associated with two different y values one positive and one negative the only exception is that again is with this kind of zero here and that's because if i plug in a 0 for x well i know that what squared would give me 0 well there's only 0 that could do that okay but for every other situation let's say i choose an x value of 9 okay so that's here and then also here so i know that a y value of 3 squared would give me 9 and then also a y value of negative 3 if that's squared it would also give me 9. so that's why we have this kind of scenario going on but you can see that kind of the vertical line test would fail here right if i drew a vertical line here it hits here and it hits here so this is not the graph of a function but it is symmetric with respect to the x-axis right this guy is going to cut it in half where again if we folded kind of this top part over the x-axis it would perfectly line up with this kind of bottom part below the x-axis now how do we prove this again we plug in a negative y for y and we see if we get the same equation so it's the same kind of concept here if i plugged in a negative y and i squared it well we saw earlier if i square this guy i'm going to get positive 1. so really this ends up just being y squared equals x same thing as i have here so that's how we know that we have a graph that's symmetric with respect to the x axis all right let's take a look at another one so now we have x equals the absolute value of y again this is kind of the absolute value function we're used to just turned on its side just like we saw the parabola we're used to turned on its side so for this guy i'm plugging something in for y i take the absolute value i get something for x so in other words if i take a negative and i plug it in i get a positive if i plug in a positive i keep the positive if i plug in 0 obviously i get 0. so with the exception of this point right here which is 0 0 anytime i plug something in for con of y i could plug in the opposite thing for y and still get the same x so here's a good example here we have 4 comma 4 and then we also have 4 comma negative 4 okay so 4 comma negative 4 and this is 4 comma 4. okay so if i plug in a 4 for y take the absolute value i get 4. if i plug in a negative 4 for y take the absolute value i also get 4. so that's what's creating this kind of effect here we know this is not a function because again it would fail the vertical line test but it is going to be symmetric with respect to the x-axis because again if i plug in a negative a negative y in the place of y we know that this guy would end up being just x equals the absolute value of y right because this guy right here i could really just split it and say it's the absolute value of negative one times the absolute value of y the absolute value of negative one is just one so i end up with this guy which is exactly what i have right there all right let's take a look at one more of these so this guy is x plus two quantity squared plus y squared equals 25. so we see this is the graph of a circle and it's not symmetric with respect to the y-axis we can see that right if i took kind of this guy and highlighted the y-axis we can see that it's got less on the right side than it does on the left side so if i folded this over the y-axis those graphs would not match up okay so it's not symmetric with respect to the y-axis but it is symmetric with respect to the x-axis you can see if i folded kind of the top part over the x-axis i would end up with something that perfectly matches up with the bottom part again we can prove this by just plugging in a negative y there but you can see because of this exponent here that this guy if it got squared it would give me the exact same thing so the equation wouldn't be changed so that's how we know that it's symmetric with respect to the x axis all right now the last one we're going to look at is when a graph is symmetric with respect to the origin so this one's a lot more difficult to kind of look at and see visually but essentially if we replace x with negative x and y with negative y we end up with the same equation we have symmetry with respect to the origin so what you're going to find is that on a given graph if you have an x comma y then you're going to have the negative of that x and the negative of that y on the graph so let's look at an example so we have f of x equals x cubed or our cubing function this is the typical example you see when you first talk about kind of an odd function or function that's symmetric with respect to the origin or just a graph that's symmetric with respect to the origin in general so we can really write this as y equals x cubed and what i can show you is that again if i replace x with negative x and y with negative y i end up with the same thing so if i plugged in a negative y and i plugged in a negative x when i cube a negative 1 right because this is really negative 1 times x if i cube the negative 1 part i get negative 1. so i can erase the parentheses i don't need them and i can erase these parentheses here and all i want to do here is just divide both sides by negative 1 and this will cancel and so will this and i end up with my y is equal to y is equal to let me make that better x cubed again right so these are the same so that's how we know it's symmetric with respect to the origin so that's kind of the easy way to look at it the harder way is when we start involving this kind of function notation this is where it gets super confusing so we define this as f of x is equal to the negative of f of negative x so this is one way to define it and then another way is the negative of f of x is equal to f of negative x so if your function has this property that the original function f of x is equal to the negative of f of negative x or if it's the negative of f of x equals f of negative x then it's symmetric with respect to the origin and we could say it's an odd function now these two kind of pieces of notation that you see are super confusing what in the world does this mean all it means is that for a given x comma y you have the negative of that x and the negative of that y so let me explain let's just take a given point let me just kind of start with this one and then i'll work on the other one in a second so let's say i take a point like let's say 2 comma 8 we can see that here so 2 comma 8 is our given point now according to what i just said it should be true that the negative of 2 and the negative of 8 should be on this graph so negative 2 comma negative 8 should be there if we look negative 2 comma negative 8 is there so we know that's true but we can use function notation to kind of show this all we're saying is that if i pick something for x and plug in it remember if i have something like f of 2 it's just the function's value when x is 2. so f of 2 we know is 8. so what is the negative of f of two well i know f of two is eight so the negative of that would be negative eight okay so that's this part right here that's this part right here okay so we know that's negative eight and then highlighting kind of maybe makes that not show up so clearly so let me not do that so then the other thing is what is f of negative x well if x was 2 then negative 2 is negative 2. so f of negative 2 we already know that if x is negative 2 y is negative 8. so these two are equal or the same okay that's all that notation is just a fancy way to say that hey for a given x comma y you're going to have the negative of x and the negative of y on that you can also do it with that other notation that i showed you and this is much less common but you can also say that if it's kind of symmetric with respect to the origin then f of x is equal to the negative of f of negative x again just using that 2 comma 8 and the negative 2 comma 8 as a reference f of 2 we know is 8. f of 2 is 8. now what is the negative of f of negative 2 let's think about this f of negative 2 is negative 8 but then i want the negative of that okay i want the negative of that so the negative of negative 8 is positive 8 so again those two are equal now graphically we have to think about this a little bit differently let me go to the next page it's not as simple to notice if something is symmetric with respect to kind of the origin as it is to notice if something is symmetric with respect to the y-axis the x-axis it's very clear that something kind of is symmetric with respect to the y-axis because the y-axis again it cuts it in half same thing for the x-axis right the x-axis would cut the graph in half but with something that's symmetric with respect to the origin one way you could do it is you could kind of flip your graph upside down right so this orange guy is the original this is my f of x equals x cubed if i flip that upside down if i took my graph paper and just flipped it upside down if i end up with the same graph okay then it's symmetric with respect to the origin another way you can do this okay if i think about the negative of f of negative x we said this is equal to f of x and i might lose some of you with this definition but essentially if we remember from the last lesson when we talked about reflecting graphs if i had something like let me erase this for a second if i had something like g of x is equal to f of negative x we know that g of x can be graphed by just reflecting f of x across the y axis so that's the graph i have here in light blue so let's just say this is g of x this is g of x and it has this definition okay so i'm going to reflect it across the y axis and so for every given point that's on f of x this orange graph it gets reflected across the y axis so for a given y value now every x is going to be the negative right so this guy right here was two comma eight now it's going to be negative two comma eight this guy was one comma one now it's negative one comma one right so on and so forth so from this point right here we go to this point right here from this point right here we go to this point right here so we end up with this kind of blue graph so that's the first one now let me further confuse you and let's say that we had h of x which was equal to the negative of g of x what would this be this would be what if i want to graph the negative of g of x well then i just take g of x and i reflect it across the x-axis well if we do that what's going to happen let me kind of erase all this let me erase all this and let me draw that line back well now i'm starting with the blue graph and i'm reflecting across the x axis so this point right here we're going to have the same x value now which is going to be negative 2 but the y value is going to be different so we're going to have instead of negative 2 8 we're going to have negative 2 negative 8 okay so it gets reflected across the x axis instead of negative 1 1 we're going to have negative 1 negative 1 right so on and so forth so down here instead of 2 comma negative 8 i'm going to have 2 comma positive 8. instead of 1 comma negative 1 i'm going to have 1 comma 1. so what happens is all my points after i do the second reflection go back to the original graph of f of x equals x cubed so now we see that this h of x ends up just being the original function f of x and i know this gets confusing so let me kind of erase all these different function names and just go back to what we really kind of started with and saying that hey if i took this guy this f of x and i reflected it across the y axis so i had f of negative x and then i reflected it across the x axis so the negative of that those two would be the same graph they would be equal to each other so that's how again we know that it's symmetric with respect to the origin all right let's look at another example of something that's symmetric with respect to the origin so we have f of x equals 1 over x and again we can show this graphically or we can also show this kind of with function notation or we can show it just by saying that you know y equals 1 over x again by definition i can replace y with negative y and x with negative x because for a given x comma y negative x comma negative y is going to be on that graph so if i said i had negative y is equal to 1 over negative x of course i know that this guy if i have a positive over negative it's just a negative overall right so if i multiply this by negative 1 and this by negative 1 the negative cancels over here and the negative cancels over here and i'm back to my y equals 1 over x that i started with so that's how i know this is an odd function or again it's a graph that is symmetric with respect to kind of the origin okay now another way you can show this is again with that function notation you can say that f of negative x is equal to the negative of f of x or you could say the negative of f of negative x equals f of x i'm going to stay away from this definition for now this is not something you really see in your textbook this is just something i gave you so that you can see that if you kind of reflect the graph twice you end up with the same thing okay so i'm just going to use this one so f of negative x would be 1 over negative x and if i had the negative of f of x this would be what i would just take the right side of this and just multiply it by negative 1 so it's the negative of 1 over x these are the same right if i have a positive over a negative it's a negative overall i could really just say this is negative 1 over x so because these two are equal or the same we can say that we have an odd function or again we can say that we have a graph that is symmetric with respect to the origin now again if i thought about kind of reflecting this guy twice if i reflected it once across the y axis and then again across the x-axis you can do that on your own you'll see that you end up with the same graph another thing you can do is just take your graph paper if you if you sketch this take your graph paper and flip it upside down you're going to end up with the same shape okay it's the same graph everything's upside down but it's the same shape but essentially that tells you that it's symmetric with respect to the origin all right so we can officially just say that an even function is where f of negative x is just equal to f of x okay so if i plug in a negative x everywhere i have an x in that function and i end up with the same thing as if i just had f of x we have an even function or again a graph that's symmetric with respect to the y axis now for an odd function you're typically going to use this definition this is what's given to your textbook the negative of f of negative x is equal to f of x or f of negative x is equal to the negative of f of x now for the odd function again this is something that's symmetric with respect to the origin now you might be saying at this point you know hey where's the thing that's symmetric with respect to the x-axis remember when we deal with that we don't have a function so we don't have kind of a name for that because we're dealing with kind of functions here we have an even function that's symmetric with respect to the y axis and then we have an odd function that's symmetric with respect to the origin but we don't have anything that's a function that's symmetric with respect to the x-axis because those are not functions right they fail the vertical line test all right so let's look at these so we have f of x equals seven x to the fourth power minus two x squared minus one a lot of students will say okay i've got even exponents everywhere so i know it's even that's not always going to be true okay certainly if you have just you know f of x equals x to some power okay let's just call that a if a is even then you have an even function if a is odd then you have an odd function but there's all kinds of other things that can happen right you're not usually going to get something that simple so you do have to go through and do the work so for this guy i'm going to start with f of negative x and see what i get so i would have 7 times we'd have negative x to the fourth power minus 2 times you'd have negative x squared and then minus 1. so i know in each case that this negative is raised to an even power so this guy is going to be the same right so f of negative x would be 7 x to the fourth power minus 2x squared minus 1. so these two are the same so this guy is going to be even okay and clearly it's not odd because again for something to be odd f of negative x is equal to the negative of f of x well f of negative x is this right this is f of negative x if i took the negative of f of x this would be negative this would be positive and this would be positive these two are not the same so this function is even only right it is not odd all right let's take a look at another so we have f of x equals x cubed minus x squared minus 5. so i'm going to start by seeing if it's even so f of negative x is what so i have a negative x and this is cubed and then minus i have a negative x and this is squared and then minus 5. so the problem here is that this guy is going to change signs right this negative that's being cubed that's going to change signs but this one isn't going to change signs right because i have a negative that's now squared so f of negative x is going to be what it's going to be negative x cubed okay then minus x squared then minus 5. so this is different from this right here we don't have a negative out in front here we do right the rest of the signs are the same so we did change this guy so it's not even is it odd well for it to be odd again f of negative x which we have right here should be equal to the negative of f of x so in other words if i multiply the right side of this by negative 1 well i'd have a negative a positive and a positive those two don't match up either so this guy isn't odd and it's not even either so it's going to be neither all right what about this one we have f of x equals x squared plus the absolute value of x minus 7. okay well what do we see here if i had f of negative x if i plugged in a negative x there and squared it i know i just get x squared then plus if i plugged in a negative x there right inside the absolute value bars i know i just have my x and then minus 7. so really this guy is going to be even right so this would be x squared this would be the absolute value of x and then minus 7. so these two are the same so this guy is even okay again you can quickly verify this is not odd because this guy right here is not going to be the same as if you had a negative f of x okay these are not going to be equal let me kind of slide this down let me make this in a different color so i'll put that this is even like this so the negative of f of x would need to be equal to f of negative x again we have f of negative x here it's the same as this if i multiply the right side here by negative 1 this would be negative this would be negative and this would be positive those two don't match up so this guy is definitely not an odd function all right what about this one we have f of x equals 5 over x cubed minus x so what is f of negative x this would be 5 over you'd have your negative x this would be cubed and then minus you would have your negative x okay so first and foremost we know minus a negative is plus a positive so i can go ahead and just say this is plus this is plus and then if i take this guy this negative and cubit i end up with a negative so i can really just say this is negative x cubed like this all right so i know it's not even because this is different from this but i can check to see if it's odd so f of negative x should be equal to the negative of f of x so i would multiply the right side of this guy by a negative so really i would have what i would have the negative of f of x would be the negative of 5 over x cubed and then plus x well that's the same thing i have here the negative's down here now but i could move it up right a positive over a negative is just a negative overall so these are the same so we can say this function is odd all right here's a common one that trips students up so you have f of x equals the kind of quantity x minus three squared plus seven if you see an even exponent a lot of people just assume that hey this is going to be an even function okay but in this particular case it doesn't work out that way before we plug anything in just think about this if i plugged in a 5 there 5 minus 3 is 2 2 squared is 4. if i plug in a negative 5 there negative 5 minus 3 is negative 8 if i squared negative 8 i get 64. so that should tell you something's wrong okay so let's go ahead and erase this and i'll officially do it so f of negative x is what well i'd have a negative x there minus a three quantity squared plus seven well if i went through and kind of expanded this i could use my special products formula so i could say this is what this first guy gets squared so now i've got to square the negative and the x so that would just be x squared then i've got minus 2 times this guy times this guy now when we use this formula we usually don't have a minus out in front so you've got to account for that now okay usually we just bank on this sign being whatever this is okay but in this particular case it's 2 times negative x times 3 those negatives are going to become a positive so 2 times 3 is 6. so you're going to have 6x there then you get 3 squared and that's not okay so then plus 7. so 9 plus 7 is obviously 16. so with this guy you get x squared plus 6x plus 16. okay but if i expanded this one we know that let me kind of do this over here you would get the first guy squared which is x squared you would get 2 times this guy times this guy so 2 times 3 is 6 6 times x is 6x but it would be negative so minus 6x and then plus this guy squared 3 squared is 9 and then your 7 comes along for the ride 9 plus 7 again is 16. so these guys have a different sign right this is x squared minus 6x plus 16. this is x squared plus 6x plus 16. so they are not the same okay so this guy is not even is it odd okay is it odd well if i go back to what i had if i multiplied this guy right here in its expanded form by a negative one so if i had the negative of f of x it would really be equal to what it would be this guy right here let me erase this equals from here and let me drag this down here it would be the negative of x squared plus 6x minus 16. so this guy's going to be neither it's not even and it's definitely not odd because the negative of f of x is not equal to f of a negative x all right let's look at one more of these so now we have f of x is equal to x times the square root of 1 minus x squared okay so if i did f of negative x this would be negative x times the square root of 1 minus you'd have negative x squared so you know inside here this isn't going to matter right negative x squared that's just x squared but it does change things because i have a negative here and no negative there okay so these are not the same so it's not even is it odd well you can see that it would be right because the negative of f of x would be what i would multiply this thing by negative one so if i multiply it by negative one i would just have negative x times the square root of one minus x squared which is exactly what i have there so this guy is going to be odd in this lesson we want to continue to talk about graphing techniques and here we're going to focus on horizontal and vertical shifts all right so the last type of function transformation that we're going to talk about involves horizontally or vertically shifting the graph so these horizontal or vertical shifts are also referred to as translations so to begin our lesson let's start out again with our squaring function so we have this f of x is equal to x squared so we all know this function at this point we're plugging something for x we are squaring it and we get our output our f of x or you could say y so we already know what the graph looks like and we already know the points that are associated but just one more time we know that if we plug in something like a negative 2 right for x we square it we get 4. if we plug in a negative 1 we square that we get 1. we plug in a 0 and square that we get 0 plug in a 1 we get a 1 plug in a 2 we get a 4 right so we should have committed these points to kind of memory at this point and we also know stuff like 3 comma 9 and negative 3 comma 9 and a lot of you will know 4 comma 16 and negative 4 comma 16 as well all right let's look at the next kind of part and what we're going to do now is just kind of compare tables we're going to look at this other function that's very close to the original so we have f of x equals x squared we're going to call this the parent function okay and then g of x is going to be kind of this guy but just a little bit different okay so g of x is x squared minus 3. so this part right here and this part those are the same right you have that x squared but now you have this extra -3 that's involved okay and the way your book is going to show this to you they're going to say that g of x can be defined as f of x which is just x squared so f of x and then minus this 3. okay so minus this 3. so let's look at these kind of two tables real quick and i'll show you what's really going on here so again we know these by heart so negative 2 4 negative 1 1 0 0 1 1 and 2 4. okay but what's going to happen over here we know that we got these values by plugging in something for x and squaring it but over here i plug something in for x and i square it but then i subtract away 3 okay so whatever i got over here for y okay for a given x value now we're here for y it's going to be 3 less because of this minus 3. so if i copy these same x values so negative 2 negative 1 0 1 and 2 the y values associated will now just be 3 units less right so instead of four it would be one instead of one it would be negative two instead of zero would be negative three and then instead of one again it would be negative two and again instead of four it would be one what effect do you think this will have graphically think about this real quick for a given x value now the y values have been decreased by three units remember the y axis is a vertical axis so what's going to happen is this is going to shift the graph down okay by 3 units let's take a look at that real quick so the orange graph is my original my original f of x is equal to x squared and the graph in light blue okay the graph in light blue is my g of x is equal to x squared minus three okay so if you wanted to graph g of x if you got this on a test and they said hey graph x squared minus three you know that you just need to graph x squared and then just shift it down by 3 units the easiest way to kind of do this is just to pick these points that you already know something like let's say 0 comma 0 and say okay now it's going to be at 0 comma negative 3. so that's my lowest point there and then other points are just going to be shifted down by 3 units so something like 1 comma 1 is shifted down to 1 comma negative 2 right and so on and so forth and that's how you would go about sketching your graph but again the reason this is happening is for a given x value you now have a y value that's 3 units less so again for example here if i plug in a 0 i square it and i get 0. here if i plug in a 0 i square it and i get 0 but now i've got to subtract away 3. i've got this extra step here so i end up getting a y value of negative 3 or a y value that is 3 units less for that given x value so that's why all these points are being shifted down by three units all right now let's consider something that is a little bit more challenging what i just showed you was very straightforward what i'm going to show you here i'm going to pretty much tell you that everyone that sees this for the first time does struggle with it including myself it's something that takes a lot of practice to kind of get used to okay and then the light bulb will go off for you i promise but when you first see it it is a bit confusing so suppose we have f of x equals x squared and g of x is now equal to the quantity x minus 2 squared so if we really think about it this right here could be plugged in for x in f of x okay so i could really define g of x as being equal to f of x minus 2 if i wanted to right because if i plugged in an x minus 2 in for x and f of x i would have this g of x which is again the quantity x minus 2 squared okay so now that we know how to define that let's think about what's going to happen so let's start with these points here we have kind of negative 2 4 we have negative 1 1 we have 0 0 1 1 and we have 2 4. now what do you think's going to happen a lot of people think that this minus 2 here okay this minus 2 here because it's associated with x they think that kind of everything's going to shift to the left by two units right because the x values are occurring on the horizontal axis so if we think about minus two horizontally we intuitively feel like this should go to the left two units but what's going to happen is it's actually going to go to the right by two units okay so let me fill in this table real quick and then i'll explain why so for a given y value okay for a given y value so 4 1 0 1 and 4 these x values are now going to be 2 units larger okay so instead of negative 2 it would be 0. instead of negative 1 it would be positive 1. instead of 0 it would be two instead of positive one it would be three and then instead of positive two it would be four now if you wanna check this you can zero minus two is negative two negative two squared is four so that one's good 1 minus 2 is negative 1 negative 1 squared is 1 that's good 2 minus 2 is 0 0 squared is 0 that's good 3 minus 2 is 1 1 squared is 1 that's good and 4 minus 2 is 2 2 squared is 4. that's good so for a given y value right these y values are the same the x values have been increased by two units okay so let's think about what's going on here well what's happening is i'm plugging in something for x and then i'm subtracting away 2 before i square it so what that means is that in order to obtain the same y value my x value now needs to be 2 units larger right because i have to undo what's being done to x and that's going to be a key thing that comes up over and over in this lesson so you want to remember that you need to think about what undoes what's being done to x because i'm subtracting away 2 from x for a given x value i subtract away 2 so i'm really getting a y value that would have occurred with an x value kind of 2 units ago if i'm thinking about in the f x function if i'm thinking about maintaining the same y value then x now has to be two units larger to undo that minus two and again i know this is super confusing when you first hear it but let's go ahead and look at the graph and see if it makes a little bit more sense so i've graphed these already kind of side by side so once again the orange one the orange one is my original f of x equals x squared and my guy in light blue here is my g of x which is the quantity x minus two squared so again what you can see if i focus on this point zero comma zero it's been shifted two units to the right and now it's going to be at two comma zero okay let's think about y here i could plug in a 0 for x and i would get a 0 for y or f of x here if i plug in a 0 for x what happens well essentially i have negative 2 right and then negative 2 squared is going to give me 4. so that no longer works that doesn't give me a y value of 0 anymore in order to get a y value of 0 x now has to be 2 units larger than it was before to undo this so now i need to plug in a 2 there okay 2 minus 2 would give me 0 0 squared is 0. so what happens is for that same vertical value that same y value of 0 x now had to be moved 2 units to the right so what happens is everywhere on that graph is going to be shifted two units to the right because again for that same y value x now has to be two units larger again to undo this minus two right here alright let's go through kind of the textbook definition of this this is something you're going to get in your course and it's confusing a lot of times so i want to break this down so we have g of x equals f of x plus k and the first situation is we say if k is greater than zero so if k is some positive real number that's all we're saying so if this guy is a positive real number then g of x is just the graph of f of x shifted up by k units this is completely logical it should make sense for you for an example let's just say f of x is again x squared okay something really easy well let's say we had g of x and this is x squared so this part's the same and then plus some positive number let's just call it 13. okay so we should know by looking at that instantly that g of x is essentially just the graph of f of x it's x squared just shift it up by 13 units right because for any given x value you're going to plug it in you're going to square it and then you're going to add 13 to the result so the y value is going to increase by 13. so every point on that graph of f of x moves up 13 units if you're going to graph g of x okay that's all this plus 13 is doing so later on you're going to hear this inside the function outside the function this is happening outside the function okay you have your f of x nothing's being done to that it's outside so you have the plus k happening on the outside let me write that so this is the outside of the function if something happens to the outside of the function it's a vertical shift okay what happens inside it's a horizontal shift now you also have the case where you're shifting down okay and we also saw an example of this earlier so suppose g of x is equal to f of x plus k now we're going to say that k is less than zero so k is some negative number now the way this is written is a bit confusing so let me explain that in a minute for now let's just say that g of x is the graph of f of x shifted down by the absolute value of k units so they just want to give you something that's a generic rule they don't cover kind of every scenario generally we don't write stuff as plus negative we just write minus a number okay so for example earlier i gave you g of x was equal to x squared minus 3 okay so following this rule i don't have plus some number i have minus some number so i would have to transform it and say this is plus a negative three okay so we would say that the graph is shifted down the absolute value of k in this case the absolute value of negative three so it shifted down by three units but that's a bit of an overkill what i would just say is if you're subtracting something away so if you're subtracting away a positive number so minus 3 here you should be able to look at that and say well i know that could be written as plus negative 3 so i know i'm going to shift down by 3 units so if it's minus a positive then you're shifting down by this number of units so if it was let's say h of x is equal to x squared minus 21 i know that again i'm just going to shift down by this number of units okay that's all it is all right let's talk about the more difficult the more challenging scenario the one that everyone seems to you know not understand at first again including myself the first time i saw this i was really confused so let's say g of x equals f of x minus h so now what's happening is our action is inside the function i'm plugging something in for x okay so i'm plugging in the quantity x minus h in for x and f of x to get my g of x okay so if it's going on the inside let me write that down you've got to be thinking horizontal shift okay so now if h is greater than zero again this is the confusing part g of x is the graph of f of x shifted right shifted right by h units okay so let's say that we had g of x was equal to the quantity let's say x minus 7 and this is squared again a lot of people look at that and they say okay well if my f of x is x squared well my g of x they'll say okay it's x minus seven so i'm shifting to the left seven units but that's wrong okay you've got to shift to the right by seven units and the reason for that once again i've got to always think about what's going to undo what's being done to x that's really how you get your shift you can use this rule if you want you can say okay well i know in this case h is going to be 7 right if you want to match that up perfectly and my rule said hey it shifts to the right by h units so it shifts into the right 7 units you can get it that way but the easiest way to think about this is just to say what undoes what's being done to x since what's being done to x is i'm subtracting away 7 to undo it i've got to add 7 so that tells me if i'm adding 7 on the horizontal axis i'm moving to the right by 7 units because to get the same y value okay now x has to be seven units larger okay so to maintain that vertical position i've got to move seven units to the right on the x-axis because i've got to undo that minus seven okay that's what's going on all right now let's look at the more confusing rule so we have g of x equals f of the quantity x minus h again notice how we are plugging in for x and f of x so this is on the inside again this is on the inside and now we say if h is less than zero well what's going to happen is g of x is the graph of f of x shifted left by the absolute value of h units now this can be quite confusing so let's break this down so suppose we had g of x and it was equal to the quantity let's say x plus 7 squared well we know at this point that minus a negative is plus a positive so you've got to be thinking this way that to write it like this and follow this rule i would say g of x is equal to the quantity x minus a negative 7 and that amount squared so following this let me just kind of erase this real quick following this i've got my minus and i've got my minus so i've written it perfectly in line with the formula they've given me so i know that h now is going to be negative seven okay h is negative seven so just following the rule g of x is the graph of f of x shifted left by the absolute value of h units again because h here is less than zero h is negative so the absolute value of negative seven is seven so it would shift to the left by seven units but i don't recommend doing this this is something you can do when you first start the way i want you to think about this is what do i need to do to undo what's being done to x in this case we're adding 7 okay we're adding 7 to x to undo that i need to subtract 7 okay i need to subtract 7 and if i think about that okay for the same y value now x needs to be 7 units less okay if i think about that on the horizontal axis i would go 7 units to the left right that's how i decrease on the x axis so just figure out what you need to do to undo what's being done to x and then consider your shift that way so let's just look at some examples real quick we have g of x is equal to the quantity x plus 4 squared and then minus 1. so our parent function here would be f of x is equal to x squared and what you'll notice is that this part right here is happening inside the function right i could plug in an x plus 4 kind of in for x in f of x and i can obtain that part then i have this minus 1. this is happening outside the function if i wanted to think about this i could say that g of x is f of x plus 4 and then minus 1 right that would be the notation if you wanted to describe it but essentially once we have that down right we don't need to write that each time we just think about what's going on so this part is just inside the function this part's outside now if you want to use those rules that i gave you earlier feel free you can do that when you first start but the quickest way when you're looking at this stuff is just to think about okay for the inside of the function part what's going to undo what's being done to x i have a plus 4 there so to undo that i need to subtract 4 away okay so if i'm subtracting 4 away what that's telling me is that for a given y value x now needs to be 4 units less so that's going to shift the graph left by 4 units so i can say that we're going to shift or really just say shifts left by four units okay and again this is kind of counterintuitive right you see this x plus four you're thinking oh i'm shifting to the right by 4 units but it ends up being that you're shifting to the left by 4 units now additionally you have this minus 1 that happens outside the function this one's very straightforward right if i'm subtracting away 1 then i'm just shifting down by one unit if this was something like plus one i would just shift up by one unit okay but it's minus one in this case so additionally we're also going to say it shifts down by 1 unit okay so those are the two things that are going to occur so based on that original graph f of x equals x squared we would shift every point to the left by four units and down by one unit so graphically we can see this our orange graph is f of x equals x squared our blue graph our light blue graph is g of x equals the quantity x plus four squared minus one so to kind of graph the blue one quickly just take these points that you already know and love like 0 comma 0 shifted 4 units left and 1 unit down so 1 2 3 4 1 unit down right you can do as many of these as you need to so let's say i take this point i'm going to go 1 2 3 4 1 unit down okay so on and so forth that's how you could quickly sketch your graph all right let's take a look at another one so now we have g of x is equal to the quantity x minus 3 cubed and then plus 1. so this guy is based on f of x equals x cubed right that would be your parent function so this part right here is the inside the function part right i could plug in an x minus 3 kind of in for x in this original function so if i had f of x minus 3 i would have the quantity x minus 3 cubed okay so that would give you this part kind of right here and then i have that plus 1. that happens outside of the function right so really we could say that g of x is equal to f of x minus 3 and then plus 1 right if you wanted to define it that way all right so what's going on here is that again from the inside the function part i think about what i need to do to undo what's being done to x in this case i'm subtracting 3 away from x so to undo that i would add 3 right so if i add 3 then for a given y value x now needs to be 3 units larger okay so again i'm adding three so that means i'm shifting to the right by three units so this shifts right by three units okay now this plus one here again that's outside the function it's just telling me we're shifting up by one unit so we're just going to say shifts up by one unit all right so graphically again our graph in orange is our original f of x equals x cubed and then this guy right here in light blue is my g of x equals x minus 3 that quantity cubed and then plus 1. so every given point let's just take this one for example is going to shift to the right by three units and it's going to shift up by one unit so this would go one two three units to the right one unit up you could take this point right here you go one two three to the right one unit up this guy right here 1 2 3 is to the right 1 unit up right so on and so forth if you already know the graph of kind of f of x equals x cubed then it's easy to graph this kind of g of x graph all right let's take a look at another one so suppose we have g of x equals the square root of x plus 1 and then plus 3. so this part right here is what's going on kind of the inside this part right here is the outside okay so in other words let me just write this one more time so f of x equals the square root of x if i plugged in an x plus 1 4x in this function again that's what's happening on the inside of the function can i replace x with that so in this case can i replace x with the quantity x plus 1 well yeah that's how i get that okay so now we need to think about for the inside part how can i undo what's being done to x well i have this plus one okay i'm adding one so i need to subtract one away to undo that so again for a given y value x needs to be one unit less so i would just say we're going to shift left or we'll say shifts left by one unit and then this plus 3 on the outside of the function that just tells me we shift up by 3 units and then we'll say shifts up by 3 units so graphically the orange graph is f of x equals the square root of x and then the light blue graph is g of x is the square root of x plus one and then you have plus three on the outside so again every given point is going to be shifted one unit to the left and three units up so let's just take this point zero comma zero on the original it goes up one two three and then one unit to the left right so on and so forth so this point up one two three one unit to the left so again if you know this graph and you should memorize the graph of f of x equals the square root of x then you can quickly graph this kind of g of x once you figure out kind of what the shifts are you can graph this guy if you need to real quickly and then shift the points and then sketch your graph of the other one so for the last example we want to look at this kind of purple graph and we want to determine kind of the function so the orange graph is kind of a reference it's our kind of parent function so you should recognize this at this point it has this v shape that's a very signature graph it's f of x is equal to the absolute value of x right so this purple graph again the parent function will be f of x equals the absolute value of x how can we define this graph of this purple function let's call it g of x and let's just look at what happens you just look at the points we know this is the lowest point here and that gets transferred to there so i would go what i would go down one two and to the left one so we went down by two and we went left by one okay so let's see if we can figure that out so down by 2 means i subtract away 2 on the outside of the function so let's say we have the absolute value of x and i'll just leave some space again if i go down by 2 i minus 2 outside the function or you could put plus negative two if you want to it doesn't matter so what about this left by one again a lot of us want to just write minus one in there but again that's not correct right you wanna put plus one why well if the graph is shifting to the left by one unit i need to put plus one in there because the shift left is really i need to subtract one away okay that's what's creating that kind of shift i need to subtract one away to undo what's being done to x so that's why this would be g of x as equal to the absolute value of x plus 1 and then we'd have minus 2 outside of the absolute value operation so the purple graph that would be my g of x function in this lesson we want to review operations on functions all right so whenever we study functions we're going to come across this topic known as operations with functions so we start seeing this in algebra 1 and we kind of carry this through to algebra 2 and college algebra and you even see it in precalculus so it's a very common topic to kind of work through and essentially it's very easy we're just asked to find the sum difference product and quotient of two functions which in the end gives us a new function the only thing that's kind of challenging here is just understanding the notation understanding what you're being asked to kind of find but once you see a few examples it's very very easy all right so let's start off by just going through kind of the textbook definition that you're going to encounter in your course essentially given two functions and for today's purposes we're just going to call it f and g so those are our two functions and for all values of x where f of x and g of x are going to be defined we're going to talk about the functions f plus g f minus g f times g and f divided by g so we have these definitions already written down essentially if you see this notation f plus g of x this means we have f of x plus g of x okay so these two are the same or equal same thing goes for f minus g of x this is equal to f of x minus g of x and then f times g of x this is equal to f of x times g of x and then f over g of x this is equal to f of x divided by g of x and i have a restriction here remember you can't divide by zero so g of x cannot be equal to zero right because division by zero is undefined now let's briefly talk a little bit about finding the domain again we've been talking about that for a while now and this always comes up when you're in this section so for our functions f and g the domains of f plus g f minus g and f times g are going to include all real numbers in the intersection of the domains of f and g when we say intersection we mean that the domain of the new function will only include values that are in the domain of both original functions so it must be valid for both functions to be valid in the new function that is created and we'll see this as we kind of go through the examples we'll talk about the domain of each scenario we see lastly we want to talk about f divided by g of x so when we come across the scenario the domain also includes all real numbers in the intersection of the domains of f and g but with an additional statement again where we say that g of x is not allowed to be equal to zero because again division by zero is not allowed okay so very easy topic let's just jump in and look at some examples all right so let's say we have these two functions we have f of x equals x minus 3 and we have g of x equals 2x minus 5. so the first thing i'm going to ask you to do is find f plus g of x and then tell me what the domain is so let's start with that and then we'll move on to kind of this other one we'll talk about that in a minute so f plus g of x we said was what it's f of x plus g of x okay now before i kind of move on think about the domain of each one because we want the intersection of the two so the domain here for x minus three if i think about is there anything i can't plug in for x is there anything i can't plug in for x and subtract the way 3 no no restrictions there it'd be all real numbers same thing goes for 2x minus 5. there's nothing i can't plug in for x there so when i have f of x plus g of x when i add those two together that new function the domain will just be all real numbers okay so we can list that down here if we want we can say the domain again i'm going to use interval notation so from negative infinity to positive infinity and then we could just sum these f of x plus g of x i would just replace f of x with this x minus 3. so x minus 3. i would replace g of x with this 2x minus 5. and it's really just as simple as that all i'm doing is just kind of adding two polynomials together so x plus 2x is 3x let me kind of do that down here so this is 3x and then negative 3 minus 5 is going to be negative 8. so essentially f plus g of x is again f of x plus g of x this gives us three x minus eight let's say you're asked to find f plus g of negative three what in the world does this mean you might be scratching your head and saying i don't know what this means let me erase this so i have a little room and i'm going to keep this 3x minus 8 because we're going to use this in a second and i'm just going to drag this up here so it's out of our way and just put a visible kind of border there let's just use a little line okay so f plus g of negative 3. what does this mean we already know what function notation is if i have f of x equals x minus 3 and i say what's f of 2 you plug in a 2 where you have x so essentially whatever's inside the parentheses you plug it in for x so in this case 2 minus 3 would give me negative 1. so the function's value when x is 2 is negative 1. okay that's all it's really doing so let's erase this and think about this now so if i have f plus g of negative 3 all i really want to do is find out what is f of x plus g of x and then plug in a negative 3 for x okay that's all i'm doing so kind of following the fact that i already know that it's 3x minus 8 all i really have to do is sub a negative 3 in there 3 times negative 3 is negative 9 and then minus 8 would be negative 17 okay so f plus g of negative 3 would be negative 17. now there's another way you can do that you can either go through this process and then sub in or you could also do it this way and you'll get the same answer you could say this is f of negative 3 plus g of negative 3 perfectly valid so i would go through and say okay f of negative 3 is what if i plugged in a negative 3 here then i'd have a negative 3 minus a 3 up here so negative 3 minus 3 is negative 6 and then g of negative 3 is what so if i plugged in a negative 3 there i'd have 2 times negative 3 which is negative 6 negative 6 minus 5 is negative 11. so what happens when i do f of negative 3 plus g of negative 3. well let's just replace i'd have a negative 6 for that and i have a negative 11 there so i'd have a negative 6 plus a negative 11 or you could say negative 6 minus 11 if you wanted to it doesn't matter and you get negative 17 either way right so if i plug in a negative 3 here i get negative 17 if i kind of go through and say f of negative 3 plus g of negative 3 i also get negative 17. so again either way i'm going to get that negative 17 as my result all right let's look at subtraction now with the same two functions so we have f of x equals x minus 3 and g of x equals 2x minus 5 again same two functions we have f minus g of x so again all i'm doing here is i'm just saying that this is f of x minus g of x okay that's all i'm doing so f of x is what it's going to be x minus 3 and g of x is 2x minus 5. you've got to be careful with subtraction i am subtracting away g of x i'm subtracting this whole thing away so the best thing to do is just wrap this in parentheses to remind yourself that you've got to distribute this negative to each term right so essentially this becomes negative 2x plus 5. so this becomes minus 2x plus 5 and so we can go through and just combine like terms now x minus 2x is negative x so i'll just get rid of that and negative 3 plus 5 is positive 2. so this becomes negative x plus 2 okay so that's the result there now before i go any further the fact that we're doing subtraction here it doesn't matter the domain is still all real numbers if i look at f of x and i look at g of x again all real numbers here all real numbers here when you get this result here it's going to be all real numbers it's the intersection of the two domains so the domain i can just kind of put a bar again here and just say the domain is from negative infinity to positive infinity all right what about f minus g of 12 again you could do the same thing you could do f of x minus g of x like we did here and then plug in a 12 for x so essentially if i plugged in a 12 i'd have the opposite of 12 plus 2 which would be what you have negative 12 plus 2 which is negative 10. so that's the result the other way to do it is again to do f of 12 minus g of 12 that's perfectly valid f of 12 would be what 12 minus 3 would be 9 then minus g of 12 you'd have 2 times 12 which is 24 and then 24 minus 5 would be 19. so you'd have 9 minus 19 which again is negative 10. so either way you do it you're going to get the same result all right let's look at one that's slightly more challenging so we have f of x equals the square root of 3x minus 5 we have g of x equals 7x minus 2. so before we get started let's think about our domain here okay remember we're going to work in the real number system so we want values of x that we plug in that are going to produce real number results okay so with a square root you have a restriction because in the real number system we can't take the square root of a negative number of course we can do this with the complex number system but when we're talking about domain unless somebody specifically tells you to include the complex number system we're working in the real number system okay so what we want to say here is that this radicand this 3x minus 5 needs to be non-negative so 3x minus 5 needs to be greater than or equal to 0. we solve this inequality to find out what the domain needs to be we get 3x is greater than or equal to 5 we divide both sides by 3 and we get that x is greater than or equal to 5 3. so x here has to be greater than or equal to 5 thirds so the domain could be written in set builder notation by saying what the set of all x such that x is greater than or equal to five thirds or again you could use interval notation which is generally what i prefer so let's say that the domain the domain is going to be from five thirds and i'll put a hard bracket there because it's included and then out to positive infinity okay so that's my domain all right so now that we've gotten that out of the way let's think about f plus g of x here so in this particular case i'm just going to drag this down here so f plus g of x f of x is what it's the square root of 3x minus 5 and g of x is 7x minus 2. so plus 7x minus 2 i can't combine this at all right so all i can really do and let me put my equal sign here and i'm just going to kind of move this down so this makes more sense so let's move this down here and let's move this up here there we go that's much better so f plus g of x is going to be equal to the square root of 3x minus 5 and then outside the square root symbol we have plus 7x minus 2. okay so that's just summing f of x plus g of x again nothing we can really combine you can't combine this with this so we're done now if i have f plus g of negative 2 again all i want to do is i can just take this guy and i can substitute a negative 2 in everywhere i see an x so that's really simple so i'll just have the square root of 3 times negative 2 and then minus 5 and then plus 7 times negative 2 and the minus 2. now before i go any further do you see a problem well yeah we already said that the domain was from 5 3 out to positive infinity and i just plugged in a negative 2 there for x here it doesn't really matter but over here is where you get the problem right 3 times negative 2 is what it's negative 6 negative 6 minus 5 is negative 11. so you get negative 11 here again in the real number system you can't take the square root of negative 11. so we would say that this guy is undefined right this is undefined okay now again if you're working with complex numbers and use your imaginary unit i you can write a solution but again we're not doing that here all right let's look at the same two functions and do subtraction now so we have f of x equals the square root of 3x minus 5 and g of x equals 7x minus 2. so we want f minus g of x and we want f minus g of 7. again my domain is going to be the same no change in that so my domain again is going to be from hard bracket five thirds out to positive infinity okay so 5 3 is included that's why i have a bracket there instead of a parenthesis so for f minus g of x i'm going to take f of x which is the square root of 3x minus 5. and i'm going to subtract away g of x which is 7x minus 2. now again when you work with subtraction be very very careful because you can make a sign mistake pretty easily wrap this in parentheses and then distribute your negative to each term so you'd have minus 7x and then plus 2. very important you do that otherwise you will make a sign mistake you can't combine this part with this part okay but you want to make sure that you have the correct sign because if you just put plus 7x minus 2 this is wrong so this becomes minus and this becomes plus okay so very important you do that all right so for f minus g of 7 again all i'm going to do is plug in a 7 for x and again because x can be anything that's five thirds or larger it's going to be valid in this situation so we'd have the square root of 3 times 7 which is 21 minus 5 which is 16. so the square root of 16 minus 7 times 7 plus 2. square root of 16 is obviously 4. and we have negative 7 times 7 which is going to be negative 49. so what we have now is 4 minus 49 which is going to give me negative 45 then plus 2 which is going to give me negative 43 right as a result so this would be negative 43. all right let's move on and look at some examples with multiplication and division now so for the process it's essentially the same with multiplication division it does become a little bit more tedious so for f of x we have six x squared minus three x minus nine for g of x we have x plus one so the domain here again i have x squared i have x here i have x there's nothing it's not inside of a radical it's not in a denominator it's not in one of those things where you'd say okay there's a domain restriction so the domain is just going to be all real numbers so the domain again it's going to be from negative infinity to positive infinity right nothing i can't plug in for x in either one of those so f times g of x again i'm just multiplying these two together so we would say what it's 6x squared minus 3x minus 9 multiplied by x plus 1. you cannot use foil here because these are not binomials one's a binomial one's a trinomial so you have to distribute each term from the binomial to each term of the trinomial so i would have x times 6x squared which is going to give me 6x cubed i would have x times negative 3x which is minus 3x squared i would have x times negative 9 which is minus 9x and then 1 times anything is just itself so i can just put plus 6x squared minus 3x minus 9. okay so now let me erase this and i can just kind of scooch this up here let's see if we can combine any like terms so 6x cubed nothing i can do with that negative 3x squared plus 6x squared would be positive 3x squared so let's get rid of that negative 9x minus 3x is going to be negative 12x so this is minus 12x and then you have your minus 9. so that's going to be our answer here 6x cubed plus let me make that a little cleaner plus 3x squared let me make that a little cleaner minus 12x and then minus 9. now if i want f times g of negative 1 again all i've got to do is plug in a negative 1 everywhere i see an x so what i would have is 6 times you'd have your negative 1 plugged in for x and that's cubed plus 3 times negative 1 squared minus 12 times negative 1 and then minus 9. negative 1 cubed is negative 1 so this just becomes negative 6. negative 1 squared is 1 so you can just basically get rid of that because 3 times 1 is just 3. negative 12 times negative 1 would be plus 12. so this would be plus 12 here and then you have minus 9. so negative 6 and negative 9 would be negative 15. so this would be negative 15. 3 plus 12 would be positive 15 so what we see is that we're going to end up with 0 okay so f times g of negative 1 is 0. and again you can prove this to yourself you can go back and say what's f of negative 1 what's g of negative one you can multiply those two together and you will end up with zero and it's obvious because if you plug in a negative one there negative one plus one is zero zero times anything is zero okay so that's why that happens all right so let's move on and look at the same two functions with division now so what we have is f divided by g of x and again all i want to do is take f of x and divide it by g of x but again when you work with division you're going to have an extra restriction so if we're doing addition subtraction or multiplication with these guys the domain is all real numbers but when i start thinking about the division g of x here this guy cannot be equal to zero so g of x cannot be zero so what that means is that this guy x plus one cannot be equal to zero well where is it equal to zero it's equal to zero if i subtract one away from each side of the equation it's equal to 0 where x is negative 1. so that's my domain restriction okay so i can say that the domain the domain is again the set of all x such that x does not equal negative one again that's your set builder notation or as i prefer and this is going to be a little bit longer to write it this way you can write it in interval notation so from negative infinity up to but not including negative one and the union with again from negative 1 anything larger than that out to positive infinity so all real numbers except for negative 1. negative 1 is not included here and negative 1 is not included here all right so now let's think about f divided by g of x so all i'm going to do is grab f of x which is 6x squared minus 3x minus 9 and put this over x plus 1. okay now we already know how to simplify something like this we can either use polynomial long division if we want or in this particular case we can think about this by kind of factoring the numerator and canceling with the denominator okay but there's going to be one thing that we need to think about here let me copy this real quick we're going to go to a fresh sheet and we're just going to work on this real fast i'll get rid of this equal sign and i want you to think about how we can factor this numerator we can pull out a 3. so i'd have 2x squared minus x minus 3. this would be over x plus 1. and what can i do here can i factor two x squared minus x minus three well this guy right here is a prime number so it's perfect to use reverse foil with so i can say this is two x and this is x now i know this last term here negative 3 so this negative 3 can only come from a negative 1 and a positive 3 or a positive 1 and a negative 3. so those are our only two possibilities if we can factor this so let's think about this if i had 2x multiplied by negative 1 so let's say i had a minus 1 here and i had positive 3 multiplied by x would that get me what i needed no because this guy right here in the middle is negative 1x what i would need to do is make this a plus 1 and this a minus 3 because the outer now would be plus 2x and the inner would be negative 3x so that would be perfect right plus 2x added to minus 3x would give me my negative x and then negative 3 times positive 1 would give me my negative 3. so this is the factorization we're looking for now we have this over our x plus 1. so look how easy this is now this is going to cancel with this and i'm just left with this 3 times the quantity 2x minus 3 which i can also write as 6x minus 9. so let's go back up and i'm going to write this as 6x minus 9. so 6x minus 9. now here's where you really got to pay attention because this is where the teacher is going to throw a fastball at we have talked about rational expressions before and the same rules are going to apply when you have a rational function like we have with this f over g of x before we simplified we noticed that a value of negative 1 we restricted that from the domain made the denominator 0. so after we simplify we see that with 6x minus 9 there's no restriction but you've got to carry that restriction over okay it's still undefined when x is equal to negative 1. so in this next problem where it says f over g of negative one a lot of students will go okay six times you have negative one minus nine and we'll say this is equal to negative fifteen that is wrong this is wrong don't do that this guy is undefined okay this is on defined this is a very common trap question you've got to say this is 6x minus 9 and you can list the domain restriction that x is not allowed to be equal to negative 1. this gets carried over even though you simplified and it disappears all right let's look at another one so we have f of x equals the square root of 4x minus 1 and g of x equals 1 over x so f times g of x again we're just multiplying these two functions together so you'd have the square root of 4x minus 1 times 1 over x you could really just write this as over x now before we go any further let's think about domain restrictions because here you've really got to think about things with f of x you have a domain restriction where the radicand 4x minus 1 has to be greater than or equal to 0. so if i add 1 to both sides of the inequality i get 4x is greater than or equal to 1. divide both sides by 4 i get x is greater than or equal to 1 4. so that's one domain restriction the other domain restriction is what it's that this guy right here is one over x well i can't have an x value of zero but this guy is more restrictive than this guy so i can just take x is greater than or equal to one fourth and i'm good to go right because one-fourth is already larger than zero so zero is already not included in the domain when i restrict the domain to these values so i can just say that my domain okay for this new guy right here is going to be what it's going to be from 1 4 again including 1 4. that's why i have the bracket there out to positive infinity okay so that's my domain and this is as simple as i can make this i really can't do anything else here if you want f times g of 3 just plug in a 3 everywhere you see an x so for this guy i'll just kind of do this over here you'd have the square root of 4 times 3 which is 12. 12 minus 1 is 11. so square root of 11 over just plug in a 3 for x so that's all that would be equal to so f times g of 3 is just square root of 11 over 3. all right let's look at the same two functions with division so f of x equals square root of 4x minus 1 g of x equals 1 over x so we want f divided by g of x so again all i'm going to do is take square root of 4x minus 1 my f of x and divide it by 1 over x my g of x so divided by 1 over x the quickest way to do this is just to say what this is equal to square root of 4x minus 1 times the reciprocal of this which is x over 1 or just x so let me put the x out in front because that's going to be a little bit cleaner so i'm going to put it in a different color so x and then multiplied by the square root of 4x minus 1. so let's just erase this and just kind of slide this down now let's think about our domain when we look at division here we have to be extra careful that we're not dividing by zero okay but we pretty much covered that already because we said the domain was going to be from 1 4 and including that out to positive infinity so in this situation we're covered here and we're covered here we're good to go and even when we had it set up in the format of the square root of 4x minus 1 divided by 1 over x we're still good to go if i plug in a 1 4 there or anything larger i'm ok the only value that's excluded there is going to be 0 and that's covered by this domain restriction here so we are ok all right so let's talk about f divided by g of one so all i would do is take this guy here this x i plug in a one there and then square root of four times one minus one four times one is four four minus one is three so all i have here is what just the square root of 3. all right so let's look at one more kind of concept here and then we'll just wrap things up all right so we have g of x equals 2x plus 4 h of x equals 3x cubed minus 4. this can be a little confusing when you first see it suppose you see negative 2g plus 3h of x now it doesn't matter if you have addition here subtraction multiplication or division it's the same process all i want to do if i see negative 2 times g it's just like if i saw negative 2 times g of x just multiply negative 2 by each part of this function okay that's all i'm doing if i see 3 times h just multiply 3 times each part of this function over here it seems a little bit challenging but it's really not so really all i'm saying here is that i have negative 2 times g of x plus 3 times h of x that's all this is asking for okay now for the domain here 2x plus 4 there's nothing i can't do with x 3x cubed minus 4 nothing i can't do with x the domain is all real numbers so the domain let's write that up here we'll put from negative infinity to positive infinity and let's just crank this out real quick so if i have negative 2 times g of x well essentially i'm doing 1 i'll multiply negative 2 by 2x so that's going to give me negative 4x and then i'm going to have negative 2 times 4 which is minus 8. and then i'm adding this to and i'm just going to wrap this in parentheses just to kind of separate things i'm adding this to 3 multiplied by h of x so 3 times 3 x cubed is going to be 9 x cubed let me write that down here so 9 x cubed and then 3 multiplied by negative 4 would be minus 12 okay so let me erase this notation let me erase the parentheses we're just using that to separate things and essentially what can i do well my 9x cubed i don't have anything to combine with that then i have my negative 4x nothing to combine with that and i just have negative 8 minus 12 which is going to be negative 20. so i end up with 9x cubed minus 4x minus 20. so it's just that simple when you see these numbers out in front of kind of your function's name either it's f or g or h or whatever you're using just multiply that number by each part of that function and then proceed with the operation you're given okay it's very very simple in this lesson we want to talk about the difference quotient all right so when we talk about the difference quotient it's going to be something we need to understand before we kind of get into our calculus course you're going to pretty much use this right away in calculus when you start talking about derivatives so what is the difference quotient well essentially this is just the slope formula that we're used to working with with some fancy notation okay that's all it really is if you want to boil it down so let's say we revisit our slope formula let's say that we have a line and on that line we have two points so one point is p another point is q and the point p has the coordinates x sub one y sub 1 the point q has the coordinates x sub 2 y sub 2. we already know that the slope m now specifically say lowercase m here is equal to what you could say it's the change in y over the change in x or you could say it's the rise over the run or you could say it's the difference in y values over the difference in x values lots of different ways to kind of remember this for the slope formula it's just y sub 2 minus y sub 1 over x sub 2 minus x sub 1 okay and then we say x sub 2 minus x sub 1 can't be 0 because again division by 0 is not defined all right so generically if we look at this on a graph we can kind of use this to get into our difference quotient we'll see this is helpful in a minute so let's say this is point p this guy right here and this is point q right here so this is x sub one y sub one this is x sub two y sub two so we know again the slope m is equal to what the change in y values over the change in x values or you could say y sub 2 minus y sub 1 so the difference in y values over x sub 2 minus x sub 1 the difference in x values okay very very easy formula to remember something you should know at this point now let's say we start using function notation okay if we have this guy as again point p and this guy again as point q instead of saying that this is x sub 1 y sub 1 using function notation we can say it's x comma f of x we know f of x is just a fancy way to write y okay so don't let that fool you it's just it's just y right it's just x comma y we're writing x comma f of x for this guy right here it might kind of trip you up four minutes you have this h here and this h here so what does that mean well if you look at this graph here the horizontal distance from point p to point q we've said this is h right it's h units away so if this is x then this guy right here this x coordinate since it's h units away is going to be x plus h okay that's all it is this just took the place of my x sub 2 comma y sub 2 okay we just change it with some fancy notation and then we say comma f of x plus h so again this is just taking the place of y that's all it is nothing that's too fancy now we can take this kind of new notation and we can get our difference quotient so point p again is x comma f of x again this just took the place of x sub 1 y sub 1 and then q that point here x plus h is the x coordinate and f of x plus h is the y coordinate so this guy again this was x sub 2 comma y sub 2. so if we plug this in this way in our slope formula m equals what again we used to have y sub 2 minus y sub 1 well now we just say f of x plus h minus f of x so it's just the difference in y values it's the same thing just fancy notation then down here we used to say x sub 2 minus x sub 1 again it's just the difference in x values so you have x plus h this guy right here minus x this guy up here okay so that's all it is and we say h can't be zero because what happens is this x minus this x leaves me with just an h in the denominator so really i can say m is equal to we have f of x plus h minus f of x okay so this is just the difference in y values over your h right this is going to be the difference in x values they're going to be different by h units right because that was the horizontal distance between point p and point q okay and again we say that h this guy in the denominator cannot be zero because division by 0 is undefined all right so when you encounter this in your course what you're going to end up doing is you're going to be given a function you're going to be told to find the difference quotient okay so all this really is is just kind of plugging things in and simplifying once you see a few examples it becomes very very easy it's just something that you can make mistakes on because it is a little bit tedious so suppose we see y equals 5x plus 4 okay the first thing is if you don't have it in a format of f of x equals something go ahead and put it in that format first so we'll say this is f of x is equal to 5 x plus 4. now if i find the difference quotient for this guy what do i expect it to be remember this is in slope intercept form i already know that the slope is 5 right so i would expect that if i find the difference quotient for this function i would get a result of 5 okay so let's go ahead and try this out and let me just write this formula out again so it's f of x plus h minus f of x and this is all over h now what i'm going to do is again if i see something like f of 3 what does that mean it means to plug in a 3 everywhere i see an x if i see f of a what does that mean it means i'm plugging in a everywhere i see an x okay so if i see f of x plus h i just plug in an x plus h everywhere i see an x so for this guy f of x equals 5x plus 4 if i had f of x plus h that would be what that would be 5 times where i have an x here i'm just going to plug in an x plus h that's all i'm doing nothing fancy so x plus h gets plugged in there and then you have your plus 4 okay then the second part of this is to subtract away the original function which is just this 5x plus 4. now be very careful here because this is where people make these sign mistakes you've got to subtract the whole thing away so don't just put minus 5x plus 4 that's going to give you a sign mistake you want to subtract the whole thing away so wrap it in some parentheses that's going to remind you to distribute so this becomes negative 5x minus 4 okay all right so once we have this we can write it over h and then we can start simplifying so let me scroll down so let's go ahead and say this is what 5 times x is going to be 5x then plus 5 times h is 5h then we have plus 4 we have minus 5x we have minus 4 this whole thing is over h what you're going to see with these is a lot of stuff's going to cancel so 5x minus 5x is gone we have 4 minus 4 is gone so you end up with 5h over h and we know that this is going to cancel with this and just give me 5. this is exactly what we expected to get because again the line was in slope intercept form and we knew that the slope of that line was 5 to begin with okay so we got the expected result all right let's look at something a little bit more challenging it's just more tedious not necessarily more challenging so we have 3x squared minus 5x minus y equals 1. again if you don't have f of x if you don't see that this guy is in implicit form so it's not even solved for y so you have an additional step here what you have to do is solve this for y first and then replace y with f of x and then you can go through and find the difference quotient okay so i would just add y to both sides and i would subtract one away from each side and that would give me one so i would say this is equal to y we know that this cancels over here i would have my 3x squared minus 5x we know that canceled so minus 1. i like the dependent variable to be on the left so let's go ahead and write it that way let's say that y is equal to 3x squared minus 5x minus 1. now i'm going to replace y with f of x so let's just erase that and put f of x and now we can find the difference quotient we're going to do the same thing we just did it's just going to be more tedious so we're going to say that we want f of x plus h minus f of x all over h okay so again if i have x plus h inside the parentheses that's what i'm going to plug in here and here that's all i'm doing there's no trick to it so what i'm going to have here is 3 times inside the parentheses i'm going to put x plus h and this guy squared again i just plugged in for x so i put x plus h there and then i squared that result then i'd have minus five times again i've got an x so plug in an x plus h and then i have minus one okay so that's the first part you've still got to do minus f of x so i've got to subtract away and i'm going to put parentheses around this i've got to subtract away this original function f of x so this is 3x squared minus 5x minus 1. and then once that's done i want to put the whole thing over h all right so let's see what this gives me it's a lot of stuff to simplify here so i would have three times if i expand this remember i could use my special products formula for that this would be x squared plus two x h plus h squared and then over here i would have negative five times x which is negative five x and then negative five times h which is minus five h then i have minus one and then i have to distribute this negative to each term so i'd have minus three x squared i'd have plus five x and i'd have plus one and this is all over h okay so a lot of stuff there all right so let's keep going here all right so i'm going to distribute this to each term so i'm going to have 3x squared plus 6xh plus 3h squared okay just distributing the three to each term there then i'd have minus five x minus five h minus one minus three x squared plus five x plus one all over h okay and i know this gets kind of tedious okay so what can we cancel this will cancel with this this will cancel with this this will cancel with this and that looks like it's about it right so we're going to be left with what we put my equals here because i forgot to do it and let me put my equals here so i'm going to have 6 x h plus 3 h squared minus 5 h over h now can i make this any simpler well the answer to that is yes we have an h here we have an h squared here we have an h here we have an h in the denominator so what i can do is i can factor so let me kind of copy this i'm going to bring it to a fresh sheet and let's just factor this guy so in the numerator i'm going to write this as h times the quantity i'll have 6 x plus 3 h minus 5. okay all i did was i factored out an h from the numerator this is over my h in the denominator this is going to cancel and leave me with what so you'll have your 6x plus 3h minus 5. okay so this is going to be your result all right let's take a look at an example with a fraction involved so now we have our f of x equals 3 over x minus 2. again if we want to find the difference quotient for the difference quotient we have f of x plus h minus f of x and then this whole thing's over h okay so in this first part here for the f of x plus h again where i see an x i just plug in an x plus h that's all i'm doing so i'm going to have 3 over just x plus h and minus two and then you have your minus your f of x so minus this thing right here so three over x minus two what i'm going to do is just write this as plus negative just to make it a little bit easier to keep track of what's going on and then i want to write this over h so i can just write it over h like this or if i wanted to i could do this numerator part first and then write it over h kind of later on it's up to you it's the same result either way all right so looking at this guy we have a complex fraction and we already know how to simplify a complex fraction we want to find the lcd of all the denominators involved so what's the lcd of this guy and this guy okay so it would just be the product of the two so what i would do is i would multiply kind of this top part up here by x plus h minus two that quantity times x minus two that quantity and then this bottom part here by the same thing so x plus h minus two again that quantity times x minus two that quantity so again this over this would give me one so i'm not doing anything illegal here okay so what happens is let me put equals here this guy when it gets multiplied by this guy what's going to happen is that this denominator is going to cancel with this kind of guy right there so i would just have x minus 2 times 3. so the first part of this is 3 times the quantity x minus 2. and then i have my plus here let me kind of undo this now so now what we're going to do is we're going to multiply this times this so we're going to see that this is going to cancel with this now and i'll have my negative 3 times this quantity so i'm going to put negative 3 times this quantity x plus h minus 2 and then this whole thing is going to be over h times the quantity x plus h minus 2 times the quantity x minus 2. all right so just working with this now let me put equals here so i'm going to use my distributive property 3 times x is 3x then minus 3 times 2 is 6. and then you're going to have negative 3 times x which is minus 3x then negative 3 times h is minus 3h and then negative 3 times negative 2 is plus 6 and again this is all over h times the quantity x plus h minus 2 times the quantity x minus 2. so don't go ahead and multiply that out just leave that in its kind of factored form you don't want to do that because you want to end up with this in the end okay so what you're going to see now is that you can cancel this with this right 3x minus 3x that's gone you can cancel this with this negative 6 plus 6 is gone so you're just left with this negative 3h here in the numerator so let's go ahead and write that so we'll say we have this negative 3h over you have this h times the quantity x plus h minus 2 times the quantity x minus 2. okay so now i can cancel this h with this h and i'm going to get a final result so we're going to have negative 3 over we're going to have the quantity x plus h minus 2 times the quantity x minus 2. okay let me make this fraction bar a little bit longer and so that is my result here negative 3 over you have x plus h minus 2 that quantity times the quantity x minus 2. all right let's take a look at this one here and you're going to see that when you work with kind of the square root operation you have to do a little bit of extra work so f of x equals you have the square root of 7 x minus 1. so what i want to do again find the difference quotient so f of x plus h minus f of x and this is over h okay so i'm going to plug in for x here again if i want f of x plus h i'm just going to plug in an x plus h so i would have the square root of seven times the quantity x plus h and then i'm subtracting away 1 and then minus for f of x i'm going to have the square root of 7x minus 1 and then this whole thing is over h now at this point you might stop and say well really i can't do much to simplify i can distribute the 7 right let's go ahead and do that real fast and i'll show you what else you need to do so the square root of you'd have 7 times x which is 7x and then plus 7 times h which is 7h and then minus 1. and then minus you have the square root of 7x minus 1. this whole thing is over h so you might say okay well i can't combine those two radicals there and i can't really do anything with the h in the denominator so i'm done but in fact what you need to do here is kind of rationalize the numerator which doesn't make a whole lot of sense right at this point but when you get to calculus it's going to make more sense for you essentially what you want to do is get rid of the h in the denominator so that's the expression we're going to be looking for okay so what i'm going to do is i'm going to again rationalize the numerator i'm going to end up with radicals in the denominator but that's okay in this situation what i would want to do to do that remember if we have a plus b if i multiply by a minus b remember these are conjugates i end up with a squared minus b squared okay that's exactly what i want to do here okay so what i want to do is i want to take this guy plus this guy okay so i want to multiply top and bottom by that so let me kind of make this a little cleaner so this is minus the square root of seven x minus one so times i don't know if i can fit this all on the screen we're going to try so the square root of 7x plus 7h minus 1. instead of a minus there i'm going to use a plus and then we have the square root of 7x minus 1. so again same two terms here square root of seven x plus seven h minus one and square root of seven x minus one those two are the same it's just a difference in sign okay so then this is over i'm gonna do the same thing so square root of seven x plus seven h minus one and plus square root of 7x minus 1. okay so let me wrap this in parentheses here and you can wrap this in parentheses to multiply the numerator you would use foil right so essentially what you're doing is the first terms and the last terms only because the outer and the inner are going to cancel okay so the first terms we know that it's this guy times this guy so it basically becomes this the square root of 7x plus 7h minus 1 squared so this cancels with this and i'm just left with the radicand okay so that's all i have let me kind of slide that down so we have room and then you're going to have minus right because you have a negative and a positive so you have alternate signs here the last times the last so again you get in this situation where you have the square root of 7x minus 1 quantity squared so this cancels with this i'm left with the radicand okay now again you've got to be careful here because you're subtracting away this whole thing so make sure you wrap it in parentheses and you'll end up with negative seven x plus one so negative seven x plus one all right then in the denominator we have h times this quantity here so we have this quantity square root of seven x plus seven h minus one and then plus you have the square root of seven x minus one okay so what we're gonna find is that in the numerator this cancels this cancels so what i'm left with is 7h over this h times this quantity here the important thing is that the h here will cancel with the h there and you're just left with 7 in the numerator okay so let's scroll down and write that so the numerator is now seven and this is over just this part here this square root of seven x plus seven h minus one and then plus you have the square root of seven x minus one okay so this is going to be your difference quotient for this guy again the reason we did that is so we could get rid of the h in the denominator when you get to calculus it's going to make more sense for right now just take it as a given and whenever you have a square root involved in one of these problems take the extra step to go ahead and do this in this lesson we want to talk about composition of functions so at some point when you study functions you're going to come across this topic known as function composition and essentially this topic just boils down to plugging one function in as the input for another that's all it really is so if you understand function notation this is really no more difficult it's just a little bit tedious because you've got to plug some things in now related to this topic we're also going to talk about domain and we're also going to talk about decomposing functions which is essentially kind of going in reverse but we'll get to that at the end of the lesson so let's just start out by talking about this simple example here and then we'll get into some stuff that's more complicated so we have f of x equals two x minus five we have g of x equals three x squared minus seven so the first thing you'll come across in your book is this kind of weird notation some students call it fog right so we have f composed with g okay so that's how that's red so f composed with g so this symbol right here is not multiplication it means composed with okay so f composed with g so let me write that out this is composed with is what we could say let me make that e a little bit better so f composed with g so essentially if i wanted to write this in a cleaner format i can say this is f of g of x okay and most of you have already seen this in algebra 2. essentially i'm just working from the innermost set out so i'm going to take g of x and plug it in for x in my function f of x that's all i'm doing so let's go ahead and just try that out real quick the process is really really simple so what i'm going to do is i'm going to say that f of g of x is equal to what so here's my f of x here it's 2 times x minus 5 and what i want to do is i want to plug in for x i want to plug in the function g of x so g of x is equal to 3x squared minus 7 so that's what i'm going to plug in right there for x okay that's all i'm going to do so let me scroll down and get a little bit of room going so 3x squared minus 7 that's what we're going to plug in let me just put this up here so 3x squared minus 7. so what i'll have is f of g of x is equal to two times i'm going to wrap this in parentheses because the 2 is multiplying the whole thing so 3x squared minus 7 and then minus 5. so all we have to do is simplify this pretty simple overall you're just going to multiply the 2 by the 3x squared so let's say this is equal to 2 times 3x squared is 6x squared and then minus 2 times 7 is of course 14 and then you have minus 5. to kind of finish this up all we can do here we know the 6x squared is not going to kind of simplify or combine with anything the negative 14 minus 5 is going to be negative 19. so f of g of x is 6x squared minus 19. so very very simple very very easy to get that now you might also see them kind of reverse things on you okay so you might see g composed with f so essentially all this is is now g is going to be kind of on the outside and the f is going to be on the inside so we're taking f of x and now plugging it in for x and g of x so this is g of f of x okay so you've got to think about this as starting on the left side here this guy's going to be your outer one and going to the right that's going to be your inside okay so this is g of f x so the order here does matter so again all i'm going to do is i'm going to take my f of x which is 2x minus 5 and i'm going to be plugging that in for x and g of x so this is going to go in there so what we'll have here i'll just put equals here we're going to have 3 times and again for x i'm plugging in so let me wrap this in parentheses i'm plugging in my 2x minus 5 and this guy is squared so let's make sure we square this guy and then we're going to subtract away 7. okay so let's work on this real quick now all of you know at this point this deep into the course that we have to expand this right we can't just say it's you know 2x that quantity squared minus 5 squared that is wrong we need to expand this so we're going to say three times we're going to use our special products formula to make this quicker so 2x that amount squared 2 would be squared that's 4 x would be squared that's x squared then we have a minus here so we'd have a minus you have 2 times the first term times the second term okay so 2 times 2x is 4x and then 4x times 5 is 20x so minus 20x here and then plus the last chi squared 5 squared is 25. close the parentheses and then you have minus 7 here okay so what do we have so we're just going to multiply the 3 by kind of each term 3 times 4x squared is 12x squared then minus 3 times 20x is 60x then plus 3 times 25 is 75 and then minus 7. so what does this equal here what is this going to be equal to well we have 12x squared can't do anything with that you have minus 60x can't do anything with that but then you have 75 minus 7 which is 68 so then plus 68 okay so g of f of x is 12x squared minus 60x plus 68. all right so i'm just going to do one example with this because this is a very easy concept and something you would have covered in algebra two you might also see a number given as the kind of inner set of parentheses so in this example we have f composed with g and then it's of three right so essentially what this means is that we have one we have f of g of three okay again just start with the innermost set and work your way out so this is g of three that's being plugged in for x in f of x now there's multiple ways to do this you can find f of g of x which we already know what that is and then you can plug in a 3 for x in that new function or what you can do is you can find g of 3 and then plug that in for x and f of x so two different ways to do it let's just do it both ways so you see that it's the same so let's find g of 3 first so what is g of 3 just plug in a 3 for x there 3 squared is going to be 9 so you'd have 3 times 9 minus 7 3 times 9 is 27 and 27 minus 7 is 20. okay so g of 3 here is 20. so really what i have is f of 20 okay because g of 3 the result is 20. so what is f of 20 well i go to my f of x my function here and i just plug in a 20 for x so it would be 2 times 20 okay 2 times 20 minus 5 2 times 20 is 40 40 minus 5 is 35 okay so that's my answer f of 20 is 35 so that tells me f of g of 3 is 35 as well now so we already found that f of g of x was 6x squared minus 19. so if i said what is f of g of 3 so i can just take this f of g of x i already know what this is and i could just plug in a 3 for x so i can say this is equal to 6 times plug in a 3 for x that's going to be squared then minus 19. we should get 35 if i got the right answer right so 3 squared is 9. so i'd have 6 times 9 minus 19. 6 times 9 is of course 54. so you'd have 54 minus 19 which is 35. okay so you get the same answer either way so you can either do it this way find f of g of x first and then plug in kind of the value you're given in this case it's three so we got a result of 35 or we can do it this other way where we find g of 3 first and then we plug that result in which in this case was 20 for f in the original function okay so two ways to kind of do that i'm not going to do another example of this because it's really easy we want to move on to something more complicated known as finding the domain so let's look at finding the domain for a composite function so this is kind of a really tricky thing so we're going to spend most of our time in this lesson on this because the other parts are really really easy first and foremost when we consider the domain of a composite function we need to do two things okay the first thing is we need to consider the domain of the inside function so this is the function that's being plugged in as the input for the outer function okay so that's very important a lot of students mess that up they forget to do it so that's the first thing then we need to think about the domain of the new composite function that is built okay so we're going to look at both of those and then we're going to state the domain of this composite function that we're building has the intersection of the domain of the inside function and our composite function we don't need to think about the outside function and i'll show you that as we kind of move on okay so remember when i say intersection i mean the overlap between the two so this means it must be valid in the domain of the inside function and also valid in the domain of the new composite function that we've just created i just want to say this again finding the domain for a composite function is really tricky and it's something that you'll probably struggle with when you first see it okay so in this particular case if we think about the domain whether i have f composed with g or g composed with f it's going to be all real numbers in each case and the reason for this is if we think about the inside function well for f composed with g this guy is f of g of x well the inside function here is g of x and it's all real numbers right so i can plug in anything here for x and square it multiply by 3 and then subtract away 7 from the result there's no problem there when i plug this in here okay i know that i end up with a result that's what it's 6x squared minus 19. in this new function that we've created there's no restrictions on x as well so the domain here for this guy is all real numbers and it's going to be the same thing with this guy as well when we look at g of f of x remember this was 12x squared minus 60x plus 68 okay so when i look at f of x it's 2x minus 5 that's all real numbers then when i look at the composite function we create the 12x squared minus 60x plus 68 that's all real numbers as well so the domain in each case here we could just say the set of all x such that x is a real number okay you could do it like that or again you could do it in interval notation again for each one it's the same so from negative infinity to positive infinity all real numbers all right let's look at an example that's a little bit more complicated now so we have f of x equals 1 over x and we have g of x equals the square root of 5 minus x so let's do f of g of x and then let's determine the domain and then we'll flip it around and look at g of f of x and we'll look at the domain of that as well so when we think about this f of g of x is quite simple to find so g of x is here it's the square root of 5 minus x f of x is here so i'm just going to plug g of x in for x and f of x it's all i'm doing so i would have 1 over again i'm plugging in for x and plugging in the square root of 5 minus x so that part is very very simple something you've done again in algebra 2 some of you even would have done an algebra 1. but finding the domain is a little bit tricky so we want the intersection again for the domains of the kind of inside function and then of the newly created function so this is f of g of x you could even name it as something new you could say this is h of x if you wanted to and i'll kind of get to that in another example i'm going to like leave that off for now because i don't want to confuse you so what's the domain for the inside function the g of x so g of x what's the domain well if i think about the square root of 5 minus x what can i do i can't take the square root of a negative so i would just take this 5 minus x this radicand and i would say that it has to be greater than it has to be greater than or equal to 0. pretty easy to solve this just add x to both sides of the inequality and you would have 5 is greater than or equal to x okay or you could say x is less than or equal to 5. so let me erase this and say that we know that x needs to be less than or equal to 5. so that's my restriction here on the domain for the inside function now when we look at the newly created function the f of g of x we have 1 over the square root of 5 minus x so we have a radical the same one so we have the same restriction here that x needs to be less than or equal to 5. so that's still there so for the domain of f of g of x we have that same restriction that x needs to be less than or equal to 5. but we have something else that we need to consider we have division here and we have a variable in the denominator so we need to make sure that something that gets plugged in for x doesn't create a zero denominator right we don't want to divide by zero that's undefined so where would the denominator be equal to zero well what i can do is i can say where is the square root of 5 minus x equal to 0 well there's a lot of ways to think about this we know the square root of 0 equals 0 if this becomes 0. so i could just say 5 minus x equals 0 just solve that i would just add x to both sides of the equation and find that x equals 5. so if i plugged in a 5 there 5 minus 5 is obviously 0 squared of 0 is 0. so 1 over 0 will be undefined so what i have to do here is further restrict the domain for the composite function the newly created function that we just built and i have to say that not only does x need to be less than or equal to five x needs to be strictly less than five okay and so what happens is i want the intersection of these two this one right here is more restrictive right so this contains all the same numbers as this one except for 5. so i've got to go with this one as the domain for this f of g of x so again the intersection of the two so let me erase this and we'll just state our domain here our domain is going to be from negative infinity up to but not including 5 okay so that's the domain you could also say that it's the set of all x such that x is less than 5 right and it's strictly less than if you put less than or equal to that would be raw okay so now that we understand that again we've got to consider the inside function and the newly created composite function so this whole thing this guy that we're looking at here that's what we're looking at now a common mistake is to look at this f of x function this is a different function okay so students will start going okay well x can't be zero well x can be zero right if i plugged in a zero there i'm fine five minus zero is 5. i can take square root of 5 that's not an issue also i can plug in a 0 here right it's the same thing so don't look at this f of x function this outer function in this kind of composite thing that we're building you need to look at the composite function as a new function okay something that we are creating based on taking the inside function and plugging it in for x and the outside function okay so i know that can be confusing at first but after you work enough examples it just becomes something that is kind of second nature for you okay so let's look at the other way around now let's think about g of f of x so what is this now so now i'm going to take this f of x which is 1 over x and i'm going to plug it in for x and this g of x again square root of 5 minus x so this guy is going in right there now this one will be a little bit more complicated so we will have the square root of 5 minus again for x i'm plugging in a 1 over x okay so now let's think about the domain and again this one will be slightly more complicated so my domain now the inside function is 1 over x right it's the f x so now i'm going to restrict this to where x cannot be 0 because 1 over x we know that the domain restriction there would be x can't be 0 because we can't divide by 0. so x cannot be 0 okay that's my domain restriction now in this particular one i have that 5 minus 1 over x is underneath the radical okay so essentially i know that i want this guy to be non-negative and so what i'd have to do here is say that 5 minus 1 over x needs to be greater than or equal to zero so this is a rational inequality it's an easy one but we do need to solve it and figure out what our restrictions are going to be so i'm going to write this as a single fraction so essentially i'm just going to multiply this guy right here by x over x and so i'll have 5 x minus 1 over the common denominator of x this is greater than or equal to 0. okay so now what i want to do is find the critical values you guys all should remember this so i know that one of them would be what makes the denominator zero so here it would just be x set that equal to zero so i know zero is a critical value and i know because it makes the denominator zero it's going to be excluded from the domain and we already know that because it was excluded from the domain before okay so nothing new there then we have our numerator so we have our five x minus one set that equal to zero get our other critical value so add 1 to both sides we get 5x equals 1 divide both sides by 5. you get x equals 1 5. okay so this critical value will actually satisfy this guy right because if i plugged in a 1 5 there 5 times 1 5 is going to be 1 1 minus 1 is 0. 0 over a non-zero number would be 0 0 is greater than or equal to 0 is true okay so 1 5 is okay now what we want to do is take these two numbers so we have x equals 0 and then one-fifth we want to kind of think about this again you've got to kind of test values in the intervals that you create with this so one interval is going to be from negative infinity up to but not including zero the other interval is going to be between zero and one-fifth and then the other interval is going to be greater than one-fifth out to positive infinity okay so those are the integers so if i look at the inequality we started with it was 5 minus 1 over x is greater than or equal to 0. okay so let's choose something that's less than 0 let's just choose negative 1. it's pretty easy to work with so if i did 5 minus 1 over negative 1 i could really write this as plus negative if i want is this greater than or equal to 0 yes it is right negative 1 over negative 1 is positive 1. so this would end up being 6 6 is obviously greater than or equal to 0. so this would be true right so values that are less than 0 would work between 0 and 1 5 we need to find something that we can plug in there let's just go ahead and use 1 10 so if i plugged in a 1 10 there i'd have 5 minus 1 over 1 10. is that greater than or equal to 0 1 divided by 1 10 is the same thing as 1 times 10. so this would be 5 minus 10 you know that would be negative 5. negative five is not greater than zero it's not equal to zero so this would be false okay nothing would work in that interval in the last interval i could just choose let's say positive one so let's put five minus one over one this would of course be one five minus one is four four is greater than zero so this is true okay so now we know what our domain is going to be we know that zero is excluded right so zero is excluded we know that values that are less than 0 would work so from negative infinity up to 0 we know that values between 0 and 1 5 do not work and then we know that values greater than one-fifth and also including one-fifth would work so i put a bracket there one-fifth out to positive infinity so let me just copy this so for my domain i can just list it out like this we'll say from negative infinity up to but not including zero right zero's not included then the union with we have that one-fifth is included and then anything larger all right let's take a look at another one so we have f of x equals 2 over x minus 1 and then g of x equals 3 over x okay so let's start with f of g of x and see what we get so again i'm just going to plug g of x which is this 3 over x in for x and f of x that's all i'm doing so 2 over i would have 3 over x and then minus 1. okay so you can simplify this guy if you want you can multiply it's a complex fraction you just multiply the numerator here by x in the denominator here by x and let me scroll down and get some room going this would give me what it would give me 2x in the numerator over x times 3 over x would be 3 and then minus x times 1 is just x so you get 2x over 3 minus x now if i want to think about the domain of this guy again if i go back up the inside function g of x g of x equals 3 over x so that's not defined where x is 0. okay so x can't be zero when i think about the kind of new function that i've created so let me erase this and just kind of drag this up so i can fit all this on the screen when i think about this guy right here this 2x over 3 minus x well i can say that 3 minus x can't be 0 right so i could just solve 3 minus x equals 0 add x to each side of the equation and i find that x is equal to 3 there right so x can't be 3 as well so x can't be zero coming from the inside function and x can't be three coming from our composite function the one that we just formed so let me erase this and i'll officially write that real quick for you so the domain the domain you could write it using this format if you want you could say the set of all x such that x does not equal zero or positive three okay set builder notation that's fine or if you want you can use interval notation which is a bit longer here so from negative infinity up to but not including zero the union with you'd have anything larger than zero up to but not including three and then the union with anything larger than three okay so it kind of gets really long with interval notation for this problem all right so two common mistakes again for this type of problem the first one is forgetting to find the domain of the inside function so g of x was the inside function if we forgot about the domain here you wouldn't have thought about zero being restricted from the domain so that's the common mistake then the second mistake again is to think about the outer function the one we plugged into the f of x a lot of students will say okay i've got f of x equals 2 over x minus 1 you'll see them say okay well x can't be 1 there and then they restrict that from the domain as well that's not restricted from the domain here we're only considering the new composite function that we formed and the inside function right the one that we plugged in okay that's all we need to do now let's look at g of f of x so let's look at g of f of x now okay so again all i'm going to do is i'm going to take f of x which is 2 over x minus 1 this guy right here and i'm just going to plug it in for that guy right there so you'd have 3 over you'll have 2 over x minus 1. okay so how can we simplify this again we've got a complex fraction so what we can do is we can multiply by x minus 1 over x minus 1. so what we're going to get is what let's scroll down and get a little bit of room going we'll have 3 times the quantity x minus 1 in the numerator over in the denominator the x minus 1's are going to cancel so you would just have your 2 there okay so you get 3 times the quantity x minus 1 over 2. so let me erase this real quick and let me kind of scooch this up and i want to think about the domain with you and i'm going to show you a little trick on this one because a lot of people get confused with this so let me kind of go through this this way and then i'll go through it a different way all right so the first way i'll say that the domain is what what did we plug in we plugged in f of x f of x is 2 over x minus 1. we know there that x minus 1 if we set that equal to 0 and we solved it we would get that x equals 1. so we know that 1 is restricted from the domain of this inside function so x can't be 1. now in the new function that we've created are there any restrictions on x well even if i multiply this out and say it's 3x minus 3 3x minus 3 no right this is a linear equation right so we have 3x minus three over two there's nothing i can't plug in for x there so the domain for this guy is just going to be all real numbers except for one so again i can say that it's the set of all x such that x doesn't equal one or you can use your interval notation again doesn't matter from negative infinity up to but not including one and then union with anything larger than one okay so you can do that as well now let me think about a different way to kind of show this to you your book might show you kind of this technique we already know the domain so we don't need it up there anymore let's say i take my outer function my g of x and i write it so i have my 3 over and then i have x so instead of writing x i'm just going to put f of x here i'm just going to put f of x here okay so i have 3 over f of x now let's consider the fact that this guy was plugged in okay this guy was plugged in so we know that we first need to find where f x is undefined right so we already know that it's undefined where x is 1. so we've already restricted that from the domain so x can't be 1. okay so now we move on and say that in this form we know that f of x can't be 0 because it's in the denominator okay so let's think about that f of x is 2 over x minus 1 we would say where is this equal to 0 okay where is this equal to 0 well if i think about that for a while i think about the fact that if i'm dividing here how can i get a result of 0 well i've got to have a 0 in the numerator and i've got to have a denominator that's not 0. so it is actually not possible to get a solution here you can go through and try you will not because 2 divided by x minus 1 will never give you 0. okay you can't plug in a 1 for x we already know that but you can plug in anything else that you want and it doesn't matter what you plug in you will never get a 0 over here because again you need 0 in the numerator and a non-zero denominator to get a result of 0. so what that tells you is that there's no restrictions here on this kind of newly created function which will be 3 over f of x and again f of x was 2 over x minus 1 and again we multiplied the numerator by x minus 1 and the denominator by x minus 1 and we found that this was 3 times the quantity x minus 1. over just 2. right so that's what we got and again you can write that as 3x minus 3 if you want over 2 doesn't matter but again the domain we know that this is from negative infinity up to but not including 1 and then the union with anything larger than one okay so either way you want to look at that i know that the way that i just showed you where you kind of say you know the f of x there which is being plugged in for x can't be equal to zero if you look at it that way that's another way to kind of go about it all right let's look at one more of these and then we're going to get into decomposing functions which is really easy okay so this is kind of the hard part if you're good to go on the domain you're good to go with everything with this topic because it's really really simple and the domain just takes a few tries to kind of wrap your head around what you need to do but again if you just remember you want the inside function you want the newly created function you want to look at those two domains and you want the intersection or the overlap between those two domains okay that's how you get the domain of your composite function so here we have f of x equals 1 over x squared plus 2x minus 35. we have g of x equals 1 over x so in the interest of time i'm just going to do f of g of x so i'm just going to do f of g of x so let's just start with that so we have 1 over and i'm plugging in a g of x everywhere there's an x so i'm plugging that in here and here okay so i would have what i would have 1 over x that quantity squared plus 2 times 1 over x that's basically 2 over x and then minus 35 okay so that's what i'd have there now if you want to simplify this quickly you can kind of square this first and say that this is 1 squared which is 1 over x squared you can go through and you can multiply the numerator and denominator by x squared right because this is just a complex fraction and you have x squared x no denominator there so i'm just going to multiply this by x squared and this by x squared and what i'm going to get is x squared in the numerator over in the denominator the x squareds would cancel so you'd have 1 there then plus x squared would cancel with x and give me x x times 2 is 2x and then minus you have x squared times 35 which is 35x squared so this is our f of g of x so let me erase this let me just kind of move this up here this is our f of g of x now you can think about the domain again in a few different ways you can first look at g of x we know that it's 1 over x so obviously x can't be zero and then you can look at this composite function here and you can say okay well 1 plus 2x minus 35x squared that can't be 0 because that's in the denominator okay so you can do that as well so let's do it that way so let's go through and say that 1 plus 2x minus 35x squared where is that equal to zero we're going to restrict those values to make this easier to put in a format that we're used to go ahead and divide everything by negative one it's just going to change the sign of everything so this would be minus this will be minus this will be plus it has no effect on zero right so we would have 35 x squared minus 2x minus 1 equals 0. you can factor this if you want to or you can solve it with a quadratic formula whatever you want to do it's going to factor so i know it would be what it would be 35 is either going to be 35 times 1 or 5 times 7 it's not going to be 35 times 1 so we're going to have to do 7x and 5x and then for negative 1 that can only come from a positive one times a negative one so you really just have to work out whether it's plus one here and minus one there or the opposite so this way it would be minus seven x and then plus five x and that's exactly what we want because negative seven x plus 5x would be negative 2x so that's good to go there okay so now i would just set these two equal to 0 and solve so 7x plus 1 equals 0 subtract 1 away from each side you get 7x equals negative 1. so 7x equals negative one divide both sides by seven you get x equals negative one seven okay so that's one value we're going to restrict so let's erase this and we'll just kind of keep this for a second then the other guy is five x minus one equals zero add one to both sides you get five x equals one divide both sides by five you get x equals one fifth okay so x equals negative one seventh and then one-fifth all right let's erase all this okay so x can't be zero we know that from the inside function and then also x can't be negative one-seventh or one-fifth so let's write our domain so let's write our domain and we probably want to write this one down here because it can be a little bit lengthy so i'll start by writing in set builder notation the set of all x such that x cannot be equal to zero one-fifth or negative one-seventh okay so in interval notation this is going to be from negative infinity up to but not including negative one-seventh and then the union width you'd have anything larger than negative one-seventh up to but not including zero and then you would have the union with anything larger than zero up to but not including one-fifth and then you'd have another union with anything larger than one-fifth okay so this one is really long when you write that in interval notation but that's generally how i like to do it all right let's wrap up the lesson and think about something that's pretty easy overall it's called decomposing functions this is another topic you're going to see in this section and it's very important because when we get to calculus we're going to have to treat a function as a composition of two functions so in other words when we compose functions we're going to compose two functions to form a new function when we decompose functions we're going to express it as the composition of two functions you're going to see this right away in calculus okay so it's going to be very important so now here's where i'm going to kind of change the name a little bit you've seen that we created a new function but you can give this new function a name a lot of times you're going to call it h right so we'll say h of x equals f of g of x and this is equal to 1 over 2 x minus 3. so essentially we've already done the function composition and we have the result of that and we want to decompose it okay so we want to find f of x and g of x so very very simple process so h of x is equal to f of g of x this is equal to one over two x minus 3. okay so lots of different ways to kind of think about this but essentially if you really think about what could be the outer function what could be the inner function well i could have plugged this guy right here in for x in 1 over x okay so the outer function the f of x would be 1 over x and the inner function the g of x would be what it would be 2x minus 3. okay so if i took 2x minus 3 and i plugged it in for x and this 1 over x i would get 1 over 2x minus 3. so that would be an answer here so find f of x and g of x f of x would be 1 over x g of x would be 2x minus 3. all right let's look at another one of these and then we'll just kind of wrap it up very easy topic overall so a lot of times they give you these they're very obvious so h of x equals f of g of x this is the cube root of x squared minus 2x minus 1. find f of x and g of x well we know that this guy right here would be the inside what's being plugged in for x and the outer would just be this kind of cube root operation so essentially the f of x the outside guy would be the cube root of x okay so that's my outer one the inner one the g of x what i'm plugging in for this x here is just this right here this x squared minus two x minus one okay so this topic is very very simple in this lesson we want to talk about the vertex form of a parabola so we've already spent a lot of time working with quadratic functions but we want to specifically learn how to graph a quadratic function and in order to do that efficiently we need to know about this vertex form okay so we're going to talk about that today in case you're clueless about what a vertex is we'll talk more about it in the next lesson when we're graphing but basically the vertex of a parabola is the lowest point if it's an upward facing parabola right you get that standard u shape and then it's the highest point if it's a downward facing parabola so if it's a u turn upside down again we'll talk more about this in the next lesson for now we need to know how to convert between this form and this form so something like f of x equals a times x squared plus bx plus c where a is the coefficient for x squared b is the coefficient for x to the first power and c is your constant term now most textbooks most people will say this is standard form this is going to vary based on which textbook you're using some textbooks will say this is standard form so i don't want to confuse you but i'm going to refer to this as standard form and this is vertex form okay so this guy right here this f of x equals you have your a times the quantity x minus h squared plus k the reason most people say this is vertex form is that you can immediately upon inspection find out where the vertex is so the vertex will occur at h comma k so this is my x-coordinate okay and this is my y-coordinate right so this would have correct h comma k this would be my vertex for the parabola so what we need to understand for today's lesson is how to go from this format here this f of x equals x squared plus 12 x plus 34 into vertex form right this is what i call standard form we want to go into that vertex form that f of x is equal to a times the quantity x minus h and this is squared and then plus k so i want you to notice something this is a binomial that's squared so we know earlier in the course we talked about the fact that if you have a binomial squared and you go through the fourth process you get a perfect square trinomial or a three term polynomial that factors into this guy right here a binomial squared remember we can create this guy from the completing the square process i know a lot of you guys are probably rusty on this so let me just walk you through it the first thing you need to make sure of is that the coefficient on the x squared here is a one so in this case that's already done force okay so you don't need to do anything there we'll see an example in a minute where it's not but for right now it's super easy so the next thing you want to do is group your x squared term and your x to the first power term together so i'm going to say i have f of x is equal to x squared plus i have this 12x so i'm just going to put parentheses around this right now to say this is just in a group and then plus my 34 okay so what i want to do now is complete the square and let me just scroll down and get some room going so i'm going to say i have f of x is equal to i'm going to make this a little bit bigger so x squared plus i have my 12x and then i'm going to add something here to complete the square and then i'm going to subtract something here and i'll explain this in a minute and then i'll put plus 34 like this okay so do you remember how to get this missing term right here remember you take the variable that's raised to the first power and you take the coefficient of that variable so in this case it's 12 right that's the coefficient you cut it in half meaning you multiply by half and you square it so 12 if i cut that in half or if i multiply by half that's six 12 times one half is six if i squared that i would get what i would get 36 okay so this is what i need to create my perfect square trinomial so i'm going to add 36 right here but i can't just do that normally i would add 36 to both sides to make that legal but in this case i want this f of x over here on the left i don't want to mess with that so what i'm going to do is i'm going to subtract 36 as well so 36 minus 36 is 0. i haven't changed anything mathematically i can always add 0 to something it's not going to change anything so what i want to do now is close the parentheses down on this okay but normally you have something out here that's multiplying this and this creates an issue here i just have parentheses here so i can at this point just close the parentheses off like this and remove this from here and i haven't done anything mathematically illegal you have to be careful though if you have something out here that's multiplying this then you've got to distribute and we'll see an example of that in a moment so for right now what i've got is a perfect square trinomial here so f of x equals i can factor this into what well it would be x plus again this guy was six okay so x plus six that quantity squared and then over here i just do negative 36 plus 34 that's going to give me negative 2. so i've found my vertex form it's a little bit off okay but you can still write it this way remember we said the vertex form was f of x equals a times the quantity x minus h squared plus k so the reason i say it's a little bit off here is because of the signs here you have a plus and here you have a minus here you have a minus and here you have a plus so if you wanted to find the vertex you'd have to write it in this format or just know how to convert it off the top of your head which most people do but essentially you would just take this and say okay i'm going to match the format so a is 1. so you don't need to write anything there you can just say x minus a negative 6 is the same thing as plus 6. so now i've matched that minus sign there okay so minus a negative 6 and then this is squared and then i'm going to do plus a negative 2 okay so from here now i can just grab this guy right here and this guy right here and i know my vertex occurs at negative 6 comma negative 2. okay so that would be my vertex for this parabola all right let's look at another one so this one will be a little bit more challenging because the coefficient on x squared is not one so we have f of x equals 2x squared plus 32x plus 123. so again the first thing you have to do is make sure that this guy is a 1. now again when you work with this when you're solving quadratic equations by completing the square you just divided everything by 2. but you don't want to do that again because you have this f of x over here you don't want to mess with that so what you're going to do now is you're going to factor it okay so i'm going to say that i have f of x is equal to i'm going to factor a 2 out from this again first group i'm going to treat these as a group and i'm going to say i have 2 times the quantity i'd have x squared there and then plus if i pulled out a 2 from there i'd have 16x okay so i've taken care of that okay let me just kind of scooch this down i know i've got to add something and subtract something and let me put my parentheses there and let me put plus 123 there okay so what are these two missing values here again i just take this guy right here this coefficient look at the variable raised to the first power we want that that coefficient cut it in half and square it okay if you just say that a few times you'll remember forever cut it in half and square it my ninth grade algebra teacher taught that to me and i never forgot it so if i cut 16 and a half i get 8 if i square 8 i get 64. so if i say 64 here and 64 here okay i have completed the square but this is where it gets a little tricky and a lot of people make mistakes normally if this guy isn't here if this isn't a two here i can just regroup this okay but because there's a two here you gotta realize that to remove the parentheses we have to use the distributive property and multiply 2 by every term there okay but i don't want to do that i just want to get rid of this part right here so what i'm going to do is say that i have f of x is equal to i'm going to keep my 2 out in front so times the quantity x squared plus 16x plus 64. and then for this part right here i'm just going to distribute the 2 to this so that i can remove it out of there okay and i'm going to say i have plus and i'll do 2 times this negative 64. and then i'll wrap this up by saying it's plus 123 okay so that's what you got to be careful of okay if you have something outside the parentheses like this 2 or whatever the number is if it's not 1 you've got to do this extra step it's a very common mistake to miss that all right so now what we want to do is write this we'll say f of x equals two times the quantity okay this guy would factor into what it's x plus we know that 64 came from us taking 8 and squaring it so this would be 8 and this would be squared and then we would have 2 times negative 64 which is negative 128 and then plus 123 we know that a negative 128 plus 123 would be negative 5. so this would be your vertex form you have f of x equals 2 times the quantity x plus 8 squared and then minus 5. again if you want to match things up perfectly you would write this as minus negative 8 okay and you would write this as plus a negative 5. so i could just write that like that and say f of x is equal to 2 times the quantity x minus a negative 8 again squared and then plus a negative 5. so the vertex here is going to occur at negative 8 and then comma negative 5. so negative 8 comma negative 5 again this is your vertex okay let's just do one more and then i'll go into the super simple vertex formula which is what you're going to use but you might need to know how to do the completing the square method it's always good to look at it because it comes up every so often alright so we have f of x equals 11x squared plus 66x plus 90. i'll kind of go through this very quickly we know we want to factor that 11 out so what i'm going to do is say this is f of x is equal to i'm going to treat this again first two as a group so 11 times the quantity x squared plus 6x and then plus 90. again i'm just factoring the 11 out from the first two terms all right so let's go ahead and say this guy right here i want to cut that in half okay again you're looking at the variable raised to the first power look at the coefficient cut it in half and square it cut it in half and square it so 6 divided by 2 or 6 times a half is 3 and then 3 squared is 9. so what i'm going to do is i'm going to add 9 okay i'm going to add 9 inside of the parentheses [Music] now if you wanted to get really advanced i recommend just putting -9 inside of here so you remember to distribute but if you can remember this on your own you can just say that okay i have 11 times negative 9 okay so it would be minus 99 out here okay so i just kind of skipped a step there once you do this a few times you're going to skip that step you just have to make sure that you multiply 11 by the negative 9 so that it's the correct number so then we have our plus 90 here and we're ready to go so what i can do is i can very quickly factor this and say this is f of x is equal to 11 times the quantity this guy factors into what x plus the 9 came from 3 squared so 3 this quantity squared and then negative 99 plus 90 would be minus 9. okay so that's your vertex form again if you wanted to match it up perfectly you would want this to be a minus and this to be a plus so you'd say f of x is equal to 11 times the quantity x minus a negative 3. this is squared and then plus a negative 9. so your vertex occurs where negative 3 comma negative 9. so the vertex okay the vertex is going to be at negative 3 comma negative 9. all right so now let's talk about the vertex formula and this is something that you're basically going to use you're not going to complete the square every time you're just going to do that if you're asked to do it so we have this f of x equals ax squared plus bx plus c okay again you've got to know that a is the coefficient for x squared b is a coefficient for x and c is the constant we already know this from the quadratic formula so it should be something that's familiar to you and if you go through the process of completing the square to put this in vertex form you would find that the vertex occurs at negative b over 2a so this is your x coordinate and then your y coordinate would be f of negative b over 2a so in other words find out what x is and then once you have that plug it in for x in your function and you're going to get y as the result so let's look at an example all right so here we have f of x equals 5x squared minus 20x plus 15. so if i wanted to find the vertex what i would do is i would say that the x-coordinate is negative b over 2a so what is b and what is a we know that a is the coefficient for x squared and b is the coefficient for x make sure you include any signs so if it's minus you want to put plus negative so you don't make a sign mistake so the negative of negative 20 is 20 okay and then 2 times a a is 5 so this is 10 so this is 20 over 10 which is 2. so the x coordinate is 2 and then what we want to do is plug this in for x to find out our y coordinate so in other words we want to just know what is f of 2 so we would have 5 times 2 squared is 4 and then plus you'd have negative 20 times 2 which is negative 40 and then plus 15. so i'll do 20 plus 15 which is 35 and then 35 plus negative 40 is negative 5. so this would be negative 5. so your vertex occurs at 2 comma negative 5. okay if you wanted to write this in vertex form again just match things up it's f of x equals a times the quantity x minus h squared plus k so h again is going to be your x coordinate so this is the x coordinate and k is your y coordinate and i'm going to run off the screen so let me kind of scooch that down okay so what we want to do here is say that we have what we have f of x equals what's my a what's this right it's the 5. so 5 times the quantity x minus what's my h it's the x coordinate so from the vertex my x coordinate is 2 and this quantity squared and then plus we have my y coordinate which is k okay so this is negative 5 in this particular case so i can put plus negative 5 if i want or i could just put minus 5. it doesn't really matter so the vertex form here is f of x equals 5 times the quantity x minus 2 squared and then minus 5. all right let's just do one more of these it's good to get a lot of practice because when you start graphing these things you know how to do this off the top of your head alright so we have f of x equals negative 4x squared minus 64x minus 261. again the vertex will occur at negative b b in this case is negative 64. so the negative of negative 64 is positive 64 over 2 times a a is negative 4. 2 times negative 4 is negative 8. so 64 over negative 8 is negative 8. so the x coordinate is negative 8 and then again to get the y coordinate i want to do f of negative 8. so what is that well negative 8 squared is 64. 64 times negative 4 is negative 256 and then you would have negative 64 times negative 8 which is positive 512 and then lastly you have minus 261. so if you do negative 256 plus negative 261 you get negative 517 and then negative 517 plus 512 would of course be negative 5. so let's just go ahead and say this is negative 5. so we'll put negative 5 there and i'll just go ahead and erase this and again if you wanted to write this in vertex form you'd say f of x is equal to your a which in this case is negative 4 times the quantity x minus whatever your x coordinate is from the vertex in this case it's negative 8 so you could put minus a negative 8 or you could put plus eight it really doesn't matter so this quantity is squared and then you're going to have your plus k k again is the y coordinate from the vertex so this is negative five so you can put plus negative five or minus five it doesn't really matter so your vertex form is going to be f of x equals negative 4 times the quantity x plus 8 squared and then minus 5. in this lesson we want to talk about graphing parabolas all right so in the last lesson we talked about the standard form and the vertex form of a parabola we learned how to convert back and forth between the two so again i'm going to define this first one here f of x equals a times x squared plus b times x plus c i'm going to say this is standard form and i'm going to say this guy right here f of x equals the a times the quantity x minus h squared plus k i'll say that's vertex form based on your textbook it might be different they might call this guy right here standard form but i'm going to stick to calling that vertex form so i don't want you to be confused by that now when we talk about the vertex of a parabola once again it's the lowest point on an upward facing parabola or it's the highest point on a downward facing parabola now when we have our quadratic function in vertex form what's special here is that we can get the vertex by just looking at the equation so the vertex will occur at h comma k so this is your x coordinate and this is your y coordinate you can just pull it straight from here so as long as it's in the minus form here in the plus form here you just take your h and you take your k now if you have different signs there you have to make an adjustment for that and we'll see that later on in the lesson now let's kick things off by looking at a parabola that we've already seen before so this guy is the simplest type of quadratic equation that you're going to come across it is the f of x equals x squared that we already saw when we're talking about graphing transformations okay so we already know the points we know that there's zero comma zero there's one comma one negative one comma one you know so on and so forth two comma four negative two comma four and then three comma nine negative three comma nine you could do four comma 16 negative four comma 16 again so on and so forth now a couple of things to kind of pay attention to here you can see that the vertex here occurs at zero comma zero okay and if you wanted to write this in vertex form you could play around with and just say it's f of x is equal to well the a here the coefficient of x squared is one so you don't have to put anything there you would just say x in this case minus zero that quantity squared and then plus zero okay and you're just doing this to show that this guy would be the x coordinate and this guy would be the y coordinate so the vertex would occur at again zero comma zero and that is your lowest point now let's move on and talk about something else that's important when you look at a parabola this vertical line that i've drawn in here that's basically going over the y-axis is known as the axis of symmetry or you might also hear people say the line of symmetry okay so basically what this is it's a line and it's a vertical line to be specific that intersects the vertex so this guy is always going to be a line that is x equals whatever h is okay so again h is the x coordinate of the vertex in this case it's zero so this line here would be x equals zero which is just the y axis in this particular case but basically what this guy does is it divides our parabola into two perfect halves so i could take a point on one side of the parabola let's just say we took this point here which is 3 comma 9 and i could fold it across that axis of symmetry and i would touch this point exactly which would be negative 3 comma 9. and we'll see this comes in handy when we're graphing because we get points on one side we can kind of reflect them across that axis of symmetry and get points on the other side all right so now let's get into graphing a quadratic function or you can say graphing a parabola basically this is a very simple process it only gets hard if the coefficient of this guy right here is a fraction okay if it's a whole number or an integer or something like that it's very very easy i'm going to show you three different ways you can do this the second method i'm going to show you is the quickest but let's start out with kind of one of the more challenging methods which is using graphing transformations so in order to do this we want to first write this in vertex form so how do we do that quickly again vertex form is f of x equals you've got your a times the quantity x minus h and this is squared plus k well again a is the coefficient for x squared in this case that's one so if this is a one you can just get rid of that okay so what's h and what's k well in the last lesson we talked about a shortcut to figure that out right we saw that we could complete the square to get this or there's a quick and easy formula we can use so the vertex formula tells us that h occurs at negative b over 2a again this is your a the coefficient for x squared this is your b the coefficient for x to the first power and this is your c okay you don't need c here in this particular case but i'm just labeling it for reference sake so what is negative b over 2a well b is 6 so i would have negative 6 over 4a i have 1. so negative 6 over 2 times 1 which is 2. so this is negative 3. so this is the x-coordinate okay the x-coordinate for the vertex so my vertex here will occur at negative 3 comma and let me just put this in here if i put minus a negative 3 that would be fine or i can put plus 3 if you want as well it doesn't matter okay as long as you know this is negative 3 for that coordinate okay so the way we get k is we plug a negative 3 in for x in our function okay so in other words k is going to be equal to f of negative 3 and so we say we'd have x squared so we'd plug in a negative 3 there negative 3 squared would be 9 and then plus 6 times negative 3 which would be negative 18 okay and then plus eight well nine plus eight is seventeen and seventeen plus negative eighteen is negative one so my vertex is going to occur at negative three comma negative one so let's put minus one out here okay so this is my quadratic function in vertex form okay so we know the vertex is here we know this is here but in this particular case we don't even need the vertex we just need this guy here because we're going to use our graphing transformations okay so we start out by comparing it to f of x equals x squared okay so what do we have here what happens inside well we add 3. remember when you talk about what happens inside the function you do the opposite so if it's plus three you're shifting to the left by three units so we would say that this guy is going to shift left three units okay and then this guy right here is happening outside the function so this is your vertical shift okay and that's straightforward if you have minus one i'm just shifting down by one unit so we'll say shift down one unit so we shift left three units and we shift down one unit okay so we can take that graph f of x equals x squared we can do these shifts and we'll have our new graph so let's go ahead and see that real quick so i've already drawn these side by side for you so again you can take a point like this this is zero comma zero you can go one two three units to the left and one unit down i'm gonna use a different color there let me use like a light or a dark blue how about that and let's just take this point here so you go one two three is the left when you're down take this one one two three to the left one unit down so on and so forth you can get this graph by again just applying shifts to the graph that you already know so that's the first method all right let's talk about the easy method this is the one that you're probably going to use in almost every case unless you have a fraction as your value for a in which case it becomes a little bit more challenging to use this guy you might want to use a different method all right so again we have this f of x equals x squared plus 6x plus 8. we already know that the vertex form is f of x equals 1 times the quantity x minus a negative 3 or you could say x plus 3 that quantity squared and then plus negative 1 or again you should say minus 1 if you want so with this method you basically want to find your vertex which we already know it's negative 3 comma negative 1. so your pattern is 1 times a 3 times a 5 times a and 7 times a now in this particular case again a is the coefficient for x squared it's right here and it's right there it's implied there okay so we know a is one so one times one is one three times one is three five times one is five and seven times one is seven so this is the pattern so first i would plot negative three comma negative 1 so i'm going to go to negative 3 comma negative 1 so that's right here okay and then i'm going to use that pattern that 1 3 5 and 7. so what does this mean basically what's going to happen here is i'm going to as i move to the right one unit i'm going to go up by this unit so i'm going to go to the right one up one okay now the next time i go to the right one i'm going to use this guy right here so to the right one up one two three now the next time i do this i'm gonna go up five so to the right one up one two three four five okay so you can continue i'm gonna go off my particular graph but if you wanted to go to the right one and up seven you could do that as well if you had enough room on your graph paper so the idea here is that you can use that to get one side of the parabola and again your axis of symmetry here is x equals again what's the x coordinate for the vertex well in this case it's negative 3 so it's x equals negative three so that's your axis of symmetry so i can use that to get these points on the other side if i fold it across this line just think about the fact from right here to right here is what it's one two three units so we just go one two three and it's over here there's my point right so here it's going to be zero comma eight and over here it's going to be negative six comma eight again you could do the same thing with this point you could take this and fold it across the line and get here take this fold across the line and get here and then once you have enough points you can sketch a pretty accurate graph let's talk about the third and final method that you can use and then we'll just burst through a couple of problems so again we have the f of x equals x squared plus six x plus eight and in vertex form the f of x equals 1 times the quantity x minus a negative 3 squared and then plus a negative 1. vertex occurring at negative 3 comma negative 1 we already know what this is now in most textbooks they're going to teach you to first find the vertex which we have and then they're going to tell you to find the x and y intercepts and then they'll tell you to just grab additional points as needed so you can use this method if you want again it's in most textbooks i found it to be pretty slow and inefficient overall but we're going to go ahead and go through this just for the sake of completeness because your teacher might ask you to do this so the first thing is how do you get the y-intercept do you remember that's where you basically plug in a 0 for x right because at the y-intercept x is going to be 0 because you're crossing the y axis so in the standard form here on the top i can just say what is f of 0 well these 2 would be gone it would just be 8. so a point here the y-intercept would be 0 8 okay so that's something i would want to plot then i would find my x intercepts how do we do that will we make y or f of x 0 now because we're crossing the x axis so i would solve the equation x squared plus 6 x plus 8 is equal to zero okay again if you just replace this with y and then you replace y with zero this is nothing more than what we've been doing all along throughout all of algebra and we come down here and we solve this by factoring it's very very easy to solve this guy so this guy would factor into what x here and x here i've got two integers that sum to six and give me a product of eight that would be four and two so plus four and plus two so the solutions here would be x equals negative 4 and negative 2. okay so that tells me that the x intercepts are going to be at negative 4 comma 0 and then also at negative 2 comma 0. so let's erase all this and again we could just plot these points and let me just paste this in here real quick and i'll write this more neatly so we have 0 comma 8 we have negative 4 comma 0. let me make that better and of course we have negative 2 comma 0. okay so let's erase this all right so we know the vertex occurs at negative 3 comma negative 1. so negative 3 comma negative 1 we have our axis of symmetry again that's x equals negative 3. and then we just plot these points so we have 0 comma 8. so 0 comma 8 is here and you can reflect this across the axis of symmetry to get this point over here this negative 6 comma 8. then you have your two x-intercepts so you have your negative 4 comma 0 and your negative 2 comma 0. so you don't get these two points from this method okay so you'd be sketching the graph with just these five points which can still give you a pretty accurate sketch of what's going on in most cases these things your teacher is not asking you to be perfect just give them an accurate sketch of what's going on we know in this day and age that we can pull up any graph from an online calculator in a millisecond so the idea is just to show that you know how to perform this task okay let's look at two more problems and i'm going to do these using the quick and easy method because in most cases you're just going to do this again unless you get a fraction for this value here for a right you can use this method it's very very effective so we have f of x equals 2x squared minus 8x plus 6. so the first thing i want to do is put this in vertex form and grab my vertex and then i want to think about that pattern okay so i would have f of x is equal to again a times the quantity x minus h this is squared and then plus k so what is h what is k we know a is 2 right it's the coefficient of x squared so let's put a 2 in there and again for my h h is equal to negative b over 2a well what is negative b well b is this guy right here include the sign so negative 8. so the negative of negative 8. so make sure you don't make a slime mistake you've got a negative there you got a negative 8 there so the negative of negative eight which would be positive eight over two times a a in this case is two so the negative of negative eight is eight so that's eight over two times two is four so this is two okay so this guy would be two and that would be the x coordinate for our vertex so let me erase this real quick and let me write this over here so the vertex we already know the x coordinate is at 2. and then what about my y coordinate again to find that i just plug a 2 in for x in my function so what is f of 2 right this would give me my k this is what it's going to be 2 squared which is 4 times 2 which is 8 minus 8 times 2 which is 16. so let me just go ahead and write that let me just put minus 16. and then plus 6. so negative 16 plus 6 is going to be negative 10 and if i did 8 minus 10 that's negative 2. so this is negative 2 this is negative two and again i can put plus negative two or i can just put minus two it doesn't really matter now what i wanna do now is just grab my vertex so two comma negative two so my vertex is two let me make that better so 2 comma negative 2. so 2 negative 2 that's going to be right there my axis of symmetry is x equals 2 so you see that line is already drawn and basically what i want to do now is just use my pattern okay so remember it's 1a 3a 5a 7a you know so on and so forth now in this particular case a is not one so you have to be careful because you can kind of think about this as one three five seven for the easy examples when a is one but when a changes you've got to pay attention so if we come back up here remember a is two it's the coefficient for x squared so when we come back down here the pattern is going to change one times two is two three times two is six five times two is ten seven times two is fourteen so what this is telling me is that when i move to the right one unit i've gotta go up two units so to the right one up two okay so that's a point then if i move to the right one unit i've now got to go up six units so to the right one up six that puts me there now again i'm going to run off the graph but if i went to the right one unit i would need to go up 10 units now to get to my next point on the graph now we don't need this because again i'm off my particular graph but if you were using some kind of long graph paper then you can go through and get as many additional points as you want using this method so i've got three points here and i'm just going to reflect across this axis of symmetry to get this point here right because again the distance from here to here this is one two units away so this is going to be one two units away from here to here this is one unit away so this will be one unit away again you just reflect across that axis of symmetry to get those other points and then you can draw a sketch of that graph let's do one more of these i think it's useful to look at an upside down parabola so we have f of x equals negative 2x squared minus 12x minus 16. so again write it in vertex form so f of x is equal to a times the quantity x minus h squared and then plus k all right a is negative 2 here so let's go ahead and write that in so we'll say this is negative 2 and then what's my h what's my k we already know that h is what it's negative b over 2a so what's b b is negative 12. so the negative of negative 12 is 12 over 2 times negative 2 which is negative 4. so this is going to be negative 3 okay so again minus a negative 3 is plus 3. so you can put plus 3 in there if you want but remember when you write the vertex when you say this is the vertex you've got to put it in as negative 3. so negative 3. all right so let's erase this what is k k is equal to what it's f of whatever this guy was again you're not putting in a positive 3 you're using this coordinate here which is negative three so f of negative three these are just common mistakes things that you'll find yourself doing when you first start when you're not really paying attention so you gotta pay close attention to the signs so f of negative three is what well we'd have negative three squared which is nine 9 times negative 2 is negative 18 and then you'd have a negative 12 times a negative 3 which is plus 36 and then you'd have minus 16. okay well 36 minus 16 is 20. and then if you did negative 18 plus 20 that's positive 2. okay so this is going to be 2 here so let's write this down here so it's negative 3 comma 2. so negative 3 comma 2. again that's my vertex and let's go ahead and write this as a 2 right there like that so let's go ahead and plot negative 3 comma 2 so we would have negative 3 comma 2 is right there so that's your vertex then again your axis of symmetry is going to go through that vertex so we would say this is x equals the x coordinate is negative 3 so x equals negative 3 is the axis of symmetry okay so again the step pattern is what you want to have so you have 1a 3a 5a in this particular case you don't need to keep writing these out because you're only going to get two additional points out of this because my graph only goes down to negative 10. okay so we know a if i go back is negative 2. again it's always the coefficient of x squared so 1 times negative 2 is negative 2 3 times negative 2 is negative 6 and 5 times negative 2 is negative 10. okay so what you want to do is say okay if i go to the right one i'm going to go down by 2 because it's a negative 2. so to the right one down 2. now to the right one the next one is negative 6 to the right one down 6 units okay so that puts me there again if you wanted to go to the right one again you would fall by 10 units to get to the next point for myself on this graph i can't do that i would run out of room okay so i can only put these two points in so it's very important to remember with this kind of pattern that again it's 1a 3a 5a 7a so on and so forth make sure you do the multiplication by a otherwise you're going to get the wrong answer if a is not equal to one okay so you've got to pay close attention to that we get lul to sleep by these kind of easier examples where a is equal to one then we get the harder examples and we don't know what to do or we just get the wrong answer now once you have these points over here again you can reflect them across this axis of symmetry again this is one unit away so this will be one unit away this is two units away so this will be two units away and from that you can have your graph in this lesson we want to talk about the remainder theorem all right so when you talk about the remainder theorem it's basically just a shortcut for evaluating a function at a given value so let me start out with this easier example and then i'll explain what's going on so we have something like f of x equals x cubed plus x squared minus 5 x minus 12. and we want to start by asking what is f of 3 so we know what this notation is we've been working with functions forever we know that this is just asking for us to plug in a 3 for each occurrence of x and find out what is the value of the function when that happens so what is f of 3 i'm just going to plug in a 3 here and here and here and so i would have what 3 cubed is 27 and then plus 3 squared is 9 and then minus 5 times 3 is 15 and then minus 12 so this would give us what 27 plus 9 is 36 36 minus 15 is 21 and 21 minus 12 is not okay so this result here we already know how to do this by plugging in but there's another way that you can do this and it comes from this remainder theorem so let me go to a fresh sheet real quick and i want to rewrite my polynomial so we have f of x is equal to we have our x cubed and then we have our plus x squared and then we have our minus 5x and then we have minus 12. okay so with synthetic division that we talked about earlier in the course you can take a polynomial like this and you can divide it by a polynomial of the form x minus k so remember we had something like x minus k as our divisor where the coefficient of x is 1 and it's raised to the first power so it's got to be of this form so this guy right here when we talk about the remainder theorem is what we're going to be plugging in for x in our function in our example we use 3. so for this guy we would be dividing by x minus 3. so let me put this as the divisor so let's just say g of x is equal to x minus 3 as an example so if i do f of x divided by g of x what would that look like well i could do a polynomial long division or again because it's in this format i can do a synthetic division which is quicker i just grab whatever this is okay and i put it over here in the top left and then i want to put kind of a division symbol or you can use whatever symbol you want it doesn't really matter and i'm going to grab just the coefficients from this dividend here so this is a 1 this is a 1 this is a negative 5 and this is a negative 12. now you've got to make sure that these are in order okay in terms of the degree so it's x cubed x squared x to the first power you got to make sure it's lined up that way and if you're missing a power of x you put a zero as a placeholder okay it's very important so i'm going to take my 1 here and write it in and then i've got a 1 i've got a negative 5 and i've got a negative 12. so i want to do the synthetic division here and the idea is that when i do the synthetic division my remainder is going to be 9 because that's what i found when i plugged a 3 in for x in my f x earlier we saw that already so to do the synthetic division i'm going to drop this down and then i'm going to go through again we're going to multiply then add multiply then at so on and so forth so we say 3 times 1 is 3 then we add 1 plus 3 is 4. then we say 3 times 4 is 12 then we add negative 5 plus 12 is 7. then we multiply 3 times 7 is 21 and then we add negative 12 plus 21 is 9. so remember how this answer works this guy right here this leftmost position is the coefficient for x raised to a power that is one degree less than this kind of polynomial that you're working with here as your dividend so this guy has a degree of three right that's the highest power so this guy will have a degree of two so this is basically going to be an answer of x squared plus 4x plus 7 and then your remainder is 9. so typically we'd say plus 9 over the divisor which was x minus 3. okay so let me erase this already we don't need this anymore we already see that we got the same answer here but i want to explain where this comes from so let me kind of drag this up and before we get into this i want to just break this down as simply as i can so i want to use an example first from just whole numbers so when you were in elementary school you did something like this so 7 divided by let's say 3 so we know this result would not be just a whole number right 7 divided by 3 would be 2 with a remainder of 1. okay and we know that we could write this as 2 and 1 3 or we could punch it up on a calculator we get a decimal form but that's not important we know the remainder is 1. so let's write it that way now what we can do is we have this guy right here which is the dividend we have this guy right here which is the divisor we have this guy right here which is the quotient we have this guy which is the remainder so if i take the quotient which is 2 and i multiply it by the divisor which is 3 and i add the remainder to that result i will get back my dividend two times three is six six plus one is seven okay so it's very important you understand that process with whole numbers before we do this with polynomials because again once you understand it in its most simplest form you can apply it to something that's more complicated we're going to do the same thing here so this guy right here is my divisor okay so i have x minus 3 and then i multiply by the quotient which is this part right here okay so the quotient is x squared plus 4x plus 7 okay then i add my remainder to that in this case that's 9. okay so that's my remainder so you have your divisor you have your quotient and you have your remainder so it's the same thing that we looked at with the whole numbers a second ago now you can pause the video and you can see that if you go through and simplify here you will get this back exactly okay but i also want you to notice that if you plugged in a 3 in this case what would you get well if i plugged in a 3 well what happens is because i have the minus 3 there okay 3 minus 3 would be 0 and 0 times this whatever it is x squared plus 4x plus 7 doesn't matter i know i'm plugging in a 3 but forget about that for a second 0 times this would just be 0. so this is gone this is 0. so i'd have 0 plus 9 which is 9 okay again that's exactly what we found from plugging a three in for x here and also through the synthetic division so that's why this works to get a more general view something you'll see in your textbook they'll write this out and say okay we have f of x and this is equal to we'll have x minus k which is your divisor okay times q of x which is your quotient plus r which is your remainder again if i want to find f of k well if i'm plugging in a k there k minus k is zero zero times whatever this is zero so i'm just left with my remainder r so this is why this works okay if you want to find f of some number well then you can just do synthetic division okay so let's look at some more examples and so for this one we have f of x equals negative 5x to the fourth power plus 15x cubed minus 11x squared plus 6x minus 15. we want to find f of 2. and again you could plug a 2 in there and do it that way but this is going to be a bit quicker so i'm going to put 2 here and i'm going to set this up so i'm just going to grab my coefficients so negative 5 15 you've got negative 11 you've got 6 and you've got negative 15. okay so let me put a bar down here i'm going to drop my first number and we're ready to roll so 2 times negative 5 is negative 10 15 plus negative 10 is positive 5. 2 times 5 is 10 negative 11 plus 10 is negative 1. 2 times negative one is negative two six plus negative two is four two times four is eight negative fifteen plus eight is negative seven so you just take this remainder forget about everything else and you say f of two is just negative seven all right let's go ahead and look at another so we have f of x equals negative 3x to the fourth power plus 9x cubed plus 4x plus 3. we want to find f of 3. now this is very important okay people forget this all the time and they get the wrong answer so if you're missing a power of x so i started with the fourth power then i have cubed i don't have squared okay got to write a 0 in as a placeholder so pay attention so i want to put something like 3 and then i'm going to put negative 3 9 i'm going to put 0 here as a place holder because i don't have an x squared then i'm going to put 4 and then i'm going to put 3. so negative 3 9 0 because i'm missing x squared 4 and then 3. okay so make sure you do that otherwise you will not get the right answer all right let's drop this down so we have negative three here three times negative three is negative nine nine plus negative nine is zero three times zero is zero zero plus zero is zero three times zero is zero four plus zero is four 3 times 4 is 12. 3 plus 12 is 15. okay so my remainder here is 15 so f of 3 is 15. so you might also get some examples with complex numbers these are more tedious so typically you see a degree of 2 maybe a degree of 3 if your teacher is you know wanting you to spend some time on it this is no more difficult but you do need to understand how to work with complex numbers we've already seen this earlier in the course so i'm confident that you can get through this so suppose we had f of x equals x squared minus 2x minus 3 we want to find f of 2 plus i okay so again you just set this up the same way so you put 2 plus i over here grab your coefficients so you have a 1 you have a negative 2 and a negative 3. so let me scroll down we need some room for this so let's drop this down here so if i had 2 plus i times 1 i would have 2 plus i so here's where it gets a little tricky so you need to add these two numbers so let's say we have negative 2 plus you have this 2 plus i so you're just going to add the real parts so negative 2 plus 2 is 0 so i'm left with just i okay so this is just going to be a result of i so now what i have is multiplication so 2 plus i times i so i'll just do i on front times 2 plus i so i times 2 is 2i and then plus i times i is i squared now remember by definition i squared is negative 1 so we're just going to write minus 1 here okay and i'm going to flip the order because that's the way you write it so negative 1 and then plus my 2i okay so let me put this up here so negative 1 plus 2i now my last step is to add and what i'm going to have is negative 3 and then we have plus this negative 1 plus 2i again add the real parts negative 3 plus negative 1 is negative 4 and then this plus 2i would come along for the ride so this is my remainder this negative 4 plus 2i so that would be my f of 2 plus i all right so let's talk about a related concept here and to get an idea of what's going on let's just start with something simple like let's say f of x equals let's just do x squared minus not so we already know that when we work with functions if i replace f of x with let's say y and i replace y with 0 what am i doing here i'm finding the x intercept right the x intercept occurs when y is set equal to 0. we already know that so i'm solving for x this is where we would cross through the x axis okay so to solve this is very easy right so i would say 0 is equal to i'm going to factor this into x plus 3 times x minus 3. we know this is the difference of squares we know this is x equals 3 and negative 3. okay we already know that but basically what i want you to understand is that if i go back to this and it's f of x form so let's say we have f of x now if i plugged in a 3 there or a negative 3 for x the result would be zero right so if i said what is f of three well this is zero what is f of negative three well this is also zero so if i plug something in for x and i get a result of zero for the function's value meaning i've found the x-intercept we call this a zero a root a solution there's a lot of different names for but i'm just going to stick to calling it a zero for now so if we plug something in for x in our function and it gives us a zero as a result we call this a 0. we'll talk more about this in the next lesson all right so let's just look at one example of this it's a very easy concept so we have f of x equals x to the fourth power minus 4x cubed minus 4x squared plus 25x minus 12. we're asking is 3 a 0. in other words would the point 3 comma 0 be on the graph of this function would it be an x-intercept one of them so what we would do is just use synthetic division to see if this is true or not so we put a 3 here and then we set this up as we normally do so we'd have a 1 we'd have a negative 4 we'd have another negative 4. we'd have a 25 and then we'd have a negative 12. okay drop this down and then we can get started so i have a 1 here and 3 times 1 is 3. negative 4 plus 3 is going to be negative 1. then 3 times negative 1 is negative 3 negative 4 plus negative 3 is negative 7. 3 times negative 7 is negative 21. 25 plus negative 21 is 4 3 times 4 is 12 negative 12 plus 12 is zero so there's no remainder here so yes we can say that 3 is a zero or a solution or root for this function in this lesson we want to talk about the factor theorem all right so before we get into talking about the factor theorem i want to make sure that you understand the remainder theorem which was what we talked about in the last lesson so if we have some polynomial function let's call it f of x okay and it's divided by some linear factor so let's call this x minus k again the coefficient on x is 1 and the exponent on x is one so that's very important so with some linear factor we call it x minus k when we think about this generically now if i divide f of x by x minus k the result will be some quotient which we call q of x plus some remainder which we call r so we can rewrite our polynomial function f of x in this way so we have this divisor x minus k times our quotient q of x plus our remainder r now if this doesn't make any sense to you we'll look at an example in a minute and also let me just give you a quick example with some whole numbers so that it makes more sense so let's say that you had the number seven and you divided it by the number three so seven is your dividend three is your divisor the quotient is two okay but then you have a remainder of one okay so what we would say here is that i could write seven using this format as what well my divisor here is x minus k my divisor here is 3. okay then times my quotient here is q of x my quotient here is 2. then plus my remainder here is r my remainder here is 1. so it's quite simple when you look at this with whole numbers we have our divisor 3 times our quotient 2 which is 6 plus our remainder 1 which gives us 7. okay so that's all we're really doing it's the same thing we just got to find a way to relate this to a harder concept when we work with these polynomial functions now the reason this is important to write it in this format is that it makes it crystal clear about the remainder theorem which tells us if we evaluate a polynomial function at some number let's just call it k so f of k is how we would notate that the result is r and the reason for this is simple f of k is equal to again if i plug in a k for x you'd have k minus k times q of k plus r well this is zero so you can erase this and put zero here zero times whatever this is is zero you can erase this and zero plus r is just r so f of k equals r that's what this tells us here and that's the remainder theorem okay so to see a quick example let's look at this guy real fast we have f of x equals x cubed plus 2x squared minus 21x plus 18. if we want to find out what is f of 2 well of course you can plug a 2 in everywhere there's x evaluate you've got your answer but you could also say well what is the remainder from dividing this guy by let's say g of x is x minus this number here whatever it is so x minus 2 in this particular case so the remainder from that would be the same as f of 2. okay so the way we're going to do this is with synthetic division so i'm going to put a 2 here whatever this number is that's what you put there and then i'm going to grab these coefficients only remember if you're missing a power of x you got to put a 0 in there as a placeholder so i'm not missing any powers there so i'm good to go i'm just going to grab my coefficients this guy's a 1 this guy's a 2 this guy's a negative 21 and this guy is an 18. okay so you remember how to do this draw a little bar down here go ahead and drop this down i'm going to scroll down get some room i'll come back so drop this down so now we're going to multiply 2 times 1 is 2 then we add 2 plus 2 is 4. then we multiply 2 times 4 is 8 negative 21 plus 8 is negative 13. so then we multiply 2 times negative 13 is negative 26 and lastly we're going to add 18 plus negative 26 is negative 8. so this is your remainder and so that tells me that f of 2 is negative 8. but what we can also do is take this answer here this would be your quotient right and the way you get this is that our polynomial let me just scroll up real quick i'm just going to erase this and keep the answer part let me just bring this up here so it's all on the same screen for us and come back up here i had this guy divided by this guy so we always know what's in the form x minus k so to take these numbers and make sense of them all you do is you take this degree of the polynomial in this case the highest exponent is a 3 this would just be one less so this is the coefficient for in this case to be x squared because 3 minus 1 is 2. okay so 1 would be the coefficient for x squared then you would have plus your 4x and then minus your 13. okay so this would be your quotient so let's erase this real quick we know the remainder is negative 8. but again i could write this let me just kind of scooch this down so we can really draw this point home so f of x could be written as this times this divisor here and i can change the order around it doesn't really matter and then we can put minus 8 here okay so if you pause the video and you go through and multiply this by this and then subtract away a from that you'll get back to this exactly okay now if you think about what is f of 2 again when it's in this format with the divisor times the quotient plus the remainder you see clearly you just get the remainder because if i plug in a 2 there 2 minus 2 is 0 0 times whatever this is a 0 i'm left with just negative 8. so again we say f of 2 is negative 8. so hopefully that drives home what the remainder theorem is and how it works now let's move on and talk about the factor theorem which again if you understand the remainder theorem is very simple so if f of x is equal to x minus k again our divisor which is a linear factor times this q of x which is the quotient plus r and we find that f of k equals zero well then what it means that this times this gives me this right there's no remainder or a remainder of zero okay so that tells me that x minus k is a factor of f of x we also know from the last lesson that if f of k equals zero then k is known as a zero or you could say it's a root or a solution or really they're the x-intercepts if you're talking about k being some real number now let's look at a couple of examples let's say that we have the same f of x but we're going to change our value for k in this case it's going to be 3. so f of x again is x cubed plus 2x squared minus 21x plus 18. now we want to know what is f of 3 if f of 3 is 0 then we can say that x minus this value x minus 3 is a factor okay so let's see what f of 3 is so i'm going to say 3 here and i'll take my 1 my 2 my negative 21 and my 18. let me get some room going just so we're not jammed up on here drop this down and get started three times one is three two plus three is five three times five is fifteen negative twenty one plus fifteen is negative six and then three times negative six is negative eighteen eighteen plus negative eighteen is zero okay so the remainder here is zero so that tells me since f of three is zero that x minus again whatever this number is in this case it's three x minus three is a factor okay so x minus three is a factor of this guy right here okay so let's scroll down and look at a couple more problems here so we have f of x equals we have x cubed plus 4 x squared minus 22 x plus 20. so we have g of x equals x minus 2 and we just want to know is this a factor okay so again you just take this number right here as long as it's in the format of x minus some number if it's not in this format you've got to change it to where it's x minus some number we'll see an example of that in a moment but basically i can again just take 2 put it out here and then grab my coefficients so 1 4 negative 22 and 20 and i'm just going to see is the remainder 0 if it is then x minus 2 is a factor if it's not then x minus 2 is not a factor so let's go ahead and drop this down 2 times 1 is 2 4 plus 2 is 6. 2 times 6 is 12. negative 22 plus 12 is negative 10. 2 times negative 10 is negative 20. 20 plus negative 20 is 0. so there's no remainder here so i can say if i kind of go back up here that x minus 2 is in fact a factor right because we can say that f of 2 is equal to 0. all right let's take a look at a longer one so we have f of x equals x to the fifth power plus 2x to the fourth power minus 6x cubed minus 12x squared minus 16x minus 32 is g of x which is equal to x plus 3 is that a fact again you want this in the form of x minus k to make this work you have a plus there if you see that you've got to change it okay it's a very common mistake to just put three out there and then to get the wrong answer okay so you want to make sure you change this and just say this is equal to x minus a negative three so that's in that format and then you grab whatever this number is minus whatever this number is that's what you're going to say so is f of negative 3 equal to zero so that's a question mark again we're just going to use some synthetic division so i'm going to say that i have negative 3 and grab my coefficients i've got a 1 a 2 a negative 6 a negative 12 a negative 16 and a negative 32. and a lot of these problems are going to get quite tedious so you just have to be prepared for that okay so let's scroll down get some room going let's drop this to here negative 3 times 1 is negative 3. 2 plus negative 3 is negative 1. okay so negative 3 times negative 1 is going to be 3. negative 6 plus 3 is negative 3 negative 3 times negative 3 is 9. negative 12 plus 9 is negative 3 and then negative 3 times negative 3 is 9 negative 16 plus 9 is negative 7 and negative 3 times negative 7 is 21 negative 32 plus 21 is going to give me negative 11. so this is not zero okay this is not zero so f of negative three is negative 11 and so i can say that this g of x which is x plus 3 or i could say x minus a negative 3 is not a factor of f of x okay let's look at the same f of x but we're going to change kind of our divisor we're going to call it h of x now and now it's x plus 2. so same thing in this case it's plus so change it to x minus a negative 2. very important you do that again it's a common mistake i can't say that enough so you want to take your negative 2. and again just see if the remainder from this guy is 0. so we would have a 1 a 2 a negative 6 a negative 12 a negative 16 and a negative 32 okay so that's everything so let's go ahead and scroll down a little bit okay let's go ahead and drop this and get started so negative two times one is negative two two plus negative two is zero negative two times zero is zero negative six plus zero is negative six negative two times negative six is twelve negative twelve plus twelve is zero negative three times zero is zero negative 16 plus zero is negative 16. negative two times negative 16 is going to be 32 negative 32 plus 32 is zero so that's your remainder it's zero right i mean you don't have one so that tells me that x plus 2 is a factor right so you can say h of x is a factor of f of x all right so let's look at another common problem that occurs in this section so basically what's going to happen here is they're going to give you one factor and they expect you to find the other ones okay so we have f of x equals x cubed minus 7x squared plus 14x minus 8 and our factors x minus 2. so what that means is that if i plugged in a 2 for x so if i had f of 2 this is 0 right they're telling us that okay so what i can say is that if i plug in a 2 there and i do my synthetic division so i have a 1 a negative 7 a 14 and a negative 8. i'll do this i'll get a remainder of 0 and i'll take that quotient and then i'll try to factor that further okay so let's go ahead and crank this out so i'll drop this down 2 times 1 is 2 negative 7 plus 2 is negative 5. 2 times negative 5 is negative 10. 14 plus negative 10 is 4. 2 times 4 is 8 negative 8 plus 8 is 0. okay so i don't need this part and again when we think about this if i go back up this guy has a degree of 3. so the answer will have a degree of two so if i erase this i'll say that my answer here is just x squared and then minus five x and then plus four okay so that's how this would go so this times this would give me this right i could say f of x is equal to this right we already know that now let's think about factoring this guy so this is a quadratic very easy to factor so we have x here and x here and we need two integers that sum to negative five and have a product of positive four so we can get negative four and negative one right because negative four times negative one is four and then negative four plus negative 1 is negative 5. so negative 4 and negative 1. and let me write that in a more compact manner so i'm just going to say f of x here is equal to x minus 4 times x minus 1 times x minus 2. in this lesson we want to talk about the rational zeros theorem so when we work with polynomial functions we're often asked to find the zeros so let's say you came across f of x equals 3x cubed minus x squared minus three x plus one and let's say your teacher asks you to find the zeros so we already know that a zero or solution or root is something that when we plug in for x in our function f of x we get a result of 0. so how could we find the zeros here so there's going to be a lot of different methods you're going to come across the one we're going to focus on today is known as the rational zeros theorem or some people call this the rational roots test now let me explain what this is the rational zeros theorem is just going to give us the possible candidates okay so it's not necessarily true that they're gonna work for rational zeros of a polynomial function with integer coefficients so let me be clear here the polynomial function you're working with has to have integer coefficients and this method is only going to give you possible rational numbers to check okay so it's not necessarily true that they're going to work now let's think about what a rational number is so that we're all on the same page in pre-algebra we define these as p over q where we say q is not equal to zero and p and q are both integers so something like two thirds right something like the number five because we could say five over one so on and so forth so if i could put one integer over another integer to form a number that's a rational number so something that's not a rational number would be like the square root of 2 right in decimal form it does not terminate it does not repeat the same pattern so it's not a rational number also if you think about numbers that are involved with the imaginary unit i we think of those as a complex number so those would not be involved here as well so when you generate this list you're only looking at rational numbers so i want to be clear there now what you want to do to kind of get things going you want to make sure that your polynomial let me just erase this is in standard form so f of x equals we have 3x cubed minus x squared minus 3x plus 1. so this is in standard form because the highest exponential power is all the way to the left okay so that's how you want it then the next highest then the next size right so on and so forth so there's two terms you're going to hear the first thing is the leading coefficient and the second thing is the constant term so the leading coefficient is the coefficient on the variable that is raised to the highest power again if it's in standard form this is all the way to the left so in this case that's going to be a 3 because the highest power here is a 3. so that's my leading coefficient so it's 3. then you're also going to talk about the constant term so this is your term if it's in standard form all the way to the end and it doesn't have a variable involved so this is your constant term now the way this is going to set up if this number p over q is some rational number and let's say it's written in lowest terms so there's no common factor other than one between p and q then if it's going to be a zero of your polynomial function let's call it f of x then p is a factor of our constant term okay so p would be a factor of one so in this case one only has positive one and negative one as factors let me just say plus or minus 1 there and then for q it's going to be a factor of our leading coefficient so in this case it's 3 so 3 is a prime number so that would be plus or minus 1 and then plus or minus 3. so you've got to include all these kind of different signs so you do plus or minus 1 for the p and then for q you have plus or minus 1 and then plus or minus 3. so the way you want to do this is you want to go through the different possibilities that will occur so how i would set it up and you don't have to do it this way but i just take whatever's in the numerator okay so i would start with one a lot of times you have more than just one number here but i would just take one and then i would go over each denominator so the way i would do this is i would just take my plus or minus and i would say that i have one over one okay so that's just plus or minus one then i would take my one again make sure to put the plus and minus out in front and i would go over the next guy so it would be plus or minus one third okay so that's how you can do this really quickly if you're wondering about the signs because a lot of people get confused with these signs you're only going to have two possibilities here okay even though you really have four numbers between here and here right because you're thinking about okay i have 1 over 1 and then i have negative 1 over 1 and then i have 1 over negative 1 and then i have negative 1 over negative 1. but think about how fractions work if it's a positive over a positive or a negative over a negative like you have here and here well that's a positive so that gives me the positive 1. and then if it's a negative over a positive or positive or negative that gives me the negative right so that's why you can just shortcut this you don't have to go through everything you can just put plus or minus okay plus or minus and then just take your number part there over the number part here and you're basically going to be good to go okay so don't even worry about the signs so we've generated four possible rational solutions you've got one and negative one you've got one third and negative one third so if we wanted to at this point say what works you can use synthetic division or you can straight up just plug things in which in this case is probably faster and you can see if the result is zero okay so i can say is f of one equal to zero if it is then one is a zero is f of negative one equal to zero is f of one third equal to zero is f of negative one third equal to zero you know so on and so forth now i'm just going to pause here for a minute as we go through these examples i'm not going to do these each time because in some cases it's going to be ridiculous okay you need other tools to make this kind of method work for you we'll get to those tools in future lessons so for right now we just want to focus on understanding the rational roots theorem so very quickly let's test these and see which ones work so what i can do is say what is f of 1 well if i plugged in a 1 here here and here well 1 to any power is just 1 and if i multiply 1 by a number i just get the number so really i can just kind of mark these out and say i'd have 3 minus this would be a 1 right so 3 minus 1 is 2 and then 2 minus 3 is negative 1 and negative 1 plus 1 is 0. okay so that's 0. so we know that 1 is a 0 of this polynomial function what about negative 1 well let's just go through and say this is 3 times negative 1 cubed and we have minus you'd have negative 1 squared minus 3 times negative 1 and then plus 1. we can simplify this really quickly we know this is negative 1 so i'll just put a negative out in front this right here would be positive 1 but you have a negative out in front and then this would be positive three okay so let's go through we have negative three plus three which is zero negative one plus one which is zero so this is zero as well so f of negative one is zero let's check one third so again we would just plug in a 1 3 everywhere there's an x so you'd have 3 times you'd have 1 3 cubed which is 1 over 27. if you simplify this 3 and 27 have a common factor of 3. so this is 1 one-ninth then minus if you had one-third squared that would be one-ninth and you could just stop right now and say one-ninth minus one-ninth is zero so just erase that so basically you have negative three times one-third plus 1 we know this would cancel and give me a 1. so this is negative 1 plus 1 which is 0. so this works as well now as we move through this section you're going to find that for this polynomial because this highest exponent is a three it can only have at most three distinct zeros okay so we've already met that so we could stop we know that this guy wouldn't work but we haven't gone to that yet we're going to get to that in a future lesson so let's just check this real quick we'll see that it doesn't work so we would have 3 times negative 1 3 cubed would be negative 1 over 27 again we can cancel the 3 and the 27 so let's just say this is a 9 down here and then you'd have minus if you had one-third and it's negative and it's squared well the negative goes away one-third squared is one-ninth so you'd have minus one-ninth there okay and we can go ahead and do this and say this is going to be overall negative one minus one which is negative 2 so this is negative 2 9. and then you have your minus 3 times negative 1 3. so there you can think about the signs cancelling because you're going to have a negative 3 times a negative 1 3. so let's put plus and 1 3 times 3 is 1 and then you have plus 1. so to get a common denominator i'll say this is 2 and 2 times 9 over 9 would be 18 over 9. and so negative 2 plus 18 is 16 so this is 16 9. okay so we see this is not zero so we know that this is not a solution right this negative one third so not a solution not a zero not a root so even with this very simple example you can see how tedious it was to go through and find the zeros now other things you can do once you have one zero again you can use this with synthetic division to find the factors of this guy so you could factor this guy and then once you have it in the state where you have x minus 1 would be one factor and then you'd have another factor well this guy right here this other factor would be a quadratic you could use a quadratic form if you wanted to so there's all kind of techniques that we can use and again we'll talk about those as we progress through this section let's look at two more examples i'm not going to see if the roots work here i'm just going to go through and list the possible rational zeros so i have f of x equals 5x cubed plus 54x squared plus 24x minus 7. again i want to find all the p over q those possible rational zeros okay so p comes from the factors of your constant term so in this case it's negative seven don't worry about the sign because you're always going to do plus or minus anyway just think about what are the factors of seven well it could be one it could be seven it could be negative one it could be negative seven so really you just put plus or minus one and then plus or minus seven okay so that's what you're going to put for that then for q again that comes from the leading coefficient so the leading coefficient again if this is written in standard form it's all the way to the left the highest power in x is a three so the leading coefficient is a 5. so in this case for 5 you have plus or minus 1 and you have plus or minus 5. so i'm giving you another easy example but you're still going to see that there's a lot of possibilities here if you want to go through and check stuff so i'm going to take my plus or minus and put that first and then again i just take this number and go over this number so plus or minus 1 over 1 is 1. then i'm going to go plus or minus i'm going to take this number and i'm going to go over this number so plus or minus one over five which is plus or minus one fifth now i'm gonna move down and i'm gonna take this number and go over this number so let me write this down here so i have enough room so plus or minus don't forget that you have your seven over one which is just seven and then lastly plus or minus i'm going to shift this down and say you have your seven over five so plus or minus seven fifths okay so this gives us what we have one two three four but then times two because you have a plus or minus in each case so we have eight possible rational zeros now i'm not going to test these it's just way too much work but if you're curious one-fifth is the only rational zero that we have here all right i want to wrap up the lesson and just look at one more of these again we're just trying to get the concept down we're not really trying to get zeros at this point we'll do that as we progress so we have f of x equals 3x cubed plus 7x squared plus 15x minus 14. so what are the possible rational zeros so again just think about p over q and again when you think about p it's coming from the factors of this constant term here so just think about 14 again forget about the fact that it's negative 14 is what it's 1 times 14 or 7 times 2. so you would have plus or minus 1 you would have plus or minus 2 you'd have plus or minus 7 and then plus or minus 14. now let's think about q that comes from again factors of your leading coefficient so in this case that's going to be 3. 3 is a prime number so you're just going to have plus or minus 1 and plus or minus 3. so again i've given you easy examples and you're going to see that even with these easy examples the list becomes really really big so if you get something that's harder you're going to have a really big list and a lot of different possibilities to check so on its own this rational roots test or rational zeros theorem is not very efficient okay so you need other tools so let's go through and figure out our possibilities so again i would go ahead and say that i have plus or minus i'm going to take this number and go over these two so one over one would be one and then plus or minus one over three would be one third and then i'm just gonna move down so i'm gonna say that i have plus or minus i'm going to have 2 over 1 which is 2 and then plus or minus 2 over 3 which is 2 3. and then let me move down so i'm going to have plus or minus i'm going to have 7 over 1 which is 7. let me come down here so let me put my plus or minus and then i'll have seven over three which is seven thirds and then lastly i'm going to come down here and i'm going to have 14 over one so plus or minus 14. and then i'll have plus or minus my 14 over 3 or 14 thirds so again an easy example and you see how quickly it gets out of control you have 1 2 3 4 5 6 7 8 times 2 so 16 possible rational zeros and if you went through and checked all these you'll find that two-thirds is the only one that works and again at this point we don't have enough information to effectively use this technique in this lesson we want to talk about the fundamental theorem of algebra all right so as we continue to talk about finding zeros of polynomial functions we're going to come across the fundamental theorem of algebra which basically tells us that every polynomial function of degree n will have n complex roots but we're going to see some of these may be repeats so let me cover a few things real fast before we continue i want to make sure you understand what i just said so let's look at something like this so f of x equals let's say x cubed minus x squared plus let's say 2x minus 1. we should all know that this polynomial function is of degree three right we think about the highest exponential power on our variable in this case that's going to be a three so the degree of the polynomial is a three so that's what we're talking about when we say that now with our fundamental theorem of algebra this tells us that we would expect to have three complex zeros but again some of these might be repeated and i'll show you an example of this in a moment for right now i just want to focus on the words complex zeros what does that mean remember when we work with numbers specifically when we say complex numbers they're of the form a plus bi so if i had the number five most of you would not expect for that to be a complex number because it's just the number five remember a number like five can be a part of more than one set of numbers five is a natural number it's a whole number it's an integer it's a rational number i could say it's 5 over 1. it's a real number and it's also a complex number because i can say this is 5 plus 0 times i okay so all the real numbers that you work with are also complex numbers so when you start reading things about complex solutions you're including things that you think of as real numbers in that okay so let's erase that now that we have that point is clear now when we talk about this fundamental theorem of algebra basically what they're saying and this is the set up here you go back to the factor theorem you say if x minus k is a factor of f of x we know f of k is 0 right so there's no remainder this means if i divide f of x my polynomial function by this x minus k i get some q of x some quotient okay and that's all there is there's no plus r over here there's no remainder so these two multiplied together give me this now what this guy tells us is that let's say i have f of x and this polynomial function is of degree one or more so you have an x involved or whatever the variable is it doesn't matter and the exponent on it is a one or higher okay so it's not something like f of x just equals the number five that wouldn't work so if this guy is involved again one or higher for the degree then you'll always be able to find at least one complex zero so where you can say this is x minus i'm just going to say this is k sub 1 this is going to be my first complex zero times q sub 1 of x okay so you can write it like this then if this guy right here you just keep this process going if this guy right here is of degree one or higher then again you can always find at least one complex zero so you just keep the pattern going okay so i can just say that this would be times x minus k sub 2. and again i can just keep this going so i can end up with just x minus k sub n in the end you're going to have a quotient that's just going to be a number now that number might be one right so it might be true that these guys just multiply together to give your polynomial function or there might be some number here so we typically say this is a and this would be put out in front so let me put this out in front here and i'll change my color okay so that's how you end up with this form that you see in your book so i want you to remember this for later because we're going to use it when we write some polynomial functions i want to look at two examples to explain what we mean when we say that some of the complex zeros might be repeats let's start out with something really easy so we have f of x equals x squared plus 3x minus 10. and again we already know how to find the zeros for something like this we've been doing it forever so let's say x squared plus three x minus ten let's set this equal to zero if we factor this to solve it put an x here and an x here and we know we want two integers that sum to 3 and give me a product of negative 10. well that would be what it would be positive 5 and it would be negative 2. 5 times negative 2 is negative 10 5 plus negative 2 is positive 3. okay so the solutions here or the zeros would be x is equal to negative five and positive two so f of negative five would be zero and f of two would be zero now this is exactly what we expect because this polynomial function is of degree two right that's for every quadratic function now let's look at this case here which you might remember will only have one solution but we're going to say that it has a multiplicity of two because it's going to occur twice so we have our alpha of x equals x squared minus 8x plus 16. again i'm going to solve this so x squared minus 8x plus 16 equals 0. so you can factor this into a binomial squared but what i'm going to do just for demonstration purposes i'm going to say this is x and this is x minus and minus so two integers that sum to negative 8 and give me a product of positive 16 well it's going to be negative 4 and negative 4 right negative 4 plus negative 4 is negative 8 negative 4 times negative 4 is positive 16. okay so the solution here is obviously x would be 4 right so 4 is your only zero f of 4 would be 0 but it occurs twice right because you have two factors of it so that's what the fundamental theorem of algebra is telling us it's saying that you're going to have two zeros here but one might be repeated alright so there's not much to this section you just need to understand what the fundamental theorem of algebra is telling you additionally you might be asked to write some polynomial functions based on the information that you're given so what we're given here is three zeros so we have 5 4 and negative 4 and we have a point here we have f of 1 equals 120 or in other words you have 1 comma 120 that's on the graph of this function now how could we write this polynomial function well again if i go back and think about this form here i would take this guy and i would just start substituting information in so let me go back down so what do we know we know that there's a zero of five there's a 0 of 4 and there's a 0 of negative 4. now with these problems they're typically going to give you some instructions that say that they want a specific degree in this case we'd be looking for a degree of 3 because it gave us 3 zeros sometimes it'll tell you give them the minimum degree required to write the polynomial okay so you might see it either way but here we're looking for a third degree okay so i know that 5 is a 0 4 is a 0 and negative 4 is a 0. so 5 is a 0 meaning f of 5 equals 0 then that means x minus whatever this is so x minus 5 is my factor or my divisor okay then i do the same thing here i'm going to have x minus 4 and then i'm going to have x minus a negative 4 which is x plus 4. so i have f of x is equal to a times this so this is what i've got to figure out and then i can go through and simplify so how do i get this well i'm going to use this point here so basically i know if i plugged in a 1 everywhere there's an x the result would be 120 okay so let's scroll down 4 minutes and i'm just going to plug this in so f of 1 is going to be equal to 120 so let's just go ahead and say we have a times 1 minus 5 is negative 4 and then 1 minus 4 is negative 3 and then 1 plus 4 is 5 and this should equal 120. so negative 4 times negative 3 is 12 12 times 5 is 60 so you would have 60 a equals 120 to solve for a i would divide both sides by 60 and i would find that a is equal to 2 okay so let's erase all this we will have everything so real simple process just tedious overall so now what you want to do is go through and just simplify so if i want to kind of leave multiplying by two till the end that's fine or if you do it at first it doesn't matter just gonna multiply everything together and leave two off till the end so x times x is x squared and then my outer would be minus four x my inner would be minus 5x so together would be minus 9x and my last would be plus 20. okay so you'd have that times this and let me come down here get some room going so i would have x times x squared which is x cubed i'd have x times negative 9x which is minus 9x squared x times 20 which is plus 20x and then i have 4 times everything so you'd have plus 4x squared minus 36x and then you would have plus 80. okay so what does this give me i'll put my 2 out in front i've got x cubed nothing to combine with that negative 9x squared and 4x squared would give me minus 5x squared and then 20x minus 36x would give me minus 16x and then lastly i have plus e okay so now i'm just going to multiply everything by 2 and let's scroll down and get some more room going so i'll have 2 x cubed minus 10 x squared minus 32 x plus 160 okay so you can just box that off this is going to be your polynomial function again you can just put half of x we have it all the way up there but almost scroll all the way back up so f of x equals again there's 2x cubed minus 10x squared minus 32x plus 160. okay let's look at one more of these another easy topic that you come across so the zeros you have one but it says it's got a multiplicity of two and then you also have negative three so again this is going to be a third degree polynomial because again we're dealing with two zeros here you have the one that occurs twice and then you have another one which is negative three so you count that as three zeros so you'd have a degree of three then we're told that f of negative one is 24 okay so let's write this part out first so you'd have f of x equals you'd have your a times you'd have x minus 1 and this occurs twice so you can put squared like this or you can write it out and say x minus 1. and then you have x again if you have a minus 3 it's going to be x plus 3. okay so we've got that part done and now i just need to figure out what a is so again if i plugged in a negative 1 for x so if i had negative 1 minus 1 that would be negative 2. if i had negative 1 minus 1 that would be negative 2 again if i had negative 1 plus 3 that would be 2. so i know this is positive 8 so i can go ahead and just say that i would have 8 a is equal to the result would be 24 divide both sides by 8 and i get that a is 3. so this is 3. okay so we got that done pretty quickly and basically what we want to do is just simplify again so i'm going to say that x minus 1 times x minus 1 that's going to be what it's x squared minus 2 times this guy times this guy so 2 times x times 1 is 2x and then plus we'll just go ahead and say that's negative 1 times negative 1 is 1. okay so we've got that let me put my 3 out in front so i don't forget it and this is going to be multiplied by this okay so let's go down here and let's go ahead and continue so we have three times so x times x squared is x cubed then x times negative two x is minus two x squared then x times one is plus x and you have 3 times everything so i'm just going to go plus 3x squared minus 6x and then plus 3. okay so let me just combine like terms inside first so we have 3 times x cubed nothing to do with that then we have negative 2x squared plus 3x squared that's going to give me plus x squared we have x minus 6x which is minus 5x and then we have plus 3. okay so almost out of the woods just one more step we have 3 times all of this so we would have 3x cubed plus 3x squared minus 15x plus 9. and again the f of x part is all the way up there so let me just kind of scroll down and i'll rewrite it so we'll just say that f of x is equal to three x cubed plus three x squared minus fifteen x plus nine in this lesson we wanna talk about the conjugate zeros theorem all right so as we continue to talk about finding the zeros of a polynomial function we're going to come across this section in our book that talks about the conjugate zeros theorem so the conjugate zeros theorem which is also known as the conjugate pairs theorem just tells us that if we have a polynomial function let's just say it's f of x that has only real coefficients then if some complex number let's say a plus bi is a zero then a minus bi its conjugate is also going to be a zero so in other words complex zeros always occur in pairs okay so if i had something like let's say 2 plus 3i and that was a 0 well then it would also be true that 2 minus 3i would be a 0. okay remember for conjugates this part stays the same okay so you have 2 and 2 and then this part stays the same so you have 3i and 3i it's just that your signs are going to be different okay so you have a plus and then a minus and we already saw this when we worked with quadratic equations right we would solve them and we would find what that you either had two real solutions you had two complex solutions meaning they were imaginary solutions with the imaginary unit i involved or you would get one solution but we already learned that we could say it has a multiplicity 2 right it's the same solution that occurs twice so those were only three possibilities so what we're going to do is use this information we have two different tasks that we're going to complete in this section the first one we already talked about this in the last lesson basically what we're going to do is we're going to find a polynomial of least degree having only real coefficients and 0s that meet this criteria so the zeros given are 1 and 1 plus 2i so we know that this guy right here this 1 that's just the regular zero we've been working with that forever but when we get this complex 0 1 plus 2i well we know that it's going to have its conjugate as a 0 as well so we can go ahead and write in here that another 0 would be 1 minus 2i again the 1 stays the same the 2i stays the same just change the sign to get the conjugate that's all you're doing so now if i want to write out my polynomial function remember the form it's f of x is equal to it's a times you have your x minus k sub 1 times your x minus k sub 2 and then you know you keep going and then you have your x minus k sub n okay so in each case it's x minus whatever the 0 is so in this one i'd say x minus 1 and then for these it's going to be a little bit more complicated because i'm subtracting away a complex number so let me just kind of write this off here and i'll say that this is x minus i'll put it in parentheses 1 plus 2i and then let me close that and i'll do x minus now i'm going to do 1. let me slide this down and run out of room and now we'll have minus 2i go ahead and close that off okay now in this particular case you're not given any information i know in the last section we were given a point and then we could figure out what a is so here a is just going to be 1 so you can just erase it or put a 1 there and what happens is we're just going to go through and simplify this it's going to take a little bit of time because it is quite tedious but we'll have our polynomial function and it's a good exercise to kind of practice working with these complex numbers okay so let's go through and start simplifying so i'll say f of x is equal to i'm going to leave this for last so i'm going to have x minus 1 outside of and i'll just put some brackets here and the way i'm going to do this is i'm going to do foil with these guys as they are okay so i'm going to treat this as the first term and this as the second term in each polynomial so this is your first term and this is going to be your second term okay so i'm going to say that i have what first terms x times x is x squared and then my outer terms i would have x times negative this guy the 1 minus 2i in parentheses so remember the minus and you have x times this quantity 1 minus 2i okay and then continuing and i'm probably not going to be able to fit all this but i'll try and see if we can squeeze it down so let me erase this and put minus x times again 1 minus 2i let's kind of scrunch that down a little bit so when i go to my inside terms i'm going to have this negative times this 1 plus 2y that quantity times this x so i'll have minus x times again this quantity 1 plus 2i and then for the last terms i have this guy times this guy now it's negative times negative so i know it's going to be a positive okay and then what i want to do is say that i have what 1 plus 2y times 1 minus 2i well i know my formula for that remember if you have a plus b times a minus b this is a squared minus b squared okay so in this particular case i'm just going to write this down here because i'm just going to run out of room i'm going to say that i have what the first guy squared so 1 squared which is 1 minus the second guy squared so if i take 2 and squared i get 4. if i take i and square it remember i squared by definition is negative 1. so this would be negative 1. so really you have negative 4 times negative 1 which is positive 4 so 1 plus 4 is just going to give me 5. so i can write this as 5 here and that allows me to fit everything on my screen okay so now we're going to simplify let me scroll down and get some room going and see where we are so we have f of x equals i've still got this x minus 1 out in front and inside i'm just going to see what i can do to simplify so let me put my brackets here so i have my x squared and then i have my negative x times 1 which just minus x and then negative x times negative 2 i would be plus we'll do 2 i x then moving down here i have negative x times 1 which is minus x and then negative x times 2i is going to be negative 2ix and then lastly i have my plus 5. okay so what we see now is that this will cancel with this and negative x minus another x would be negative 2x so let's go ahead and write that we'd have f of x equals the quantity x minus 1 times i'll have x squared again negative x minus x is minus 2x and then we're going to have plus 5. so starting to look pretty good so let's scroll down a little bit i know this is tedious but you have to do some examples of these so we'll have f of x is equal to i'll go through and multiply x by everything so x times x squared is x cubed and then minus x times 2x is 2x squared and then plus we have 5 times x which is 5x and then i have a negative 1 times everything so basically i'm just going to change the sign of everything here so i would have minus x squared and then i would have plus 2x and then i would have minus 5. okay so let's continue now all we have to do is just combine like terms so x cubed nothing to combine with that i have let me get some room going here i have negative 2x squared and minus x squared so that would be negative 3x squared i have 5x and 2x so that would be plus 7x and then i have minus 5. so we end up with our polynomial function f of x equals x cubed minus 3x squared plus 7x minus 5. all right the next type of problem you're going to see is also pretty tedious so suppose you're given some polynomial function let's say f of x equals x to the fourth power minus three x cubed plus six x squared plus two x minus sixty and they say find all the zeros and they give you a complex zero so let's think about what we know at this point at this point we know that the degree of the polynomial is four so i know that i'm going to have four solutions right at most it would be four distinct solutions but i know that i'm going to have four solutions even if some of them are repeats okay the other thing we know is that we're given a complex 0 of 1 plus 3i so there also has to be a 0 of 1 minus 3i because these come in conjugate pairs okay so i know if i have two zeros here i can get this guy down to a quadratic and at that point i can use the quadratic formula or if i'm lucky i can factor or complete the square or whatever you want to use it doesn't matter we know that once we get it down to a quadratic we're good to go so there's two ways that you could attack this problem the first thing is you could write this out as x minus this 1 plus 3i okay that's your factor let me make that clear and then x minus you have your 1 minus 3i that's another factor so you could multiply these two together get that polynomial okay and then you could divide synthetically this guy by whatever that is and that would give you two different quadratics that you could solve with the quadratic formula so that's one way you could do it another way you can do it and the way that i'm going to do it probably would take a little bit longer you can use synthetic division okay with this guy and with this guy to get it down to a quadratic so that's the way i'm going to do it just to get a little practice of working with these guys and doing some synthetic division so i'm going to start with this one right here okay and what i'm going to do is go 1 plus 3i and then over here i'm just going to take my coefficients so 1 negative 3 we've got 6 we've got 2 and we've got negative 60. and i made a little bit of a mistake here because when you work with this you want to make sure you have a lot of rum so let me actually redo this so let me come back up here and say that we have negative 3. i'll push that over here i'll push my 6 over here my 2 over here and my negative 60 all the way over here you need to run because these complex numbers they take up a lot of space so let me kind of move this down and then we'll be good to go okay so let me put a bar down here drop this down and we'll start so we'll come back up in a minute we'll go back and forth so basically i drop this down 1 plus 3i times 1 is 1 plus 3i okay if i had negative 3 plus 1 plus 3i i just add the negative 3 and the positive 1 which gives me negative 2 and the plus 3i comes along for the ride okay now where this gets tedious is that you have to multiply now you multiply two complex numbers you basically have to stop and do four okay so we have this 1 plus 3i times this negative 2 plus 3i okay so 1 times negative 2 is negative 2. the outer would be 1 times 3i which is plus 3i the inside would be 3i times negative 2 which is minus 6i and then the last 3i times 3i is 9i squared now remember i squared is negative 1 so you can really just say this is negative not okay so you have negative 2 minus 9 which is negative 11 and then on top of that you're going to have 3i minus 6i which is minus 3i okay so let's erase this we don't need that scratch work anymore and so if i said 6 plus negative 11 minus 3i what is that again just work with the real parts 6 minus 11 is negative 5. so this guy's going to be negative 5 minus 3i okay so here comes again the tedious part we've got to multiply so we have 1 plus 3i times you have negative 5 minus 3i okay and then we're going to do 1 times negative 5 that's negative 5. we're going to do 1 times negative 3i that's minus 3i we're going to do 3i times negative 5 that's minus 15i if you want to combine these at this point it would be negative 18i let's just go ahead and do that and then lastly you have 3i times negative 3i so that's negative 9i squared but again i squared is negative 1. so you're just going to change the sign of this to plus so positive 9 plus negative 5 or you can think about this as 9 minus 5 that's just going to be 4. so this would be 4 minus 18 i okay so now we want to add 2 plus 4 minus 18i well 2 plus 4 is 6 that's all you're going to be able to do there and then you bring the minus 18i along for the ride all right last multiplication again i know this is tedious but we have 1 plus 3i times you've got your 6 minus 18i so 1 times 6 is 6 and then i'm going to do my 1 times negative 18i which is minus 18i and then my 3i times 6 which is plus 18i okay then we have 3i times negative 18i so we know that's a minus 3 times 18 is 54 and then you'd have i times i which is i squared i squared is negative 1 so just change the sign so it's just plus 54. now 54 plus 6 is 60 okay we know that negative 18i and positive 18i those are going to cancel so we're just left with 60. we already knew this was a zero so we knew we wouldn't have a remainder by definition so we know negative 60 plus 60 is zero and that is what we would expect right to have no remainder or a remainder of zero now once we're done here we already know that one plus three is a zero we don't need it and we know that one minus three i is a zero so we're going to use this information to further get this down so remember if you divided if we go back up here if you divided this synthetically by x minus this number you would get this guy right here these are your coefficients for the quotient okay so we're just going to use that i'm going to erase this i'm just going to take these guys okay so i'm going to take my 1 i'm going to take my negative 2 plus 3i i'm going to take my negative 5 let me put that as a minus so minus 3i and then i'm going to take my 6 minus 18i so what we're going to do is now take the other 0 the 1 minus 3i because we know that's a 0. and basically when i take this quotient and divide it by this i'm going to have a quadratic that i can then solve again with the quadratic formula we're factoring or something like that and i'll have all my zeros okay so let's bring this down and say we have a 1 so again 1 minus 3i times 1 would be 1 minus 3i okay so when we add these you've got negative 2 plus 1 which is negative 1 and then positive 3i minus 3i is 0. okay so it's just negative 1. then 1 minus 3i times negative 1 just change the sign of everything so this would be negative 1 and then plus 3i okay we add negative 5 plus negative 1 or negative 5 minus 1 is negative 6 negative 3i plus 3i that's 0. okay so now lastly we have 1 minus 3i times negative 6. again i can just multiply each term by negative six so this would be negative six and this would be plus 18i so again we knew this was a zero so this comes out to be zero exactly okay and these would be the coefficients for your polynomial function okay that's your quotient now this is going to be of degree two now we started with a fourth degree we did one synthetic division so it was a third degree now we did another one so it's a second degree so this guy is going to be the coefficient for x squared then this is x to the first power then this is your constant term okay so this guy right here is what we're looking for so i'll drag this up here and we'll come here and we'll just say okay where does this guy equal zero okay and we can solve this with factoring put an x here and an x here give me two integers that sum to negative one and give me a product of negative six well you could do negative three and positive two you could do negative 3 and positive 2. so obviously if you solve this you would get x is equal to 3 or negative 2. so i can just add this to my list here and say that another 0 would be 3 and another would be negative 2. i know i said zero here this should be zeros but when i started it was singular we just had one so let me just mark this out and say these are the zeros okay plural so we have our 1 plus 3i which was given to us or 1 minus 3i which is the conjugate and then we have 3 and negative 2 which we found in this lesson we want to talk about descartes rule of signs so as we continue to talk about finding the zeros for a polynomial function we're going to come across the topic of descartes rule of signs so basically what this is going to be is just another tool for us it's going to help us to narrow things down a bit so it's going to tell us the possible number of positive and also negative real zeros for our polynomial function given that the polynomial function we're working with meets certain criteria so basically you want to make sure that the function you're working with has real coefficients and also has a non-zero constant term so if you have that met then basically you want to just start out by writing your polynomial function in standard form so for this example we have f of x equals we have 2x to the fifth power plus 3x to the fourth power plus 7x cubed plus 5x squared plus x plus 15. so this is already in standard form our term with the highest exponential power is all the way to the left then you have the next highest exponential power you know so on and so forth all right so once you've done all that you're ready to calculate the number of possible positive real solutions okay so the way you do this is you're going to count the sign changes so i'm going to put a plus there for emphasis and i'm going to say from here to here there's no sign change right positive to the positive sign's the same same same same same right so there's no sign changes at all right it doesn't change so that means there's zero sign changes so we're going to put a zero here and that means there's zero possible positive real solutions okay so that's what that means now if it's not zero okay then what you want to do is you want to say you can have that number or you could subtract away 2 from that number until you get to 0 or you can't subtract away 2 anymore we'll see an example in a moment of what i'm talking about there but for right now there's zero sign changes so there's zero possible positive real solutions okay now let's talk about negative so for negative real solutions what you're going to do is plug in a negative x everywhere there's an x in your function now i'm going to give you a little shortcut for this don't go through and actually start plugging things in because the sign is only going to change when you have an odd exponent right so here here and then this is the first power so here if i think about it i'm really just plugging in negative 1 times x so just think about the negative 1 part negative 1 raised to an odd power is going to give you back negative 1 so it's going to change your sign okay but if i had negative 1 raised to an even power like negative 1 raised to the 4th power it becomes 1 right so no sign change so i'm going to go through and change the signs of again the terms with the exponents that are odd okay so this would be negative 2x to the fifth power no change here and this would be negative 7x cubed no change here and this would be minus x and then plus 15. okay so now we have a lot of sign changes so we go from here to here so that's one here to here that's one here to here that's another one here to here that's another one and here to here that's another one so we have from negative to positive so that's one positive negative that's another one negative the positive that's another one positive negative that's another one and negative is a positive that's another so you've got one two three four five so we're gonna put a five here for the possible negative real solutions now here's where this gets a little bit tricky okay when you think about the imaginary solutions that are possible or the non-real complex solutions remember they come in conjugate pairs so right now if this was the scenario you would have zero because this is a fifth degree polynomial right so you're going to have five solutions again some of those could be repeats but you're going to have five solutions or five roots or five zeros however you want to say that now there's other scenarios that could be possible again when i talk about this you're going to decrease this by 2 whatever these numbers are because again the complex solutions are the non-real complex solutions the imaginary solutions are going to come in conjugate pairs so i'm always going to have 0 here but this could be reduced by 2 so i'd put 3 here and now i would put 2 here because again if i have one solution there i've got to have 2. it can only be 0 2 4 6 8. it's got to be some even number now if i decrease it again this would still be zero this would be one and now this would be four okay so notice how the total in the row is always five right the total is always five so that's what you're looking for so i started with five i went down to three then i went down to one started with zero went up to two went up to four okay now notice how i can't decrease this guy by two again because that would tell me that i have negative one negative real solutions that's nonsense right that's not possible so you need to keep going and reducing by 2 until you hit 0 or something that's not possible like in this case right here okay so for this example we have one scenario where you have 0 positive real solutions five negative real solutions and zero imaginary solutions or you could have zero positive real solutions three negative real solutions and two imaginary and then lastly you could have zero positive real solutions one negative real solution and then four imaginary solutions let's take a look at another example once you do these a few times it's very very easy so again if i want to find the possible number of positive real solutions i just look at f of x i make sure that again all my coefficients are real and then i have a non-zero constant term which i have you'll notice here you're missing x to the third power and x to the first power that's okay you don't have to worry about that just as long as you have this guy right here your non-zero constant term all right so it's in standard form so i'm just going to count my sign changes so i go from plus to plus no change plus to minus one change so that tells me this could be one right here okay now this is a fourth degree polynomial so i know there's more stuff coming okay so let's erase this and let's think about f of negative x would anything change no because this is even and this is even so a negative one to the fourth power or a negative one squared okay would give me positive one no sign change so this is two x to the fourth power plus 7x squared minus 4. so it's going to be one sine change again you go from positive to positive no change positive negative one change so really what we find here is that there's only one possibility because again this is a fourth degree polynomial this row here has to add up to four so this has to be two right again these come in conjugate pairs so the only possibility is that you have one positive real solution one negative real solution and two imaginary solutions all right let's look at another one so we have f of x equals 2x to the fourth power minus 13x squared plus 6. so again i have a non-zero constant term and then i have real coefficients so i know i can use this so again just count the sign changes to get the possible positive real solutions so this is plus so going from here to here that's 1 going from here to here that's 2. so i'm going to put a 2 there and again i know i can decrease that by 2 and go to 0. so i can either have 2 positive real solutions or 0 positive real solutions for the negative real solutions remember if you plugged in a negative x there if i plugged in a negative x there nothing would change because this is even and this is even so you'd have two sine changes again so let's go ahead and put 2 here and 0 here but you have another two scenarios that you have to consider so you could have two positive real and two negative real and of course because it's a fourth degree polynomial you'd have zero imaginary you could have zero positive real zero negative real and then you'd have four imaginary again the row has to always sum to four then the other scenario would be you could have two and zero and this would be two so two positive real zero negative real and two imaginary and then the final scenario would be that you have zero positive real two negative real and two imaginary so again when you go through the different possibilities you want to think about the imaginary as being what's left over okay so in this case you have two positive real two negative real so there's nothing left over because you're gonna have four solutions because again this is a fourth degree polynomial okay so this guy zero zero and then you get four right because there's nothing here it's gotta sum to four two zero and then this has to be two two plus two is four zero two and then again this is two because two plus two is four all right let's just look at one final example so we have f of x equals six x to the fifth power plus three x to the fourth power minus 50 x cubed minus 25x squared plus 16x plus 8. again i've got my non-zero constant term all my coefficients are real numbers so i'm good to go so how many sign changes do i have so no sign change then i've got one no sign change i've got another one and no sign change so this is going to be two so let's put two here and then we know it's also possible to have zero now if i change this to a negative okay think about what would swap so this would swap because it's odd this would swap because it's odd so let's put a plus there this would not change this would swap because it's odd and this would not change okay so now sign changes you have one you don't change there then you'd have another one then you don't change there then you'd have another one so you have three or you could have one so to go through the different possibilities again this is going to be a 5th degree polynomial so i know that my solutions have to add up to 5. so 2 plus 3 is 5 already so this would be 0 right 2 positive real three negative real so zero imaginary if i had zero and one well this would have to be four okay because again it's got a sum to five now other possibilities i could have two positive real i could have one negative real and then i would have to have two imaginary again two plus one is three three plus two is five and then the last possibility is that i could have zero positive real solutions i could have three negative real solutions and then i would have to have two imaginary solutions in this lesson we want to talk about the intermediate value theorem we also want to look at the upper and lower bound rules all right let's start with something that's very simple it's known as the intermediate value theorem and basically this can help us to determine if a zero lies between two numbers so what we do is if we have f of x again some polynomial function with only real coefficients and let's say we have two real numbers for simplicity let's just call them a and b for right now so if the values f of a and f of b are opposite in sign then it must be true that at least one real zero occurs between a and b now if you're scratching your head when you hear that or when you read that in your textbook i promise you in a second you're going to understand this is a very very intuitive concept so let's say we start with something like f of x equals x cubed plus two x squared minus five x minus six now i'm just going to tell you that there is going to be a zero that occurs between negative two and zero so between these two numbers between negative two and zero there will be a zero meaning there's some value that i can plug in for x and the result of the function or the function's value if you want to say that would be zero now to prove this in this particular case it's going to work out to where you're going to have one of these guys going to be negative and one's going to be positive so in other words f of negative 2 if i did that what would we have well this would be negative 2 cubed plus 2 times you'd have negative two and that would be squared minus five times negative two and then minus six okay so we know negative two cubed is negative eight we know that negative two squared is four four times two is eight negative 8 plus 8 is 0. so you can just erase this basically what you have here is negative 5 times negative 2 which is 10 10 minus 6 is 4. so this is a positive 4 right so the function's value when x is negative 2 is 4 okay i can also write this as an ordered pair and say that negative 2 comma 4 is on the graph of this function now we also want to look at 0. so f of 0 is equal to what well i know if i plug in a zero this is gone this is gone this is gone i'm just left with negative six so on the graph of this function would be zero comma negative six okay so what i told you was that if f of negative two is a positive number and f of zero is a negative number well then there must be a zero between these two numbers so there must be a zero between negative two and zero so graphically we can prove this if we go back and look at this point so we had negative 2 comma 4 so this is my negative 2 comma 4 and then also i had 0 comma negative 6 so that's my 0 comma negative 6. well think about this logically the value of the function here is 4 and let me use a different color because that doesn't show up very well so this is 4 right and then the value of the function down here is negative 6 so negative 6 there okay well in order for me to get from here to here as i move to the right notice that every y value between 4 and negative 6 is going to occur and that y value includes a y value of zero right where we cross the x axis and when we cross the x axis that's our zero for the polynomial function it is going to end up happening right there at negative one comma zero so that's how this guy works if you plug in one number and you get a negative and then you plug in another number and you get a positive well then there's a zero between those two numbers now what you don't want to think here and this is a common source of confusion let's say for example you plug in and you get a positive and then you plug in and you get a positive again with a different number well that doesn't necessarily mean there's no zero between them okay it's just true that if it's positive and then it's negative or you get negative and positive whatever it is if there's a sign change then it must be true that at some point you're going through the x-axis so you will have at least one zero let's look at one more quick example of this it's a very easy concept usually the problems you see with this they'll just give you two values and say hey prove that there's a zero between them so all you need to do is say what is f of negative six and this is going to be what negative six cubed is negative 216 then plus you would have negative 6 squared which is 36 36 times 14 is 504 then plus you'd have 59 times negative 6 which is negative 354 so i just put minus 354 and then plus 70. so if you go through and do the addition here negative 216 plus 504 is 288. if i subtract away 354 i get negative 66. then if i add 70 i get 4. so let's just say the value here is 4 right so one point would be negative 6 comma 4. okay so the other one let's look at f of negative 4 so what would that be negative 4 cubed is negative 64. then you'd have plus negative 4 squared is 16 16 times 14 is 224 then plus you'd have 59 times negative 4 which is negative 236 so just put minus 236 and then you have plus 70. okay to do this quickly i know that 70 minus 64 is going to be 6. so let's just get rid of this and put a 6 here and i know that negative 236 plus 224 is negative 12 okay is negative 12. and i can quickly do negative 12 plus 6 that's going to be negative 6. so this guy is positive this guy is negative right so one point would be negative 4 comma negative 6. so because we go from positive to negative i know that there's going to be a zero between those two numbers again my y value here is positive four my y value here is negative six so at some point between an x value of negative six and an x value of negative four i'm going to have to cross the x axis or get a y value of 0. all right so now let's move on and talk about the upper and lower bound rules these rules are helpful again to just narrow things down when you're working with finding the zeros for a polynomial function you have to have as many tools as you can get especially if you don't have a graph or you don't have you know a calculator or something to use to help you out you've got to just narrow things down as quickly as you can in order to not spend hours and hours and hours trying to find zeros all right so if we have a polynomial function f of x with real coefficients and a positive leading coefficient okay let me say that again a positive leading coefficient then if we divide f of x by x minus k and your k that you're working with is positive if each number in the last row of your synthetic division is either positive or zero then you can say that k is an upper bound this just tells you that you're not going to have a zero that's above that okay you can stop looking so if you had a list of values like you generate with the rational roots test and you test something in the middle you take a positive value in the middle of the range and let's say you find out it's an upper bound or you can cut off a lot of numbers you can start working down now all right so let's see this with this guy real quick and then i'll show you about the lower bound in a minute so basically what we want to do is just do some synthetic division so i put a 5 here and i'm just going to take my coefficients so 1 negative 4 negative 2 negative 3 negative 1 and negative 1. okay so let's scroll down to get some room and basically what we're looking for is that this entire bottom row has to be either positive or zero okay if you have any negatives involved you have to stop okay it didn't work so this guy's going to come down so you have 5 times 1 that's 5 negative four plus five is going to be positive one and then five times one is five negative two plus five is positive three five times three is fifteen negative three plus fifteen is twelve five times 12 is 60 negative 1 plus 60 is 59 5 times 59 is 295 negative 1 plus 295 is 294 okay so that's your remainder there but again all you're looking for here is the fact that you have a positive number here okay and you had a positive leading coefficient so if all the numbers down here are positive in the final row here your results well then you know that 5 is an upper bound okay it tells you that you don't need to look at anything larger than 5 if you're looking for the zeros of a polynomial function you would want to start going the other way okay so let me tell you about the rule for the lower bound so if your k your number you're doing the synthetic division with is negative now and the numbers in the last row are alternating in signs so it goes positive negative or negative positive if it alternates like that zero in this case can be treated as whatever you want it can be positive or negative whatever you need okay but if this pattern occurs then we can say k is a lower bound so none of the real zeros will be below that point so you can kind of stop your search all right so let's go ahead and crank this one out and i'll show you that this one does work so this is negative three we'll have three five negative four negative four again 3 and negative 2. okay so this is a negative value we have a positive leading coefficient so let's go ahead and try this so i'll go ahead and drop this down we'll put a 3 here negative 3 times 3 is negative 9. 5 plus negative 9 is negative four negative three times negative four is twelve negative four plus twelve is going to be positive eight negative three times eight is negative twenty-four negative four plus negative twenty-four is negative twenty-eight negative three times negative twenty-eight is 84 3 plus 84 is 87 and then lastly negative 3 times 87 is negative 261 and then negative 2 plus negative 261 is negative 263. so if you had done the rational roots test and you were just checking negative 3 well immediately you would say okay this is not a 0 but i know negative 3 is a lower bound because this guy is negative okay and if i look at these guys this is positive negative positive negative positive negative so it alternates here so i know that negative three is a lower bound so i'm not going to be looking for any real zeros that are less than negative three okay again the upper bound scenario is where this guy is positive and your whole row down here is positive in that case i'm not looking for any real zeros that are above that number in this lesson we want to talk about finding the zeros of a polynomial function all right so now it's time to take all the things we've been talking about over the last few lessons and apply it to a real world example so we have f of x equals 5x cubed plus 12x squared plus 54x plus 20. so if you got this problem on your test and your teacher said find the zeros for this polynomial function what would you do well there's a lot of different strategies that you can employ here but the first thing that i would do is i would look at the degree of the polynomial and we see that it's a 3. so immediately i know that there's three solutions now there could be repeated solutions but i know there's three solutions now because the imaginary solutions or again the non-real complex solutions come in conjugate pairs meaning there could be zero there could be two there could be four so on and so forth well that tells me that there's either one real solution and two imaginary solutions or there's going to be three real solutions now we can further break this down by thinking about again descartes rule of signs remember you look at the sign changes with f of x so this is positive positive positive positive no sign changes in f of x so there's zero real solutions that are positive okay so that tells me that all the real solutions are going to have to be negative you can verify this by doing f of negative x okay and so again when you do this don't worry about the x because that part stays the same it's the negative here that you're thinking about because it's a negative one that gets plugged in that's going to change the sign of that term if the exponent is odd so there and also there right because negative 1 raised to an odd power is negative 1 that changes the sign so in this particular case you'd have negative 5x cubed this wouldn't change because this is an even exponent so plus 12x squared this would change it would be minus 54x and then plus 20. okay so we see that we have three sign changes here you go from negative to positive positive to negative and negative to positive so one two three sign changes all right so that means one scenario could be zero positive real solutions three negative real solutions and zero imaginary and then again we decrease that number by two okay so we can have zero positive real solutions decrease this by two you get to one so negative real solutions you'd have one in that case and then this right here imaginary would have to be two now right because the row always sums to three because you've got a degree of three for that polynomial all right so this already narrows things down for me i know that in terms of real solutions i'm only looking for negative ones okay so if i do my rational roots test i can throw out all the positives i don't even need to look at those so now let's go to a fresh sheet and let's think about this rational roots test so it's been a while since we talked about it so let me refresh your memory remember the polynomial needs to be in standard form and you want to look at the leading coefficient so this guy here is a 5 that's the leading coefficient because it's on x cubed the third power is the highest power for this polynomial right this one's the second power and this one's the first power so this is the leading coefficient because again it's the coefficient on the variable raised to the highest power so this is where i get my cues from right all the factors of 5. then this the constant term is where i get my p's from okay remember all the factors of 20. so think about the factors of 20 so i'm just going to put p here 20 is going to be what it's 1 2 4 5 10 and also 20. and then for q again all the factors of 5 you've got 1 and 5. and i know there's plus or minus in each case but again i'm just kind of trying to speed this up we're going to be throwing out the positives anyway so i'm just going to write everything and then i'm going to put a negative sign in front of everything just to make it quicker because in this case we don't need plus or minus we already know we don't have positive real solutions okay so i've got to come up with all the possible p over q's so i'm going to start with the number one one over one would be one and then i'm just going to slap a negative on it and then we'd have one over five which would be one-fifth again i'm gonna put a negative on it and then we'd have what i moved down so now you'd have 2 over 1 which is 2 so it'd be negative 2 and then 2 over 5 would be 2 fifths so negative 2 fifths and then moving down we have 4 over 1 which is 4 so i'll put negative 4 and then 4 over 5 is 4 fifths so negative four-fifths and then let's talk about we have five so we have five over one so that's negative five and then five over five is one right so we already have negative one we don't need that so then you'd have ten over one which is ten so i'll put negative 10 and then 10 over 5 is 2. again if i think about this in terms of negative 2 i already have that don't worry about it you can just move on so we have 20 20 over 1 is 20. so i'll put negative 20. and then 20 over 5 is 4. we already have that again we're changing the sign of negative 4 we already have that so again just so you're not lost i didn't put plus or minus because we already know that there's no positive real solution so that's why i only have negatives there okay so let me erase all of this and now we have a list of potential candidates okay so here's what i'm going to tell you to do and this sometimes works sometimes it doesn't we talked about the upper and lower bound rules okay in the case where the leading coefficient is positive which it is here and you have a negative k a negative value that you're doing your synthetic division with okay we know that if the bottom row of that synthetic division alternates in sine we've found a lower bound right so there won't be a zero that is below that value or less than that value so what i'm going to do when i think about the integers here forget about the fractions i've got negative 1 negative 2 negative 4 negative 5 negative 10 negative 20. so i'm going to pick something kind of in between i'm just going to go with negative 4 and i'm going to see if it's a lower bound and sometimes you do this and you end up with a 0. so i'm going to take negative 4 and just do some synthetic division so i'm going to take 5 i'm going to take 12 i'm going to take 54 and i'm going to take 20. okay so let me come back up in a minute i'm just going to drop this down and see what we get so negative 4 times 5 is negative 20. 12 plus negative 20 is negative 8. negative 4 times negative 8 is 32 54 plus 32 well 4 plus 2 is 6 and 5 plus 3 is 8 so that's 86 and then negative 4 times 86 is going to be negative 344 20 plus negative 344 is negative 324 okay so although we didn't find a 0 here we did figure out that negative 4 is a lower bound right because the signs here this is positive this is negative this is positive this is negative so they alternate so again that tells you if this guy is negative and this guy is positive and these guys alternate you've found a lower bound now i want to stop here and i want to caution against a common mistake if you had done this with negative 4 and you didn't find that it was a lower bound that doesn't mean that it's not okay it only says that conclusively you can say if these situations present themselves so again a negative a positive and alternating signs then you do have a lower bound for sure okay so let's erase this and let's go back up now i can take negative 4 off i don't need it i can take negative 5 off negative 10 often negative 20 because those values are less than negative 4 okay so everything that i have now would be greater than negative four now the next thing i would think about here negative one is something easy to just plug in for so i would just check that real quick so all we would see is that again if you were plugging in a 1 you would just erase all those x's and you would just straight add the coefficients right so if it's negative 1 the sign of the coefficient would change if it's an odd exponent so here and here okay so if we do this let me kind of go up a little bit here this would be negative and you can get rid of this get rid of this this would be negative okay so negative 5 plus 12 is 7 7 minus 54 is negative 47 and negative 47 plus 20 is negative 27 so i know negative 1 is not a 0. so f of negative 1 let me write that down it's negative 27 so i know this doesn't work but something you can check really quickly is f of 0. so f of 0 is what this is all gone it's just 20. okay so that takes a millisecond now the reason i did that is because of the intermediate value theorem all these fractions here negative one-fifth negative two-fifths and negative four-fifths they're between zero and negative one on the number line so because this guy is positive and this guy is negative again the intermediate value theorem tells us that for sure we're going to have a real zero between zero and negative one okay so what i'm going to do now is just clean up a little bit so let me rewrite these fractions so we have negative one-fifth we have negative two-fifths and we have negative four-fifths and then we have negative two okay but i'm going to work with these fractions because again i know that there's a zero between zero and negative one okay so let's start with negative one-fifth so if this is a zero again we use our synthetic division so the remainder would be zero if this is a zero so i'm going to put negative one-fifth like this this is a little bit harder because you're working with a fraction but if you have a calculator it's not too bad so we've got five we've got 12 we got 54 and we have 20. okay so let me scroll down and get some room i'll pop back up in a moment so let's bring this down and so negative one-fifth times five we know that the denominator of five here would cancel with this guy which is a 5 okay so i'd be left with just negative 1 in that case so this is going to be negative 1. 12 plus negative 1 is 11. then negative 1 5 times 11 this would be negative okay we know 11 times 1 is 11 and this is over our denominator of 5. so here's where it gets a little tricky you've got to get a common denominator going because you have 54 plus negative 11 fifths so let's do that over here so 54 plus you've got negative 11 fifths and you can just put minus if you want it doesn't matter so for the 54 i need a common denominator so i'm going to times it by 5 over 5. 54 times 5 is 270 right 50 times 5 is 250 and 4 times 5 is 20. so this would be 270. okay so if i subtract 270 minus 11 that's 259 so this would be 259 fifths so 259 fifths okay so now we want to multiply negative one-fifth times 259 fifths so what is that so negative one-fifth times 259 fifths five times five is 25 and negative one times 259 is negative 259. so negative 259 over 25. okay so negative 259 over 25 all right so now to do this final addition we want to multiply this by 25 over 25 so this would be 500 over 25 okay so 500 over 25 and 500 minus 259 is 241 so this would be 241 over 25. now this is not zero and so we can say that this negative one-fifth here is not a zero right so it doesn't work so let me just erase this let me put this back to 20 and let me erase this and let me scooch up here for a second and let me get rid of this one and let's go to the next one which is negative two-fifths all right so we're gonna repeat this process again drop this down negative two-fifths times five the fives would cancel you'd have a negative two twelve plus negative two is ten okay so now negative two-fifths times ten if you think about five and ten the ten would cancel the five and give you two two times negative two is negative four so 54 plus negative four is fifty fifty times negative two-fifths again 50 divided by 5 is 10 10 times negative 2 is negative 20 aha we found it 20 plus negative 20 is zero okay so we know that negative two-fifths is a zero okay so we want to erase this keep those numbers down there you need them i'm going to erase this we don't really need this because we have one zero and we're taking it down to a quadratic so we'll quickly find the other ones so i'm going to go ahead and say that for the zeros one of them right now we know is negative two fifths okay so that's one of them so the next thing i'm going to think about is the fact that synthetically i divided this by x plus two-fifths right x minus a negative two-fifths is x plus two-fifths so the result here is going to be a quadratic so this would be 5x squared plus 10x plus 50. so where does this equal zero now to make this a little bit easier if you're just trying to find out where this is equal to zero i could divide everything by five so i'm working with smaller numbers because this is divisible by five so is this so is this and if i divide this by five i get zero anyway right so it's fine it's not going to affect anything it's going to make your your life easier so what i'm going to say is we have x squared plus this would be 2x plus 10 equals 0. so this right here again we're just going to use the quadratic formula so my a is going to be 1 my b is going to be 2 my c is going to be 10. if i think about this it's what x is equal to negative b in this case b is 2 so negative 2 plus or minus the square root of b squared so b is 2 2 squared is 4 minus 4 times a a is 1 times c c is 10. so 4 times 10 is 40. this is all over 2 times a again a is 1 so just 2. so 4 minus 40 is negative and we know if we pull that negative out it would be i right the imaginary unit so i can say this is i times the square root of 36 which is 6. so you can put 6i here and basically you could factor out a 2. let me erase this we don't need this anymore you could factor out a 2 and so you could say x is equal to put a 2 out here and then inside the parentheses you would have negative 1 plus or minus you'd have 3i and then over 2 and you can cancel this with this okay so because this is divisible by 2 and this is divisible by 2. again you can factor that out and cancel and make this more simple so let's erase these and so we found our other two solutions our other two zeros so one of them would be negative 1 plus 3i and the other is going to be negative 1 minus 3i so at this point if you want to stop the video and check these solutions by plugging them in be my guest but i can promise you they work so i know this was a bit tedious but it's good to go through a harder example that way if you get something like this on a test you know what to do in this lesson we want to talk about one-to-one functions and the horizontal line test so in this section of our course we're going to be talking about exponential and logarithmic functions but before we can get to these topics we need to first discuss one-to-one functions and also inverse functions so today we're going to talk about one-to-one functions in our horizontal line test and then in the next lesson we'll kind of follow up and we'll look at some algebraic methods and then we'll move on to inverse functions and then exponential functions and logarithmic functions all right so let's start out with just a brief review of functions in general this is something we covered in depth earlier in the course so we already know that a function is a special type of relation where each member of our domain or x value corresponds to one and only one member of the range or y value so for example if we look at this f here this is a relation and it's also going to be a function and it just consists of these four ordered pairs so we have 3 comma 2 1 comma negative 9 7 comma 4 and 2 comma 2. so how do we know if this is a function again with this type of simple example you would go through and you would look for a duplicate x value right because if you had a duplicate x value that means that that given x value is associated with more than one y value now we can also look at the mapping diagram and we can see that an x value of 3 is associated with a y value of 2 an x value of one is associated with the y value of negative nine an x value of seven is associated with y value of four and then an x value of two is associated with y value of two now the reason i chose this example is because it's gonna relate to something in a minute and also we need to remember that this part right here does not break or you can say does not violate the definition of a function i can have a y that is associated with two different x values that's not an issue with a function if i give you something as an input or as an x value it has to give me as an output or as a y value a unique value so if i said hey what is y when x is 3 well we know it's going to be 2. right if i go to this ordered pair if x is 3 y is 2. if x is 3 y is 2. then additionally if i said hey what is y when x is 2 well it's going to be 2 right if x is 2 y is 2. so there's a clear association there there's an answer if i said hey if x is this what's y there's a clear answer now if we look at something that's not a function you're going to have a situation where there's not a clear answer not a clear association between an x value and a given y value so let's look at g g is a relation and it's not a function and again the easy way to look at this is you look for duplicate x values you have negative one comma three two comma negative five negative one comma seven and four comma six you've got that negative one that occurs twice so that's a red flag that this is not a function and again when you look at the mapping diagram you see that negative 1 this x value corresponds to 3 that y value and 7 that y value so if i said hey what is y when x is negative one you don't know whether answer three or to answer seven so this guy is not a function now when we talk about a one-to-one function this is a stricter definition so something can be a function and not a one-to-one function so here for each x there's one y but also for each y there's one x so our definition here is that each x value corresponds to only let me highlight that only one y value so that's always been the case with functions but now and each y value corresponds to only one x value okay so the takeaway here is that for each x there's one y and for each y there's one x again if we go back to this example here this is a function but it's not a one to one function because this y value of two is associated with an x value of two and then also an x value of three so how can we determine if we have a one-to-one function well previously we learned that we could use a vertical line test to determine if a relation was a function right this is a graphical test so you have to first graph your relation and then draw vertical lines and if any vertical line impacts the graph in more than one location it's not the graph of a function so why does this work well remember a vertical line is something like x equals some real number let's just call it a it could be x equals negative 2 it could be x equals 7 it could be x equals again any real number it could be a trillion what happens is for any point on that line x is always the same value so if that line impacts the graph in more than one location well that given x value is now associated with more than one y value so that breaks or violates the definition of a function so you don't have a function now similarly we can use a horizontal line to test for a one-to-one function so a horizontal line is y equals some real number let's just call it k so y equals 7 or y equals negative 100 any of those would work it's y equals some real number what we need to know is on that horizontal line the y value is never going to change so if it impacts the graph in more than one location that tells us that that given y value is going to be associated with more than one x value so it's not a one to one function so let's just blow through some examples real quick and again in the next lesson i'll show you an algebraic method to kind of do this if you don't feel like graphing stuff so first we have f of x equals the quantity x minus 2 squared so immediately you would see that this is going to fail the horizontal line test i'm just going to draw one horizontal line and i'm going to do y equals 4. so what you can see is that this horizontal line that i drew again this would be y is equal to 4 okay and what happens is it impacts the graph here and also here so this is where an x value is 4 so this would be 4 comma 4 and this is where the x value is 0 so this would be 0 comma 4. so let's write these over here we don't need this anymore so you'd have 0 comma 4 and you would have 4 comma 4. so again think about this you have a y value of 4 in each case and it's associated with two different x values it's associated with an x value of 4 and it's also associated with an x value of 0. now you can clearly see what's causing this it's because of this exponent here that's a 2. it's an even exponent so what happens is when you have a 2 squared you get 4 and then also if you get a negative 2 squared you get 4 as well if i plugged in a 0 for x there you say f of 0 is equal to 0 minus 2 that would be squared this is going to be negative 2 and it's squared which is 4. then also if i did f of 4 this is what it's 4 minus 2 that quantity squared 4 minus 2 is 2 2 squared is 4. so either way i get 4 and it's because of again that squaring operation that takes a negative and makes it positive and takes a positive and keeps it positive so that's why for that given y value you've got two different x values that's kind of associated with it all right let's look at one that is a one-to-one function so we have f of x equals the square root of x plus 1 and then you have plus 4 and this is outside of the square root symbol okay so what we have here is a one-to-one function you can draw as many horizontal lines through this thing as you want you're never going to impact the graph in more than one location and i'll just draw two lines so just two lines you can draw as many as you want or as few as you want you can really just observe this thing and see that no horizontal line is going to impact the graph of more than one location so this is going to be a one-to-one function all right let's look at another one that's not a one-to-one function and again as soon as i pull the graphs up you can immediately tell it's not you can see that a horizontal line is going to impact the graph in more than one location there so f of x equals the absolute value of x plus one plus one so it's the absolute value operation that's giving us the problem here because again that takes a negative and makes it positive it takes a positive and keeps it positive so let's just draw one horizontal line here so again what you can see here is that it impacts the graph there and there so that's three comma five three comma five and it's also going to be negative five comma five so negative five comma five so let's just see this real quick if i did f of three as an example this would be the absolute value of three plus one and then plus one we know the absolute value of three plus one is four right three plus one is four absolute value of 4 is 4. so this would be 4 plus 1 which is 5. so that's simple enough then f of negative 5 is going to give me the same thing it's the absolute value of negative 5 plus 1 we know negative 5 plus 1 is negative 4 the absolute value of negative 4 is 4 so it's 4 plus 1 again which is 5. so it's that absolute value operation that's creating this kind of effect where this y value of 5 is associated with two different x values in this case three and negative five in this lesson we want to continue to talk about one-to-one functions and what we're going to do here specifically is we're going to focus on an algebraic method that's going to allow us to determine if a function is a one-to-one function all right so in our last lesson we talked about one-to-one functions and we also talked about how to determine if a function was a one-to-one function using the horizontal line test now a one-to-one function just to kind of recap is where each x-value corresponds to only one y-value and then also each y-value corresponds to only one x-value now the quick way to kind of recap that is to say for each x there's one y and then also for each y there's one x remember when we had a function that wasn't one to one it was okay for two different x values to correspond to the same y value that's okay that doesn't break the definition of a function but it does break the definition of a one-to-one function so in the last lesson we talked about this horizontal line test basically with a horizontal line the y value is always the same so if that horizontal line representing that kind of single y value impacts the graph in more than one location that function is not going to be a one to one function because that single y value is associated with more than one x value now additionally we have an algebraic method that we can use in some cases this is faster in some cases it gets a bit complex so it just depends on the given scenario but if you're in a pre-calculus course or college algebra course a lot of times they'll tell you that you need to show this algebraically and they'll give you reasonable examples to work on all right so this is what we want to start with here it's just basically breaking down the definition we just saw but into a more kind of algebraic format so f is said to be 1 to 1 if and only if f of x sub 1 is equal to f of x sub 2 implies that x sub 1 equals x sub 2. so what does this mean if you read this in your textbook you might get confused by it the only thing you need to focus on is the fact that when we have f of x it doesn't matter this is f of x sub 1 f of x in general is what it's just a replacement for our variable y that we're so used to right it's the output of a function so f of x sub 1 is just the output when i plug in the x sub 1 for each occurrence of x in that function it's the same thing for f of x sub 2. it's just the output or again the y value when i plug in x sub 2 for each occurrence of x in the function now if these two y values are equal or the same well if the function is one to one that can only be true if the two x values are the same so that's all this notation is really saying here it's written in a format that is confusing but once you kind of break it down it's not so bad so what i want to do now is just show you a method that relies on this kind of notation here and basically we'll just jump right into an example what you want to do if we have f of x equals negative 3x minus 1 you want to say okay if i had f of you can use x sub 1. i'm going to use a and b here so i'm going to say f of a this is negative 3a minus one all i did here was i plugged in an a where there was an x here okay that's all we've done and we've seen this before in previous lessons now what i also want to do is i want to look at f of b so what would that be so it would be negative 3b minus 1. again all i did was i plugged in a b where there was an x in that function okay so so far so good so from our notation let's just go back to it again we have f of x sub 1 is equal to f of x sub 2 implies that x sub 1 equals x sub 2. well instead of using x sub 1 and x sub 2 we're using a and b just to keep it simpler and so what i want to do here is say okay if it's true that f of a is equal to f of b okay then it has to be the case that a is equal to b okay otherwise this function is not going to be one to one so what we can do is we can set these two equal to each other remember f of a is actually equal to this and f of b is actually equal to this okay so what i'm going to do is i'm going to set these two equal to each other so i'm going to say negative 3a minus 1 is equal to negative 3b minus 1. and if it's a one-to-one function it should be true that a is equal to b always so let's go through and kind of simplify this is really simple we're just going to you could say you're solving for a you could say you're solving for b i'm just going to go through and add 1 to both sides and what i'm going to do here is i'm just going to cancel this and that's going to give us what let me kind of scroll down a little bit that's going to give us negative 3a is equal to negative 3b if i divide both sides by negative 3 what am i going to get well we know that this is going to cancel with this and this is going to cancel with this so i get that a is equal to b so we've proved that this function is 1 to 1. let me erase this real quick and let me just drag this up here and let me just go back up so we can recap this real quick we said that again if f of a is equal to f of b and this function is one to one then it has to be true that a equals b well we have f of a defined as negative three a minus one we have f of b defined as negative three b minus 1. so if i set these two equal to each other and i go through a process of either you could say i'm trying to solve for a or b it doesn't really matter i end up with this a equals b so i know this function is 1 to 1. now if you want to look at a graph of this guy again i don't have to draw horizontal lines here you can see that this graph here there's no horizontal line that would impact it in more than one location so we know that this function is going to be one to one when you first start doing this if you want to kind of graph it to check and make sure that you got the right answer that's probably a good idea all right let's look at another example so we have f of x equals the absolute value of x plus two again if i use f of a this is what just plug in an a where there's an x that's all you have to do so a plus 2 then for f of b okay i'm just going to now plug in a b where there's an x so we'd have b plus 2. nothing that's too fancy nothing that's too confusing again if f of a is equal to f of b then it's got to be true that a equals b and that's got to be true all the time so what can we do here again this is f of a this is f of b so if these two are equal set them equal to each other and let's see so we have the absolute value of a plus 2 is equal to the absolute value of b plus 2. now we've solved absolute value equations before and we know that if we have an absolute value expression equal to another absolute value expression well these two can only be equal if what the first scenario is you drop the absolute value bars so the first scenario would just be a plus 2 is equal to b plus 2. but because you have absolute value involved here there's going to be an or right so it could also be true that the negative of one of these expressions is equal to one of the expressions so in other words i would drop the absolute value bars so i put a plus two is equal to b plus two and then i would wrap one of these in parentheses it doesn't matter which one and i would put a negative sign out in front now you can distribute that and say this is negative a minus 2 and what's going to happen is you're going to run into a problem okay so over here it's fine i subtract 2 away from each side and i get a is equal to b but this is where it gets messed up because you have this or and over here i have negative a minus 2 equals b plus 2. let's just solve it for a so in that particular case i would add 2 to both sides and i would have negative a is equal to this cancels b plus 4 and then to get a by itself let's divide both sides by negative 1 or you could multiply by negative 1 it doesn't matter so let's divide both sides by negative one and let me scroll down to get a little bit more room so we know that this is going to cancel and i'll have a is equal to negative b minus four now the problem lies in this one right here okay it's okay this one right here on the left is fine but you have this or situation and that's what throws a wrench in this so this function is not going to be one to one so let's look at this graphically and we can clearly see this function is not one to one if i drew a horizontal line through if i go up to y equals four so let's say i had a horizontal line y equals four let's just draw that in real quick and not perfectly straight but you get the idea and when i look at this guy again it's going to impact the graph here and also here so that's an x value of 2 and a y value of 4. so 2 comma 4 works and then also an x value of negative six and a y value of four so an x value of negative six and a y value of four so this y value of four is associated with two different x values two and negative six so of course this is not a one to one function all right let's take a look at another one so now we have f of x equals we have 1 over x minus 2 and then plus 2. so again i'm going to go f of a is equal to 1 over a minus 2 then plus 2. then i'm going to do f of b is equal to 1 over b minus 2 and then plus 2. okay so again what i want to do is set these two equal to each other so this is f of a this is f of b set them equal to each other and show that a equals b if this guy is one two one so one over a minus two plus two is equal to one over b minus two plus two well i have 2 on each side so i can just get rid of that right if i subtract 2 away from this side and i subtract 2 away from that side i'm just subtracting the same thing away from both sides of the equation so it's going to cancel over here and cancel over here so let's scroll down now and what we have is 1 over a minus 2 is equal to 1 over b minus 2. so the quickest way to solve this is just to cross multiply okay so what we would have is what we would have b minus 2 times 1 which is just b minus 2 is equal to a minus 2 times 1 which is just a minus 2. and of course now i can just add 2 to both sides of the equation this is gone this is gone and what i get is that b is equal to a or you could say a equals b it doesn't really matter but you've proved that this is a one to one function so here's the graph and this one is a bit confusing because you do have that horizontal asymptote at y equals 2. but all we're doing here is this graph if i'm going to the left it's approaching a y value of 2 but it's never going to actually touch it this guy is constantly getting closer and closer to 2 but the value isn't always the same right it's getting closer and closer it looks like it graphically because it gets infinitely close to 2 in our eyes when we look at that we say oh that looks like a horizontal line would hit that in more than one location but it wouldn't right so you just have to understand the fact that this is an asymptote and the idea is that it's getting closer and closer and closer it's not going to actually touch it but the values aren't the same right so those y values are changing all right let's look at another one so now we have f of x equals the quantity x minus 3 cubed plus 1. so we already know what to do so we're going to go f of a is equal to the quantity a minus 3 cubed plus 1 and then f of b is equal to the quantity b minus 3 cubed plus 1. okay so we're going to set these two equal to each other so i'm going to say the quantity a minus 3 cubed plus 1 is equal to the quantity b minus 3 cubed plus 1. so we know that we have 1 on this side and 1 on this side so i can subtract 1 away from both sides and that's going to go away in each case so now what i'm looking at is a minus 3 cubed is equal to b minus 3 cubed okay and what i can do to get rid of this kind of exponent of 3 i can use kind of the inverse operation i can take the cube root of both sides of the equation or i could really raise both sides to the 1 3 power it doesn't really matter how you think about that i'm just going to say i'm going to take the cube root of both sides of the equation and so we know that this index is going to cancel with this exponent in each case and that's going to leave me with just this part right here okay so i'm going to have that a minus 3 is equal to b minus 3 and let's scroll down a little bit and we can add 3 to both sides of the equation and we will have that a is equal to b so a is equal to b here so graphically when we look at this guy we can see that no horizontal line would impact this in more than one location you can see this kind of spot here looks like it might but it's not this guy is constantly changing in terms of the y value so this is going to be a one to one function in this lesson we want to talk about how to find the inverse of a function over the course of the last two lessons we talked about how to determine if a function was a one-to-one function meaning for each x it had one y and also for each y it had one x so once we understand that concept if we have a one-to-one function we can then find something known as the inverse of that function and when we talk about a function and its inverse they basically reverse each other so the x and the y values are going to be swapped so to see that with a very simple example let's start out with f and f is just a very simple relation and it's also a function and it's also a one-to-one function so it just contains three ordered pairs so 3 comma 2 negative 1 comma 7 and 5 comma 1. if we say that g is going to be the inverse of f then all we need to do to find this is just swap the x and y values so this ordered pair again this is x comma y this ordered pair is 3 comma 2. so down here when i form this ordered pair i just swap them so the 2 that's the y value will become the x value and the 3 that's the x value will become the y value now moving on to this one i'm going to do the same thing and i can just kind of draw this like this i'm just going to swap so this will become 7 and this will become negative 1. so a y value of 7 became the x value of 7 in the inverse and then the x value of negative one became a y value of negative one in the inverse and again we're just going to do the same thing one more time so this will be a one and this will be a five so we're just swapping the x and y values it's a pretty simple thing overall the next thing you're going to come across in this section is they're going to ask you to find the inverse of a one-to-one function so we already know something like f of x equals 3x plus 5 is a one-to-one function again we know how to prove that graphically and also algebraically so once you've determined that it's a one-to-one function how can you find the inverse well basically what you want to do is replace f of x with y so you're going to go y equals 3x plus 5. so that's the very first step then the next thing you want to do is you want to swap x and y remember these guys are going to be reversed the x values become y values and the y values become x values so if i swap x and y you can kind of do it like this i'm going to say this is x and this is y and now what i'm going to do is i'm just going to solve it for y so i can just erase this so i have enough room to do this let me just kind of slide this up and to solve it for y is very easy we subtract 5 away from each side of the equation we know that this is going to cancel and i'll have x minus 5 is equal to 3y and let me erase this because i like my y on the left so i'm going to put 3y is equal to we have x minus five i'll erase this now so how can i solve for y we have this three that's multiplying y divide both sides by three and of course we get that y is equal to x minus five over three so this guy and this guy are inverses the notation seems a bit different because we have f of x and then y so to fix that i'm going to change this and i'm going to put some special notation in here i'm going to say f inverse of x okay so that's how that's read it's not f to the power of negative one as it appears to be so if we think about this the common mistake is you see this notation and you're thinking about exponents and you're saying okay f inverse of x is 1 over f of x no no no this is completely wrong okay so this is just the way that we notate the inverse okay so it's not one over f of x now let's think about this for a moment and see how these guys reverse each other if i think about what f of x does if i plug something in here for x well what happens to it we multiply by three so step one is we multiply by three and then what do we do in step two we add five so then we add five okay so that's the first one what happens in the inverse well the first thing is when you plug something in for x what happens the first thing is i subtract five i subtract 5 okay away from whatever x is so you see how this is going to reverse what this is if i add 5 to something and then i subtract 5 away from something i'm basically reversing that and getting back to where i started if i started with 3 and i added 5 and then i subtracted 5 well i'm back to 3 okay so if i look at this here the second step is what it's going to reverse this multiplication by 3 and we see that once we've subtracted away 5 we divide by 3 down there so we can put divide by 3. so what this is telling me is that when i plug something in for x and i go through the multiply by 3 step in the add 5 step if i take that result and then plug it back in here it's going to completely undo what i've done it's going to first subtract away 5 okay undoing this and then it's going to divide by 3 undoing this and i'm going to get back when i started to show you this let's just erase this so we have enough room let's just pick one number and we could just choose something easy like two so f of two would be what so i'm just plugging in a two for x here and that would be three times two plus five again the first thing is i'm multiplying by three 3 times 2 is 6. then i add 5 to the result 6 plus 5 is 11. okay so if i take this result and i run it back through the inverse as the inputs i'm going to go back to 2. okay that's all it's saying so then f inverse of 11 should be equal to 2. so what we would have we would have 11 minus 5 over 3. again the first thing i'm doing is i'm subtracting away 5. i'm undoing this part right here 11 minus 5 is 6. so now i'm back to that and then the next thing i'm doing is i'm dividing by three okay so i'm undoing this part right here six divided by three is two so what you see is that in the first one we have an ordered pair that's two comma eleven in the second one we reverse that ordered pair if i plug in 11 i get back to 2. so these guys again they reverse each other so this is the main concept of an inverse all right let's take a look at another example again when you get these problems they're very simple you just want to start out with taking this f of x and making it y [Music] and then you're just going to swap x and y so we can just erase it if you want and say this is y and this is x and then just solve for y okay so pretty easy so we're going to add 2 to both sides and i'll just do this over here we're going to have x plus 2 is equal to 2 y cubed now let me erase this and i'm going to flip the order because i like y on the left so 2 y cubed is equal to x plus 2. and then from here how can we solve for y well we have two steps we have to do we first are going to divide both sides by 2 and that's just going to clear this out so i can just erase it and then the last thing is this is raised to the third power i can undo this by taking both sides to the kind of 1 3 power or we could say we take the cube root of each side either way you want to show that okay and i didn't make that kind of long enough so let me redo that so what's going to happen is this will cancel and i can really make this a lot better so let me kind of slide this down so we have some room and i'll do that properly so y is equal to the cube root of x plus 2 over 2. and really you just change this at this point say this is f inverse of x and you're done right you've found your inverse so these guys are inverses again you can pick a number plug it in and run it through here and then what happens is let's just do one for example if i do 2 well 2 cubed is 8 8 times 2 is 16 16 minus 2 is 14. so this would be 2 and 14. if i take 14 and plug it in here i'm going to get a 2. okay so 14 plus 2 is 16 16 divided by 2 is 8 the cube root of 8 is 2. so again you get 14 comma 2 as an ordered pair here so again these guys are just reversing each other the x and the y's are just going to be swapped let's take a look at one that's a little bit harder to wrap things up so we have f of x equals we have x minus 5 over 2x plus 3. so in this case you have two x's in your problem so you're still going to do the same thing so i'm still going to change this one into a y and this one into a y so i'll say y is equal to x minus 5 over 2 x plus 3. again just swap these so i'm just going to let me kind of do this down here we're going to need a lot of space for this i don't want to erase anything because people get confused on this one so i'm going to be switching this and this into a y [Music] and this one i'm going to change into an x okay so we'll have x is equal to y minus five over two y plus three so so far not too bad now let me scroll down and get lots of room you want to solve this guy for y but you have two y's involved so what you're gonna have to do is you're to have to get all the y's to one side okay you're going to have two terms that are not going to be like terms with y's in it and then you have to factor the y out okay we've already done this earlier in the course but it's been a while so you probably have forgotten so what i'm going to do let me just slide this down a little bit i'm going to multiply this by this denominator so 2y plus 3 and i'm going to multiply by the denominator over here so it's going to cancel from here and what i'm going to have is this 2y plus 3 times x which is basically x times 2y so let's put 2xy and then plus x times 3 is 3x and this equals my y minus 5. so i've got a y here and i've got a y here so i've got to get all those terms to the left everything else is going to go to the right so let me subtract 3x away from each side and let me subtract y away from each side and what is that going to give me well we know that this cancels and this cancels so don't worry about that we're going to have 2xy minus y is equal to negative 3x minus 5. okay so now what i want to do i want to factor this y out okay so if i do that i'll have y and then inside of parentheses i'll have 2x minus 1 and this equals we have this negative 3x minus 5. so i'm going to run out of room here so let me copy this and i'm going to start a new sheet and we'll just paste this in so we can keep going so now all i need to do think about this y is being multiplied by this right here okay i know this doesn't look like if we had something like 6y equals 30 and we divide both sides by 6 to say y equals 5 it's the same thing here y is being multiplied by this quantity 2x minus 1. so to get y by itself i just divide both sides by that quantity and this cancels and so i find that y is equal to negative 3x minus 5 over 2x minus 1. and let's just paste this in here and i'm going to erase the y and i'm going to put f inverse of x so we have our correct notation and again you can choose some point arbitrarily and show that it reverses itself in the next lesson i'll show you how to prove that you have inverses but for now we'll just show this with a simple point so let's just go with an x value of 1 again so f of 1 is what plug in a 1 here 1 minus 5 is going to be negative 4 and this will be over 2 times 1 is 2 plus three is five so you get negative four-fifths so the ordered pair here would be one comma negative four-fifths now if i take this negative four-fifths and plug it in for x i should get one back so i should get the ordered pair negative four-fifths comma one so let's try that so if we had f inverse of negative four-fifths this would be negative three times negative four-fifths minus five over two times negative four-fifths minus one so let's do this in the numerator first so we know that nothing is going to cancel here you would have negative three times negative four which is positive twelve so you'd have twelve-fifths then minus five you can multiply this by five over five and basically what you'd have here is twenty-five so twelve minus twenty-five is negative thirteen so this would be negative thirteen-fifths [Music] then down here two times negative four-fifths would be negative eight-fifths so this is negative eight-fifths and then you're subtracting away one so you can say minus five over five now negative eight minus 5 is negative 13. so this would be negative 13 fifths so what do we get negative 13 fifths divided by negative 13 fifths the same nonzero number divided by itself is 1. we know that this would be negative 13 fifths times the reciprocal of this which is 5 over 13 and it's negative and again these cancel and become 1 which is exactly what we said right we plugged this in we got a 1. so these guys are inverses of each other again in the next lesson i'll show you exactly the process you can use to prove this in this lesson we want to talk about how to determine if two functions are inverses so in the last lesson we learned how to find the inverse of a to one function basically we swap the x and y values okay and then we just solve for y and then we use our special notation f inverse of x that's all there is to it really so in this kind of section here what i want to do is show you how to determine if two functions are inverses so they give you two functions one f one g in most cases and they say hey determine if these two are inverses so the rule for this is very very simple if they're inverses then f of g of x is equal to x so this is f composed with g we already saw this earlier in the course and then also we'll say and here g of f of x is equal to x as well so both of these conditions have to be met don't just check one you have to check both so if we have f of g of x so f of g of x what is this equal to well again we already know how to do this we take f of x which is 2x minus 9 and everywhere there's an x there's one right there we're going to plug in g of x so g of x is this guy right here so what i'm going to do is i'm going to plug this in for x there i'm just going to simplify so i would have 2 this guy right here times x plug this in for x so times x plus 9 over 2 and then i'm subtracting away 9. so what does this give me let me scroll down well we know that 2 over 2 is 1 so that's gone and we basically have what you'd have this numerator here of x plus 9 and then minus 9 and this equals what 9 minus 9 is 0 so this is just x so in the first case it's true right we have f of g of x is x so let me erase everything and let's just put a check mark here and now let's do g of f of x so g of f of x and what's this going to be equal to we'll take g of x and everywhere there's an x there's just one we're going to plug in f of x so again this gets plugged in right there that's all we're doing so you'd have 2 x minus 9 okay that's for x there then plus 9. then over your denominator is going to be two so let me scroll down so we have some room to work and what do we have first we have minus nine plus nine so you basically say this is plus negative nine so negative nine plus nine is zero you can basically cancel those out and say this is two x over 2 which is just x right this cancels so we've proved the other condition and let me erase this also and i'll just put another check mark here so f of g of x is equal to x and g of f of x is equal to x so if both of those are true they're inverses so if you wanted to you could really kind of scratch this out and say this is f inverse of x like this if you wanted to i'm just going to leave it as g of x for now but you might also see this notation as f of f inverse of x so in some textbooks they'll do that and say this has to be equal to x and then also the and okay the and here would be f inverse of f of x would be equal to x so again both of these have to be true so you might see it this way or the other way all right so let's erase this and let me show you why this kind of works out to get a good intuition into this because this is generic notation let's just pick a point that works here so let's just say f of 2 f of 2 is equal to what you'd have 2 times 2 which is 4 minus 9 which is negative 5. so an ordered pair here that works is 2 comma negative 5. so 2 comma negative 5. again with the inverse those are going to be flipped so the y value becomes the x value the x value becomes the y value so it should be true that an ordered pair that satisfies g of x would be negative 5 comma 2. well let's see if that's the case so g of negative 5 is equal to what well you'd plug in a negative 5 here and then plus 9 that's 4 4 divided by 2 is 2. so these two are kind of the reverse of each other the x value became the y value and the y value became the x value now let's think about this a little bit more deeply if i think about let's just take the first scenario f of g of negative five okay what is g of negative five g of negative five is the y value that you get when you plug in a negative five in for x in g of x okay so we already know that if i plug in a negative 5 here i get a 2 okay so what's going to happen is what this turns out to be f of 2 because this this right here g of negative 5 we already know that's a 2. so now i have f of 2 and i want to say what is f of 2 well i go back up here i know this is negative 5. so f of g of negative 5 i can just erase this is negative 5. okay and you can play around with this a little bit more but basically that's all we're saying f of g of x is equal to x for any x in that domain and then g of f of x is equal to x again for any x in the domain so again just play around with this for a little while and this will start to make some sense for you again when you work with generic notation a lot of times it's confusing okay let's go ahead and jump in and look at another one so we have f of x equals the cube root of x minus three and g of x equals the quantity x plus three cubed so are they inverses so let's check f of g of x first and see what we get so this is my f of x i'm going to plug this in here for x so again this whole thing gets plugged in there so you would have the cube root of the quantity x plus 3 cubed and then you'd be subtracting away 3 out here so what does this give me let me scroll down and get some room we know that this kind of index would cancel with this exponent here and i'd just be left with this x plus 3. so i would have x plus 3 minus 3 which is just equal to x so the first condition is true f of g of x is x so what about g of f x so g of f of x is equal to y so let's scroll back up and again i'm starting with g of x and i'm just going to plug in for x there and this is what i'm going to plug in so i would have inside the parentheses this cube root of x minus three and then you'd have plus three and the whole thing is cubed okay so what do we have here well we have this minus three plus three that can be done so minus three plus 3 is obviously 0. so i can say this is the cube root of x and then this is cubed again this is going to cancel with this and i'm just left with x so in both cases the f of g of x is x and g of f of x is x so because both of those conditions were met we can say these guys are inverses all right let's take a look at another one so now we have f of x equals we have 2 over x minus 1 and then minus 3. and we have g of x equals we have negative 3 over x minus 1 and then minus 2. so again i'm just going to start with f of g of x okay this equals what well i'll have 2 over where i have an x here i got to plug this whole thing in so be really careful with this so i'm going to do negative 3 over x minus 1 and then minus 2. so this whole thing right here got plugged in right there okay you see that so if i just kind of x this out now what's left well i have minus 1 okay that's got to be done and again that's in the denominator down here and then outside of this i've got minus 3. so this one's really really easy to make a mistake on so be very very careful when you're plugging stuff in so let's scroll down so we have a lot of room to work so the first thing i'm going to do is do negative 2 minus 1 and i'm just going to do that in line and say this is -3 and now i would need a common denominator here so i would multiply this and before i do this let me put plus negative 3 so i don't make a silly sign mistake so i'm going to multiply this by the quantity x minus 1 and this is all kind of running over over the quantity x minus one okay so let me fix this real quick so it doesn't run over here and get confusing this is going to be negative 3x plus 3 so negative 3x negative 3x plus 3 and you're going to have a common denominator so you can just get rid of this and you can just kind of extend this over we'll say this is x minus 1 here now before we deal with anything over here realize that you have a complex fraction here so really what i can say is that i have 2 over 1 divided by and if i simplify this negative 3 plus 3 those are going to cancel okay so what i'd have is 2 over 1 divided by negative 3x over x minus 1. so this equals what 2 over 1 times the reciprocal of this which is x minus 1 over negative 3x okay so now that we know that let's go ahead and multiply 2 times x that's 2x and then 2 times negative 1 is minus 2 and this is over negative 3x so this is over negative 3x now i have this minus 3 over here that i have to deal with i want to get a common denominator going so what i'm going to do is i'm going to multiply this by negative 3x over negative 3x and again to not make a sign mistake i'm going to put plus negative here really easy to make a sign mistake and blow the whole thing so i'm going to say that this is equal to 2x minus 2 and then you have negative 3 times negative 3 times x so this would be plus 9x and then this is all over the common denominator of negative 3x so what does this give me 2x plus 9x is 11x and then you have minus 2 and this is over negative 3x so obviously this is not equal to x so you can say these two are not inverses of each other you don't have to go back and check g of f x if one of them fails they're not inverses in this lesson we want to talk about how to find the inverse of a function with a restricted domain so in a lot of cases you're going to come across a one-to-one function that has an inverse but that also has a domain restriction involved so a lot of these that you're going to see in your book are going to involve kind of square roots so something like f of x equals the square root of x plus 1. so we're going to see this is a one-to-one function and it does have an inverse but we have to be really careful when we calculate okay so let me explain why first and foremost we know this has a domain restriction because when we work with the set of real numbers we cannot take the square root of a negative unless somebody specifically tells you otherwise you're working with a set of real numbers so we have the square root of x plus one we know that this part right here cannot be less than zero so x plus one has to be greater than or equal to zero it can be zero that's fine square root of zero is zero you're okay but it can't be negative if you get square root of negative one you get into complex numbers or you say there's no real solution all right so from here i can solve this by just subtracting one away from each side cancel that out and say x is greater than or equal to negative 1. so let's just list this here and just say that this is valid for x that is greater than or equal to negative 1. okay now if we look at this graphically we can clearly see that this guy is a one-to-one function right it passes the horizontal line test with flying colors but we can also see that our domain restriction is here right you can see this point right there which is the point negative 1 comma 0 okay that's the leftmost point on this graph so for any x value to the left of that it doesn't exist on this graph it's going to be either negative 1 for the x value or anything to the right or anything larger going out to positive infinity now let's think about if we found the inverse of this guy without considering the domain restriction what's going to happen so let's go ahead and say this is y equals the square root of x plus 1. let's swap x and y so i'm going to swap this and i'm going to swap this so i'm going to say this is x equals the square root of y plus 1. and then to get rid of this kind of radical here i'm just going to square both sides so i'm going to have that x squared is equal to this will cancel with this and i'll have this y plus 1 here so just y plus 1. because i want my y what i'm solving for on the left that's just how i like it i'm just going to go ahead and flip this guy it's completely legal so i'm going to say y plus 1 is equal to x squared and then to solve for y all i need to do is subtract and i meant to use a marker there subtract 1 away from each side of the equation this will cancel of course and you get y is equal to x squared minus 1. now in most cases we would stop and say this is the inverse right so we would erase this and say f and let me kind of slide this down because i'm going to run out of room so we'll say f inverse of x is equal to x squared minus 1. but then you'd scratch your head and say hmm i don't think that's true because this if i had g of x equals x squared minus 1 that's not a 1 to 1 function so what's going on so let's take a look at that graphically this is let's just say g of x equals x squared minus one right so if i had g of x equals x squared and then i shifted it down one unit that's this graph here so this thing would fail the horizontal line test like crazy right you can make a bunch of different horizontal lines that would fail all over the place so what do we need to do to fix this well we need to impose a domain restriction so let's go down here and consider these two graphs side by side i want to think about the domain of this one and then what should be the domain of this one okay if we consider that this guy is going to end up being the inverse well let me kind of go down and isolate this one for a moment and we can think about this further when we find the inverse of a function we swap the x and y values so what that tells me is that the domain okay from the first one let's just say f of x becomes the range in the inverse f inverse of x so if i consider my f of x function and i just think about my domain well we already know that it's the set of all x such that x is greater than or equal to negative 1. we found that earlier but what about my range we can look at the graph and see this or you can go back and play around the equation but you can easily see from the graph that the lowest point is again here which is at negative one comma zero every other point on that graph is going up and it's going to go out to positive infinity so the range we could say is the set of all y let me make that a little better so we'll say the set of all y such that y is greater than or equal to zero right it can be zero or anything larger now because the x and the y's get swapped when we think about the inverse so let me do this in a different color if i do f inverse of x well now what's going to happen is the domain takes on the range so it would be the set of all x such that x is greater than or equal to 0 and the range would take on the domain so it would be the set of all y such that y is greater than or equal to negative 1. and if we want to see this graphically i've already plotted the function f of x which is in blue so this is f of x and i've plotted the inverse which is this guy in i guess that's orange not really sure so f inverse of x is right there and you can see that we've kind of cut the left side of the parabola off we got rid of it completely we only considered x values that were greater than or equal to zero so this is kind of the start of this thing when you consider the x values it's the x value of zero and then it continues out to positive infinity and a couple of things you can notice about the graph and its inverse the first thing is if you graph the line y equals x which is this guy in red maybe pink not sure so let's say y equals x but it's this line right here what happens is when you have a graph and its inverse you can take a given point and reflect it across this line y equals x and you'll come over here so this point right here is eight comma three so this is eight comma three if i fold it across the line y equals x i'll come over here to three comma eight so again you can see that the x and the y values just swapped okay that's all you're really doing and you could take other points if i took this one right here this is three comma two folded across the line i get to two comma three okay so on and so forth you can take this point fold it across the line get to there this point folds across the line get to there again so on and so forth so if we go back to right here where we wrote this we need to put a domain restriction in for this to be valid we can't say that this is true right now okay so let's copy this and slide it over here and we're just going to put a comma here and say that x needs to be greater than or equal to 0 and for this one x needs to be greater than or equal to negative 1. so this is how you would want to list this this function does have an inverse it's this but with a restricted domain now i want to prove this to you even further let's go down to a fresh sheet and let's just write f of x is equal to the square root of x plus one and i'm going to write the other one as just g of x just so i can save a little bit of that notation that makes everyone crazy so i'm going to say this is x squared minus 1 okay so if i don't consider any domain restrictions what happens is when you do the check remember you have this f of g of x is equal to x and remember that g of f of x is equal to x as well one of them is going to fail and it's going to fail into a lot of people because they don't remember the kind of rules that we talked about with basic algebra they're going to think it didn't fail so let me show you this real quick so that you understand what's going on so let's say i try f of g of x so f of g of x so i would take this function here f of g of x and i would plug in right there i'm plugging this in okay so i would take square root of this is going to be x squared minus one and then plus one okay so so far everything looks fine and dandy i know that minus 1 plus 1 that's going to go away so i have square root of x squared and a lot of people at this point go oh that worked so this cancels with this and it equals x it doesn't though so if you remember back let me kind of erase this from kind of the basic stuff we did at the very beginning of the course we know that the square root of x squared is equal to the absolute value of x to prove this real quick let me scroll down i'll come back up let's say i took the square root of something like negative 4 squared okay i have a negative 4. so this is not going to be equal to negative 4 it's going to be equal to positive 4. why because negative 4 squared is done first so negative 4 squared that quantity is 16. so let me just erase this and say now you have the square root of 16 which equals positive 4. so this became positive so you have to be careful about this okay so it would only be true let me kind of erase this real quick it would only be true that if x was greater than zero or equal to zero that the square root of x squared would be equal to x because the absolute value of x is equal to x if x is greater than or equal to zero and then the absolute value of x is equal to the negative of x if x is less than zero okay but because we put a domain restriction in here we can throw this case out and we can say that okay the absolute value of x just equals x because we have x restricted to be greater than or equal to zero so then we can say that this is actually equal to x because we're following this domain restriction so that's how we make this work out again if you're not paying attention and you forget that the square root of x squared is equal to the absolute value of x you'll mistakenly just cross those out and say this is x it works out what's this guy talking about okay so you have to be very very careful about this so we're going to come back up here and we're going to make sure that when we list this we again we kind of slide this down put our domain restriction in and say x needs to be greater than or equal to zero for this to be true okay it's true only when that happens now if we erase this and do the other one there's no issue no problem so we're going to say g of f of x is going to be equal to so instead of x squared remember i'm plugging this in for x so i'm going to have the square root of x plus 1 this quantity would be squared and then minus 1. no issues here you can go through and just cancel this away and say this is what it's x plus 1 minus 1 which equals x okay so that's why it's very important to check both of these okay that's why we told you that earlier where a lot of people just want to check one and be done if you get something like this and you just kind of fly through it you're not paying attention you're going to get the wrong answer and especially something like the sats or the acts they know these are common things that you'll make a mistake on and they're good trap questions let's take a look at one more so suppose we have f of x equals 1 over the square root of x so we already know that this guy would have a domain restriction right we would say that first and foremost you can't take the square root of a negative okay so we know that x would need to be greater than or equal to zero and then also it's in the denominator so x can't be zero so you can just erase that so that's my domain restriction what's my range here okay because i got to think about that when i do the inverse i've got to know what the range is because that's going to be my domain and the inverse i know that i can't plug in a 0 there but if i plug in something that is between 0 and 1 okay something in that range so let's say x was greater than 0 and less than 1 what's going to happen well the closer i get to 0 remember this is going to become a really really really small number infinitely small if you want it to be and 1 divided by some infinitely small number is going to approach infinity so if x is greater than zero and less than one we know that the y value is approaching positive infinity okay as we get closer and closer to zero now if x is one we know y is just 1. and then if x is some larger number let's say you start getting something like let's say you plug in 4 okay well square root of 4 is 2 you get 1 half and then as you increase this it approaches a y value of zero but it's never going to actually touch it so the range here the range here would be the set of all y such that y is greater than zero so the domain and the range are actually the same okay so the domain is the set of all x such that x is greater than zero the range is the set of all y such that y is greater than zero so when i find the inverse this range here will be the domain so let's go to a fresh sheet and we'll say that f of x is equal to 1 over the square root of x and i'll just go ahead and say that we'll have y is equal to 1 over the square root of x i'll interchange or swap x and y so let me do that over here i'll say now that we have x is equal to 1 over square root of y and what i can do now is square both sides but actually what i'm going to do is i'm just going to take this guy we know that y right now is raised to the one half power so i can just say this is raised to the one half power and if i wanted to drag it into the numerator i could say it was y raised to the power of negative one half well how could i cancel that well i could raise both sides to the power of negative 2. so let me raise this side to the power of negative 2 and this side to the power of negative 2 and so i get x to the power of negative 2 which is 1 over x squared is equal to y right because power to power rule those guys are going to cancel and become 1. now let me erase this over here and erase this and let's state this as our inverse we'll say f inverse of x is equal to we'll have 1 over x squared but again we have to put our domain restriction in here so we say that x needs to be greater than 0. and again my f of x is equal to 1 over the square root of x and we'll say again for the domain here x is also greater than zero now i'll leave it up to you to prove this you can go back and do the f of g of x or in this case because we have this notation you could say f of f inverse of x is x and then f inverse of f of x is also x and you need to do this with the domain restriction if you don't in one of the cases you're going to run into that problem where you have the square root of x squared and that becomes the absolute value of x so you're going to have to restrict that domain to kind of clear that up in this lesson we want to talk about graphing exponential functions so most of you have worked with exponential functions before they're of the form f of x equals a which is our base raised to the power of x which is our exponent so notice that you have a variable now in the exponent and basically you need to pay attention to these two restrictions on the base we say here that a again the base is strictly greater than zero and also a the base is not allowed to be equal to one i'll talk more about those restrictions in a moment for right now i want to talk about the kind of general expectations of what you're going to see when you graph one of these guys first and foremost it's not that easy to do because of the kind of limitations of a hand-drawn graph if you have a smaller base like two or three or even four it's doable but when you get into higher bases let's say you use 12 or something like that well you're going to have a problem kind of doing this by hand because if i had something like f of x equals 12 to the power of x well 12 to the first power is 12 but then 12 squared is already 144. okay so can you plot 2 comma 144 with a sheet of graph paper it's going to be really hard to do so that's why these are pretty difficult to graph overall and your textbook is usually going to use these kind of smaller bases as i am going to do as well today so on your graph as long as your base a is greater than zero and not equal to one you're gonna have these three points so negative one comma one over a will be one point zero comma one will be another and one comma a will be another so you can use these three points to kind of get things started and then find additional points as needed in order to sketch your graph let's say you have something like f of x is equal to let's just use 2 to the power of x very common example well if i plugged in a negative 1 for x f of negative 1 here would be 2 to the power of negative 1 which is 1 over 2 to the power of 1 which is just 1 over 2 okay and this is always going to be true again as long as a is greater than 0 and not equal to 1. if i had something like g of x is equal to let's say 5 to the power of x well if i had g of negative 1 well this is what it's 5 to the power of negative 1 this is just 1 over 5 to the power of 1 which is 1 over 5. so it's always going to be 1 over whatever the base is okay in this case it's five so we get one over five in this case it was two so we got one over two all right so we understand that one so let's go to the next one and the next two are very easy to understand so you have zero comma one as a point again if you raise something to the power of 0 as long as it's not 0 itself you get 1. and we know that we're not going to deal with 0 because this base here is restricted it can't be 1 and it can't be something that's 0 or less so we're good to go if i raise 2 to the power of 0 i get 1 if i raise 5 to the power of 0 i get 1 so on and so forth then the last one is also very easy we get 1 comma a which is the base so if i raise 2 to the power of 1 i get 2 right if i raise 5 to the power of 1 i get 5. you raise something to the power of 1 you just get the number so using this kind of function here i would have negative 1 comma 1 over 2 i would have 0 comma 1 and i would have 1 comma 2. now these aren't enough points to get the full graph but it's enough to kind of get things started now before we get into some graphs i want to talk about the restrictions involved with the base and i want to tie this into the domain and the range and explain why there's an asymptote at the kind of x-axis so you'll have a horizontal asymptote there so let's start out by just listing this guy again so we have f of x is equal to a which is our base raised to the power of x and the restrictions are that a the base has to be greater than zero and also a the base cannot be equal to the number one so first and foremost we're given the fact that the domain is the set of real numbers so from negative infinity to positive infinity i can plug in whatever i want for x there's no restriction but that comes at a cost to us because we have to now restrict a because we're not restricting what gets plugged in for x let's say for example we allowed a the base to be something that was negative so let's say we had negative 2 for example well if i plugged in something like one-half or 1 4 you'd run into an issue in the real number system negative 2 raised to the power of one-half is the square root of negative 2 which in the real number system we don't have that okay same thing goes if i raise this to the 1 4 power or the 1 8 power or something like that so because we want the domain to be all real numbers we've got to restrict the base if we think about the number 0 itself and the number number one then basically we restrict these because you wouldn't have an exponential function okay if you think about one let's say you had f of x is equal to one to the power of x well no matter what you plug in for x there if it's negative 0 or 1 you just get 1 here so this is basically going to go down to y equals 1 which is just a horizontal line right that's not an exponential function if you had something like let's say 0 to the power of x well this leads to all kinds of problems first off it's only defined where x is greater than 0 okay because 0 to the power of 0 is undefined and then if x was negative let's say it was negative 1 well 0 to the power of negative 1 is 1 over 0 that's undefined as well and again if you use negative exponents that's going to keep happening so it's only defined where x is greater than 0 and if you do plug something in that range well anything you plug in there let's say you plug 1 or 2 or a million you're just going to get 0 right 0 to the power of a million is just 0. so that's why that's restricted okay so that takes care of this and this so now let's think about the range why does it have to be greater than 0 well the reason is is because if you think about this and let's just write this as y equals a to the x right now and let me just use a practical example just say 2 to the power of x okay so let's say i was plugging different things in for x and getting a value for y let's say i started with a positive number like 2. 2 squared is 4 okay so that's positive if i kept going i would just get positive numbers if i'm plugging in positive numbers for this now if i plugged in 0 i would get 1. what happens when i plug in negative numbers well if i plug in something like let's say negative 2 okay then over here this is going to be what it's 1 over 2 squared which is 1 4. if i plugged in let's say negative 5 this is 1 over 2 to the fifth power which is 1 over 32. so by increasing the negative here so the way to think about this because as you move to the left on the number line the negatives are getting bigger and bigger right bigger in terms of being more negative that's what i'm saying they're actually smaller because we're moving left on the number line but as we get bigger and bigger negative numbers this guy gets closer and closer to zero but it will never actually get there okay so you can think about two to some incredibly large negative numbers so 2 to the power of negative let's say 100 000 for example something like that that would be really close to zero you can get as close as you want you could do two to the power of negative one trillion if you wanted to okay and see what that is it can be really really close to zero but you'll never actually touch it so because of this fact your y values in this function will approach zero but they will never touch it and so it forms a horizontal asymptote okay where y is zero so at the x axis now if you have a problem where you have some shifting involved this may change okay but generally speaking if you get the f of x equals just some number raised to the power of x you're going to have the horizontal asymptote that occurs at the x axis now let's actually look at the first example i've already pre-drawn all of these because i'm a terrible drawer and these things are hard to draw so i'm going to start with f of x is equal to 2 to the power of x this is a typical example you'll see in your textbook and the basic way to do this is just start out with kind of making a table of values just like you've always done so you know those three points already you know that there's negative one comma you're going to have 1 over the base the base is 2 so 1 half so that's a point and we can go ahead and plot that so negative 1 for the x and then one half of the y so that's there and then you know you have 0 and then 1 right if i raise 2 to the power of 0 i get 1. so the x value is 0 and the y value is 1 so that's right there and then we know we have a value of 1 for x that gives me a value of 2 for y right to the first power gives me 2. so an x value of 1 and a y value of 2 so that's right there and then i would just need to get some additional points going to make the graph clear we know going to the left that it's going to approach the x-axis but never touch it so i know if i chose some values let's say i picked negative 2 for example well this is 1 over 2 squared so that's 1 4. if i picked negative 4 this is 1 over 2 to the fourth power so that's 1 over 16. again so on and so forth so as this goes to the left and i'm picking bigger and bigger negative values for x the y values are getting closer and closer to 0 but they'll never touch it you don't need this you can get rid of this you're just going to draw it to where it approaches 0 but doesn't touch it and basically what i would do is get these points here so just think about the integers that are left if i go from 1 and go to 2 well 2 squared is 4 and then i would go to 3 2 to the 3rd power is 8. so that's enough for this guy you can go to an x value of 2 and a y value of 4 and then an x value of 3 and a y value of 8. so that would give you enough points to sketch this graph again it takes off very quickly and with f of x equals 2 to the power of x it's one that you can graph with your kind of graph paper let's look at one that takes off more quickly so now we have f of x is equal to 3 to the power of x and you'll notice that it takes off more quickly okay so it's harder to get points here because typically you're going to work with a coordinate plane that the max y value might be 10 or 15 or something like that if you think about 3 to the power of x well if i take 3 squared that's 9 but 3 cubed is already 27. so that's way off the page right if you do 3 to the fourth pi it's 81 so that's just accelerating very very quickly right so we know that again going to the left is the same phenomenon it's going to approach a y value of 0 but never touch it so we know how to draw that part and then for the ordered pairs again i would just start with what we know negative 1 and then 1 over the base so 1 3. 0 and then 1 1 and then 3 and then you could do 2 and 9 okay but that's basically where you'd stop because again if you do an x value of 3 you're going to get a y value of 27 you could maybe do that on a graph paper if you started really really low okay and you didn't really include this then maybe you can get to that but it's still pretty hard to do we have negative one for the x value and one third for the y value and we have zero comma one so zero for the x value one for the y value and then 1 for the x value of 3 for the y value and then 2 for the x value and 9 for the y value so that's enough points to get an idea of what this guy looks like and sketch your graph all right let's talk about some reflections now so we understand what a reflection is we get that if you have a graph that's f of x and then g of x is defined as the negative of f of x this is a reflection across the x-axis and then if g of x is equal to f of negative x this is a reflection across the y-axis so in this case we have a reflection across the y-axis we can see that and so what's going to happen is i'm going to define f of x as 2 to the power of x and then g of x is going to be f of negative x so basically we could say g of x is equal to 2 to the power of i'm just plugging in a negative x in the place of x so this is going to be a negative x up here now there's another way to look at this and i'll talk about this i'm going to graph this separately in a moment but right now let's think just think about it as a reflection so we already know these points because we already graphed this before so you have this three comma eight you have this two comma four you have this down here which is one comma two and then over here you have this point here which is zero comma one that's going to be on both of them and then lastly you have this one which is going to be negative 1 comma 1 half so to reflect this i'm going to keep the y value the same so over here the y value is the same but the x value will be its opposite so you'd have negative 3 right just change 3 into negative 3 and then the y value is the same then this point over here so the y value is the same so it's 4 and then i'm going to change the x value to its opposite so just negative 2. and then down here so on and so forth it would be negative one comma two and it's kind of hard to fit that in there so let me just try so negative one comma two and then if i go down here this guy gets flipped over so this would be positive one and then one half okay and then you have this point here that's on both so you have that 0 comma 1 that would be on both so if you wanted to graph this without reflecting it it would be fine you could do f of x is equal to 2 to the power of negative x and realize that this guy could be graphed in its current state or you could kind of manipulate this with exponents and you could write this as 1 over 2 to the power of x and 1 to the power of x would still be 1 so i could really say this is 1 to the power of x over 2 to the power of x and i can now write this as 1 over 2 all raised to the power of x okay so you can look at it this way or this way let me just kind of erase all this this is just getting us there whichever way you find to be easier i'm just going to work with it in this format so kind of getting some ordered pairs going i know my three points are going to be available to me so my negative one and then one over the base the base is one half so one over half is two i know my 0 1 is there right 0 for x 1 for y and i know my 1 and then the base the base is one half so that's there so those are my three points now this guy is an example of where in nature we would say it's a decay graph okay so it starts out high and goes low the other ones are a growth graph right they start out low and they go high so in this case what i want to do if i want to get some points up here kind of going up i need to go in the negative direction in terms of x so i would go something like negative 2 and negative 3 and of course we know that if we plug in a negative 2 there one half raised to the power of negative 2 through the rules of exponents we know that would be 2 squared right that's all that is which is 4. so this would be four and the same thing one half raised to the power of negative three through the rules of exponents we know that would be two to the power of three or eight so we saw this by reflecting it but we see it by kind of plugging in the points here it's the same thing so if you want to kind of plot these let's start with the 0 for x and 1 for y and then we have that an x value of 1 would give me a y value of one half and let's kind of go to the left now so we would have an x value of negative one which would give us a y value of two and then an x value of negative two will give us a y value of four and an x value of negative three will give us a y value of eight so again that's another way you could sketch the graph of this guy once you have your points plotted you could sketch this down like this and of course it's going to approach a y value of zero but again not touch it so these are always going to approach the x axis but not touch them in the case of a decay graph like this it's going to approach it coming to the right if you go back to your example with the growth graph you see that it approaches it as you go to the left so it starts out small if you're coming from the left and going to the right and then it grows like this so this is a growth graph and again because this one starts out high and comes down this is a decay graph all right let's look at another reflection so this guy is going to be reflected across the x-axis so we know that again if we were to find something like f of x to be equal to in this case it's going to be 3 to the power of x well then my g of x here if it's reflected across the x axis it's defined as the negative of f of x so in this case it would be the negative of this so the negative of 3 to the power of x so let me just erase this and put the negative of 3 to the power of x and of course now we're reflecting the points across the x-axis so what happens is the x values stay the same but the y values are now made into their opposites so if i have this point here which is 2 comma 9 and i reflect it across the x axis it becomes again 2 comma negative 9. so the x value now stays the same the y value gets turned into its opposite similarly this point right here which is 1 comma 3 is going to become what it's going to become 1 x 5 stays the same and then negative 3. then this point right here which is 0 comma 1. so this is 0 comma 1 is going to get reflected across and it's going to become 0 comma negative 1. and then this last point here which would be negative one let me write this over here so negative one comma one third would be reflected across so it's still an x value of negative one but now the y value is going to be negative one third okay so on and so forth so you can see that's a good way to kind of reflect it and sketch the graph you could also graph this guy on its own you don't need to reflect it so let's say g of x so you could say f of x it doesn't matter it's the negative of 3 to the power of x now you've got to be careful here because basically you're taking in terms of the y values that you're used to and you've got to make them negative so those three points that you're kind of working with all the time the negative 1 for x and then the 1 over the base well now you've got to make that negative so instead of 1 over 3 it's got to be negative 1 over 3. okay so you got to keep that in mind if it's a reflection and then if i had 0 comma 1 well now it's 0 comma negative 1. if i had 1 comma 3 well now it's 1 comma negative 3. you know so on and so forth so i can do my 2 here but now instead of getting 9 i get negative 9. okay so just keep in mind that you can work with the normal x values but you've got to make the y values into the opposite so again just taking these four points if i do negative 1 comma negative 1 3 that's there 0 comma negative 1 is right there 1 comma negative 3 is right there and then 2 comma negative 9 is right there so again that's enough to sketch your graph you know this guy going to the left is going to approach the x-axis but not touch it and coming to the right we're just going to go through those points and again you just draw a smooth curve put an arrow at the end all right so i want to do one more of these this is pretty typical in this section we look at some reflections but you're also going to have horizontal and vertical shifts like we talked about earlier so let's say f of x which is my guy in yellow is going to be 2 to the power of x we already know that graph we're comfortable with it now let's say we have g of x and this is equal to two to the power of x plus one and then minus one over here okay so we're going to have a horizontal shift one unit to the left that's what that plus one is going to do right there let me highlight that and you're going to have a vertical shift down by 1 unit that's what that minus 1 is going to do remember how you think about this we could say that g of x is equal to f of in this case it would be x plus 1 and then minus 1. okay so what's happening is this plus 1 here again you've got to think about the opposite of that if it's getting plugged in for x so to undo that this is the one that confuses everybody so if it's associated with x you undo it so the plus one you got to go minus one to undo it so it's a shift to the left let me just do this shift to the left by one unit okay and then this one that's outside of the function this minus one this just directly affects the y value so we're just shifting down by one unit so we'll say down by one unit okay so if you've got a graph of f of x equals 2 to the x already drawn and you just want to kind of shift this guy down by one unit and to the left by one unit it's pretty easy to do so just get your points going so this point right here which is three for the x value and eight for the y value gets shifted one unit to the left and one unit down so now what is this going to be let me write this over here well the x value is left by a unit so it's one units less so it's two and the y value is one unit less so it's seven then if i take this point right here which is 2 comma 4 again if i shift it it goes 1 unit to the left and 1 unit down to get me right there and so that is going to be 1 comma 3. then if i take this point right here which is one comma two again if i shift this one unit to the left and one unit down it ends up being what zero comma one now that ends up intersecting with another point that's there so that's a little bit confusing but this point is already there on the original one and that point of zero comma one gets shifted down so let me write this twice so let me put zero comma one over here so this zero comma one that's on the graph of f of x equals two to the power of x would get shifted one unit to the left and one unit down so that's this guy right here okay that's crossing through the x-axis and basically that's going to be what it's going to be negative 1 comma 0. and then lastly if you can see i have this last point here which is negative one and let me just draw an arrow for that one so it's negative one comma one half well that's going to go one unit to the left and one unit down so that's going to be what negative two comma and then if i took one half and subtracted away one i'd be at negative one half okay so i know this is messy but again if you're trying to show these kind of reflections or shifts or whatever it is you just want to kind of mark things up and draw your other graph it doesn't have to be perfect you just want to show the teacher that you understand what's going on in this lesson we want to talk about solving exponential equations with like bases so we just got done talking about exponential functions in general and now the next topic is to discuss how we can solve exponential equations when you have like bases so we're not going to involve kind of logarithms yet we'll see that in a few lessons so for now we just want to use this rule to solve some problems so if a to the power of x equals a to the power of y then we can say x equals y so what does this mean well as long as a this base here is greater than 0 and it's not equal to 1 then basically if these bases are the same meaning i have the same number here as i have here well then i can just set x equal to y and i can solve so if you have the same base then you're just going to set the exponents equal to each other and solve the resulting equation so it's easy enough let's just kind of jump right in and look at an example so we have 2 to the power of 5x minus 7 is equal to 64. so when you see problems like this again a lot of it is just making you kind of think through your properties of exponents you should know at this point that 64 is 2 to the sixth power if you didn't always look at this number over here the one that's kind of simpler if you have something that's a prime number like 2 or 5 or 7 check to see if the number on the other side is divisible by that so i know 64 is divisible by 2. if i set up a little factor tree then i'm going to go through and find out that it's 2 to the 6th power so what happens is i can now rewrite my problem and say this is 2 to the power of 5x minus 7 and this equals instead of 64 i'm just going to write 2 to the 6th power so now through a little bit of manipulation i have like bases so i have 2 and i have a 2. so what i want to do now is just set these exponents equal to each other and solve very simple process so 5 x minus 7 is equal to 6. we're just going to add 7 to both sides of the equation and of course this is going to cancel and i'll have 5 x is equal to 13. divide both sides by 5 and you get x is equal to 13 fifths all right so let's go ahead and look at another one so we have 1 7 raised to the power of 6x equals 343. again if you're in this section the idea here is that you're going to be able to rewrite things in such a way that you're going to have like bases so 7 is a prime number so that should be a little hint for you that hey is 343 divisible by 7. you checked it on a calculator it is 343 divided by 7 is 49 divided by 7 again you get 7. so that tells me 343 is 7 cubed so i know that on the right side i could just say this is 7 cubed now on the left side i don't have 7 i have 1 over 7. but again here's where you got to use your rules of exponents i know that 1 over 7 is equal to what it's 7 to the power of negative 1. remember that little rule so the first thing i'm going to have to do this is going to be two steps i'm going to say that 1 over 7 is 7 to the power of negative 1 but then it's raised to the power of 6x now what i want to do is i want to use my power to power rule and i want to say okay if 7 is raised to the power of negative 1 and then it's raised to the power of 6x the base here stays the same i multiply exponents so all this is going to be is negative 1 times 6x is my exponent and that'll just be negative 6x so most of the problem here is just setting things up okay after you've got that done a lot of these are really really simple equations to go through so i'm going to set this equal to this and just solve it you can almost do that mentally so negative 6x equals 3. divide both sides by negative 6 and we're going to get that x is equal to what this cancels and 3 over negative 6 is a half and you've got that negative there so it's negative one half as your answer all right let's look at one that involves a little bit more work so we have 216 raised to the power of negative 4x plus 7 equals 36 raised to the power of 2x plus 1. so if i look at this now neither of the bases are a prime number so i've got to start to think in my head how i can relate these two what would the like base be okay so 36 i know off the top of my head is 6 times 6 right so once i figure that out then i can look at 216 and say okay is that divisible by 6 yeah 216 divided by 6 is 36 well then okay 216 is 6 cubed so you're going to have 6 cubed and then this is raised to the power of negative 4x plus 7 and this equals you're going to have 6 squared and this is raised to the power of 2x plus 1. so here's where people start to make mistakes make sure that this 3 when you do your power to power rule gets multiplied by the whole thing make sure the 2 over here gets multiplied by the whole thing so let me kind of break this down so you don't get lost so power to power rule tells me that i'm going to multiply these two exponents so 3 gets multiplied by this quantity so 3 times the quantity negative 4x plus 7. make sure you use parentheses because if you don't you're going to make a silly mistake this equals 6 to the power of what use parentheses 2 times the quantity 2x plus 1. okay so once we've done this now we're just going to simplify i've got 6 to the power of 3 times negative 4x which is negative 12x and then plus 3 times 7 is 21 this is equal to 6 to the power of what 2 times 2x is 4x and then plus you've got 2 times 1 which is 2. okay so we've done all that work to get in the format where we can use our little rule we've got six and we've got six so we've got like bases so now what i want to do is just set my exponents equal to each other so i'm going to take this one set it equal to this one very simple so negative 12x plus 21 equals 4x plus 2. i'm going to go ahead and subtract 4x away from both sides of the equation and this is going to cancel i'm going to subtract 21 away from both sides of the equation and this is going to cancel let me get more room so now what i'm going to have is negative 12x minus 4x which is negative 16x and this equals 2 minus 21 which is negative 19. to finish this up let's divide both sides by negative 16. and let's just go ahead and cancel this out and we'll say i'll just write this up here that x is equal to negative 19 over negative 16 that's just going to be 19 over 16 right 19 is a prime number so nothing you can really do to simplify there unless you want to write it as a decimal okay let's look at the next one right now we have one that is a little bit more complicated so each of these is kind of grown in complexity so we have 32 raised to the power of 5 minus 3x times 64 raised to the power of negative 2x minus 1 and this equals 1 over 8. so if you look at 32 you look at 64 and you look at 8 you should be thinking immediately about a base of 2 right because 2 cubed is 8 to the 6 power 64 and 2 to the fifth power is 32. so if i just start with this 32 and say this is 2 to the fifth power then raised to this 5 minus 3x let's just go ahead and crank this out real quick again power to power rule this is going to get multiplied by each one of these so remember to do this you would multiply with this guy in parentheses so 5 times 5 is 25 and then 5 times negative 3x is minus 15x so i'm just going to erase all of this and use this as my exponent so 25 minus 15x okay so this one 64 is 2 to the sixth power so 2 to the 6th power again this is raised to this guy negative 2x minus 1. power to power rule 6 is going to get multiplied by each of these so what i'm going to do i'll just do this down here you wrap this in parenthesis so 6 times the quantity negative 2x minus 1 this equals 1. this is going to be 6 times negative 2x that's negative 12x and then 6 times negative 1 is minus 6. so let's go ahead and erase all of this and i'll say this is to the power of negative 12x and then minus 6. okay so let's just stop for a minute we'll figure out the right side in a moment remember our rule for kind of multiplying when exponents are involved if i have a to the power of m times a to the power of n this is a to the power of m plus n well it's the same thing here i have like bases i have a 2 and i have a 2 i'm multiplying so all i'm going to do is i'm going to say that this is 2 okay this is 2 raised to the power of i'm going to add these exponents together so let me write this as plus negative 15x and i'll say negative 15x minus 12x is negative 27x and then i'm going to do 25 minus 6 which is 19. okay so that takes care of that and you can even erase this at this point so this is the left side as simple as we can make it now for the right side i have 1 over 8 and remember i can say this is 1 over 2 cubed but we know that this is not 2 this is 1 over 2 cubed now i want you to think about the fact that you could write 1 over 2 cubed using a negative exponent by saying this is 2 to the power of negative 3. again if you drag this into the denominator everything stays the same we just change the sign of the exponent okay so that's our rule from earlier in the course so i'm going to write this as 2 to the power of negative 3. and again the main part here is just getting things set up so now this is going to be equal to this because i do have like bases so i would say negative 27 x plus 19 equals negative 3 and then from here let me scroll down get a little room going i would just subtract 19 away from each side of the equation cancel this and say negative 27 x is equal to negative 22 divide both sides by negative 27 and what's going to happen is this cancels with this and i'll have x is equal to i'll have 22 over 27 because again negative over negative is positive if you think about 22 it's 11 times 2 and 27 is 3 cubed so no way to really make that any simpler all right let's wrap up the lesson with one that seems kind of easy but does trip up a lot of students so we have 4 is equal to x to the power of 2 3. so you might also see this in this section what in the world do you do to solve this there's no way to really get like bases right because this is 4 and this is x so what you want to do here is think about this using kind of a radical notation that we learned earlier so i'm just going to go ahead and say this is x and i want the cube root of that and then this whole thing is going to be squared and i'm going to set this equal to 4. okay so this is just another way to write the x to the power of two-thirds this is going to be what it's raised to and this is going to be your index on your radical okay so what can i do to clear this 2 up here well i could take the square root of both sides so i could take the square root of this side and then over here let me kind of slide this down remember if you're taking the square root over here over here you've got to go plus or minus the square root so now i know this would cancel with this and i'm just left with that and so what i would have is just the cube root of x is equal to plus or minus 2. so let's erase this let's think about what this means we have the cube root of x is equal to 2 or we have the cube root of x is equal to negative 2. so there's two scenarios now what i can do from here is just cube both sides in each case to get rid of this kind of radical so i'm going to do that over here as well and what i'm going to find is that what well here i just have x is equal to 8 and then over here i would have x is equal to negative 8 so i can simplify that and just say x is equal to plus or minus 8. and this is a good one to check because it's kind of one that you have to think about for a while so if you plug in an 8 here and i say what is 8 to the power of 2 3 well let's think about that what is the cube root let me make that better what is the cube root of 8 well that's 2 right and then if i square that i'm going to get 4. so this becomes 2 and then if i square it i get 4. so that one checks out what if this was negative 8 okay what if it was negative 8 well what i would have is the cube root of negative eight which is negative two and we would put parentheses here because this amount this quantity of negative two would be squared okay and so we'd have negative two squared and that would also give me four okay so keep in mind you have two solutions here it would be x equals positive eight and also x equals negative eight and i'll just write that one more time so x equals plus or minus eight in this lesson we wanna talk about the compound interest formulas so when you talk about kind of exponential functions somewhere in this section you're going to run into this topic of compound interest now we talked about the simple interest formula earlier in the course now we're going to be able to talk about compound interest which is where you earn interest on interest so the formula goes like this we have a is equal to p times the quantity 1 plus r over n and this guy is raised to the power of t times n so this process here is very very straightforward okay you just need to understand what each letter in this formula represents so a is the future value so what we're saying here is this is the account balance in the future if i have an investment for five years at the end of the five year period a this here would be my future value or the account balance at that time p is the principal or you could say the present value it's how much the kind of investment is worth right now today how much am i putting into the bank or the cd or whatever it is then r is the rate as a decimal so if you have 10 percent it's 0.1 5 percent is 0.05 so on and so forth t is the time in years that's really straightforward and then n is the number of compounding periods per year so this is one that you do have to kind of know off the top of your head so if they say something like it's compounding monthly well we know that in a year there's 12 months so that number is going to be 12. if it was semi-annually well that means twice in a year so we would put that number as 2. you know so on and so forth something like quarterly would be four right four times per year so let's use this formula to kind of calculate this example so we have jamie invests three thousand one hundred fifty five dollars in a cd which is a certificate of deposit with an annual interest rate of four percent compounded quarterly let me go and highlight that it's quarterly so four times a year what is the account balance after five years so basically with this you're just plugging some stuff in and then you're going to use your calculator to get an answer you just have to make sure to follow the order of operations as long as you understand what you need to plug in this is super simple so again our formula is that a is equal to p times the quantity we have one plus you have r over and then we're going to have n and then this guy right here is raised to the power of n times t so jamie invests this amount right here this 3155 dollars that's the current value or the present value of the investment or you could say the principal so that's your p so i'm just going to instead of erasing it let me just kind of draw an arrow and say this is going to be 3155 and then when we go over here to r that's the rate as a decimal so my rate is right here it's this four percent as a decimal we can draw an arrow this is going to be zero point zero four and then we say it's compounded quarterly again think about a quarter as being what it's a fourth of a dollar okay so if you had four quarters you'd make a dollar well it's the same thing you have four quarters in a year okay so when we say the n is the number of compounding periods here that's going to be 4 and i'm just going to plug that in here as well and then t is the number of years that's here right it's 5 years so really once you understand what each letter is for you just write your formula out and you plug in and then you just do the calculation it's really really simple so let's come down here and we'll say a is equal to now i'm just going to plug in so i'm going to say 3155 times the quantity i'm going to have 1 plus i'm going to have 0.04 divided by 4 and this is raised to the power of 4 times 5. we'll go ahead and just write that as 20 for now and then let's come down here and see if we can simplify this a bit you could plug this into your calculator right now but i'm going to tell you if you don't know the order of operations you need to really revisit that topic okay because if you just plug it straight into the calculator a lot of times you're not going to get the right answer okay so let's just go ahead and simplify this as much as we can so i'm going to say a is equal to 3155 times what is 0.04 divided by 4 that's 0.01 so this would be 1 plus 0 0.01 and i'll just go ahead and say that's 1.01 so this is 1.01 and this is raised to the power of 20. okay so now it's very straightforward all i have is multiplication and i have an exponent operation again we know our pemdas or order of operations we raise to the power first we take that result and multiply it here but if you punch this into your calculator at this point it's not going to make an error it knows that if you do 3155 times this raised to this it'll do the exponent first and then it'll multiply so let's go ahead and crank that out and see what we get so on my calculator i get 3849 and then i get point six nine nine five seven six now with these problems you're usually going to round to kind of the hundredths place so what i want to do is think about this number here it's a 9 so i know i'm going to round up so i'm just going to make this a 70. so i'm going to just say that this is going to be 3849. and 70 cents okay and yeah that's an approximate amount but in real life this is what they would do you don't get an amount past this when you work with real money right you would just stop at 70 cents and so we're just going to copy this and i'll just paste this in here we can just say the account balance [Music] is we'll say three thousand eight hundred forty nine dollars and seventy cents and you can even put after five years and keep in mind we did round this but on this type of problem you're expected to round so this is definitely the correct answer all right let's take a look at another one and we can blow through this one now so robin invests twelve thousand dollars in a bond fund with an annual interest rate of six percent compounded quarterly what is the account balance after 11 quarters well again i want the formula a is equal to p times the quantity 1 plus r over n raised to the power of t times n again what is all this stuff a is the kind of future value of the investment so after the time period in this case it's going to be 11 quarters so we don't know that that's what we're trying to solve for p is the principal or amount invested in this case robin invests 12 000 so twelve thousand dollars okay times this quantity here we have one plus r which is the rate okay so that six percent is put in as a decimal so point zero six or you can put 0.06 if you want and then this is over the number of compounding periods again in a year it tells us that it's compounded quarterly so this is going to be again 4. now here's where it gets a little tricky you have this t times n t is the time period in years we're told this guy in quarters okay so let's get to that in a second we know n is already four so if i have 11 quarters how many years is that well the quick way to relate this in your mind is to think about okay if i had 11 quarters in money how much would that be in dollars it'd be two dollars and 75 cents right you do that pretty quickly because you know 12 quarters is three dollars but if you didn't do that the official way is to set up a little unit fraction and say okay well i know that one year is equal to four quarters and so i could say one year over let me just erase this four quarters has a value of one right because these two things are equal i can write the same non-zero quantity over itself okay it's going to give me 1. so i can then multiply this by 11 quarters and what's going to happen is these units that i don't want are going to cancel so i would have 11 divided by 4 which would give me 2.75 so if i come back up here i can just say this is 4 times 2.75 and if you do that on a calculator you're going to get 11. okay so let's bust this out real quick just pull out your calculator again to not make a mistake i always recommend to simplify what's inside the parentheses first because if you punch this in on your calculator you're not paying attention you're going to get the wrong answer so .06 divided by 4 is .015 so let's erase this and just temporarily put .015 and of course if i add 1 plus 0.015 it's just 1.015 so from this point you can just raise this to the 11th power in your calculator and multiply that result by twelve thousand you're going to get fourteen thousand one hundred thirty-five point again you get a long decimal so three eight seven two five but we're going to go ahead and round this again i'm just going to round it to the nearest hundredths so i'm going to think about this guy okay and so since this guy is a 7 i'm going to round this to a 39 there okay so i'm just going to erase all this and i'm going to erase this and i'm going to make this into a 9 and i'll put my dollar sign out in front and again you can make a nice little sentence and just say that the account balance is and i'll just drag this up here fourteen thousand one hundred thirty five dollars and thirty nine cents after you could say eleven quarters or two point seven five years all right so another type of problem you're going to see in this section involves finding the present value we have that ned savings account has a current balance of six thousand dollars if ned opened the account five years ago and received an annual interest rate of 1.8 percent compounded monthly what was the opening balance of the account so this is very very simple you use the same formula again a equals you've got p times the quantity 1 plus r over n raised to the power of t times n okay so let's think about this for a second we don't have p anymore that's what we want to solve for we want to solve for that now at this point you could divide both sides by what's multiplying p but i would just hold off on that for one second and i would just plug some stuff in so this is not all going to fit on the screen so i'm just going to go back and forth so the future value here or the account balance in the future or when the investment is done is six thousand dollars that's given to me right there so let's scroll down for a minute and i'm just going to write six thousand dollars or six thousand and this will be equal to again we don't know what p is that's what we need to solve for so then times we have one plus what's the interest rate let's go up so we have 1.8 percent and it's compounded monthly my r is going to be 1.8 as a decimal which is you can put 0.018 and then when we say compounded monthly this n here is going to be 12 right because there's 12 months in a year so let's put a 12 there and then the sky is going to be raised to the power of t times n we know n is 12 already what's t let's go back up and it says five years ago so he had this guy for five years so let's go back down and put a five here and now we're basically ready to go so all we're going to do is kind of simplify this inside so if i divide 0.018 by 12 i'm going to get .0015 so i'm going to get .0015 if i add that to 1 i'm just going to put a 1 out in front so that's pretty simple so i know 12 times 5 is 60. so let's take care of that and now all i need to do let me kind of scroll down and get a little room going i'm going to divide both sides by this amount so i'm going to divide this side by 1.0015 raised to the power of 60 and let me just kind of slide this down and slide this down so i have enough room and then i'm going to divide this side by that as well so 1.0015 raised to the power of 60. we know that the same non-zero number divided by itself is going to cancel become 1. so p will be by itself now okay and it's equal to this 6 000 divided by this number here 1.0015 raised to the power of 60. so if you punch that up on a calculator you're going to get that p is equal to 5483 you get .956896 so we're going to round this again if i look at this digit here it's a 6. so i'm going to cut this off and just put this as a 6 here and you can put an approximate symbol here if you want because it's not exactly equal when you round but we just want to answer the question with this number here so let me copy this and i'm just going to erase all this again you can make a little sentence if you want to you can say the opening balance let me drag this out the opening balance and we'll just say was and we have this five thousand four hundred eighty three dollars and ninety six cents and you can just put a period after that's good enough the opening balance was five thousand four hundred eighty three dollars and ninety six cents and if you wanted to check this you could pause the video and take this number run it back through okay the formula we've been using for the other two examples and you'll see you get pretty much this amount in this lesson we want to talk about the continuous compound interest formula the a equals p times our special number e raised to the power of r times t so we've already worked with the compound interest formula and we've already used it to solve some simple word problems let's just look at this real quick we have a the future value is equal to p the principal or present value of your investment times the quantity 1 plus r which is the rate as a decimal divided by n the number of compounding periods per year and this guy inside the parentheses is raised to the power of t which is the time in years times n again the number of compounding periods per year so if we want to kind of build up to this idea of continuously compounding interest let's start out with what we know so i have a basic example here so we can compare them we have a thousand dollars is invested at an annual interest rate of ten percent give the account balance after one year for the different compounding periods so we've got quarterly monthly daily and continuously we can just blow through this real quick we know the formula let me just write it again it's a equals you've got p times the quantity 1 plus r over n raised to the power of t times n okay so what you want to do is just plug in and use your calculator it's really simple so for quarterly this is going to be compounded four times per year so four times per year they're going to give you your interest deposit it in there and then you're gonna earn interest on that interest okay so for the p the principal it's a thousand dollars it's the amount invested and this is times one plus if you have r that's the rate as a decimal so that's point one then over n which in this case is going to be four right the number of compounding periods it's raised to the power of t times n now in this case our t our number of years is one so you can just forget about that because multiplying by one is going to leave something unchanged so we just put n there which is four so if you punch this up on a calculator you're going to get approximately and i'm just rounding this number one thousand one hundred three dollars and eighty one cents again i rounded this so let's just put this i'm just gonna kind of slide this down let's put this over here we'll say this is one thousand one hundred three dollars and eighty one cents now for a given interest rate and a given time period if i increase the number of compounding periods or how often i give you your interest so that interest is earning interest i should have more money right at the end of that kind of time period so let's see this so for monthly all i want to do is put a 12 here and a 12 here everything else is the same if you punch this up on a calculator you're going to get about 1104.71 cents again it's an estimate we're just rounding so let's go ahead and put this as 1104.71 cents again we increase the number of compounding periods from 4 to 12 and we saw that we ended up with more money at the end of the year now if we go with daily that would be 365 days for a normal year if it was a leap year would be 366. so in this case if you punch that up on a calculator you're going to get again if you round about one thousand one hundred five dollars and sixteen cents so let's erase this formula we don't need it anymore so we have this one thousand one hundred five dollars and 16 cents now there is a theoretical maximum number of compounding periods that you can kind of get to in other words if we took this n here and we kept increasing it right instead of 365 we went to a thousand then to a million then to a trillion we just went towards infinity what happens is as we're continuously getting interest deposited into our account our money is continuously compounding then the formula changes a bit okay and it involves our special number e so we say a is equal to p times the special number e remember that's about 2.718 raised to the power of r times t so the r again is the rate as a decimal the t is the time in years so this is a really simple formula to use in this case p again is the principle so it's 1000 times e okay and that's going to be on your calculator raised to the power of r which is point one and then times t which is one so really you can just get rid of this part and say we have one thousand times e raised to the power of point one and if you punch that up on a calculator you get about one thousand one hundred five dollars and seventeen cents so you can see it's very close to what we got with daily and if you continued out if you went with a thousand and then two thousand and three thousand as you keep increasing the number of compounding periods you get closer and closer to this kind of number that you get from this formula here all right let's look at another example this is a very easy concept just memorize the formula and plug into it that's all you have to do so we have that james invests two thousand dollars in a savings account paying three percent annual interest compounded continuously for seven years what is the account balance at the end of the term very easy again a equals p times e raised to the power of r times t most of this is just memorizing the formula hopefully your teacher lets you write it down on a piece of scratch paper but basically you're just going to plug in so for p i have 2000 then times e raised to the power of r times t in this case the rate is 3 percent so as a decimal it's 0.03 and then the time in years is seven so it's times seven so basically .03 times seven just think about three times seven that's 21 you'd have two decimal places there so it would be 0.21 and if you punch that up on a calculator you're going to get approximately 2467.36 so if you wanted to make a nice little sentence you can let me just erase this real quick we'll use that number so we'll just say that james has and i'll just kind of drag this up here so 2467.36 at the end of the term [Music] all right let's take a look at one more problem again this is super simple so susan wants an investment with an annual interest rate of six percent which is compounded continuously a zero coupon bond can be redeemed in five years for ten thousand dollars how much should susan pay for this bond so first and foremost let me explain this zero coupon bond in case you've never taken a finance course before basically these trade at a discount so they don't pay any interest you're not getting any payments while you own the bond you basically just get a discount to the kind of price that you're going to get at the end so it would be like if you loaned me 50 and at the end of let's say two years i gave you back a hundred but during that two year period you didn't see any money so that's what this is like so we know that let me kind of write the formula down again it's a is equal to p times e raised to the rt we know that she wants to earn an interest rate of 6 percent so i can go ahead and erase this and put .06 and we also know that this investment is for five years so times five if you want to go ahead and do that would be 0.3 or 0.30 if you wanted to write that so you can just put 0.3 here e we don't need to really worry about that that's built into our calculator but the p the principal is what well that's what we want to solve for because we want to know how much should susan pay for this bond how much should susan pay for the bond we don't know but we have the amount she's going to get at the end right at the end of this in the five years she's going to get 10 000 so i'm just going to erase this and put my 10 000 here put my comma in and basically i just want to solve for p so i want to divide both sides by e raised to the power of 0.3 and i'll have my answer so let me just kind of erase it from over here and you can punch that up on a calculator and what you're going to get let me just kind of erase this and i'll just do this over here it's going to be approximately four hundred eight dollars and eighteen cents so that's how much she should pay for the bond again she wants a six percent annual interest rate where it's being compounded continuously and it's a zero coupon bond so she pays this amount and five years later she gets this amount so let's go ahead and write that down as our answer and just say that susan should pay and i'm gonna put this eight dollars and 7408.80 cents just for the bond in this lesson we wanna talk about logarithmic functions so something like f of x equals a to the power of x is our exponential function where again a your base has to be greater than zero and also not equal to one we already talked about why those restrictions are important but basically from the lesson on graphing an exponential function you saw that it would pass the horizontal line test and therefore it's a one-to-one function and every one-to-one function has an inverse so basically we would start by saying that f x is y and it would be equal to a to the power of x now if you want to find the inverse you swap x and y so i would now say that i have x is equal to a to the power of y and then from here you would solve it for y and say you have f inverse of x but there's a problem here we don't really have a way to solve for an exponent until you introduce something known as a logarithm okay so a logarithm is an exponent that's something you need to remember if you've never worked with logarithms before so what i can do is i can write this in a different format in such a way that it allows me to solve for this exponent y okay so i'm going to put y out in front so y will be equal to and then i'm going to have my logarithm here so i'm going to have log with a base of a and it's the same base for my exponential expression there so the base is a and then of you're going to have this answer over here so that's going to be x okay so basically the way this works this is your base and it came from this base here okay then this guy right here your exponent is what you're solving for so it's going to be out in front so y equals this guy right here again when you say it equals something it means it's the same as so y is the same as this log base a of x that's why we're saying that a logarithm is just a fancy way to write an exponent then lastly this x right here when you work with it in a logarithm you call it the argument but really to try to make this simpler i'm just going to say it's the answer okay so the answer so a and s so this is your answer from this original problem so if you write that down it's easy to kind of go back and forth between the forms and if you get confused write out this generically and then you can kind of practice going back and forth that way when you get to the problems that don't involve kind of the same letters you can go back and forth between your exponential form and your logarithmic form now there's one thing that i want to note here okay i want to just erase this real quick if we look at this in kind of exponential form this x equals a to the power of y you'll notice that you need another restriction so these restrictions are going to hold so you're going to have that a is greater than 0 you're going to have that a is not allowed to be equal to 1 but also you're going to see that x needs to be greater than 0 as well so this is another restriction you have to have when you're working with logarithms and the reason for this is simple if you take some number a and again we're working with something greater than zero not equal to one and you raise it to some number y and y can be any real number it can be from negative infinity to positive infinity so negative zero positive whatever you want well what happens is the result of this can never be 0 and it can never be negative and it just has to do with the rules of exponents if i have let's say 2 as an example as a and let's say y is some negative number let's say it's negative 2 we've talked about this previously this is 1 over 2 squared which is 1 4. so if you increase the negative okay meaning i go to the left on the number line or i make it a bigger negative this guy right here this result just gets closer and closer to zero but it's never going to actually touch it okay so that's why we have this restriction here that x is greater than zero so if you ever see a problem with the argument here as zero or something that's negative then you know they're trying to throw you a trap question okay so that's not going to have an answer all right so let's look at this one more time so we have our log form y equals log base a of x in our exponent form x equals a to the power of y again this is your base this is your base okay easy to remember this is your exponent over here it's solved for the exponent that's how i keep track of it i know that it's solved for the exponent in logarithmic form and then this right here again your argument you could really say it's the answer so a to the power of y will give me an answer of x a to the power of y will give me an answer of x all right so let's jump in now and look at some basic examples of kind of going back and forth between the exponential form and the logarithmic form so i already have this laid out for us the log base a of x equals y just as a reminder again this is a to the power of y is equal to x so a to the power of y is equal to x okay so here we have log base 2 of 4 equals question mark so how do we figure this out well we can convert it into exponential form again using the same process so 2 to this question mark here 2 to something is equal to again this guy right here so 4. so we don't need to do anything fancy here we know that 2 squared is 4 okay that's pretty obvious so the question mark here would just be 2 okay so 2 squared equals 4 and we can write this in logarithmic form by just saying log base 2 of 4 is equal to 2. again it's solved for the exponent see how this exponent here matches what it's solved for there all right look at one more and for this one i'm going to first put it into exponential form and then put it into logarithmic form so 3 to the power of negative 2 we know is what so 3 to the power of negative 2 is 1 over 3 squared which is 1 9. okay so if i have 3 to the power of negative 2 equals 1 9 how can i put this in logarithmic form again this is the base okay this is the base so you would have log base 3 of what's my answer it's this one right it's 1 9. so that comes in here that's going to be the argument of the logarithm so 1 9 and this equals what it equals this guy remember it's solved for the exponent so the exponent is negative 2 so that goes on one side by itself so log base 3 of 1 9 equals negative 2 3 to the power of negative 2 equals 1 9. again different ways to kind of write the same thing all right let's wrap things up by looking at some very simple equations as we move throughout this chapter we'll see much more complicated examples but we need to talk about the properties of logarithms first before we can get into that so suppose we have log base a of x equals k and then we transform this into what a to the power of k is going to be equal to x a to the power of k is equal to x so we're going to use this in order to solve some of these problems and we start with log base 3 of 81 equals x so we're just going to put this in exponential form so again the base is 3 the exponent is x so 3 to the power of x equals the answer is 81 okay now some of you are saying hey you need logarithms to solve that but remember you don't because 81 can be written with a base of 3. i can say that 81 is 3 to the 4th power and earlier in the course we talked about the fact that if you have the same base okay if you have the same base so 3 and 3 then the exponents you would set those equal to each other and solve in this case you would just get x is equal to 4 okay you can prove to yourself that that's true if i had log base 3 of 81 is equal to 4 we got 3 to the fourth power equals 81. 3 to the 4th power equals 81. what about log base 4 of x equals 7 halves well again let's set this up in exponential form so you've got your base 4 raised to this power again solve for this so raised to the power of 7 halves and this equals x right it equals this guy right here now we don't need to do anything fancy here we just need to figure out what is 4 to the power of 7 halves and remember this guy right here can be written as what 4 to the power of 7 halves is really 4 to the one-half power raised to the seventh power okay and four to the one half power is the square root of four which is two so this is two to the seventh power which is a hundred twenty eight so again i could write this as log base four of 128 and this equals 7 halves because 4 to the 7 has power gives me 128. all right look at another one so we have log base x of 1 8 equals 3 again we'd have x which is the base to the third power equals 1 8. okay so for this one i do need to do a bit of work i need to take the cube root of each side so i'll take the cube root of this side and the cube root of this side and so this whole thing is going to cancel with this kind of exponent of 3. i'm just left with x and that'll be equal to what the cube root of 1 8 is one-half because the cube root of 1 is 1 and the cube root of 8 is 2. so it's one-half so we can erase this and we can basically say x equals one-half because log base one half of one eighth is equal to three right if i took one half and i raised it to the third power i would get one eighth right one cubed is one and two cubed is 8. all right let's take a look at a common trap question so we have log base x of negative 5 equals 10. so whenever you see a negative or a 0 as your argument you need to stop immediately okay this is a very common trap question there will not be a solution here remember your base for this guy is going to be restricted it's got to be greater than zero and it can't be equal to one okay so given those restrictions if i set this up as x to the power of 10 is equal to negative 5 there's no such x right there's no solution here that's not possible not possible if i plug in something that's positive there and again not equal to 1 well i'm always going to get a result that's positive so there's no solution here all right so now let's kind of look at two harder examples or you could really just say more tedious examples they're not necessarily harder so we have the negative of log base 5 of 7 minus 6x and then outside of the parentheses we have minus 2 and this is equal to negative 1. so if you get a problem like this the first thing you want to do is write it in a familiar form so we know that we can solve something like log base a of let's say x is equal to just something like b okay we know that a to the power of b is equal to x and we can solve stuff given that we only have one unknown right so for this one i'm just going to put it into that format so i want to take this part right here and set it equal to some number over here so how can i do that well what i want to do is i want to add 2 to both sides of the equation and this will cancel and that will leave me with this negative here i'm just going to write that as negative 1 times log base 5 of you've got the 7 minus 6x and this will be equal to negative 1 plus 2 is positive 1. okay from here it's pretty simple because we just have the negative 1 that we've got to get rid of because it's multiplying this guy over here so i can either divide both sides by negative 1 or multiply both sides by negative 1. it doesn't really matter if i multiply the left side by negative 1 then i would just clear that it would just become 1. so i would have log base 5 of 7 minus 6x and then over here if i multiply this side by negative 1. let me just kind of do this so let me do this over here and i'll put times there so these two would cancel just give me one and over here one times negative one is negative one okay so from this point on it's pretty simple okay once you kind of figure out how to get in this format then basically just do what we've been doing 5 to the power of negative 1 is equal to this guy right here okay so you'd have negative 6x plus 7 that's how i'm going to write that and basically 5 to the power of negative 1 is one-fifth so one-fifth is equal to negative 6x plus 7. if you want to clear that denominator you can multiply everything by 5. so if i multiplied one-fifth by 5 i'd get 1. if i multiplied negative 6x by 5 i would get negative 30x if i multiplied 7 times 5 i would get 35. so at this point i can solve for x i'm just going to subtract 35 away from each side of the equation this will of course cancel 1 minus 35 is negative 34. this equals negative 30x and if i want to get x by itself i divide both sides by negative 30. and what's going to happen is this cancels with this i'm going to get that x is equal to negative 34 over negative 30 the negatives cancel each is divisible by 2. 34 divided by 2 is 17. 30 divided by 2 is 15. so let me write this over here i'm going to say x is equal to 17 over 15. okay let's take a look at one more of these again this is very very simple you're just using some basic basic algebra so we have 5 times log base 6 of negative 8x minus 6 then outside we have minus 4 and this equals 6. again you want to get it to the point where it's this guy right here is equal to one number over here okay so to do that what do i need to do i need to get rid of this 4 over here so add 4 to both sides and that cancels and then what i'm going to have is 5 times log base 6 of negative 8 x minus 6 and this equals 6 plus 4 is obviously 10. now to get rid of this 5 over here that's multiplying this log base 6 of negative 8x minus 6 then i'm just going to divide both sides by 5. okay and that'll take care of that so this cancels and this over here is going to be a 2. so now i have in the format that i want so log base 6 of negative 8 x minus 6 is equal to 2. so don't let it scare you when you see something long like that it's not that big of a deal you just need to go through a few steps to get it into a format that you can do something with so now we can put this in exponential form to get a solution so my base is 6 and it's raised to the exponent of 2 and it's equal to what it's equal to this guy right here the argument of the logarithm so negative 8x minus 6 6 squared is 36 and let me flip this around so let me say negative 8x minus 6 equals 36. you scroll down get a little bit of room going and let's add 6 to both sides of the equation that's going to cancel we get negative 8x is equal to 36 plus 6 is 42. so this is 42 here and let me divide both sides by negative 8 to finish this off and so this cancels we get x is equal to 42 over negative 8. now i know that's going to be negative each is divisible by 2 42 divided by 2 is 21 and 8 divided by 2 is 4. so you get x equals negative 21 4. in this lesson we want to talk about the properties of logarithms so the first property i'm going to cover is known as the product property for logarithms or you might hear called the product rule for logarithms and what we'll say here is that x is greater than zero y is greater than zero a is greater than zero and a is not allowed to be equal to 1. and so this is our property log base a of x times y so this is multiplication here this is equal to log base a of x plus log base a of y so this is a really important property something we're going to use all the time when you start out with the form on the left and we go to the form on the right it's known as expanding okay and if we start out with a form on the right and we go to the form on the left it's known as condensing and we'll see examples of each at the end of the lesson after we go through the properties now really quickly i just want to show you some common misconceptions some trap things that you might be thinking that work but that don't work so the first one is if you say something like log base a of let's say x plus y so a lot of people get confused okay and they say okay this is log base a of x times y like this this is not the same okay this is not the same as this what i wrote up here notice that you have log base a of x plus y like this over here you have log base a of x plus log base a of y so don't do this this does not work out this is wrong okay so that's the first misconception the second misconception is to take this and think that you can use some sort of distributive property here so a lot of people will say okay this is log base a of x plus log base a of y this is also wrong okay we know that log base a of x plus log base a of y is equal to log base a of x times y it's not log base a of x plus y all right so next we're going to talk about this quotient property where again you could say the quotient rule for logarithms and we have the same restrictions so x is greater than 0 y is greater than 0 a is greater than 0 and a is not allowed to be equal to 1. so we have log base a of x over y is equal to log base a of x minus log base a of y so here you've got to pay attention because this guy right here in the numerator goes right here okay so it's in the first position so it's log base a of whatever's here first in the numerator then minus log base a of whatever's here in the denominator in this case it's y okay so this is how this sets up all right the last one i'm going to cover today and there's more properties but i'm going to do them in the next video again if x is greater than 0 y is greater than 0 a is greater than 0 and a is not equal to 1 and r is going to be any real number well this one is where log base a of x to the power of r is equal to r times log base a of x so basically what this property allows you to do is take your exponent out in front okay so here it's out in front it's just been pulled down so this is very useful for us when we're solving equations and we'll see that in the upcoming lessons all right so now that we know the properties let's jump in and look at some examples in most cases these are very very simple you'll get a few that are kind of difficult but most of them are very easy so we have log base 9 of we have x to the sixth power times y cubed times z so we want to expand this okay so we're going to use our product property for this or again the product rule however you want to think about that and the first thing i'm going to do is realize that again if this is multiplication in here i can say that this is log base 9 of x to the sixth power plus let me change colors here log base 9 of this y cubed then plus let me change colors again so we'll have log base 9 of your z okay so that's pretty simple the only thing you want to do in addition to this if you have these powers here you want to bring them out in front if you're expanding okay so let's go ahead and do that real quick so i'm going to have 6 times this log base 9 of x now and then i'm going to have plus my 3's coming out in front so 3 times log base 9 of this y and then plus let me change colors once more and just have my log base 9 of z okay so log base 9 of x to the sixth power times y cubed times z turns into 6 times log base 9 of x plus 3 times log base 9 of y plus log base 9 of z all right let's try another one so we have log base 7 of and then for the argument here we have this zx that's raised to the fourth power over y and then this whole thing is raised to the fourth power so you just have to be careful here to set this up correctly again we want to expand this so the first thing i'm going to do and you can do this multiple ways but the first thing i'm going to do is apply this exponent here to the z and the x so i'm going to say that i have log base 7 of z to the fourth power x to the fourth power over y and this whole thing will be raised to the fourth power okay so at this point what i'm going to do is i'm going to apply the exponent here this 4 to everything so i would have log base 7 of you would have z to the 16th power right you just multiply exponents here z stays the same then x to the 16th power same thing multiply exponents x stays the same this is over y to the fourth power okay so once it's in this format it's pretty simple overall and let me just put equals here now you have division involved and you also have multiplication involved so what you can do and this is helpful so you don't make a sign mistake and we'll see some examples where you can clearly make a sign mistake in the next example but what you can do here is just go ahead and say okay well this is my numerator this is my denominator so based on our rules i can say that this is log base 7 of the numerator in this case that's z to the 16th power times x to the 16th power okay and then i'm going to subtract away log base 7 of the denominator which in this case is y to the fourth power okay so i'm just going to start with this you can do it a different way but that's how i'm going to do it again this is going to keep you from making a sign mistake later on when you get to tougher problems so over here what i'm going to do is expand this now because i have this multiplication okay so to do that i'm going to say i have log base 7 of z to the 16th power plus log base 7 of x to the 16th power and then minus my log base 7 of y to the fourth power okay so we're almost there now let me put my equals in here and all i'm going to do is i'm just going to bring these out in front in each case and we will be done so the 16 comes out in front we have 16 times log base 7 of z then plus 16 comes out in front so we have 16 times log base 7 of x then minus my 4 comes out in front so 4 times log base 7 of y and that's it so we have 16 times log base 7 of z plus 16 times log base 7 of x minus 4 times log base 7 of y all right let's look at another example and this one's a little bit more challenging and you'll run into an issue here where you can easily make a sign mistake so i'll show you how to guard against that so we have log base 5 of you have the square root of x squared plus 1 over x cubed times let me put that multiplication symbol in there so it's clear the quantity x plus 1 to the fourth power okay so if we want to expand this let me start by thinking about first off if you have a radical with these problems go ahead and write it in exponential form so you know if you have the square root of something it's raised to the one half power if you have the cube root it's raised to the one third power fourth root raised to the one fourth power so on and so forth so i'm going to first say this is log base five of i'm just gonna say that this is and let me kind of make some parentheses here and then some inner parentheses with x squared plus one this is raised to the one half power and then down here i have my x cubed times again this x plus one this quantity raised to the fourth power so that's my set up here okay so what i'm going to do is i'm going to think about my quotient property first that always keeps me from making a sign mistake so what's in the numerator this guy right here so let's start with that so we'd have log base five of what's in the numerator which is x squared plus one again this is raised to the one half power and then i'm going to subtract away again you're going to have that log base 5 so log base 5 of what's in the denominator okay so in this case it's going to be x cubed times this quantity x plus one to the fourth power here's where you've got to again pay attention you're subtracting this whole thing away this can be further expanded because you have this multiplication involved so to not make a sign mistake what i'm going to do after i set this up i'm going to put some brackets around this okay and i advise you to do the same i think you should always work with your quotient property first so you can set it up this way because if you don't it's very very easy to forget about the sign and turn in the wrong answer so let's scroll down a little bit and let's put equals here and what i want to do now i'm just going to pull this out in front so this comes down here i'm going to have one half times log base 5 of you've got this x squared plus 1. so this part is done i can't do anything else then you've got minus i'm still going to have my brackets here with this i know that i can expand it because i have my multiplication so i'm going to have log base 5 of this x cubed and then plus you have your log base 5 and let me kind of scroll this over so it fits so then of this x plus one okay and this is raised to the fourth power and then i'm going to close these brackets now i'm not done yet so let's put equals and let's keep going i want to pull this down and i want to pull this down and i want to distribute the negative to finish this off so let's first pull those exponents down so we're going to have that one-half times log base 5 of x squared plus 1 and then minus i'm still going to have my brackets i'll remove them in a second so we have this 3 that comes down so 3 times log base 5 of x now the 4 comes down so plus 4 times log base 5 of x plus 1. okay so now if you want to remove the brackets this is what's going to remind you again you're subtracting the whole thing away so remember if you do this if you have minus in front of parentheses or in front of brackets or whatever it is you've got to distribute that to each guy each term okay so if you get rid of this everything becomes negative so this gets distributed to this and this and now you're not going to make a silly sign mistake okay and what would have happened is if you don't use brackets the common thing is you just put minus the first thing and you have plus the second thing and that's where you get messed up and you turn in the wrong answer okay so let me put those brackets back so basically you want to set it up this way so you end up with a nice clean answer and you don't make a silly mistake you get points off on your test all right let's take a look at some examples with condensing now so in my opinion these are a bit trickier they're not that bad overall but they are a little bit harder than when you expand so we have log base eight of x plus log base eight of y plus log base eight of z plus log base eight of w this is all over two so i want you to realize first and foremost that this is a common denominator okay so i could have log base 8 of x over 2 plus log base 8 of y over 2 you know so on and so forth so what i'm going to do is consider this as if i had one half multiplied by each one of these guys and then what i could do is i could just pull it out right so i can say that i have one half times inside of brackets or parentheses or whatever you want to use i would have my log base 8 of x and then plus my log base 8 of y and then plus my log base 8 of z and then plus my log base 8 of w okay so once that's done what you can do from here is use your product property on the inside so i can condense this all down to i'm just going to start with my one half times then i'll have log base 8 and again i can condense this down to x y z w or if you want to write that in alphabetical order just do w x y z okay like that now this guy can come back up here so we'll say this equals log base 8 of you'll have your w x y z and this whole thing so each one of these guys can be raised to the one half power but instead of doing that i'm just going to write it with the square root so i'm just going to say we have log base 8 of and i'll say the square root of w x y z all right let's take a look at one more and this one's quite tricky just because of the signs so we have 24 times log base 7 of a minus 4 times log base 7 of b minus 4 times log base 7 of c okay so there's a lot of different ways to do this if you see this situation this is how i do it if you want to choose to make these guys plus negative 4 and plus negative 4 and then treat them as negative exponents that's fine that'll get you there too but what i'm going to do is i'm going to factor out the negative in each case i'm going to just treat that as negative 1 pull it out so what i'm going to say i have here is 24 times log base 7 of a i'm going to pull this minus out so i'm going to put a minus here and then inside of parentheses or brackets or whatever you want to use i'm going to have 4 times log base 7 of b this will be plus now because remember i pulled the sign out the negative so 4 times log base 7 of c and so inside of here i know that i can do some additional work and i can do some additional work over here so let's do that real quick and then we'll use our quotient property because we see we have a minus sign so over here i know i can bring this back up so i'm going to start by saying i have log base 7 of a to the 24th power okay you can put parentheses around that if you want to make that cleaner and then minus again inside of these brackets i'm going to go ahead and pull this up so i'm going to have log base 7 of b to the fourth power then plus i'm going to pull this up so i'm going to have log base 7 of c to the fourth power okay so from here it's not too bad i'm going to keep this one unchanged so log base 7 of you've got your a to the 24th power then minus over here what i'm going to do i have my product property i can use so i've got this addition i'm going to condense it down and say i have log base 7 of you've got b to the fourth power and you've got c to the fourth power that are not going to be multiplied together so i'm going to put b to the fourth power times c to the fourth power like that okay and now i have my quotient property i can use because again i have this minus here so i have the same log base 7 and log base 7 so that is what i'm going to write first so i'm going to have my log base 7 and then of so this guy's going to go in the numerator this guy's going to go in the dominator so i'm going to go a to the power of 24 in the numerator and then over in the denominator i'm going to have my b to the fourth power times c to the fourth power okay so this is how you want to write it make sure you're not putting log base 7 of a to the power of 24 over log base 7 of b to the power 4 times c to the power 4. that's another common mistake please don't do that this is how you write this log base 7 of this guy in the numerator and then this guy in the denominator that's all you need to do in this lesson we want to continue to talk about the properties of logarithms and also we're going to move into finding the inverse of an exponential or logarithmic function all right so before we can get into finding the inverse of an exponential or logarithmic function in some cases you're going to need some additional properties and we need to develop those and i just want to show you where they come from instead of just kind of giving them to you so i'm going to start with an exponential and logarithmic function we know they're inverses i'm going to prove that they're inverses through function composition and from that process we're going to get two new properties that we can use in our lesson today so we have f of x equals a to the power of x and we know that a is greater than zero and not equal to one and we have g of x equals log base a of x and we know that a is not equal to one greater than zero and x the argument of the logarithm is greater than 0. all right so from this we already know they're inverses so we can do our function composition so f of x equals my a to the power of x and g of x equals log base a of x we know these guys are inverses so if they're inverses we know that f of g of x equals x and g of f of x equals x let's start with this one so i would take my f of x which is a to the power of x and where i have this x here i'm going to be plugging in g of x g of x equals this so this is going to get plugged in right there so up in the top for my exponent i have log base a of x now let's stop for a minute and think about this why would this equal x well if i take this logarithm on its own and just kind of write it separately so we can think about this log base a of x what happens is this is the base a is the base we already know that let me kind of highlight this or circle it or put a square around it whatever you want to say there so that's my base and it's raised to a certain power and i get an answer of x so really i can say that log base a of x is the exponent to which this base a needs to be raised to in order to obtain again this answer of x well i have a base of a and it's raised to the exponent here because this is really just an exponent required to give me an answer of x so i just get x to see an example of this because i know when you work with this generically it's quite confusing to see an example of this let's suppose that we had 5 raised to the power of log base 5 of 25 what's this going to be equal to well again think about this separately well log base 5 of 25 is the exponent to which the base 5 again this base 5 must be raised to in order to obtain an answer of 25 so this is an exponent it's a value we know it's 2 but let's pretend we didn't know that well i have 5. it's raised to the power required to give me an answer of 25 so the answer is just 25 okay so that's all it is so in general when you see let me just kind of write this as a generic statement over here when you see a raised to the power of log base a of x it's just equal to x so let's now erase this and let's think about the other one so now i want to do g of f of x so g of f of x and remember this is going to give me x as well so now i'm going to plug this thing right here the f of x and for x in g of x so this equals log base a of and let me change colors here so you're going to have a to the power of x as the argument of the logarithm now for this one it's a lot easier to wrap your head around you already know that from the power property you can bring this out in front legally so what i can do is say this is x multiplied by log base a of a now this guy right here represents one why does it represent one well this is a raised to what power gives me a right that's the exponent that this guy represents a raised to what power gives me a well it's 1. so really i can just say that this is x times 1 which is just x so that's another property we're going to be using we can basically just say and i'm just going to put over here this is just x we can say that log base a of a is equal to 1 and then also we can say log base a of a to the x power is equal to x all right so now that we have those special properties under our belt let's move on and look at how to find the inverse of an exponential or logarithmic function so these problems are pretty easy overall i'm going to start out with f of x equals this log base 2 of 4x and again the first thing you want to do is just replace f of x with y so i'm going to say y is equal to my log base 2 of this 4x and then i'm going to swap x and y so instead of y here i'm going to have x and this is equal to log base 2 of again i have 4x so now i'm going to put 4y okay so to solve this for y which is our next step what i want to do is put it in exponential form so remember how to do this this is your base so your base is 2 then it's raised to this power remember it's solved for the exponent when it's in logarithmic form so 2 to the power of x is going to be equal to this guy right here the argument of the logarithm that's your answer okay so this is the 4y here so to solve this for y is pretty simple i'm just going to divide both sides by 4 and let me kind of scroll down and get some room going so i'm going to divide both sides by 4 and what i'm going to end up with and i'm going to flip the sides around so that y is on the left so i'm going to say y is equal to this 2 to the power of x over 4 okay so the only thing you want to do now is just replace y with your f inverse of x so f inverse of x is equal to 2 to the power of x over 4. be careful here the tendency for a lot of students is to think this is 2x over 4 it's not this is 2 to the power of x over 4 so you can't go through there and cancel a common factor of 2 in case you were trying to do that let's take a look at it all so i'm going to replace my f of x with y this equals log base 4 of my x cubed and again i'm going to swap x and y so i'm going to put x is equal to log base 4 of y cubed and now i just want to solve for y so again what i'm going to do here is i'm going to put it in exponential form so i'm going to say that 4 to the power of x is equal to y cubed so 4 to the power of x is equal to y cubed now at this point if i want to get y by itself i just take the cube root of each side and let me just kind of slide this stuff down so i have enough room to do this so i'm just going to take the cube root of this side and the cube root of this side and again you really can't do anything with this over here other than to say that you have y is equal to the cube root of 4 to the power of x you could also say you have 4 to the power of x over 3 if you wanted to do that so you could write this with exponents and say it's 4 to the power of x over 3. that's equally valid but again you're going to change this notation and say you have f inverse of x is equal to 4 again raised to the power of x over 3 or you could say the cube root of 4 to the power of x all right let's take a look at another one so i'm going to replace again my f of x with y so this equals my log base 4 of you've got your 5 to the power of x minus 6 all over 4. and again i'm going to swap x and y so to do that let me kind of scroll down a little bit i'm going to say i have x equals my log base 4 of 5 to the power of y minus 6 over 4 and then from here again i'm just going to put this in exponential form so i'm going to say 4 the base here raised to the power of x so 4 to the power of x is equal to this guy right here so you have 5 to the power of y minus 6 over 4. now i want to solve for y and so to do that i can clear the denominator here by just multiplying both sides by 4. but you have to be careful here this is 4 to the power of x so it's not going to be 16 to the power of x you just want to say i'm just going to multiply this side by 4 and this side by 4. it's basically just 4 times 4 to the power of x okay so don't say that's 16 to the power of x that's a common mistake so we want to cancel this out and i'm just going to say this is equal to 5 raised to the power of y minus 6 and then basically what i want to do now is add 6 to both sides of the equation let me kind of scroll down a little bit more and so from here i'm going to have 4 times my 4 raised to the power of x and then plus my 6 and this equals over here this is going to cancel i have 5 and i'll write this over 5 to the power of y so now here's where we can use one of our properties we learned earlier if i want to get this y to come down i can basically use any log to do that but if i use log with a base of 5 you do this log base 5 of 5 raised to the power of y we learned earlier this is just going to be y okay so i'm going to do that to both sides let me scroll this a little bit down to fit this so i'm going to say i have log base 5 of this whole thing and we know that log base 5 of 5 is 1 so this is just y so i'm going to flip the side that this is on it's going to come to the left now so i'm going to say that this is y is equal to over here i'm going to have log base 5 of i'm going to clean this up a little bit i want you to realize that this is 4 to the first power times 4 to the power of x so by the rules of exponents i know that if i have the same base and i'm multiplying i can add exponents right i keep the base the same so 4 would stay the same and i would add the exponent of 1 to the exponent of x so let's say this is 4 to the power of 1 plus x or x plus 1 however you want to write that i'm just going to put x plus 1 and then we would have plus 6 like this so this is going to be your answer i'm just going to take the y away and say this is f inverse of x all right let's look at another one so again i'm just going to change this to y so y equals 10 to the power of x and then minus 5 and again i just want to swap x and y so this would be x and this up here will be y and then from here i just want to solve for y let me scroll down and get some room going so i want to say that i have x is equal to i'm going to add 5 to both sides of the equation go ahead and get rid of this so i'll say x plus 5 is equal to 10 raised to the power of y now i want to get this y on one side by itself right now it's as an exponent so to get rid of that again i'm just going to use a log with a base of 10. and in the next lesson we'll learn this is the common log so i can just write log but for right now i'm going to go log base 10 so i'm going to go log base 10 of this left side so x plus 5 is equal to you're going to have the log base 10 of 10 okay raised to the power of y let me make that more clear what happens is log base 10 of 10 raised to the power of y is just y let me make that better so this is a y okay so if i scroll down get some more room going what i have is log base 10 of x plus 5 is equal to y and i can just replace this at this point and say this is my f inverse of x all right i saved the best one for last this is an example of one that gives students a lot of trouble there's a lot of different techniques to solve it okay and if you attempt this one on your own i advise you to pause the video and try it on your own at first and struggle with it and if you can get it on your own that's great if not we can go through the solution right now so again the first thing i'm going to do same as always i'm going to say that y equals 4 to the power of x plus 1 plus 1 okay over 4 to the power of x i'm going to swap x and y so let me scroll down a little bit i need a lot of room for this so i'm going to put x is equal to i'm going to have 4 raised to the power of y plus 1 then plus 1 then over 4 to the power of y so i've got a y here and i've got a y here okay so the first part of this that's tricky is if you're not used to working with exponents remember what this represents here okay if i have 4 to the power of y plus 1 this is really let me kind of write this off to the side if i had 4 to the power of y times 4 to the power of 1 we saw this earlier this is 4 to the power of y plus one okay so you've got to be thinking this way when you see problems like this because you're expected to know this and to be able to kind of break it back up so what i'm going to do let me kind of scroll down get a little room going i'm going to use that so x equals i'm going to have 4 raised to the power of y and then i'm going to have times 4 raised to the power of 1 and then this is plus 1 and this is over 4 to the power of y now at this point you have a lot of different roads or a lot of different paths that you can take the path that i'm going to take the one that i find the easiest is i'm going to make a substitution you don't need to do this but i am most comfortable doing this so i'm going to show you this and i think it'll help you so i'm going to over here say that we're going to let u be equal to 4 to the power of y okay so u equals 4 to the power of y what i'm going to do is say i have x is equal to u times 4 plus 1 and this is over u so now i'm just going to solve for u that's very easy to do it's not complicated at all we've been doing it forever so let's go down and i'm just going to multiply both sides by u to get this thing started and so i multiply by u over here and by u over here this cancels so you get x u or u x however you want to think about that is equal to you're going to have 4 u plus 1. okay if i want to solve for u i've got to get all the terms with u on one side so let me subtract 4u away from both sides of the equation and i'm going to run out of room so let me copy this and go to a fresh sheet and just paste this in and what i want to do here i want to have minus 4u and then plus xu is equal to this cancels the number 1. okay so i'm going to factor the u out so i'm going to have u and then in the parentheses i can write negative 4 plus x where i can switch that around and say x minus 4 which is what i want to do and this is equal to 1. okay so i'm going to divide both sides by x minus 4 okay and now i have this in terms of u okay so i have u is equal to 1 over x minus 4. now we're not done because remember i made a substitution i said u is equal to 4 to the power of y so at this point i can erase this and just say that this is 4 to the power of y from here i know i can just use log base 4 on both sides and i'll be done so let me just move this over and i'll say that i have log base 4 of this guy and let me slide this over and i'll say i have log base 4 of this guy okay so we know that this is just going to come down and this part's 1. so i get that y is equal to log base 4 of 1 over x minus 4 okay and this i can replace with f inverse of x and i'm done okay so i get f inverse of x equals log base 4 of 1 over x minus 4. in this lesson we want to talk about common and natural logarithms additionally we're going to talk about the change of base formula for logarithms now most of you already know that the log key on your calculator corresponds to a logarithm with a base of 10. this type of logarithm is known as the common logarithm and when we write the common logarithm we can use a little shorthand and not include the base so something like log base 10 of x can be just written as log of x so when you use this key on your calculator again you have to understand that the base is 10. so something like let's say log of 100 okay if you type that into your calculator log and then for the argument just put 100 this is going to give you a result of 2. and again the reason is because this is log base 10 of 100 and 10 to the second power or 10 squared gives me an answer of 100 so this guy has a value of 2. all right so additionally most calculators also come with this ln key so this ln key represents the natural logarithm so this is going to be a logarithm with a base that's our special number e so this guy right here so you remember the special number e about 2.718 we talked about this earlier in the chapter when we're looking at different compound interest scenarios so it turns out that the natural logarithm comes up a lot when you have problems that involve growth or decay so you'll see this a lot in science class you also see it a lot in math class economics class things like that so log base e of x we can write this in a shorthand notation by saying ln for the natural log of x and if you wanted to calculate the natural log of some number let's say 15 for example you'd hit the ln key on your calculator and then when the parentheses come up just put 15 in okay and hit enter and in this case you get approximately because this is an irrational number 2.708 so 2.708 and again i'm rounding this in a lot of situations you're going to have to round things because you're going to get an irrational number all right so now what i want to do is go into this change of base formula for logarithms and this is useful because again most calculators just have the log key and the ln key i know that nowadays some of the ti 84s and the ti-89s and you know the upper end calculators do let you punch this in directly and get an answer but it's so useful to know the change of base formula so let's say i start with log base 2 of 3 and i don't know what it is so i just say it's equal to x okay from here i can put this in exponential form so 2 to the power of x is going to give me 3. 2 to the power of x is going to give me 3. so i have the log key available to me on my calculator so let's say i take the log of both sides and again this is a logarithm with a base of 10. so log of this side and log of this side so this comes down from our power property so we get x times log of 2 is equal to log of 3. okay so how do we solve for x well we already know how to do this we just divide both sides by what's multiplying x so in this case it's log of 2 okay nothing's changed so log of 2 over here and basically this is going to cancel right that's gone so i get that x or the value of this guy is going to be equal to log of 3 over log of 2 and i can calculate this using my calculator right so this is approximately because i'm going to have to round here it's an irrational number i would say 1.585 so 1.585 with that being said let's take this process and dive into a general formula so we don't need to set it equal to x each time and go through this complicated thing what you'll realize is that it works out the same way each time so what you can do if you have log base a of x where x is the argument and a is the base we can say this is equal to log of sum base b of x over log of some base b of a well in our case i just chose a base of 10 here okay and so i said log of x over log of a remember this was 3 and this was 2. so this was 3 and this was 2. right we found our answer by saying log of 3 over log of 2. so log base 10 of 3 divided by log base 10 of 2 gave us our answer and again that was about 1.585 again generally speaking that's all you need to do and your restrictions here are as expected we have that a is greater than 0 and not equal to 1. well a is the base here and the argument there then b is greater than 0 and not equal to 1 that's your base here and here again that can be whatever you want usually you're going to use a base of 10 or a base of e so you can use those two kind of buttons on your calculator the ln key or the log key and then x here and here that's got to be greater than zero because in each case it's just an argument so let's just blow through a few of these is a very easy process log base 4 of 12 is equal to what again just take i'm just going to use the common logarithm so i'm going to go log of 12 okay so log of the argument here that's what comes first and then over log of the base okay that's what comes second so the base is 4 and this would be approximately because again you've got an irrational number here i'll say about 1.792 all right let's look at another one so we have log base square root of 5 of square root of 8. so again all i'm going to do is this looks kind of hard but it's really the same thing you just punch it into a calculator and i'm going to use the natural logarithm so i'm going to go the natural log of again this argument here so square root of 8 and then over you're going to have the natural log of again square root of 5. so all you need to do is hit the ln key square root of 8 divided by ln key square root of 5 and you're going to get approximately 1.292 all right let's look at one more of these and then i'll show you something else so we have log base one-fifth of 20. so again to set this up i'm going to say this is what you can use ln the natural logarithm or you can use log the common logarithm either one of those keys will be available so i'm going to use the common logarithm so log of again the argument here so 20 over log of again i'm going to use this base here so one-fifth and then so this guy right here would be approximately when you type that into your calculator you're going to get about negative 1.861 all right so in addition to those easy problems we also get these kind of multiplication scenarios that come up and these problems typically involve you need to realize that you can simplify in some way so let me go through one of these as a typical example so we have log base 3 of 4 times log base 5 of 9. so the very first thing you want to do is set up each with the change of base formula so i'm just going to use my common logarithm i'm going to say the first one is log of again the argument there 4 over log of again the base there is 3 then times for this one it's going to be log of again the argument is 9 over log of again the base is 5. now before you go any further and start typing this up into your calculator you need to look and see if there's anything you can simplify because a lot of times they want you to show that okay so here i know that i have a base of 3 let me just highlight that and here i have a base of 9. so i know that 9 is 3 squared so the first thing you need to realize is that you want to rewrite this as 3 squared like this and then from your properties of logarithms this guy can come out in front so basically you can erase that and just put a 2 out here now you see that this is just multiplication involved here so there's nothing stopping me from canceling this with this because those are common factors log of 3 over log of 3 is the same non-zero number so you can cancel those okay and the result will be 1. so you're meant to show that this will be simplified to basically 2 times log of 4 over log of 5. and sometimes these will work out perfectly to be a rational number but in this case it's not going to be so we'd still have to type this into a calculator and we would get approximately 1.723 but just be on the lookout for these because in a lot of cases they'll give you a problem and say show how you could simplify this and they mean that you need to go through and cancel things like we did here in this lesson we want to talk about solving exponential equations using logarithms so at this point we already have all the tools that we need to solve any exponential equation that we come across so i'm just going to start out with an easy example we'll see some harder examples as we progress through the lesson so we have 10 raised to the power of negative 2x and then we have plus the 6 and this is equal to the number 18. all right so how could we solve this let me just rewrite it so 10 raised to the power of negative 2x then we have again plus 6 and this equals 18. well the first thing you want to do if you're solving for x you want to isolate the term that contains x in it so this guy right here is the term that contains x in it so to isolate that i would just subtract 6 away from both sides of the equation so that's the very first thing you want to do so this cancels all right so now on the left i have this 10 raised to the power of negative 2x this is equal to 18 minus 6 is 12. now from this point here's where you have to think about your logarithms that you worked with you know that you have the power property for logarithms and basically that allows you to take your exponent down out in front okay so what i want to do is use that property here so that i can get this guy to come down okay so if i take the log of the left side and i also take the log of the right side to make that legal what's going to happen is from my power property this guy right here can now come out in front okay so let's scroll down and get some room going and we'll see this real fast so we're going to have our negative 2x out in front we'll make that negative just a little bit better and then times this log of 10. now this is 1 and i'll cancel that in a minute i want to explain why it's 1. i think most of you know that already but this is equal to log of 12. now this is 1. so you can just cancel and put 1. the reason it's one is remember when you have log of something you don't show a base it's because you have the common logarithm the base is understood to be 10. so if i showed you that you had log base 10 of 10 again what is this this is equal to 1 because 10 to the first power gives me 10. so if this is 1 then really this left side is just negative 2x because negative 2x times 1 is just negative 2x and now this is equal to log of 12 and once we get it to this point it's really really simple log of 12 is just a number so don't let that scare you basically all you want to do is divide both sides by negative 2 just like we've always done okay to isolate x and so we know that this right here is going to cancel and i have x by itself and it equals my log of 12 over my negative 2. now this is the answer if you don't want to give an approximation if your teacher says hey i want the exact value then you give this log of 12 over negative 2. now if you wanted to approximate this using your calculator that's fine too generally speaking you're going to have some specific decimal place that you're going to want to round to and the teacher will let you know that in your assignment so i'm going to punch this up on a calculator and i'm going to get about and i'm going to use my approximate sign there so negative and then i'm going to say 0.5396 and again i'm rounding this guy and again i'm using the approximation symbol there to show that it's a rounded amount it's an approximation it's not the exact amount but it's pretty close all right let's take a look at a harder example or i'd say a more tedious example so now we have 3 multiplied by 4 raised to the power of 3 minus 8x then we have minus 4 and this is equal to 30. so again the first thing i want to do since x is here i want to isolate this guy okay so to do that i have multiplication here and have subtraction here let me add 4 to both sides of the equation to get things started i know that this will cancel and on the left side now i have 3 multiplied by 4 raised to the power of 3 minus 8x and this equals 30 plus 4 is 34. okay so from here again if i'm trying to isolate this guy well what do i have i have multiplication here so to undo that i'm just going to divide both sides by 3. and now this is going to cancel here and what i'm left with on the left is what i wanted to isolate so i have 4 raised to the power of 3 minus 8x and this equals my 34 over 3. okay so from this point i've got to find out how i can get this variable down okay and to do that again we're going to use our power property for logarithms so i'm just going to take the log of both sides so i'm just going to go log of 4 raised to the power of 3 minus 8x and this will be equal to log of you have your 34 over 3 or 34 thirds if you want to say it that way okay so from here this can come down right from the power property so you have your 3 minus 8x times your log of 4. now i want to caution you again something it's very important that you use parentheses here because this is a quantity 3 minus 8x and it's being multiplied by log of 4 which is just a number so we'll say this is equal to log of again 34 thirds now let me make this a little better a little cleaner so 34 thirds okay so at this point you can do this two different ways you can use your distributive property here so you can multiply log of 4 by 3 and then log of 4 by negative 8x you can do it that way or you can get rid of this guy right now by dividing both sides by log of 4 and that's what i'm going to do you'll get the same answer either way so it doesn't really matter so let's go ahead and cancel this with this and let's scroll down now so i'll have 3 minus 8x and you can drop the parentheses you don't need them anymore and then this is equal to you've got your log of you've got this 34 thirds and then this is divided by log of 4. all right so from here what we need to do is again isolate this x so i'm going to start by subtracting 3 away from each side of the equation so i'm going to subtract 3 away from here that's going to cancel and subtract 3 away from over there so now i have what i have negative 8x is equal to this log of 34 thirds and this is over my log of 4 and then over here i have minus 3. okay so to finish things up i'm just going to again try to solve for x and to do that just like we've been doing forever i want to divide both sides by negative 8 but really an easier way to think about this here because we have a fraction involved i'm going to multiply both sides by negative 1 8. okay so that would be the same thing make sure you use parentheses here so times negative 1 8th this of course is going to cancel and on the left you just have your x that you're looking for on the right i'm going to distribute this to each and so what's going to happen is for the first one i'm going to have the negative of log of 34 thirds over you're going to have this 8 out in front times log of 4. okay then what do we have we basically have negative 3 times negative 1 8 so that would be what that would be plus 3 8. so this is going to be your answer again if you don't want to give an approximation you would basically say that x is equal to the negative of log of 34 thirds over 8 times log of 4 plus you have your 3 8. now if you wanted to give an approximation you're going to get about okay i'm going to bring this down here because i'm going to run out of room we're going to say 0.1561 so again here's your approximation and here's your exact value again the negative log of 34 thirds over 8 times log of 4 plus 3 8 and then as an approximation 0.1561 all right let's take a look at another one so we have 7 raised to the power of x plus 2 is equal to 2 raised to the power of 2x minus 1. so this one is a little bit quicker to do but you do have some things that kind of trip you up on this there's nothing really to do other than to take the log of both sides to start because you've got to get these guys down okay you can't solve it the way it is so you go log of 7 raised to the power of x plus 2 and this equals log of you have 2 raised to the power of 2x minus 1. let me slide this down so it's in the middle of the screen and so now we know that those exponents can come down so legally i can say that this is what this comes down this comes down so we have this x plus 2 again this is a quantity so i'm going to use parentheses times log of 7 equals this comes down so the quantity 2x minus 1 times log of 2. so again here's where i'm going to use my distributive property you've got to get everything with an x on one side so you can factor that out and solve for x so i'm going to go and distribute this to each term here so i'm going to have x times log of 7 then plus 2 times log of 7 and this equals over here i'm going to have 2x times log of 2. and let me slide this down so i don't run off the screen and then i'm going to have minus 1 times log of 2 which is just log of 2. okay so now what i want to do is get all my terms with x involved to one side okay so i'm going to have this guy right here and this guy right here everything else you want on the other side again the reason you're doing this is so you can factor the x out and then solve for it so what i'm going to do is subtract 2 times log of 7 away from each side of the equation and i'm going to subtract 2x times log of 2 away from each side of the equation so let me do that over here so minus 2x times log of 2 okay and so this guy's going to cancel and this guy's going to cancel and what do i have well my new left side is going to be x times log of 7 minus 2x times log of 2. let me scooch this down a little bit and this will be equal to my right side will be what i'm going to say that i have the negative of log of 2 and then minus 2 times my log of 7. okay so how do i solve for x again i want to just think about factoring it out so if i factor out an x from here and here what would that leave me with so this comes out so i have x out in front and inside the parentheses i have the log of 7 minus you're going to have this 2 here times this log of 2 here so i would have log of 7 minus 2 times log of 2 okay and let me put some closing parentheses there and this will be equal to again over here this isn't going to change so we'll have negative log of 2 minus 2 times log of 7. all right so now how can i get x by itself well this guy right here is just multiplying x so all i have to do is just divide both sides by this so divide by log of 7 minus 2 times log of 2 on both sides so log of 7 minus 2 times log of 2. and what do we get well this cancels so that's done we get x is equal to we'll have that negative log of 2. minus 2 times log of 7 over log of 7 minus 2 times log of 2. so again this is your exact answer if you want to give an approximation i'll just give one here as an approximate value and let me make that sign a little bit better so you punch that into your calculator you're going to get about negative three 8.1931 again that's just an approximation all right let's go ahead and take a look at another example so we have e raised to the power of x squared equals 144. so for the previous examples i used the common logarithm or again the logarithm with a base of 10 that's the log key on your calculator now i'm going to use the natural logarithm the ln key on your calculator because whenever you deal with this special number e remember if you type ln for the natural logarithm into your calculator and then you hit e as the argument you're going to get a value of 1 right because this ln this represents a log with a base of e and then if it's of e this is what e to the first power gives me e so that's why this has a value of 1. so what i'm going to do here is i'm going to use the natural log for each side so this ln is how you want to write that of e raised to the power of x squared and then again for the natural log on this side write the ln and then of 144 and so this can now come down so i'll say i have x squared times if i have the natural log of the number e this is one okay this equals again the natural log of 144. okay so from this point what do i have i have that x squared is equal to the natural log of 144 and how can i get x by itself why just take the square root of each side so take the square root of this side let me make that better so the square root of this side and then remember you got to do your plus or minus the square root of this side okay so we know that this on the left is just going to be x now and this equals plus or minus the square root of this natural log of 144. once again this is going to be the exact answer so you have plus the square root of the natural log of 144 and then you have minus or the negative of the square root of the natural logarithm of 144 so that's the exact value if you want an approximation okay you can type this into your calculator and you're going to get approximately plus or minus we'll say 2 point [Music] let me make that a little better so 2.22 again 9 3 is a good approximation for this all right what i want to do now is look at a typical word problem that you're going to see in this section and basically it goes like this we have how long will it take for money in a savings account that is compounded continuously at three percent annual interest to double so this is a very simple word problem to solve but a lot of times if you haven't seen it before you might get tripped up so the first thing is to remember the formula for the continuous compound interest so that's the a is equal to p times the special number e raised to the power of r times t so again the r is the rate as a decimal we have that here it's three percent so you're going to plug that in there so let me just do that now so let me put this as point zero and then times again that t this is your time in years your time in years okay so remember that now when we think about the rest of this we have p which is the principle and we have a which is the account balance in the future you could say the future value of the account so typically for these problems you're given an interest rate you're given a time in years and you're given a principle a starting amount you don't really have any of that here okay so that's what trips people up what you have to realize is that you're trying to figure out how long it's going to take for the money to double so you're going to be solving for t here okay and in order to get a and p to behave in this problem you really have to think about writing a in terms of p okay so again you're trying to double your money so whatever you put in for p you're going to have double that for a when this happens so this is going to be 2 times whatever p is so once you realize that once you get past that roadblock the rest of this is super easy you're just going to divide both sides by p and i made that a little too long so let's divide both sides by p and we know that this would cancel and this would cancel and from here it becomes very very simple we have 2 is equal to e to the power of .03 times this variable t okay so from here let me move this down a little bit i'm going to take the natural log of each side and the reason i'm going to do that is again i'm working with this special number e so i'm going to say the natural log or again use the ln of 2 is equal to the natural log of this side which is e raised to the power of 0.03 times t okay so from this point i know that this can come down so let's come down here and we'll say what we have the natural log of 2 is equal to my .03 times t and then times what is the natural log of e we know that this guy is 1. okay so that's going to cancel so what i have now is that t is going to be equal to what well this is multiplying t so i just divide both sides by .03 and i am done so very very simple overall we can say that t or the time in years is equal to the natural log of 2 over .03 now this was a word problem so if you give this as an answer it's basically going to be nonsense right you need to give an approximation here and depending on how precise you want to be i would round this and say that it's about 23 point let's just go ahead and say one zero that's pretty close again you can add more decimal places if you need it to be more precise but that's pretty much up to you or your teacher and how they assign the question so let's take this 23.10 and make a little sentence and i'm just going to say that the money will double in and i'm going to use about because it's an approximation so 23.1 and you can with zero if you want but 23.1 and 23.10 are the same let's just put 23.10 years in this lesson we want to talk about solving logarithmic equations all right so we're going to start out with an easy example so we have that negative 9 times log base 11 of negative 4x plus 2 plus 4 equals negative 5. so when you see something like this with just one logarithm involved you should be thinking about simplifying things and basically trying to get this into exponential form that's how you're going to solve this type of equation so what i'm going to do to kick things off i'm just going to subtract 4 away from each side of the equation so what's going to happen is that would cancel over here on the right side i would now have negative 9 because negative 5 plus negative 4 or negative 5 minus 4 if you want to think about it that way would be negative 9. on the left side i would have that negative 9 times log base 11 of this negative 4 x plus 2. okay so from here what do i want to do now well notice how you have this negative 9 here that's multiplying this log base 11 of again this negative 4x plus 2. so to get rid of that because this is multiplication i can use division so i'm going to divide both sides by negative 9 and that will clean this up further so this is going to cancel and on the left i'm just going to have that log base 11 of you've got that negative 4x plus 2. on the right what do i have negative 9 divided by negative 9 that's going to be 1. okay so let's scroll down and get some room going and we'll look at what we have here now i'm going to write this in exponential form so i'm just going to take this base here and i'm going to raise it to this guy right here this one so we have 11 to the first power and this will be equal to this guy right here this negative 4x plus 2. now 11 to the first power is just 11 so you can erase that if you want and basically you just want to solve this very simple equation to get our solution so what i'm going to do is just subtract 2 away from both sides so i'll have 11 minus 2 which is positive 9 so 9 is equal to negative 4x and what i want to do now to get x by itself again i have this negative 4 that's multiplying x so let's divide both sides by negative 4 and of course now this cancels with this i'm going to have that x is equal to you've got 9 over negative 4 which is basically negative 9 4. now let's stop for a minute when you work with these kind of logarithmic equations you always want to check your solutions and make sure they're valid in the original equation okay a lot of times when you're working with these you're going to get solutions that violate the domain okay and so you've got to reject those solutions so let's go ahead and check this real quick it's going to tell you in advance this one does work out but you'll see some in this lesson that don't work so let me erase everything real quick so again we have that x is equal to negative 9 4. so i'm just going to plug that in right here and see what happens so we have negative 9 times log base 11 of negative four times negative nine fourths let's do that off to the side negative four times negative nine fourths we know the fours would cancel the negatives would cancel and you'd have nine so this would be nine plus 2 which is 11. let's just erase this real quick and then we have plus 4 and this equals negative 5. what is log base 11 of 11 well again 11 to the first power would give me 11. so log base 11 of 11 is just 1. so let's put a 1 here now negative 9 times 1 is just negative 9 and negative 9 plus 4 would give me negative 5. so it does take a few minutes to kind of check things but you always want to make sure that your solution works out and also that you don't have a domain violation if you do if you plug something in there and you end up with taking the logarithm of a negative number as an example you want to reject that solution let's take a look at kind of a harder example and i will say harder but it's just a little bit different in the way that you solve it you have a log base 7 on each side of the equation so if i have log base a of x is equal to log base a of let's say y this implies that x is equal to y so we have a logarithm with the same base on each side of the equation they're set equal to each other so you just say that x is equal to y in this circumstance so that's what we're going to use here to solve this equation so we have that log base 7 of 3x equals log base 7 of 70 minus x squared again notice how you have the same base on each side they're set equal to each other so you don't have to do anything else other than just say that i have 3x here is equal to this guy right here which is 70 minus x squared so this is what we want to solve so to make this simpler i'm just going to move everything to the left hand side so if i add x squared to both sides i get x squared over here then plus 3x and then if i subtract 70 away from both sides i have minus 70 and this equals 0. okay so from this point i can solve this using factoring or i can use the quadratic formula or i can even complete the square if i want it doesn't really matter whatever you want to do i'm just going to solve it with factoring because it's usually the fastest so i'm just going to factor this and say this is x and it's going to be let's see plus 10 and then x minus 7 and this equals 0. okay so from here i'm going to set each of these equal to 0. so i'm going to say x plus 10 equals 0. of course the solution here is that x is equal to negative 10 and then here if i say x minus 7 equals 0 of course the solution is x equals 7. so these are the two proposed solutions x equals negative 10 and x equals 7. so here's going to be an example where we have a domain violation and we've got to reject one of the solutions so let me erase everything real quick and now let's look at these more closely so when we work with negative 10 we'll see we have a problem let's say i have 3 times negative 10 that's negative 30. so log base 7 of negative 30 again this is undefined you don't even need to plug anything else in once you see that it fails somewhere just reject the solution and move on so this is rejected so that's not a possibility so now i can try 7 and i could say okay i have log base 7 of what is 3 times 7 that's 21 and this equals you have log base 7 of 7 squared is 49 and 70 minus 49 is 21 as well so this one does work out as a solution all right let's take a look at another one so we have log base 5 of 9 plus log base 5 of 2x squared plus 3 and this equals 3. so for this type of scenario you want to turn to your properties of logarithms notice that you have the same base here in each case and you have this addition involved so we already know that we can manipulate this into one logarithm right we can condense so what we're going to do here is say that we have log base 5 of remember if it's addition you're going to turn it when you condense it into multiplication it's going to be this guy right here times this guy right here so 9 times the quantity make sure you use parentheses 2x squared plus 3 okay and let me put another parenthesis there and this equals 3. okay so let's go through and figure this out we have log base 5 of 9 times 2x squared is 18x squared and then plus 9 times 3 is 27. okay so this equals 3. so from here all i want to do is just write this in exponential form i can very easily get a solution so 5 to the third power which is 125 will be equal to the argument here this guy this 18x squared 18x squared plus 27. so what i'm going to do is i'm going to subtract 125 away from each side of the equation and what is that going to give me i'm going to have that 18x squared minus you do 27 minus 125 you're going to get negative 98 so that's why i have minus 98 there this would be equal to zero okay so from here you can use the quadratic formula if you want but it's quickest to realize that you have the difference of squares if you factor out a 2 okay so if i take a 2 out i would have 2 times the quantity from 18 i pull out a 2 i get 9 then times x squared and the minus if i pull out a 2 from 98 i get 49 of course this equals zero and once you see it like this you realize that it's the difference of two squares so we can factor this very quickly using our formula we say this is two times the quantity three x plus seven times the quantity three x minus seven of course this is still equal to 0. so from here what we're going to do is set each factor with a variable involved equal to 0. so i'm going to say that 3x plus 7 equals 0 subtract 7 away from each side of the equation you get 3x is equal to negative 7 divide both sides by 3 and one proposed solution would be that x is equal to negative 7 3. so let me erase this so we have a little room and i'll write this up here say x is equal to negative 7 3. and then for the other guy i'm going to say that 3x minus 7 is equal to 0 and i'm going to add 7 to both sides of the equation and i'll have that 3x is equal to seven divide both sides by three and i'm gonna get that x is equal to seven thirds now in the interest of time because i don't want this to go on forever and ever and ever i'm just gonna tell you in advance that both of these solutions are gonna work they're valid solutions but again when you're solving logarithmic equations you've got to stop you've got to check okay you got to make sure that you don't end up with a domain restriction i've already checked this so in the interest of time we're just going to move on and say our solutions are going to be x equals negative seven thirds or seven thirds let's go ahead and look at one more example again this is a very easy topic most of these can be solved within just a few minutes usually the checking is what takes the longest because sometimes you run into some messy things that you have to work with all right so we have log base 9 of you have negative 3x minus 3 then minus log base 9 of 4 this equals 1. so i have a minus here and again i've got log base 9 and log base 9. so again when i try to condense this into one logarithm because there's a subtraction sign i'm thinking about division so i'm going to have my log base 9 of this first guy here is going to be the numerator so negative 3 x minus 3 this will be over the second guy is my denominator and this all will be equal to 1. okay so how do we solve this we put this in exponential form so 9 to the first power which is just 9 is equal to this guy right here this negative 3x minus 3 this is over 4. all right so let's solve this real quick it's very easy we just multiply both sides by 4 to clear the denominator so you're going to get what multiply by 4 multiply by 4 this cancels you get 36 equals negative 3x minus 3. you can add 3 to both sides of the equation and so you're going to get that 39 is equal to negative 3x let's go down and get a little bit more room going and i'm going to divide both sides by negative 3. and what do i end up with this cancels i get x is equal to negative 13. let me make that a little bit better so x equals negative 13. so that's going to be our solution again i've already checked this if you want to go back pause the video you'll see that it does work out as a solution in this lesson we want to review systems of linear equations in two variables and specifically we're going to look at the graphing method the substitution method and the elimination method so in this section of our course we're going to do a deep dive into matrices and before we can really get into this topic we have to do a thorough review of the basics so over the course of the next few lessons we're going to revisit some stuff that you learned in algebra 1 and algebra 2 just to make sure you have a fundamental understanding of these kind of basic concepts before we get into this advanced material so at this point we've all seen a linear equation in two variables we talked about this earlier in the course something like x plus y equals one now one thing i immediately want to kind of draw your attention to is that this type of equation has what an infinite number of solutions so there's an infinite number of x comma y values or ordered pairs that will satisfy this equation we can find them by plugging in something for x and solving for y or plugging something in for y and solving for x now the question you always get when you do this is where did you get that number from and the answer is i made it up right so i can choose any value i want for x anything i want it could be a million if i wanted it to okay i'm not going to use that because it's a little big i'm just going to use the number 2. again just made it up plug that in for x and then solve for y so 2 plus y equals 1. so if i want to solve for y i just subtract 2 away from each side of the equation this cancels i get y is equal to 1 minus 2 is negative 1. okay so what this tells me is that i plugged in a 2 for x so when x is 2 y is negative 1. so that ordered pair 2 comma negative 1 satisfies this equation and i can do the same thing over and over and over again and generate an infinite number of these ordered pairs i could choose something for y like let's say 5. again where did i get 5 from i made it up so 5 gets plugged in for y so you get x plus 5 equals 1. so if i subtract 5 away from each side of the equation i get x is equal to negative 4. so now another ordered pair is negative four comma five okay again we can continue doing this forever and ever and ever now when we first look at linear equations in two variables we do this kind of activity to generate enough ordered pairs so that we can kind of sketch our graph but as we kind of move on we find that there's an easier way to sketch our graph and the way to do this is to solve the equation for y and put it in slope intercept form so looking at this guy if i just solve it for y if i subtract x away from each side i get y is equal to negative x plus 1. okay so this is in the format of y equals m the slope times x plus b the y-intercept so in this case my slope m is negative 1 right because you should have a negative there you could really erase that and put in negative 1 if you wanted to make that more clear and b my y-intercept is 1. so that means the y-intercept occurs at 0 right for the x coordinate comma 1 this guy right here so graphically all i really need to do is find 0 comma 1 which is right here that's my point and then i can use my slope of negative 1 to generate additional points remember slope m is rise over run so if this is negative 1 and i could say it's over 1 right because anything over 1 is just itself well really i could say okay from this point i could fall 1 go to the right one fall one go to the right one fall one go to the right one you know so on and so forth i've given you this point here and this point here but really you just need two points so if i started there i could go down one to the right one and have a point there that's really enough to graph a line but of course i've already pre-drawn this so we could have a nice clean line i made it with a computer versus just freehand drawing which generally doesn't look so good all right so we know that this is how we could graph this guy and we also know that every point on this line is a solution for this equation let's just take this point right here which is going to be 4 comma negative 3. so let's erase this real quick i just want to show you this just so you have a complete understanding so that means if i plugged in a 4 for x and i plugged in a negative 3 for y i should get a true statement and obviously that would work because 4 plus negative 3 or 4 minus 3 would give me 1. right 4 minus 3 equals 1 you get 1 equals 1 so that's true so again every point on this line is a solution for this equation now suppose we throw another equation into the mix and we consider these two equations together so this is known as a system of linear equations and in this case because we have two variables involved you could say it's a system of linear equations and two variables now in this particular case you're not going to have an infinite number of solutions because the solution now has to satisfy this equation and this equation so that means whatever i get for x and whatever i get for y i have to be able to plug in for x here and here and for y here and here okay and it has to satisfy both so one way that we can find such a solution is using the kind of graphing method it's not very efficient okay but it is a good way to teach the concept so if we graph this one x minus 2y equals negative 8 we would what we could solve this for y so x minus 2y equals negative 8. i would subtract x away from each side so let me kind of do this up here this would cancel i'd have my negative 2 y equals negative x minus 8. what i need to do now is get rid of this negative 2. so let me divide both sides by negative 2 and so this is going to cancel and so now i'm going to have y is equal to you essentially have negative over negative which is positive so this is x over 2 but we're going to just write this as what just one half times x then negative 8 over negative 2 is plus 4. so you've got y equals one half x plus four so i'm going to add that i'm gonna say this is y equals one half x plus four okay that's going to be this guy right here again i would go to my y intercept which occurs at zero comma four so that's going to be right here and let me use a different color because that didn't show up so well so let me use this color and then i could find additional points with my slope again m my slope is one half so it's rise over run so it's one over two so i would rise one go to the right one two so there's a point or i could fall one and go to the left one two so there's a point now let me draw your attention to this point right here okay so this point lies on this line y equals one half x plus four and then also this line which we already know is y equals negative x plus one again in slope intercept form or x plus y equals 1 in standard form either way you want to kind of label that but the main thing to notice here is that that point that lies on both lines or the intersection of these two it's going to solve both equations because it's on both lines right every point on the line is a solution for the equation so every point on this yellow line is a solution for this equation every point on this i guess you would say that's teal every point on this tl line is a solution for this equation that point of intersection which is negative 2 3 so let me label that i'll label that over here i'll say it's negative 2 3. that guy is on both lines so it solves both equations of the system and so it's a solution for the system if i go back we can prove this mathematically okay and so let me just write the solution here as negative 2 comma 3 and you just plug in a negative 2 for each x and a 3 for each y and you verify that it works in both equations and then you know you have your correct solution so in the first one i would have negative two plus three equals one so we know that works right because this is going to be one equals one so it works in this first one then for the second one you would have negative 2 minus 2 times plug in a 3 for y and this equals negative 8. okay so i would start with negative 2 times 3 that's going to be negative 6 and then negative two minus six is negative eight so you get negative eight equals negative eight so it works here as well so we've found our solution for the system graphically again all we really need to do there is graph each equation okay and then look for the point of intersection now this method is not practical okay it's just something we use to teach the concept because what if you have a solution like you know one half comma 2 19. something like that that's not going to be very easy to find graphically so fortunately for us we have other methods and these methods are algebraic okay so you already know that we have the substitution method and we have the elimination method so let's start out with the substitution method let's use the exact same two equations and for reference sake i've labeled the top one as equation one that's why you see this one here and i've labeled the bottom equation as equation two just so i can talk about and say okay equation one i'm talking about this guy right here if i say equation 2 i'm talking about this guy right here okay just for reference now with substitution the idea is that we want to solve one of the equations for one of the variables okay doesn't matter which one and then you want to plug in for that variable and the other equation what that's going to do is it's going to give you a linear equation in one variable okay so we can easily solve that and then we're going to substitute back into one of the original equations and get the other one okay so what i just said might sound kind of complicated if you've never seen this before if you have seen it you know it's very easy you usually want to use substitution when one of your variables has a coefficient of one or negative one in this case with equation one you have a coefficient on x of one you have a coefficient on y of one and then also in equation two you have a coefficient on x of one okay so any of these would be easy to use i'm just going to use equation 1 and i'm just going to solve it for y because we already did that so i'm just going to say that y is equal to what i would subtract x away from each side so it would be negative x and then plus 1. okay so once you've done that you've got it in terms of y so now you're going to plug in for y in the other equation so what do i mean by that here's y right here we're saying y is equal to or y is the same as this negative x plus 1. so what i'm going to do here is i'm going to plug this guy right here in for y there okay that's all i'm doing it's very very simple so i have x minus 2 you've got to use parentheses here because 2 is multiplying this whole thing so times the quantity i'm going to use a different color for this so negative x plus 1 and then i'll switch back and this is equal to negative 8. so let me scroll down and i'll come back up you see that we only have one variable now we only have x involved okay so we have x and then i'm going to distribute this negative 2 to each term inside the parentheses negative 2 times negative x is going to be positive 2x negative 2 times 1 is going to be negative 2 this equals negative 8. okay so combine like terms x plus 2x is 3x and then i can add 2 to both sides of the equation so negative 8 plus 2 is going to be negative 6. lastly i want x by itself so i'm just going to divide both sides by 3 and i find that x is equal to negative 2. we already knew that because we solved this already using graphing but we just solved it using substitution and we found that x is negative 2. so let me erase this and we can erase this as well so i'm just going to put this over here and say that x is equal to negative 2 okay now we already know because we saw this already that y is positive 3 but let's prove this what we would do once we found one of the variables we can plug into either original equation and we can find the other so i can just plug in here or i can plug in here doesn't matter so let's just plug a negative 2 in for x in equation 2. so negative 2 minus 2y equals negative 8 and let me add 2 to both sides of the equation that will cancel we'll have negative 2y is equal to negative 8 plus 2 is going to be negative 6. and let me scroll down a little bit get a little room and let me divide both sides by negative 2 so i can isolate y so this cancels and we found that again y equals positive 3. let me erase this and let's go back up so y is equal to 3 okay so negative 2 comma 3 as an ordered pair and we've already checked this so we don't need to check it again but again that's the substitution method so let's look at a harder example of this so now we have negative 9x minus 10y equals 5 and negative 7x minus 6y equals 11. so let me label this as equation 1 and this is equation 2. again just for reference sake so what i'm going to do is i'm going to solve one of these for one of the variables and then i'm going to plug in for that variable in the other equation so i can solve for x i can solve for y here it really doesn't make a difference you don't have anything that's easier i'm just going to work with equation 1 and i'm going to solve it for x okay so i'm going to solve for x so i have negative 9x minus 10y equals 5. to solve for x i want to isolate x on one side okay so that's all i want to do so i'm going to start by isolating this term so add 10y to both sides let me scroll down get some room going so this will cancel obviously we'll have negative 9x is equal to you'll have 10y plus 5. okay so how can i get x by itself well what i want to do is get rid of this negative 9 that's multiplying x so i'm going to divide both sides by negative 9 and of course this is going to cancel let me kind of drag this up here we'll have x is equal to you have 10 over negative 9. we'll just say negative 10 9 and then times y and then you'll have minus 5 9. okay so a lot messier than when we saw in the last example but this is a typical thing that you'll see as you get into kind of more challenging examples so let me erase this and now we've solved this guy for x okay so we've solved it for x let me scroll up and let me drag this up so we use the equation one so now what i want to do is use equation 2 and again i want to plug in for x there i have x now it's it's equal to this or it's the same as this so i'm going to take this and i'm going to plug it in for x there so i'm going to have negative 7 times again the quantity you need parentheses because negative 7 is multiplying this whole thing so let me do this in a different color so you have negative 10 9 times y and then minus 5 9. and let me go back to this color and then we'll have our minus 6y and then we'll have that at equals 11. so notice that i only have a y involved right there's no x so that's why this method works right it breaks it down into kind of a linear equation of one variable that you can solve then you can go back and substitute and get the other value that you're looking for so now let me kind of distribute this negative 7 to each term this would be positive 70 9 times y then this would be positive 35 9 and then we'd have minus 6y and this equals 11. now to make this a little bit quicker if you don't want to work with fractions remember we can just multiply both sides of the equation by this lcd 9 so i would have what i would have 70y plus 35 then minus if i multiplied 9 times 6 it would be 54 then times that y and this equals 9 times 11 is 90 not so from this point i can combine like terms here 70y minus 54y would be 16y and then let me say plus 35 equals 99. i'm going to subtract 35 away from each side of the equation this will cancel we'll have our 16y is equal to if i do 99 minus 35 that's going to give me 64. if i then divide both sides by 16 i'll find that y is equal to 4. okay so let's erase all this all right so at this point we know that y equals 4 and again let me put these numbers back up so i can refer to these equations and now i can just plug a 4 for y in either equation 1 or 2 doesn't matter let's just use 2 so we'd have negative 7 x minus 6 times again plug in a 4 for y equals 11. so now i can solve for x so negative 6 times 4 would be negative 24 and what i can do now is just add 24 to both sides of the equation so this is going to go away over here 11 plus 24 is 35 so this would be equals 35 and then what i can do to finish this off is just divide both sides by negative 7 and i'll find that x is equal to negative 5. okay so x is equal to negative 5. again as an ordered pair as an ordered pair we could say this is negative 5 comma 4. and if you want to check this plug a negative 5 in for each x plug a 4 in for each y verify you got the right answer so negative 9 times negative 5 minus 10 times 4 equals 5. is that true negative 9 times negative 5 is 45 negative 10 times 4 is 40 so minus 40 and this equals 5. this is true 45 minus 40 is 5. that one works out then the other one would be what it would be negative seven times you have negative five minus six times you have four this equals eleven negative seven times negative five is thirty-five minus six times 4 6 times 4 is 24. we know that 35 minus 24 is 11. so that one checks out as well so negative 5 comma 4 is the solution for this system all right so now let's move on and talk about the elimination method the elimination method is probably a little bit more preferred but if you have a situation like we just looked at it really can go either way okay but the elimination method does require a little bit of extra work so the first thing you have to do is you have to make sure that the equations are written in standard form so that's the ax plus b y equals c now when we did our lesson on this we talked about there being kind of a high school definition and then being kind of a relaxed definition as you move on a b and c here can be any real number that you want they don't have to be you know positive for a and it doesn't have to be an integer any real number will do for a b and c just write it in this format so what we would do here is we would subtract 7y away from each side of the equation so that would give me i'll say this is equation 1 i would have 6x minus 7y this would be equal to 9. right so this is ax plus b y equals c okay then over here for equation 2 we'll say i would add 2x to both sides so i would have 2x plus y is equal to negative 7. okay so this is the format that we wanted it now the idea here is that once you have each equation written in standard form you want to think about how you can make the coefficients of one pair variable terms into opposites so this is really easy to do in this particular case you'll notice that this guy right here is a negative seven the coefficient of y and this guy if i kind of erase this and make this plus one this guy is a one okay so this guy is a negative seven and this guy is a one so we know we can multiply both sides of an equation by the same nonzero number and not change the solution so what i could do is i could multiply equation two by seven so if i did that two x times seven is fourteen x one y times seven is going to be plus seven y and then negative seven times 7 is negative 49. so notice how if we kind of look at it this way this guy right here this negative 7 and this guy right here positive 7 those guys are opposites okay and we need that because one of the variables is going to drop out in the next step when we achieve this so what i'm going to do now is i'm just going to take this 6x minus 7y equals 9 and i'm going to add the two left sides and i'm going to set this equal to the sum of the two right sides so let's do this real quick and then i'll explain why this works mathematically so 14x plus 6x is going to give me 20x and then 7y minus 7y those are going to go away right that's going to be eliminated so this would be equal to if we have negative 49 plus 9 this is going to be negative 40. so at this point we see that we only have one variable involved so we can just divide both sides by 20 and find that x is equal to negative 2. okay so very very simple so let me erase this we don't even need this anymore i'm just going to write that x equals negative 2. and let me explain why this worked and then we'll find y okay so why was i able to add this to this and set this equal to this plus this well let's just think back to the addition property of equality for a moment we know that if i have a equals c and this is very generic if i had a plus b this is equal to c plus b so in other words if these two quantities started off as being equal and i add the same number to both sides it's still equal okay so in other words if i had x plus 4 equals 7 we know at this point that i could add something like three to both sides of the equation so x plus four plus three equals seven plus three and i've got the same solution here the solution is three right three plus four is seven here the solution is still three three plus four is seven seven plus three is ten over here i have seven plus three which is also ten so adding the same number to both sides did not change the solution now it's the same thing here okay it's just a little bit more complicated to understand remember that this guy is set equal to this guy so they're the same okay they're set equal to each other so what i'm doing is i'm adding this to this side and i'm adding this to this side so i'm adding the same thing or the same quantity to both sides of the equation it just looks different okay it just looks different so that's what you need to understand about this method now let's erase this and let's find y so i can just plug in a negative 2 here or i could plug it in there it doesn't matter okay and you can find y that way so y equals negative 2 times negative 2 minus 7 negative 2 times negative 2 is 4 and then 4 minus 7 is going to give you negative 3. so y here is negative 3. so as an ordered pair this is negative 2 comma and then you have negative 3. okay so let's check this real quick so 6 times negative 2 equals 7 times negative 3 plus 9. 7 times negative 3 is negative 21. negative 21 plus 9 is negative 12 and obviously 6 times negative 2 is negative 12 so this works out then for this one y which is negative 3 is equal to negative 2 times negative 2 minus 7. we already know this is true because we just did it this is 4 minus 7 which is negative 3 so this works out as well all right let's do one more of these and then we'll kind of move on to some systems that don't have a solution or an infinite number of solutions all right so we have negative 7x equals negative 6y minus 10. we have 4y equals 5x minus 8. again the first step when you want to use elimination you've got to put everything in standard form so i would just say that this guy right here is negative 7x and let me label this as equation 1 so we can keep track so again it's negative 7x i would add 6y to both sides so plus 6y and this is equal to negative 10. okay then for my equation 2 i would have what i would subtract 5x away from each side so i would have negative 5x plus 4y is equal to negative 8. okay all right so now that we have this set up we want to think about how we can eliminate a variable so to do this what i'm going to do is i'm going to look at y and notice how this is a 6 and this is a 4. well the least common multiple of 4 and 6 is going to be 12. so what i can do here is i can multiply equation 1 by negative 2 right because 6 times negative 2 would be negative 12. so negative 7 times negative 2 would be 14 then times x then 6 times negative 2 would be negative 12 then times y and this equals negative 10 times negative 2 is 20. okay when you do this make sure you multiply every part of that equation by that number okay to make it legal then for this guy right here again my least common multiple between 6 and 4 we said was 12. i made this one negative 12 so this one's going to be positive 12. so i'm going to multiply this by 3 right this equation so negative 5 times 3 is going to be negative 15 then times x and then 4 times 3 is 12. so plus 12 and then times y and this equals negative 8 times 3 is negative 24. now we have opposite coefficients on our variable y we have negative 12 and positive 12. okay so let's scooch this down here and again what we want to do is add the two left sides and set that equal to the sum of the right sides so 14x minus 15x is going to be negative x this is going to cancel right negative 12y plus 12y that's gone and this equals 20 minus 24 which is negative four divide both sides by negative one and you get that x is equal to four okay so that was pretty quick so x is equal to four and pretty easy method overall so x equals four plug this in for x in one of the original equations so let's plug that in there see what we get 4y is equal to 5 times 4 which is 20 minus 8. 20 minus 8 is 12 right and we can all see that y is 3. right divide both sides by 4 you get that y is equal to 3. so we'll put our solution as 4 comma 3. so let's check this real quick so if i plug in a 4 here in the top equation for x negative 7 times 4 is equal to negative 6 times plug in a 3 for y then minus 10. negative 7 times 4 is negative 28 this equals negative 6 times 3 which is negative 18 minus 10. obviously negative 18 minus 10 is negative 28 so this one works out then for the other one 4 times 3 is 12 and this equals 5 times 4 which is 20 minus 8 which is 12. so this one works out as well all right so at this point we've only looked at systems that have graphs that intersect at a single point when we see this type of situation you're going to have a system that's known as being consistent and the equations are said to be independent okay but most of you already know that there's some special case scenarios that you're going to encounter you're going to come across a linear system with no solution this is referred to as an inconsistent system this happens when the two lines are parallel and then we're also going to have another situation where we have an infinite number of solutions in that case the equations are said to be dependent but let's start out by looking at an example where there's no solution let's just go through this and use the substitution method and see what we end up with so we have 2y equals 10x plus 6 and negative 5x plus y equals negative 7. the reason i'm choosing substitution here is because this coefficient is a 1. so it's really easy to solve this for y i can just add 5x to both sides and my second equation here my bottom equation is going to be y is equal to five x minus seven okay so let's erase this and all i need to do now is just plug in for y in this first equation this top equation so this guy right here if i highlight that and i highlight this again this right here this is what y is equal to or the same as i'm just going to plug that in there and i'll have a linear equation in one variable okay a linear equation with just x involved so i would have 2 times again use parentheses so we're going to have 5 x minus 7 and this is equal to you're going to have your 10 x plus 6. okay so let's just go ahead and solve this 2 times 5x is 10x 2 times negative 7 is negative 14. this equals you have 10 x plus 6 and so over here you notice that i have 10x and 10x on each side of the equation so immediately you see you have some kind of problem okay you might not know what it is but let's say i subtracted 10x away from each side right if i have the same thing on each side you just get rid of it so if i erase this from each side what do i get i get nonsense right i get negative 14 equals 6 which is never true right this is a false statement so when you solve this type of system with either substitution or elimination you're going to end up with a false statement so that's how you know you can stop and say there's no solution essentially you want to write this symbol for the null or empty set meaning the solution set contains no elements now you can also see this graphically if i solved each one of these for y you would notice that the slopes are the same right because parallel lines have the same slope in different y-intercepts parallel lines are never going to intersect right that's the definition of them so if i solve this for y again we know that it's y equals 5 x minus 7. this guy if i solve it for y i have what i have 2y equals 10x plus 6 i just divide everything by 2 right so i can get y by itself and i'm going to have that y is equal to 10 over 2 is 5 times x and then plus 6 divided by 2 is 3. so if you see this and you notice that you have the same slope and different y intercepts you already know not to even work this problem just stop and say there's no solution because you have two parallel lines if we copy this kind of come here i've already graphed this to show you what this looks like so we have y equals five x plus three that's going to be this guy in orange right this is my y equals five x plus three let me erase it from here and my guy in teal again i don't know what color that is i think it's teal the guy in teal is going to be this y equals 5x minus set okay so you can see for this guy y equals 5x plus 3 you have a y intercept that occurs at zero comma three and a slope of positive five so up one two three four five to the right one for this guy y equals five x minus seven the y intercept occurs at zero comma negative seven so right there and again the slope is five so up one two three four five to the right one okay so those are the two lines you can see they have the same slope so they continue like this forever in each direction they're never going to intersect okay they're always going to be apart from each other there's no point that lies on both lines so there's never going to be a solution for that system so to show this with elimination let's just start by making this equation 1 we'll label it and again i want these in standard form this guy is going to be equation two so for the one in equation two it's already ready to go it's negative five x plus y equals negative seven again we're using a looser definition so this guy can be negative it's ax plus b y equals c again where a b and c those are just real numbers at this point okay and then for equation 1 i'm just going to subtract 10x away from each side so that would give me negative 10x and then plus 2y and this equals 6. so if you want to you can divide both sides of equation 1 by something like 2 to make it simpler so this would be negative 5. this would just be y and this would be 3. okay so now what we're going to see is we're going to have a problem right if we go through the elimination method we want to set it up to where kind of one pair of variable terms has opposite coefficients okay if i take let's say equation 1 and i say well if i multiply this by negative 1 i would have positive 5x negative 5x but also i'd have negative y and positive y so that's going to be a problem because both those variables are going to drop out right so you end up with 0 equals something and 0 equals something other than 0 okay it's going to be a false statement so let's just go ahead and do that let's multiply equation one by negative one so i would have positive five x minus y equals negative three and then this guy right here this negative five x plus y equals negative seven so we add the two left sides and they cancel so you get 0 this is equal to negative 3 plus negative 7 which is negative 10 which is always false okay so again if you solve something like this and you get a full statement you can say there's no solution right you basically have parallel lines they're never going to intersect so it's not possible to have a solution all right so let's wrap up our lesson and look at the other special case scenario so for this scenario you're basically going to be given the same equation twice right it's going to be algebraically manipulated to look different but it's the same equation okay so when you get this situation you're going to say the equations are dependent and you're going to have an infinite number of solutions because whatever works in kind of the top equation is also going to work in the bottom equation because they're the same equation okay you could see that if you look at this top equation in this bottom equation here for negative 3x minus 15y equals 45 and x plus 5y equals negative 15. well if i divided the top equation by negative 3 so negative 3x divided by negative 3 would be x negative 15y divided by negative 3 would be plus 5y this equals 45 divided by negative 3 is negative 15. so you have the same equation here and here okay so it's just algebraically manipulated to look different but essentially you have an infinite number of solutions for this type of scenario now if we solve this guy using substitution and we don't realize this or we solve it using elimination we don't realize this you're just going to end up with a true statement okay we saw when we solved an inconsistent system or a system with no solution that you ended up with a false statement now you're going to end up with a true statement okay so that's how you know the difference between the two so let's solve this with substitution real quick so i know that this is going to be easy to solve for substitution because the coefficient on that x is a 1. so i can solve this for x and say x is equal to negative 5 y minus 15 so then i can plug in for x here and i can plug this in right so this is going to go right there so let's go ahead and do that real quick i'm going to have negative 3 times again i'm plugging this in for x so negative 5 y minus 15 and then minus 15 y and this equals 45. let me scroll down and get a little room going and let's just go ahead and distribute so negative 3 times negative 5y would be 15y negative 3 times negative 15 is going to be plus 45 and then you have minus 15y and this equals 45. so you can see you have a problem right because you have this will cancel with this so you just have 45 equals 45 which is always true okay and you can subtract 45 away from each side and say you have zero equals zero but it doesn't matter you still have something that's true okay so when this scenario occurs whether you're doing substitution or elimination i'll let you do that on your own if you want to try it that way you just end up with a true statement so that's how you know you have an infinite number of solutions and i'll just write this out real quick and i'll say there's an infinite number of solutions okay so that would be our answer for this type of problem in this lesson we want to review systems of linear equations in three variables all right so in our last lesson we reviewed how to solve a system of linear equations in two variables we looked at the graphing technique we looked at substitution and we also looked at elimination so this was nothing new for you if you're in this course you saw this in algebra 1 you saw it again in algebra 2 this is just to review something that we need to understand as we kind of move forward in this chapter and we start looking at matrices so kind of the next step or the next piece of this is to review system of linear equations in three variables so this is also something you should have seen probably in algebra one if not definitely in algebra two so let's recall when we work with a linear equation in three variables we typically see the variables x y and z right with two variables you see x and y with three you see x y and z so something like let's say 2x we could say plus 3y then minus 4z let's say this is equal to 12. so notice how you have again three variables you have this x you have this y and you have this c okay so before we worked with a linear equation in two variables our solution was given as an ordered pair right you had x comma y well now because you have x y and z it's given as an ordered triple so this guy would be x comma y comma z okay now you can generate as many ordered triples as you want right there's an infinite number of order triples or kind of solutions for this type of equation just like we saw for a linear equation two variables right it's the same thing when we looked at a linear equation in two variables we could pick something for x and solve for y or we could pick something for y and solve for x here i would have to pick something for two of the three variables and i could solve for that third unknown so as an example let's just say i picked the number one and let's just say i picked the number two okay so i'm choosing one for x so this is going to be a one and i'm choosing a two for y so this is going to be a two and i don't know what z is i'm going to plug in here i'm going to figure that out okay so that's going to give me an ordered triple or one solution for this equation so 2 times 1 would be 2 then plus 3 times 2 would be 6 then minus we have 4 times z and this equals 12. so 2 plus 6 is 8. i'm just going to erase that real quick and i know that to solve this it's very simple i would subtract 8 away from each side of the equation i'm just going to erase this from over here 12 minus 8 is 4 okay and so i can quickly divide both sides by negative 4 and i'm going to find that z is equal to negative 1. okay so this is a negative 1 here so we're saying that if x is 1 and y is 2 then z is going to be negative 1. so this is an ordered triple or one solution okay amongst an infinite number of them for this equation 2x plus 3y minus 4z equals 12. now suppose we take three of these equations and we put them together we make ourselves a system how can we get a solution for this are we going to use graphing are we going to use substitution are we going to use elimination well graphing is pretty much out the window with this it wasn't very practical when we looked at a system with two variables right it it is if you have integer solutions right but if you have something that's not an integer solution it gets pretty hard right or if you have really big or really small kind of answers it also gets really hard well for this one you're not going to be graphing with a two dimensional plane now you need three dimensional graphing so this is pretty much only going to happen on a computer now the other thing you could do is you could use substitution okay but it's very very tedious so typically in your textbook you're not going to see substitution used the fastest method is actually to use a matrix but we haven't gotten to that yet so what we're going to do is we're going to use kind of an extension of the elimination method okay so the way this works we're going to take two of the equations and we're going to eliminate one of the variables so we can choose to eliminate x y or z it doesn't really matter then we're going to choose a different two equations okay it has to be a different two equations and we're going to eliminate the same variable okay so once we've made that choice in the first step we've got to stay consistent so what this is going to do is it's going to give us a linear system in two variables once we have that we can use substitution or elimination or really whatever we want because we're just working with two variables at that point okay so it'll be very very simple so to show you this in practice we have 7x plus y plus 5z equals 27. i'm going to label this as equation 1. then we have 4x plus 3y plus 5z equals 21. i'm going to label this as equation 2 and then we have 6x plus y plus 2z equals 9. i'm going to label this as equation 3. okay these numbers are just for reference sake what i'm going to do is i'm going to start with equation 1 and 2 and i'm going to eliminate one of the variables okay the easiest one to eliminate is going to be z right because i have 5z and 5z if i just multiply either equation 1 or equation 2 by a negative 1 i'll have opposite coefficients on the variable z and so if i kind of add the left sides together and add the right sides together on the left sides when i add them together z is going to drop out okay so what i'm going to do is i'm going to keep 1 as it is and i'm going to take 2 and i'm going to multiply both sides by negative 1. so let me copy equation 1 again so 7x plus y plus 5 z equals 27 then for equation 2 i'm going to multiply it by negative 1. so everything's just going to have a change of sign so instead of 4x i'd have negative 4x instead of 3y i'd have negative 3y instead of 5c i'd have negative 5z and this equals instead of 21 it'd have negative 21. and again this was equation two so let's label that so we can stay consistent all right so with elimination you're going to add the two left sides set this equal to the sum of the two right sides and for this to work you've got to have it in this format where you have ax plus b y plus c z equals d okay so you want all your variable terms on one side and you want the constant on the other okay so that's how it's going to be set up so that's already done for us so we don't need to do any additional work so when i sum 7x plus negative 4x is going to be 3x and then y plus negative 3y would be minus 2y we know that these would cancel and this would be equal to 27 plus negative 21 or 27 minus 21 would be 6. so this equation here that i've generated i'm going to call this equation 4. so let me erase these two i don't need this information anymore and i'm just going to write this off to the side and say this is my equation 4 it's going to be 3x minus 2y is equal to 6. okay now i used equation 1 let me mark this in equation two okay so now i can't use equation one and two again i can use equation one and three or i can use equation two and three but the key thing is i've gotta eliminate the variable z again okay so let's just go with two and three so let me erase these check marks and i'm just going to put a mark here and here to remind you that we're working with two and three okay so i need to eliminate the variable z this guy is a five this guy is a two so what i can do is take the lcm of 5 and 2 which is 10. so for this guy equation 2 i'm going to multiply both sides by 2 that's going to make the coefficient of z a positive 10. so i'm going to say that 2 times 4x is 8x and then plus 3y times 2 would be 6y and then plus 5z times 2 would be 10z this equals 21 times 2 is going to be 42. so again this is equation 2 and i'm not going to label this here because we're going to get rid of it in a second then for equation 3 i need to multiply this guy right here by a negative 5 right because i want to end up with a negative 10 so that it will cancel with this guy right here okay so i'm going to multiply both sides of the equation by negative 5. so 6x times negative 5 would be negative 30x and then y times negative 5 would be minus 5y and then 2z times negative 5 would be minus 10z and this equals if you had 9 times negative 5 you'd have negative 45. okay so let's add the left sides together set this equal to the sum of the right sides so 8x minus 30x is going to give me negative 22x 6y minus 5y is plus y and then we know this is going to cancel and this equals 42 minus 45 is negative 3. okay so let's erase this we don't need this anymore and let's drag this up here and we'll call this equation five okay let me change the color here so this is equation four so now if i look at equation four and five this gives me a linear system in two variables now right z is eliminated so let's work with this now we'll figure out the values for x and y and then we can go back into either equation one two or three plug in for x and y and find out the unknown z okay so for this one it looks like it would be pretty easy to use substitution because the coefficient on y here is a 1 right so i can solve this for y pretty easily i can say that y is equal to if i just add 22x to both sides of the equation 22x minus 3 okay so then what i can do is i can take this guy right here and i can plug it in for y in this equation 4. okay so that's going to get plugged in right there so what i'd have there is 3x minus 2 times again this quantity right here this is being plugged in for y so that is going to be 22x minus three okay let me change the color again and then we have that it's equal to six so let me scroll down and get some room going and we'll come back up so for this guy we have three x and then i'm gonna distribute the negative two to each term so that would be minus 44x and then plus 6 and this equals 6. so i have 6 on each side of the equation so i can just get rid of it right if i subtracted 6 away from each side this would just be a 0 and over here it would go away so 3x minus 44x would be negative 41x so negative 41x and we know this equals zero at this point but let's just complete this so divide both sides by negative 41 and we'll find that x is going to be equal to zero okay so let's erase this and let me scroll back up and let me write this over here so we've found already that x is equal to zero okay so that's the first one so i can plug in a zero for x in either equation four or five find out the value for y so let's use equation four three times zero is zero so i would have that negative two y is equal to six divide both sides by negative two and we'll get that y is equal to negative three so now i know y is equal to negative 3. okay so x is 0 and y is negative 3. okay so from here i just have to figure out what z is and to do that i can plug into either equation 1 2 or 3. doesn't really matter so let's just choose equation one so we would have seven times zero which is zero plus you have y which is negative three so negative three plus five z equals twenty-seven so let's add three to both sides of the equation let me scroll down for a minute i'll come back up so we know that this is going to cancel 5z is equal to 27 plus 3 is 30. divide both sides by 5 and we get that z is equal to 6. okay so let's erase all this and let's scroll back up and we'll say again z is equal to 6. so i would write my solution again as an ordered triple so i would say it's the x value which is 0 comma the y value which is negative 3 comma the z value which is 6. now i'll let you check this on your own i've already checked it i know it works to check this you're just going to plug in a 0 for each x a negative 3 for each y and a 6 for each z and you want to verify that in each equation that solution gives you a true statement all right let's take a look at another example now that we've kind of done one we can go a little bit quicker for this one so we have 2x plus 4y plus 5z equals 15. let's call this equation one we have negative six x plus y plus four z equals negative twenty-five let's call this equation two we have negative x minus three y plus three z equals negative sixteen let's call this equation three okay so again i want to pick two equations and with those two equations i want to eliminate one of the variables okay it doesn't matter which one i eliminate but in the second step when i eliminate the same variable from another two equations it's got to be the same one so if i look at kind of equation one and two i see that it would be pretty easy to eliminate the variable y okay so let's just go with that and i can just multiply this guy right here equation 2 by negative 4 and what's going to happen is if i had a negative 4 here and i have a negative 4y and a positive 4y so y would drop out okay so i'm going to multiply equation 2 by negative 4. so negative 4 times negative 6 is 24 so it'd be 24x negative 4 times y is minus 4y negative 4 times 4z is minus 16z and this equals negative 4 times negative 25 is 100 okay so this is equation two again i'm not going to label it because it's going to drop out in a minute and we're going to have something different then let me write equation one underneath so 2x plus 4y plus 5z equals 15. so again now i want to add the two left sides set this equal to the sum of the two right sides and again make sure the left side has all your variable terms make sure the right side has just a constant okay that's how you want to set it up so 24x plus 2x is 26x negative 4y plus 4y that's going to go away right we can cancel that negative 16z plus 5z is going to give me negative 11z and this equals 100 plus 15 is 115. so let's erase this and let's drag this up and we'll call this equation four and let me use the same color so it's consistent so this is going to be equation four okay so now we want to eliminate the same variable which was y from two other equations so i can't use one and two again so i can use one and three or 2 and 3. so if i use 1 and 3 it's going to be a little bit more challenging than if i use 2 and 3 right because if i use 2 again i have y with that coefficient of 1. it's a little bit easier to work with so let's eliminate y from 2 and 3. so for equation 2 i'm just going to multiply it by positive 3 because if i had a 3 there and a negative 3 there again i could cancel the y so if i multiply equation 2 by 3 i would have negative 18x i would have plus 3y i would have plus 12z and this would equal negative 75. okay i just multiplied each term there by three okay that's all i did then what i want to do is take this guy right here this equation 3 and write it underneath so negative x then minus 3y then plus 3z this equals negative 16. okay let me scroll down and get some room and i'll come back up so what i want to do is add the two left sides set this equal to the sum of the two right sides so negative 18x minus x is negative 19x we know that 3y minus 3y that's going to cancel we know that 12z plus 3z would be plus 15z and this equals for negative 75 minus 16 that's negative 91. okay so let's erase this and let's drag this up okay let's drag this up and let me scroll back up here so let me label this in the same color so we're consistent so this is going to be equation five okay so now with equation four and five we have a linear system with two variables now this one is not as easy to solve with substitution it's not really easy to solve with elimination either we just have to kind of pick our poison so let's just go ahead and use elimination for this one i am going to choose to eliminate z so the least common multiple between 11 and 15 would be the product of the two which would be 165 okay so all i really need to do they already have opposite signs i'm just going to multiply equation 4 by 15 and i'm going to multiply equation 5 by 11. okay again they have opposite signs already so if we just do that these guys are going to end up being negative 165z and positive 165z and we can cancel them out so i would start by multiplying equation four by 15. so 26x times 15 is 390x then minus 11 times 15 is going to be 165. so this would be 165 and then z and this equals 115 times 15 is 1725. what we're going to do for equation 5 is we're going to multiply both sides by 11. okay so negative 19x times 11 is going to be negative okay negative 209 x then we have our plus we know that 15 times 11 is 165 and then times our z and this equals we're going to have negative 91 times 11 which is negative 1001. okay let me scroll down and get some room going and we'll come back up so if we add the two left sides together this is going to cancel 390x minus 209x gives me 181x and this would be equal to 1725 minus 1001 is going to be 724. now what i want to do to isolate my x there i want to divide both sides by 181 and what i'm going to get is that x is equal to 4 right 724 divided by 181 is going to be 4. okay so at this point we know x is 4 so let me erase this let's come back up here and i'm just going to say that x is equal to 4. so we know that at this point so i can plug in a 4 for x in either equation 4 or equation 5 and i can find out what z is so let's just go ahead and use equation 5 so we would have negative 19 times plug in a 4 for x plus 15 z equals negative 91. okay so let's scroll down a little bit so negative 19 times 4 is negative 76 and then plus 15 z equals negative 91. so what i'm going to do is i'm going to add 76 to both sides of the equation so this will cancel obviously let me scroll down a little bit more and get some room negative 91 plus 76 is going to be negative 15. so what we would have is 15z equals negative 15 so we know z is going to be negative one right divide both sides by 15 we get that z is equal to negative one okay so let me put z is equal to negative one all right so we know x is four we know z is negative one we really can erase these two equations at this point because we don't need them anymore now all we need to do is figure out what y is so y equals 1. we'll just plug into either equation 1 2 or 3 okay and figure out what y is so what i'm going to do is i'm going to plug a 4 in for x and a negative 1 in for z and let's just use equation 2. so i'm going to do negative 6 times 4 plus y plus 4 times for z is a negative 1 this equals negative 25. okay so we know this would be negative 4 so let's just write that and we know this would be negative 24. so let's write that we also know that negative 24 minus 4 would be negative 28. so let's write y minus 28 like this if we add 28 to both sides of the equation we're going to find that y is equal to positive 3. so let's erase this and let's put that y is 3. so let's erase this we found our answer again as an order triple it's the x value first so 4 comma the y value second so 3 and then the z value is last so negative 1. again it's x comma y comma z okay and again you can remember this because it's in alphabetical order so the order triple is 4 comma 3 comma negative 1 again an x value of 4 a y value of 3 and a z value of negative one so one type of problem that you might encounter that is actually easier to solve but causes a lot of confusion for students is where you have a linear system in three variables but one of your equation only has two variables okay so something like this where we have 4x minus c equals three you've got only x and z and then the other two equations have x y and z so the other ones you have six x minus five y minus z equals negative twenty-five and you have x minus y minus two z equals negative 7. so let's just go ahead and label these this is going to be equation 1 this is going to be 2 and this is going to be 3. so what can we do here a lot of students will struggle and say i don't know how to get started well really a lot of the work has already been done for you you have a linear equation in two variables only with equation one so what you can do is you can take equation two and three you can eliminate the variable that this one is missing which is y okay and so you would have equation one and let's just call it equation four so those two would make up your linear system in two variables and from there you can figure out everything else okay so it just kind of saves you a step so what i'm gonna do is i'm gonna take 2 and 3 and i'm going to eliminate the variable y so to do that i would want this to be plus 5y so it would cancel with this minus 5y and i can do that by multiplying both sides of equation 3 by negative 5. okay so if i multiply x by negative 5 i get negative 5x if i multiply negative y by negative 5 i get plus 5y if i multiply negative 2z by negative 5 i get plus 10z and this equals if i multiply negative 7 by negative 5 i get plus 35. so i've got this guy and again i've got this equation 2 which stays as is 6x minus 5y minus z equals negative 25. okay so if we add the two left sides set this equal to the sum of the right sides what we're going to find is that this guy is going to cancel and negative 5x plus 6x is going to be x and then plus you've got 10 z minus z which is going to be 9z and this equals you've got 35 minus 25 which is 10. okay so let me erase this real quick and let's just label this as equation four okay so equation one and equation four give us a linear system in two variables so let me just copy this one down here again this is four x minus z equals three in this case it's easiest to use substitution because i've got a negative one as the coefficient of z in equation number one right i drug this down here this is equation one and in equation four i've got a one as the coefficient for x okay so whichever one you want to use i'm just going to go with equation 4 solve it for x plug it in for x in equation 1. okay so let me re-copy this number here so i don't get confused all right so let's scroll down and what i'm going to do again is solve equation 4 for x so i would have x is equal to just subtract 9z away from each side so negative 9z plus 10. and then the whole idea is i'm plugging this again what x is equal to i'm plugging that in for x in this equation so that's going to go in right there okay that's all we're doing so 4 times again plugging in for x it's negative 9z plus 10 and then you have minus 1 z or just minus z equals 3. so let's solve this real quick find out what z is we'll go back and figure out what x is so 4 times negative 9c is minus 36z then 4 times 10 is 40 so plus 40 then minus z and that kind of looks like a 2. so let me make that more clear and this equals 3 okay so we have negative 36z minus e that's negative 37z let me get rid of that and what i can do is i can subtract 40 away from each side of the equation so that's going to be gone 3 minus 40 is negative 37 so we can see that z is going to be positive 1 right because if i divide this side by negative 37 and this side by negative 37 we're going to get that z is equal to 1. okay so let me erase all this i'll put a line and say that z is equal to 1. now with equation 4 or equation 1 we can plug back in and figure out what x is so let me erase this from here i don't even need it anymore i'm just going to plug into equation 1 and i can just erase equation 4 because i'm not going to need that anymore either so i'm going to plug a 1n right here okay so i'm going to have 4x minus 1 equals 3. if i add 1 to both sides of the equation i get that 4x is equal to 4 obviously x there is going to be 1. right if i drag this up and i divide both sides by 4 i get x equals 1. so z is 1 and so is x now what is y well i need to plug into either equation 2 or 3 again something with y involved so let's just go ahead and plug into equation 3. it's a little bit easier so x is 1 so i'd have 1 there then minus y that's still unknown then minus 2 times z z is 1 so i can just leave that as it is and this equals negative 7. so 1 minus 2 is negative 1. so let's erase this and put this as a negative 1. and so let me add 1 to both sides of the equation this goes away negative 7 plus 1 is negative 6 and to finish this off let's divide both sides by negative 1 and we'll find that y is going to be equal to positive six so y is equal to six again as an ordered triple you would have that x is one that's the first one y is six that's the second one and z is one that's the third one again it's going to be your x value comma your y value comma your z value it's in alphabetical order so it's very easy to remember all right so now let's move on and talk about the kind of special case scenarios we're going to see so we're also going to have situations where we have no solution and situations where we have an infinite number of solutions again it's the same scenario so if you see your variables drop out and you get something that's always true you have an infinite number of solutions if you see your variables drop out and you have a false statement then you know you have no solution okay it's just that simple so let's go ahead and look at this guy right here so we have negative 3x minus 3y minus c equals twelve so let's say this is equation one we have five x plus two y plus five z equals negative three this is equation two we have negative six x minus six y minus two z equals negative thirty let's say this is equation three okay so let's go ahead and eliminate from equation one and two let's eliminate the variable z okay it looks like it's pretty easy to do so let's just multiply equation one by five so what i'd have is negative 3x times 5 is negative 15x then negative 3y times 5 is minus 15y then negative z times 5 is minus 5z and this equals 12 times 5 is 60. okay and then i have equation two i'm going to leave that alone so i'll have my 5x plus 2y plus 5z and this equals negative 3. okay so let's add the left side set this equal to the sum of the right sides so negative 15x plus 5x is negative 10x and then negative 15y plus 2y is minus 13y and then negative 5z plus 5z that's going to cancel this equals 60 minus 3 which is going to be 57 so let's call this equation 4 okay so so far no problems right everything's okay so let's call this equation 4 let me label this in green and now let's eliminate z from equation 1 and 3 okay because that's pretty easy to do equation 3 i'm going to leave alone so i'm just going to leave that as negative 6x minus 6y minus 2z equals negative 30. and for equation 1 what i'm going to do i need this to be plus 2z so it would eliminate with this minus 2z here okay so to do that i'm going to multiply both sides of the equation by negative 2. so negative 2 times negative 3x is positive 6x negative 2 times negative 3y is plus 6y you can already see a problem and then negative 2 times negative z is going to be plus 2z okay and this is going to be equal to negative 2 times 12 is going to be negative 24 okay you already see the problem because this whole left side is going to cancel it's going to be gone right so negative 24 minus 30 is negative 54. so this is basically going to turn into 0 is equal to negative 54 okay which is nonsense so this is always false so when you get something like this just stop and say there's no solution okay so you can put the symbol for the null or empty set or you can write out no solution all right let's wrap up the lesson to look at one more special case scenario so suppose we have x plus 2y minus 2z equals 1. again i'm going to label this as equation 1 and then 5x minus 3y minus 2z equals 27. i'll label this as equation 2 and then negative 2x minus 4y plus 4z equals negative 2. i'll say this is equation 3. okay so let's just go through and eliminate our variable let's just go ahead and do x okay so from equation 1 and equation 2 i can get rid of x pretty easily i can multiply equation 1 by negative 5 okay so negative 5 times x is negative 5x and then plus negative 5 times 2y would be negative 10y so let's say minus 10y and let me make that a little bit better and then we would have negative 5 times negative 2z which would be plus 10z and this equals 1 times negative 5 is negative 5. okay and then i'm just going to copy equation 2. so 5x minus 3y minus 2z equals 27. okay so add the left sides and the x variable is going to cancel right so this is basically gone negative 10 y minus 3y is going to be negative 13y so negative 13y 10z minus 2z is going to be 8c so plus 8z and this equals you have negative 5 plus 27 which is 22. so this equals 22. and again so far so good right no problems everything looks okay so let's drag this up so again we used equation one and two so now let's use equation one and three and let's again eliminate the x variable okay so to do that now i'm gonna multiply equation one by positive 2. so what i'm going to have is 2 times x which is 2x then plus 2 times 2y which is 4y then minus 2 times 2z which is 4z and this equals 2 times 1 which is 2. okay so this is a transformed version of equation 1 again we just multiplied it by 2. so then i want to take equation 3 which is negative 2x minus 4y plus 4z and this is equal to negative 2. now do you see a problem yes of course because this is going to eliminate and this is going to eliminate you're going to get 0 equals 0 which is always true okay so when it's always true when you end up with a true statement like this you have infinitely many solutions okay so let's just erase this and we can just list our answer so let's just write this out and say we have infinitely many solutions in this lesson we want to talk about curve fitting and specifically here we're going to be finding the equation of a parabola given three points on the parabola all right so in the last lesson we reviewed how to solve a linear system in three variables again if you're in this course that's nothing new for you you've been doing that you know since algebra one in most cases you definitely did it in algebra two okay so that was nothing new so when you're in this section and you're talking about systems and you're talking about matrices you're going to get a lot of different application problems kind of thrown at you so these might be word problems or something related or even something like this where we're told to find the equation of a parabola given three points on the parabola okay we want this equation in standard form so we want it in the form of y equals ax squared plus bx plus c okay and again we're given these three points that lie on the parabola so remember these are x comma y values okay that we're given these are x comma y values so this is pretty easy to do all i'm going to do is set up three equations okay so that i can solve for the three unknowns the three unknowns are going to be the a the b and the c here that i'm going to need to write this equation okay so what i'm going to do i'm going to rewrite this let's just say it's y equals we've got a x squared plus b x plus c and what we're going to do is we're going to plug in for x and y we're going to do that three times so let me erase this and let me just put some parentheses and i'm just going to be doing some plugging in and let me erase this and put some parentheses and to make this really easy i'm going to highlight where i'm plugging in x with a yellow okay and i'm going to highlight where i'm plugging in a y with a kind of blue maybe it's a light blue i'm not sure the official color on that so let me highlight this and then let's go through and do this so y gets plugged in here right because it's y equals ax squared plus bx plus c so y is all the way to the left so i'm going to plug a 5 in there then this equals we have a times there's an x that goes in here so that's going to be 1 and then that's squared then plus b times again you have an x that goes there so that's going to be a 1 and then plus c okay so that's your first equation so for this guy it's going to be 29 for y and it equals a times in my parentheses that have negative 2 and that's squared and then plus b times again i'm going to have negative 2 and then plus c and then for my third equation my final equation again i do the same thing so y is 9 so 9 is going to go out here and this equals you have a times in the parentheses i'm going to have a 2 this is squared then plus b times again i'm going to put a 2 and then plus c so let's copy this real quick and let's go to a fresh sheet so we have some room to work okay and i'm just going to rearrange this a little bit so i'm going to simplify some things so i'm going to say equation 1 i know that 1 squared is 1 so this would just be a and then plus b times 1 is just b and then plus c and this equals we have 5. okay so that's my equation 1. so let me just erase this from here i don't need it anymore and then for equation 2 i have negative 2 squared which is 4 then times a and then plus negative 2 times b so i can just put minus 2b if you want and then i have plus c and of course this equals 29. and so let's erase this we're done with that and then now for equation three i have this guy 2 squared is 4 then times a and then i have 2 times b which is 2b so plus 2b and then plus c and this equals not okay so i have my three equations three unknowns pretty easy to solve for this so let me scooch this up and let's get started all right so we know from the last lesson that all we need to do is take kind of two equations from the system pick a variable and eliminate it okay that's how we kind of get the ball rolling and what i want to do i'm going to look at equation one and two it's probably easiest here to eliminate c okay so what i'm going to do is i'm going to multiply equation 1 by negative 1 okay and then i'm going to add that to equation 2 okay because what i'm going to do if i multiply equation 1 by negative 1 this is going to end up being a negative c there okay and then negative c and positive c those are going to cancel all right so negative 1 multiplied by equation 1 would be negative a minus b minus c equals negative 5. okay i just multiplied everything by negative 1 so it just changed the sign of each term that's all it did then for equation 2 we keep that as it is so 4a minus 2b plus c equals 29 and then we're going to add these two equations together so remember when we do this we're adding the two left sides together we're setting this equal to the sum of the two right sides now in order for this to work you have to have your equations in the format where the variables you're working with are on one side and the number the constant is on the other okay so you see how i have all my variable terms my terms with the a the b and the c involved that's all on the left and the constants on the right so you want to set it up that way all right so if i add negative a plus 4a i'm going to get 3a then negative b plus negative 2b would be minus 3b then negative c plus c we know this would cancel and this equals negative 5 plus 29 would be 24. so let's call this equation 4 and i'll say this is 3a minus 3b is equal to 24. so let's erase all this we don't need any of this anymore and now let's pick two other equations it can't be one and two because we already used those so any two other equations we could do one and three or two and three and we want to eliminate the same variable so we want to eliminate c again and that's pretty easy to do so if i pick two and 3 let's just go with that if i multiply equation 2 by negative 1 i'd have negative 4a plus 2b minus c equals negative 29. and then for equation 3 i'll just keep it as it is so 4a and then we're going to have plus 2b and then plus c and this equals 9. okay so when we do this we get an even easier situation this is going to cancel and so is this so i get 2b plus 2b which is 4b and this is equal to negative 29 plus 9 is negative 20. so for equation 5 i get 4b is equal to negative 20. and let me erase this real quick this one just happens to be much easier to work with if i divide both sides by four i find that b is equal to negative five okay so i've already got one value b is negative five i can plug that back in here in equation four and i can figure out what a is so i would have 3a minus 3 times plug in a negative 5 for b so plug in a negative 5 there and this equals 24 okay so what we'll do we have 3a and then negative 3 times negative 5 is plus 15 and this equals 24 okay so if i subtract 15 away from each side of the equation let me kind of do this up here we're going to have 3a is equal to 24 minus 15 is going to be 9 divide both sides by 3 and we find that a is equal to 3. okay so my board's getting kind of busy so let me erase some stuff so at this point i know that a is equal to 3 and that b is equal to negative 5. so what is c c equals what well i can find c by just plugging back into either equation 1 2 or 3. it's easiest to work with equation 1. so let me plug in a 3 for a and a negative 5 for b what would that give me so you would have 3 plus negative 5 plus c equals 5. 3 plus negative 5 is negative 2 so you'd have negative 2 plus c equals 5. if i add 2 to both sides of the equation i get c is equal to 7. so now i have enough information to write the equation of this parabola in standard form so let me just copy this and go back up so we have that a is 3 b is negative 5 and c is 7. so just plug into your standard form we have y is equal to for a i'm going to put a 3 and then times x squared then plus then for b it's negative 5. so instead of plus let me just put minus 5 here and then times x and then plus c and c of course is 7. so the equation is y equals 3x squared minus 5x plus 7. if you wanted to check this you could plug in for x and y and verify that you get a true statement in each case i've already done this i know that it works out but again just to do one of them you could plug in a 1 for each x and a 5 for the y okay and verify you get a true statement one squared is one so one times three would be three and then five times one is five so minus five plus seven three minus five is negative two negative two plus seven is five so we would have five equals five which is true okay so we know this one works again i'll leave it up to you to check these other two but i can tell you right now i've already checked them and they do work so our equation is going to be y equals 3x squared minus 5x plus 7. all right so let's look at another example so again we have our y equals ax squared plus bx plus c and now we're given the points 1 comma 6 negative 1 comma 2 and negative 2 comma 15. so again the same process we're going to say that y is equal to a x squared plus b x plus c and we're just going to plug in for x and y again using this data from the points so for the first one again if i'm plugging in for x we have an x comma y the x is one so that's one and the y is six so let's put a six there then for the next one you would have a two for your y and then your x would be negative one okay so let's fill that out and then for the last one your y would be 15 and your x would now be negative two okay so let's copy this and go to a fresh sheet okay so let's go ahead and make sense of this real quick so 1 squared is 1 1 times a is just a 1 times b is just b so that's fine like it is so we'd have a plus b plus c and let's just put equal 6 like this and i can clean that up and make it look a little nicer so for one we have a plus b plus c equals six all right for this second one negative one squared is one one times a is just itself negative one times b would just be minus b and then you have plus c and of course you have this equals you have this 2 over here so let me move that over here we'll say that a minus b plus c a minus b plus c is equal to 2. so this would be my equation 2. okay then for equation 3 we have negative 2 squared which is 4. so this would be 4a you have negative 2 times b which is minus 2b so minus 2b and then you just have plus c so plus c then let me just move this over here so i'm going to erase this and put equals 15 like this so this will be my equation three okay so we can just pick a variable to eliminate again you can do c b or a whatever you want to do c looks like it's pretty easy you have a coefficient of one in each case b is also pretty easy as well so let's just use b to kind of change things up so if i eliminate b from using equation one and two what i would do is just add one and two together right because i have opposite coefficients on b here and here right so i have a 1b and a negative 1b there so i'm going to just add the two equations together and get equation 4. so a plus a is 2a and then 1b plus negative 1b that's going to cancel c plus c is 2c so plus 2c this equals 6 plus 2 is going to be 8. now i can simplify the equation because everything's divisible by 2. so if i divide 2a by 2 i get a if i divide 2c by 2 i get c and if i divide eight by two i get four so a plus c equals four a plus c equals four all right for equation five let's just go through and get this from equation one and equation three okay so i'm going to multiply equation 1 by 2 so i would get 2a plus 2b plus 2c equals 12. okay and then for equation 3 i'm going to leave it as is so 4a minus 2b plus c and this equals 15. okay so the reason i multiplied equation 1 by 2 is so i could have opposite coefficients here i've got a plus 2b and i've got a minus 2b okay so those are going to cancel so 2a plus 4a is going to give me 6a these again are going to cancel 2c plus c is going to be 3c and then 12 plus 15 is going to give me 27 okay now each part here is divisible by 3. 6 divided by 3 is 2 3 divided by 3 is 1 27 divided by 3 is not okay so let's erase this now we can solve this system pretty easily either using substitution or elimination let's just use substitution that looks pretty easy so i'm going to solve equation 4 for a so i'm going to say that it's a is equal to subtract c away from each side of the equation so negative c plus 4 i'm going to plug in for a in this equation 5. okay so this right here that's what a is equal to that's what i'm going to plug in for right there so i would have 2 times this quantity here negative c plus 4 okay just plug it in there then plus c equals 9. let me scroll down real quick and get some room so 2 times negative c is negative 2c plus 2 times 4 is 8 then plus c equals 9. negative 2c plus c is going to be negative c and then plus 8 equals 9. subtract 8 away from each side of the equation you can basically just erase it here and just say this is a one so to finish this up we can divide both sides by negative one and we'll find that c is equal to negative one so that's our answer for c so let me erase all this so i'm going to say that c is equal to negative one and then i can plug in for equation 4 or 5 doesn't matter figure out what a is so let me use equation 4 because it's a little bit easier if i plug in a negative 1 for c i'd have a you could basically say minus 1 equals 4 add 1 to both sides and you'll find out that a is equal to 5. so now we have c and we have a we need to find b okay so b is still unknown i can get rid of this i don't need that information anymore let me scooch this up and so what i'm going to do is plug into again either 1 2 or 3 doesn't matter figure out what b is so the easiest equation looks like equation 1. so i have an a value of 5 plus b plus a c value of negative one so let's just write minus one and this equals six so five minus one is four so we'll say that this is b plus four and i'm gonna go ahead and just subtract four away from each side of the equation and find that b is equal to two okay so b is 2. all right so let's grab these values and go back up okay so plugging back in here we have y is equal to for a we have 5 so 5 and you have your x squared and then plus for b we have two so times x and then for c we have negative one so minus one so the equation of the parabola that passes through these three points would be y equals five x squared plus two x minus one again you could check this you could take these points plug them in and see if you get a true statement you should for every scenario i advise you to pause the video and check i've already done that i know this is the correct solution all right let's look at one more problem and for this one i'm going to kind of give you a spin on this suppose you have a sideways parabola okay you know if you get something like x equals y squared this is a sideways parabola okay so we have x equals we have a y squared plus b y plus c and we have three points that are on this parabola this sideways parabola okay so we're going to solve and figure out what this equation is going to look like the parabola that goes through these points again this one's going to be a sideways parabola so i'm going to do the same thing but i've got to kind of change up what i do normally i take my y value and put it out on the left but now i'm going to take my x value and do that so i'm going to take this x value of 4 okay because x is on the left here this equals you have a times 4y i'm now going to plug in a 2 okay that's going to be squared and then plus b times for y i have a 2 again and then plus my c okay and i'm going to do this again so i'm going to have my 12 that's my x value equals a times i've got my negative 2 that's going to be squared again that's my y value then plus i've got b times my negative 2 and then plus c okay so you're just switching what you're doing normally you have your x's going here and here and your y here now that's flipped right now your x goes over here your y goes here and here okay so one more we have our negative 3 that goes here this equals a times again plug in the y here so 1 squared then plus b times 1 and then plus c so let's copy this we'll come back up get a nice fresh sheet going again 2 squared is 4 so this would be 4a so let's put 4a and then 2 times b is 2b so plus 2b plus 2b and then plus c okay so plus c and let me put equals 4 on this side i can erase it from here and let's go ahead and just label this as equation one okay let me kind of slide this down save a little room and we'll slide this down a little bit more okay so for the next one negative 2 squared is going to be 4 so we'd have 4a and then negative 2 times b is minus 2b so we'll put minus 2b and then we have plus c so let's put plus c and of course this equals 12. we'll just erase it from over here so this is going to be equation 2. let me just kind of slide this one down okay so one more to do so we have 1 squared which is 1. you just get rid of that 1 times b is just b so you get rid of that so you have a plus b plus c so a plus b plus c and this equals negative 3. so let's get rid of this let's put this as equation three and let's slide this down okay so what variable do we want to eliminate again it's pretty easy to do b it's also pretty easy to do c a looks like it's a little bit more work but even that's pretty easy so let's just go ahead and eliminate the variable b so i can go through with the first two equations and i can just add them okay because i have 2b and negative 2b so 4a plus 4a is 8a so this is 8a and then 2b plus negative 2b that's going to cancel and then c plus c is 2c and then 4 plus 12 is 16. again i can simplify this divide each part here by 2. a divided by 2 is 4 so 4a plus 2 divided by 2 is 1 so just c and this equals 16 divided by 2 is 8. okay then for equation 5 let's get that from 2 and 3 okay so what i'm going to do is i'm going to multiply equation 3 by positive 2 so that i can have opposite coefficients on that variable b okay so if i multiply equation 3 by 2 i'd have 2a plus 2b plus 2c equals negative 6. again i just multiplied everything by 2. then i keep equation 2 as is so 4a minus 2b plus c equals 12. okay so let's go ahead and add the left sides set this equal to the sum of the right sides 2a plus 4a is 6a then 2b minus 2b that's gone 2c plus c is plus 3c and then negative 6 plus 12 is 6. okay so let's erase this we don't need this anymore and again we see that we could simplify this we could divide each part by three so this would be two a plus c equals two all right so let me erase this and write this just as four a plus c equals eight and let me erase this and just kind of drag this up again it's 2a plus c equals 2. so we can just use substitution here i know that i can solve this guy for c i say c is equal to subtract 2a away from each side so negative 2a plus 2 and then plug this in for c here okay that's all i'm going to do plugging this in because c is equal to this in there so what we're going to have is that 4a plus we'll plug this in for c so negative 2a plus 2 and this is equal to 8. okay so i can just drop these parentheses they don't really do anything here so 4a minus 2a is just going to be 2a and if i subtract 2 away from each side this would go away and this would be six divide both sides by two and i'm going to find that a is equal to three okay so a is three i can erase all this and i'll just write that over here for now so a equals three plug that in in equation five plug in a three there and you'll see that 2 times 3 is 6. so 6 plus c is equal to 2. if i subtract 6 away from each side of the equation i'll find that c is equal to 2 minus 6 is negative 4. okay so let's erase these we don't need this information anymore and we're just going to write this up here and say that a is equal to 3 and c is equal to negative 4. let's erase this and now let's just go ahead and plug in and find out what b is so b is our unknown so let's plug into equation 3 that's the easiest so i'm going to plug in a 3 there i'm going to plug in a negative 4 there so i already know that 3 minus 4 is negative 1. so let me just write this as b minus 1 equals negative 3. add 1 to each side of the equation and i'll find that b is equal to negative 2. okay so let's put a negative 2 there and let's just copy this real quick and go back up so we'll copy these values okay so let's go ahead and write our equation so we have x now remember this is different this is a sideways parabola so x is going to be equal to for a we have 3 then times y squared then plus 4b we have negative 2. so let me just write minus two so minus two times y and then for c we have negative four so minus four okay so x equals three y squared minus two y minus four so you just wanna pay close attention to the problem they give you this is much less common but you might get it so just pay attention to the equation you get if you're told that the three points lie on the parabola and the equation is y equals ax squared plus bx plus c do what we did in the first two problems if you get a sideways parabola again like this one where x is equal to a y squared plus b y plus c again you want to follow this format when you write your equation in this lesson we want to review applications of linear systems all right so over the course of the last few lessons we've kind of talked about systems of linear equations in two and three variables now we're just going to look at some kind of application problems or some word problems that come up in this section and then from here we're just going to be dealing with kind of matrices all right so let's start out with the first problem so we have that a health bar is made from a corn mixture an egg mixture and a vegetable mixture one ounce of the corn mixture contains 0.25 grams of protein 0.3 grams of carbohydrates and 0.4 grams of fat the egg mixture contains 0.4 grams of protein and 0.2 grams of both carbohydrates and fat lastly the vegetable mixture contains 0.2 grams of protein 0.1 grams of carbohydrates and 0.3 grams of fat how many ounces of each mixture should be used to create a health bar that contains 22 grams of protein 18 grams of carbohydrates and 28 grams of fat so just like any other word problem that we've kind of worked on before the very first thing you always want to do is figure out what are you being asked to find a lot of times you'll struggle because you're kind of flopping around through information you're not specific on what you need to find so you need to key in on that you need to say okay i want to find how many ounces of each mixture okay so remember we had the corn mixture the egg mixture and the vegetable mixture and should be used to create a health bar that contains 22 grams of protein 18 grams of carbohydrates and 28 grams of fat okay so we have three unknowns right we need to know again how many ounces of the corn mixture we're going to use how many ounces of egg mixture we're going to use and how many ounces of the vegetable mixture we're going to use to kind of create this health bar with these specifications so let's go ahead and label these three as x y and z so let's let x be equal to the amount in ounces for the corn mixture and then we'll say y can be equal to the amount in ounces for the egg mixture and just to keep this shorter i'm just going to leave mixture off and then we'll also say then for z we're going to say the amount in ounces for the vegetable mixture so i'm just going to put veg to kind of abbreviate that okay so the best way to kind of look at this type of problem is with a table you don't need one but it definitely helps so i've already kind of pre-drawn one for us so i'm going to show you how you can fill out the information here and then get some equations going and get a solution so the first thing is let's talk about the different kind of mixtures you have you have the corn mixture you have the egg mixture and you have the vegetable mixture okay which i'm going to abbreviate veg then we're given the amounts of protein carbohydrates and fat so i'm just going to put protein just pro carbohydrates c-a-r and then for fat just fat of course that spells fat but for the other ones they're abbreviated okay then we also want to put in a total here and this will make sense once we kind of have the information going so what i can do is i can look up back in the problem the statistics for how much protein carbohydrates and fat each one of these guys have i'm just going to fill in my table so let's go back up so for the corn for every ounce it's got 0.25 grams of protein 0.3 grams of carbohydrates 0.4 grams of fat so we've got 0.25 for protein 0.3 for carbs and 0.4 for fat all right for the egg mixture we have 0.4 grams of protein and then 0.2 grams for both carbohydrates and fat so we've got 0.4 and then 0.2 and then 0.2 again then lastly for our vegetable mixture we've got 0.2 grams of protein 0.1 grams of carbohydrates and 0.3 grams of fat so we've got 0.2 for the protein 0.1 for the carbohydrates and 0.3 for the fat now i have a column here for the total and what i'm going to do here notice how going across it's all protein right it's the protein for the corn the eggs and the vegetable we want to know what the protein total is going to be from these three sources and again it's based on the requirement that's given to us when we're building this health bar if we go back we see that it needs to be 22 grams of protein 18 grams of carbohydrates and 28 grams of fat so for the total there for protein it's going to be 22 right just going across for carbohydrates going across the total needs to be 18 and for fat going across the total needs to be 28 okay so this is how we're going to get our equations because what we know is that the protein only comes from three sources right it comes from corn eggs and vegetables the number of ounces of this corn mixture i use so that's going to be x times this kind of 0.25 which is the number of grams okay of protein per 1 ounce so this amount here would tell me the amount of protein i'm getting from just the corn okay but i need to reach this amount so i'm going to add to this the amount of protein i get from the egg so that's 0.4 times y okay then plus additionally the amount of protein i'm getting from the vegetables so 0.2 times z this has to be equal to 22 okay because that's the only place the protein is coming from the same thing is going to happen with the carbohydrates and the fat we can just go across so i can say that 0.3 times x plus 0.2 times y plus 0.1 times z is equal to 18. and i'm going to run out of space here so let me scroll down i'm going to cut this off but remember that the kind of column here is for corn then this guy is for egg and this guy's for vegetables so we're just going to copy we have point four x plus point two y plus point three z equals twenty eight okay so that's all we really need to do so let's just cut this away from here we're going to take a fresh sheet and let's solve this system real quick and get our solution for our problem now you can work with decimals if you want if you have a calculator it's fine a lot of people don't like working with fractions or decimals so i'm just going to transform this into a kind of decimal free equation in each case i'm just going to go ahead and multiply kind of everything by a hundred so what i'd have is 25x plus 40y plus 20z equals 2 200. okay just multiplied everything there by 100 so that i could clear all the decimals okay so this one's gone now and then for this one i'm going to do multiplying everything by just 10 because that's all you need so this would be 3x plus 2y plus 1z equals 180. so let's erase that and then for this one i only need 10 as well so i'm going to have 4x plus 2y plus 3z equals 280. okay so let's erase this and let's scooch this up so again when you work with a linear system with three variables and you're not using a matrix the first thing you want to do is kind of look at these equations so i'm just going to label this as 1 2 and 3. you just want to pick two of them and then from the two you want to eliminate one of the variables so if i look at these equations it looks like it's pretty easy to eliminate the variable y okay because i have 2y here and 2y here so from 2 and 3 i can really quickly eliminate y by just multiplying one of the equations by negative 1. and i'm going to start with 1 and 2 and just multiply equation 2 by negative 20. so i'm going to keep equation 1 as it is so 25x plus 40y plus 20z this equals 2200. and again for equation two i'm gonna multiply both sides by negative twenty so negative twenty times three x would be negative sixty x negative twenty times two y would be minus forty y and you see that now i have opposite coefficients on y the negative 20 times z is going to be minus 20z and this equals negative 20 times 180 is going to be negative 3600 okay now again for this process to work you want to have this kind of in the format where all the variable terms are on one side and your constants on the other that way when you add the two left sides and you set it equal to the two right sides you're adding like terms so 25x minus 60x is going to be negative 35x then this is going to cancel and so is this so this would be equal to i've got 2200 minus 3600 which is negative 1400 okay so let me erase this information i don't need it anymore you could really call this equation four if you wanted to so i could say this is equation four negative thirty-five x is equal to negative fourteen hundred so from this i can just solve for x right away we kind of got lucky because two things dropped out so i'm going to divide both sides by negative 35 and i'm going to find that x is equal to 40 okay so at this point i know that x is 40. so let me erase this and just put equation 4 is saying that x is equal to 40. okay so that's one of my equations now i can't really do anything with x equals 40 because i still have two unknowns so i'm still going to eliminate the same variable but it's worth noting that when you have this situation i can eliminate either z or i can eliminate y okay from equation two and three or from equation one and three and what that's going to do is it's going to give me a linear system with kind of two variables i already know that x is 40 so i can immediately just plug in and figure out what y is or what z is again based on what i choose to eliminate what i'm going to do is i'm just going to stay consistent and i'm going to eliminate y again and i'm going to do that from 2 and 3. so what i'm going to do is i'm going to multiply equation 2 by negative 1. so i'm going to have negative 3x minus 2y minus z equals negative 180. again i just multiplied everything by negative 1. i'm going to keep equation 3 as it is so 4x plus 2y plus 3z and this equals 280 and again i have opposite coefficients on my variable y i have negative 2 and positive 2. so when i add the left sides this is going to cancel all right so negative 3x plus 4x is going to be x and then we have negative z plus 3z which is plus 2z this equals negative 180 plus 280 is 100. so let's erase this and again we already know from our equation 4 that x is going to be 40. so i'm going to call this equation 5 now and i'm going to plug a 40 in for x there so i have 40 plus 2 z equals 100 and let me subtract 40 away from each side of the equation this cancels and this will become 60. so let me kind of clean this up a little bit we essentially just have 2z equals 60 divide both sides by 2 and of course we find that z is equal to third okay so let me erase all of this now we don't need any of this information anymore i'm just going to put over here that x equals 40 and z equals 30. and now we just need to find y okay and that's pretty easy to do just plug into any of the original equations a 40 for x and a 30 for z and figure out what y is so it looks like it's pretty easy to work with this second equation or this middle equation so 3 times x 3 times 40 plus 2 y plus 1 times z z is 30 so just plus 30 equals 180 okay well 3 times 40 is 120 120 plus 30 is 150. so let's erase this and put 150. then if i subtract that away from each side this cancels you'll have 2y is equal to 30. divide both sides by 2 you get y equals 15. okay so y equals 15. all right so let's erase everything we can state our answer so let me copy this real quick and let me just paste that here so we remember that x is the amount in ounces of the corn mixture we're going to use and that's going to be 40 okay so we're going to use 40 ounces of the corn mixture and then we know that y was the amount of the egg mixture we're going to use that's going to be 15 ounces and we know that z is the amount in ounces of the vegetable mixture that's going to be 30. okay so let's write this as a nice little sentence so we will say that they should use again the corn mixture was x and x was 40. so 40 ounces of the corn mixture with you're going to have again 15 ounces remember y was the amount of kind of egg mixture and we had y equal to 15. so 15 ounces of the egg mixture and then and z was the amount of the vegetable mixture they used and that was 30. so and 30 ounces of the vegetable mixture okay so they should use 40 ounces of the corn mixture with 15 ounces of the egg mixture and 30 ounces of the vegetable mixture all right let's look at a simple interest problem so james has a total of two hundred thousand dollars to invest and wants to earn a total of forty nine hundred dollars in annual simple interest after meeting with a financial planner he decides to spread the money into three different investments so james will invest in a reit which is a real estate investment trust which pays three percent annual simple interest a bond fund which pays 2.5 annual central interest and a savings account which pays 1.5 percent annual simple interest if he invests 80 000 less in the savings account than the sum of the other two accounts how much was invested at each rate so pretty easy problem overall again when you start these you want to figure out what you're being asked to find so we just need to find how much was invested at each rate remember he had a total if we go back up of 200 000 that's invested and he wants to earn this amount right here so we want to set up kind of three variables to represent the three unknowns we need to know how much he's going to invest in the reit or again the real estate investment trust the bond fund and then also the savings account so let's go ahead and let our variables like x y and z represent those so x can be the reit okay that's the amount he invests in that and then we can say y will be equal to the bond fund okay the bond fund and then z can be equal to the savings the savings account okay so we need to get three equations to solve for this kind of three unknowns that we have here so let me go down and let's think about the first equation okay the first equation is pretty easy you know that the amount that he invests is 200 000 okay and he invests this with three different kind of investment vehicles so it must be true that x plus y plus z equals two hundred thousand so that could be my first equation x plus y plus c equals 200 000 okay that's an easy one because again the amount he invests in the reit which is x plus the amount he invests in the bond fund which is y plus the amount he invests in the savings account which is z has to be equal to 200 000 because that's what he puts in to all three now for the next one we can read the problem and it tells us specifically that he invests 80 000 less in the savings account okay then the sum of the other two accounts okay so the savings account was z and the sum of the other two accounts the other two accounts were x and y so if i sum that that's x plus y so it's 80 000 less than x plus y so for equation two z again the amount he invests in the savings account is equal to the sum of x plus y this is the amount he invests in the other two so the reit and the bond fund minus eighty thousand right because it's eighty thousand dollars less than this amount here now for the third equation we need to think about the interest okay the interest so to do that let's set up a little table so what we're going to have is our reit we're going to have our bond fund we're going to have our savings account and then we're going to have the amount invested or you could say the principal we're going to have our rate and this is as a decimal and then we're going to have our interest okay now we're working with simple interest it's not compounding okay so most of you already know because we talked about this earlier in the course the simple interest formula is i which stands for the amount of simple interest earned is equal to p which is the principal okay the amount invested times r which is the rate as a decimal times the time and generally speaking this is going to be in years okay unless they tell you it's in months or something like that but most of these problems is going to be years okay now in this case we're talking about on an annual basis so it's one year so because t would be a one you can just drop it out and you could say that the interest earned which is this column right here okay is going to be equal to the product of these two the amount invested or the principal multiplied by the rate okay as a decimal okay so let's erase this real quick and let's see if we can get this information so specifically it tells us the reit pays three percent annual simple interest the bond fund pays 2.5 percent annual simple interest and the savings account pays 1.5 annual simple interest so for the rate for the reit it's going to be 3 percent or 0.03 as a decimal and then for the bond fund it's going to be 2.5 or 0.025 as a decimal and then for the savings account is 1.5 percent or 0.015 as a decimal now for the amount invested remember this is unknown x is used to represent the amount invested in the reit y is used to represent the amount invested in the bond fund and z is used to represent the amount invested in the savings account so again the product of these two would give me the simple interest earned over the course of a year so .03 times x .03 times x is the sample interest earned from the reit .025 times y okay is the simple interest earned from the bond fund and .015 times z is the amount of simple interest earned from the savings account now what do we do with this for information how do we make an equation well if you were paying attention to the first part of the problem it tells us specifically he wants to earn 4 900 in annual simple interest now the 4 900 can only come from these kind of three sources so i know that the interest from the reit which is .03 x plus the interest from the bond fund which is .025 y plus the interest from the savings account which is .015 z has to be equal to that amount okay that 4 900 dollars so let's copy this this is going to be our third equation so again if you don't like working with decimals you can clear them for this one you've got three decimal places here and here so you've got to multiply everything by a thousand okay by a thousand so this guy would move three places to the right so it would be 30x so it would be 30x this guy would be 25y so plus 25 y then this guy would be 15 z so plus 15 z and then this guy again if this is right here let me kind of scooch this down so we have room i'm moving everything three places to the right so i put a zero here a zero here and a zero here this would go one two three places to the right so i would end up with four million nine hundred thousand okay so now that we have our three equations we can solve the system and figure out the answer now for equation two you really want to set this up into the format of you know ax plus b y plus c z equals d okay all the variables on one side the number the constant on the other so i can really just drop the parentheses from here and essentially what i have is if i move each of these over here i would just subtract them away i would have negative x minus y plus c okay so negative x negative x minus y plus z is equal to and it's going to be negative 80 000 okay so it's really easy to use elimination to get rid of either x or y from the kind of first two equations so let's just choose to get rid of x you'll notice that both of these are going to drop out so it's going to be a nice easy scenario for us so for equation 4 what i'm going to do is i'm going to add equation 1 and equation 2 together so x plus negative x that's gone y plus negative y that's gone z plus z is two z and this will be equal to two hundred thousand minus eighty thousand is a hundred twenty thousand and of course if we divide both sides by two we'd find that z is equal to 60 000. so we already know what z is right or the amount that he invests in the savings account what about y and x so we're going to pick another two equations so again we can do 1 and 3 or 2 and 3 and we're going to eliminate you can do either x or y it doesn't matter you just eliminate either x or y so that you can plug in for z and figure out what the other one is and then back substitute again so from 2 and 3 let me just go ahead and eliminate x so let me scroll down and get some room going so equation 2 it's already negative but let me multiply it by 30. so i would have negative 30x minus 30y plus 30 z and this equals if i took negative 80 000 and multiplied it by 30 negative 8 times 3 is negative 24 and then i have this many zeros so 1 2 three four and then i'd add one more zero so this would end up being negative two million four hundred thousand okay so then i'm just going to copy equation three as it is so 30x plus 25y plus 15z equals 4900 000. okay so let me scroll down and get some room going and we know that this is going to go away negative 30y plus 25y is going to be negative 5y and then plus 30 z plus 15 z is going to be 45 z and this equals if i take negative two million four hundred thousand and add that to four million nine hundred thousand i'm gonna get two million five hundred thousand okay and we'll call this equation five again i already know what z is so what i'm going to do is i'm going to plug in for z here it's going to plug that in and i'm going to find out what y is so i have negative 5 y plus 45 times 60 000 okay is equal to two million five hundred thousand okay so forty five times sixty thousand is two million seven hundred thousand so i would have negative five y plus two million seven hundred thousand okay this equals two million five hundred thousand let me scroll down a little bit so we already know if i subtract two million seven hundred thousand away from each side this is going to cancel i'll have negative five y is equal to two million five hundred thousand minus two million seven hundred thousand is going to give me negative two hundred thousand so negative two hundred thousand if we then divide both sides by negative five i'm going to find that y is equal to forty thousand okay forty thousand so what's x going to be well if i just plug into this first equation which is pretty easy to work with i can easily find out that x is going to be a hundred thousand and you can eyeball that and see 60 plus 40 is 100 right so this could be 100 grand and then if you had 200 there while missing 100 right that's how i did it in my head but again you can plug it in it doesn't matter so x plus y is 40 000 plus you have your z which is 60 000. this equals 200 thousand and again you know that this sum here forty thousand plus sixty thousand a hundred thousand so this is a hundred thousand here and if i subtract a hundred thousand away from each side this goes away from here and over here on the right it's going to be a hundred thousand okay so we get that x is a hundred thousand y is forty thousand z is sixty thousand so that means that he invests a hundred thousand in the reit he invests forty thousand in the bond fund and then sixty thousand in the savings account all right so let's write that real quick and i'm going to make a nice little concise answer and just say that he invests 100 000 dollars in the reit or again the real estate investment trust and then he put forty thousand dollars in the bond fund and he put sixty thousand dollars in the savings account all right for the last problem we have that on monday jennifer went to the store and spent a total of seven dollars and 45 cents on three one gallon containers of milk and two cartons of eggs later that day her husband jim stopped at the same store and spent 6.45 cents won two one gallon containers of milk and three cartons of eggs when jim got home he realized his error and decided to only return the milk as it would expire too quickly for them to consume how much will jim receive as a refund okay so this is what we're trying to figure out how much will jim receive as a refund now this is given the fact that he's only going to return the milk okay he's not returning the eggs he's just returning the milk and we don't know the individual prices for anything we're only told that you know jennifer goes to the store and she spends 7.45 cents on three one gallon containers of milk and two cartons of eggs and then later on jim goes to the same store okay so we're going to assume it's the same price for the milk and the eggs and he spends 6.45 cents he only buys two one gallon containers of milk and three cartons of eggs okay so we've got to figure this out with a system of equations the unknowns here are going to be the cost for a one gallon container of milk and the cost for one carton of eggs so let's let x be equal to the cost for one gallon okay and this is a one gallon container of milk and then y can be the cost for one carton of eggs okay so what can we do with this information let's go back up so we know that jennifer spent seven dollars and 45 cents on three one gallon containers of milk and two cartons of eggs so the way we can model this since x is the cost of one gallon of milk and she bought three of them then three times x would be the amount she spent on milk then plus you have the fact that she buys two cartons of eggs okay so two cartons of eggs why is the cost of one carton of eggs so two times y would give me the cost for eggs in that kind of transaction and again if i sum these two amounts together i get the amount you spent which is seven dollars and 45 cents okay again you can clear the decimal if you want you can multiply everything by a hundred so we could say this is 745 is equal to 300 x plus 200 y not necessary if you're working with a calculator but i'll just do it because your book would probably do it then for equation two i'm going to do the same thing with jim so if we go back up so for jim he spends 6.45 cents again on two one gallon containers of milk and three cartons of eggs so we would have that 6.45 or 6.45 cents is equal to again he only bought two gallons of milk so 2x plus he bought three cartons of eggs so 3y again you can multiply everything by 100 and say this is 645 is equal to 200x plus y okay so let's erase this and let's drag this up so we only have a linear system with two variables so pretty easy to work with this at this point after we've been dealing with three all right so we can solve this with substitution if we want or we can solve it with elimination i think it's probably better to solve it with elimination although substitution would probably be just as fast but let's just use elimination so what i'm going to do is i'm going to eliminate the variable x and the way i'm going to do that is i'm going to think about the least common multiple between 300 and 200 so that would be 600 right if i think about three and two okay three and two the least common multiple would be six just because it's three hundred and two hundred now it would be six hundred okay so what i'm going to do is i'm gonna multiply equation 1 by let's say negative 2. so 745 times negative 2 would be negative 1 490 and this equals 300 times negative 2 is negative 600 then of course times x and then 200 times negative 2 would be minus 400 and then times y then for this guy right here i'm going to multiply it by positive 3 okay so 3 times 645 is going to be 1935 and this equals 3 times 200 is 600 then times x then plus 3 times 300 is 900 then times y okay so if we sum the left sides together we're going to get 445 right so 1933 plus negative 1490 would be 445. on the right sides this would cancel negative 400 y plus 900 y is 500 y so let me erase this we don't need this anymore and let me just kind of slide this up so let me bring this up let me divide both sides by 500 so divide by 500. so i need to just erase this and say y equals this now this is a fraction if you want to simplify this quickly here's how i would do it i would break this up into 5 times 100 okay because it's easy to divide by 100. 445 divided by 5 would be 89. so if i cancel this with this i would have 89 okay so i have 89 over 100 now so we know if we divide by 100 just moving my decimal point two places to the left so this would be eight nine here this would go one two places to the left so point eight nine okay so that's my y so y is equal to 0.89 so that's the cost of one carton of eggs so apparently where they live it's super cheap for eggs okay so then now we need to find out x and of course i could just plug into either equation one or two so i'm just going to use equation one so 745 is equal to 300 times x plus 200 times you've got your 0.89 where again you can do this in in fractional form if you want remember this was 89 over 100 okay so that allows me to just cancel this with this and get a 2 and then 2 times 89 is 178 okay so 178. let me subtract that away from each side of the equation that will cancel 745 minus 178 is going to be 567. so this equals 300x let me erase this and we can use a similar trick again this will help you if you don't have a calculator right so this guy right here if i divide both sides by 300 i can think about three times a hundred okay so what's 567 divided by 3 well that's going to be 189 so this would be 189 and i'm going to cancel this down here over 100 right so let me erase this so x is 189 over 100 or again this moves two places to the left so x is equal to 189 move this one two places to the left so 1.89 or a dollar and 89 cents so that's the cost per one gallon of milk so apparently wherever they live they have really cheap milk and really cheap eggs okay so this does not actually answer our question okay so this is where a lot of students mess up they say their answer is it's a dollar 89 for one gallon of milk and it's 89 cents for one carton of eggs and the teacher marks it wrong and they wonder why the reason is you have to read the problem okay how much will jim receive as a refund he only returns the milk okay so how much how much milk did jim get jim got two one gallon containers of milk again the milk is a dollar eighty nine a piece so all i need to do is take the 1.89 again that was x that was the cost per one gallon of milk and multiply it by two so multiply it by two again all you have to do is 189 times two and then move the decimal point two places to the left so 189 times 2 is 378. so it's 378. again move the decimal point two places to the left you get 3.78 cents so he will get or he will receive if you want to say that three dollars and 78 cents for his refund again just for the milk make sure you read these problems because they throw kind of curveballs at you to make sure you're paying attention in this lesson we want to talk about gaussian elimination and also gauss jordan elimination all right so over the course of the last few lessons we've been talking about how to solve systems of linear equations in two variables and also three variables we basically covered everything that you would see in a typical algebra one in an algebra two course now what we're going to do is just go a little bit further and we're going to talk about these kind of matrix methods that we can use and we'll see a lot of these as we kind of progress through this chapter today we're just going to start with gaussian elimination and gauss jordan elimination now some of you saw this in algebra 2 others didn't i just want to assure you that this is an easy process most of the difficulty involved with this is just keeping track of what's going on okay so to begin i'm just going to revisit this first system using the elimination method and then we're going to go into the matrix method from there okay so i'm going to start out with 4x plus 16y equals 28 and 5x plus 8y equals negative 1. so once again i'm just going to label this as equation 1 and this is equation 2. now with the elimination method we know that our goal is to eliminate one of the variables okay so we want to take one pair of variable terms okay and we want to make the coefficients opposites so i look at the kind of coefficients for x i have 4 and 5. then the coefficients for y i have 16 and 8. so it's easier to work with y right because i can basically multiply this 8 by negative 2 and i would have negative 16 there so this would give me a negative 16y and a 16y okay so what i want to do is i want to multiply equation 2 by negative 2. so i'm going to multiply 5x by negative 2 and get negative 10x i'm going to multiply 8y by negative 2 and get negative 16y i'm going to multiply negative 1 by negative 2 and get 2. okay so that's my transformed equation again we're always allowed to multiply any equation by the same nonzero number and preserve the solution now what i'm going to do i'm going to take my equation 1 i'm going to leave it as it is so 4x plus 16y is equal to 28. okay so two things here before we kind of do the elimination i want to make sure you understand that when you do the elimination method you want to make sure that your equations are lined up properly so in other words i want to write everything in standard form so ax plus b y is equal to c okay so ax plus b y equals c ax plus b y equals c and the reason for that is is i'm adding the two left sides together and setting this equal to the sum of the two right sides so i want to have like terms on each side right so when i go to add these two equations together on the left sides i have my negative 10x plus my 4x that's going to give me negative 6x then over here i have my negative 16y plus my 16y and i'm going to go ahead and just cancel this right that's 0. and this equals 2 plus 28 which is third okay so let's erase this we already know x is going to be negative five i can complete the process by just dividing both sides by negative six and i'll just write over here that x is negative five right thirty divided by negative six is obviously negative five okay so pretty easy overall we know how to do that and we could just plug in for x in either equation one or two find out what y is right very simple so let's just take equation two because the numbers involved are smaller so five times negative five plus eight y equals negative one we know 5 times negative 5 is negative 25 so plus 8y equals negative 1. let me add 25 to both sides and if i add it over here this is going to cancel so let me just erase it and over here this would be 24. so let's just quickly do this we'll divide both sides by eight and we'll find that y is equal to three okay so very simple very easy something you've been doing since algebra one and also for the last few lessons in this course as well what we wanna do now is think about how we could solve this using a matrix so we're going to start out with the kind of gaussian elimination okay this is usually the first thing that you learn and for gaussian elimination the very first thing you need to do is you need to set up an augmented matrix okay so for an augmented matrix i'm just going to take the numerical information only but again i got to make sure that these equations are written in the correct format so i want to do ax plus b y equals c and i'll explain why this is important in a minute but essentially i'm just going to grab this information and i'm going to write it in the order that it appears so i'm going to take my 4 i'm going to take my 16 and i'm going to take my 28 okay so this row going across represents the numerical information from this equation 1 okay so that's all that is then the second row that i'm going to form which is going to be the 5 the 8 and the negative 1 that's the numerical information from equation 2. so the reason we want to write it in that ax plus b y equals c format is that i now know that this column here this leftmost column that numerical information corresponds to the coefficients of the x variable then i know that this column here is going to be the coefficients for the y variable and i know this column here is going to be the constants so writing it in that manner is what allows you to know that immediately okay so what i'm going to do now is i'm just going to put some brackets around this and i'm going to put a vertical line here that's going to separate the coefficients from the constants okay so this is our first step and i'm just going to copy this now and go to a fresh sheet okay so the goal is to use these what are known as elementary row operations or sometimes you'll hear them called matrix row transformations to put the matrix into what is known as row echelon form okay so row echelon form looks like this so we're going to have this let me put this and i'm just going to use this symbol here to represent a real number so what we're going to have is we're going to have ones starting at the upper left and going down on a diagonal okay and then below it's going to be a 0 here okay so i wanna make this a one and this a one and i wanna make this a zero okay just using those row operations that i'm about to tell you now these row operations are the same things we can do when we're solving a linear system with just equations right if we didn't have a matrix involved so the first thing is we can interchange any two rows so that means i could really write this matrix here by saying i have 5 8 and negative 1 and then i have 4 16 and 28 okay so i took the kind of second row and flipped it to the first one took the first one and flipped it down to be the second one this matrix represents the same linear system okay it's just like if we had these two equations and i chose to write this equation 2 up on top of equation 1. that doesn't make any difference to the answer okay so you can always interchange any two rows the second thing we can do we can multiply or divide the elements of any row by a non-zero real number that should make sense because again if we go back up we remember that we multiplied equation two by negative two when we're solving this right we can always multiply both sides of an equation by the same non-zero number and preserve the solution so since this is just the numerical information from the equation in other words if i took this matrix and i multiplied row 2 by let's say negative 2 this would be negative 10. this would be negative 16 okay and this would be positive 2 okay this would be positive 2. and that's exactly what we saw when we were working with this kind of system right we multiplied equation 2 by negative 2 and again we got negative 10x we got minus 16y this was equal to positive 2. now the third and final elementary row operation tells us that we can replace any row of our matrix by the sum of the elements of that row and a multiple of the elements of another row so this goes back to what we did when we looked at this remember we multiplied equation two by negative two and then we added that to equation one well it's the same thing that we are allowed to do here okay it's just that we're working with the numerical information only okay so let me move this over here so that we can keep in mind what we're trying to do and let's just go ahead and start on this this doesn't take that long when you first start doing this you want to use a lot of scratch paper and you want to really take your time because it's easy to make very simple mistakes on this that give you the wrong answer all right so the way i'm going to do this i'm going to get a 1 for this first position here right because we need a 1 there and then once that's done i'm going to get a 0 in this position here okay so you want to finish a column at a time working from left to right okay so what i'm going to do to get a 1 there you got to think about what's legal again you can multiply a row by a non-zero real number well how can i get something there that's going to be 1. 4 times what would give me 1. remember 4 times its reciprocal which is 1 4 would be 1. okay so what i can do is i can multiply this whole row row 1 okay by 1 4. now the way i'm going to show you this okay is something you might see in your textbook so you'll see this abbreviated for row one you see r with a one down there so r sub one you could just say r1 it doesn't matter and that just means row one so i'm going to say one fourth times row one is going to replace row 1. so i'm going to multiply 4 times 1 4 okay and obviously that's going to give me 1. i'm going to multiply 16 times 1 4 and that's going to give me 4 and i'm going to multiply 28 times 1 4 and that's going to give me 7. okay so i'm going to replace the top row with what i just found so let me write that right here so i'm going to have a 1 i'm going to have a 4 and i'm going to have a 7. and then this bottom row stays the same so 5 8 negative 1. let me close that off and let's erase this and we can even erase the original one we don't need it anymore let's just move this up so that we have room to work okay so for the next step i want to get a zero here right i've already got my one here so this is done right so now i want my zero so how can i get a zero well this is the tricky one it's always harder to get zeros but basically what you want to do is multiply row 1 by something and then add that result to row 2 so that this becomes a 0. okay so in other words i've got to add something to 5 to get that to be 0. so what can i add to 5 to get a result of 0 i want to add negative 5 okay and how can i get negative 5 well i would multiply 1 by negative 5 okay and then add that to 5 to get my 0. that's all i need to do is multiply this top row or row one okay so row one times negative five and then we're going to add that to row two so that's all this is saying so we're going to say that we're replacing row two with negative five times row 1 plus row 2. so again when you first see this it's a bit confusing but after you work enough examples it becomes pretty easy so if i take negative 5 and i multiply it by everything in row 1 negative 5 times 1 is negative 5 then 4 times negative 5 is negative 20 and then we have 7 times negative 5 which is negative 35 okay so i'm going to now add these values and let me just take those results so negative 5 plus 5 is 0 negative 20 plus 8 is negative 12 and negative 35 plus negative 1 is negative 36. so i can just replace these with those numbers so i'm going to have a 0 i'm going to have a negative 12 and i'm going to have a negative 36 okay all right so we just have one more thing to do okay so i'm going to just highlight this to show that it's done so now what i want is i want this to be a one okay so again getting a one is easy because i just multiply by the reciprocal of what i want to be a one okay so i would multiply row two by the reciprocal of negative 12 which is negative 112 okay so 0 0 times negative 1 12 would be y it would be 0. okay so this is unchanged this equals 0. negative 12 times negative 1 12 is obviously 1 and then negative 36 times negative 1 12 we know that this cancels with this and gives me a three negative times angle is positive so this is three okay so this is zero this is one okay and this is three now at this point at this point we have achieved row echelon form okay you have your ones down this diagonal here okay and you have a zero below now this is enough information for us to get a solution for the system remember this column is the coefficients for x this column is the coefficients for y this is the costs okay so here's where this comes into play so i'm going to take this row 2 this 0 that's the coefficient for x plus this one that's the coefficient for y and this equals this three okay so zero x plus one y equals three what does that really tell me zero times x is zero so really i have one y or just y equals three so y equals three is part of my solution right if i go back up we know y equals 3. right so we found that part and now what we can do is we can just back substitute and find x so i can take this information here this is 1x plus 4y i know y is 3 is equal to 7. 4 times 3 is 12. if i subtract 12 away from each side of the equation i'll find that x is equal to negative 5 and again if i go back up i see that x is negative 5. so again it's not really that hard it is a little bit time consuming when you first start because you have to figure out what's going on you have kind of some new notation and some new concepts to wrap your head around but once you kind of get going with this this is a very very quick process now we talked about this gaussian elimination in this row echelon form there's this other related concept which is gauss jordan elimination so what happens with gauss jordan elimination is we take this a step further instead of there just being a zero down here now i would put a zero here okay so let's do that real quick let's transform this kind of entry here this 4 into a 0 and what we'll see is that this gives us all the information from the matrix directly right we don't have to do any substituting and this is known as reduced row echelon form okay so reduced row echelon form so how can i get this to be a 0. again i can use row 2 here the fact that i have a 1 there okay and i can say okay 1 times negative 4 would be negative 4 negative 4 plus 4 is 0. so multiply row 2 by negative 4 and then add the result to row 1. okay so this is going to give me my row 1. that's what i'm going to replace it with so we know 0 times anything is 0 so this isn't going to change right i don't need to do anything for that 1 times negative 4 would be negative 4 then negative 4 plus 4 would be 0 okay so that's done and then 3 times negative 4 would be negative 12 and negative 12 plus 7 is going to give us negative 5. so you see that we have our answer now directly from the matrix okay very easy to read because again this is 1x plus 0y equals negative 5 or i can say that this is x equals negative 5 okay and then this guy right here is 0x plus 1y equals 3 or i can just say y equals 3. okay so directly from this matrix i read the solution and although it took longer for us to do it this way that's just because we're getting start okay so let's go ahead and take a look at another example so the second example here we have negative 10 equals 9y minus 10x and we have negative 9x equals negative 4y plus 32. okay so the first thing we want to do again make sure this is written as ax plus by equals c okay so for this guy over here really all i have to do is just kind of flip it i don't need to change anything i can just say this is negative 10x plus 9y equals negative 10. that's perfectly legal then for this guy right here i have my negative 9x i'm going to add 4y to both sides of the equation and this equals 32 okay so now what i want to do is just take the numerical information only and i want to make sure i write it in the order that it comes in so negative 10 9 and negative 10 then i want negative 9 4 and 32 okay so i put my bar here to separate the coefficients from the constants and we're ready to go so let me copy this we'll go to a fresh sheet and let me paste this in here and let me just rewrite this row echelon form and actually let's just use reduced row echelon form so i want ones down the diagonal and i want to zero above and below and then over here i'll just put this symbol here to represent two real numbers okay and you can use letters there if that makes you more comfortable you could say this is i and this is j it just represents two real numbers so i'll just put these symbols in here to represent that okay so we're going to start out by getting a 1 in this position here that's how i always start and then i'll get a 0 below and then i'll move to the right so again if i want to get a 1 it's really easy just think about what you're trying to make into a 1 and just multiply that row by the reciprocal of that so for negative 10 the reciprocal is negative 1 10 so i'm going to multiply the entire row 1 by negative 1 10 so negative 1 10 times row 1 that's what we're going to replace row 1 with so negative 1 10 times negative 10 would be 1. this would be 1 9 times negative 1 10 would be negative 9 10 okay and then negative 10 times negative 1 10 this would cancel give me positive 1. okay so now that i have that done the next thing i want to do and i'm just going to mark this off and say this is done the next thing i want to do is get a 0 here okay so i want that to be a 0. again not that hard because you have a 1 here ok so i can just think about i need to add a 9 a positive 9 to make negative 9 a 0. okay so the way i do that is i think about okay if i need that 9 i just multiply 9 by this whole top row okay and then i add the result to the second row so i'm going to say that i have 9 times row 1 okay the top row plus my row 2. this is going to give me my row 2. okay that's what i'm going to replace row 2 with so let's do this down here so we can keep track of what's going on so 9 times 1 is 9 and then negative 9 10 times 9 would be negative 81 tenths so negative 81 tenths and then 9 times 1 would be 9. okay so i'm going to add these to these kind of corresponding entries so 9 plus negative 9 we know that's 0. so this is 0. let's erase this if i had negative 81 10 plus 4 that's going to be a bit of work so let's do this one for a minute 9 plus 32 we know that's 41. so let's just knock this one out real fast and then this over here i got to get a common denominator going so i know this is going to be plus 40 over 10 right if i took 4 and multiplied it by 10 over 10 i'd get 40 over 10. and now this is going to simplify to negative 41 over 10. so i'll say this is negative 41 over 10. okay so that part's done pretty easy now i'm just going to check this off and say that's done and i'll move on to this one right here i always want to get my one first okay so i want to make this into a one that's super easy to do again i just multiply that whole row by the reciprocal of that the reciprocal of negative 41 over 10 is negative 10 over 41. okay that has no impact on zero if i multiply this by this i get one okay i get one and if i multiply this by 41 the 41s are going to cancel and i get negative 10. okay i get negative 10. at this point we have our row echelon form right our gaussian elimination is completed but we can take it a step further again and put it in this reduced row echelon form again that's from the gauss jordan elimination so what i'm going to do is i'm going to make this guy right here into a zero and the way i can do that again i'm going to use this one right here to my advantage i'm going to multiply row 2 by what's necessary to kind of get rid of this negative 9 10 or change it into a 0. okay so if i had negative 9 10 i would want to add 9 10 to make that 0. so i'm going to multiply row two by nine tenths so nine tenths times row two add the result to row one again i just use this symbol to say hey i'm going to replace row one with nine tenths times row two plus row one okay so let's go ahead and do that so i know that nine tenths times zero is zero and zero plus one is still one okay so that's no change and then one times nine tenths is nine tenths nine tenths plus negative nine tenths is zero so we've got that and then for this last one here i'm gonna have nine tenths times negative ten which the tens are going to cancel is going to give me a negative one so this would be negative nine okay negative 9 and then i'm going to add the result to this row 1 so negative 9 plus 1 is negative 8 okay so this is negative 8 and we don't need to do any substituting we already can just look at this and know our answer we can say that 1x plus 0y equals negative 8 or we can basically say x equals negative 8. and we could say that 0x plus 1y equals negative 10 or just y is equal to negative 10. okay so again not as quick as we're used to but just because we're getting this kind of process started once you really get going with this you can do this very very quickly all right let's take a look at one more of these and then i'll show you the kind of special case scenarios and what happens with those so we have 12x equals negative 11y plus 20 and then 12y equals 15x plus 78. so again what i want to do is i want to start by writing everything in the format of ax plus b y is equal to c okay so over here i have 12x i'm just going to add 11y to both sides and this will be equal to 20. over here i'm going to subtract 15x away from both sides and then plus 12y and this will be equal to 78. now one thing you want to do if you notice that everything in an equation is divisible by some number you want to take the opportunity to simplify it because it's going to make your calculations easier so if i look at this second equation or this bottom equation everything is divisible by 3. this would be negative 5. this would be 4 and this would be 26. right i just divided every part of that equation by 3. that's something you definitely want to do because it's going to make it easier to achieve your answer okay so let's scroll down just a little bit i'll set up this matrix so again just take the numerical information so 12 and negative 5 11 and then 4 and then you have 20 and then 26. okay so let's go ahead and copy this and again what we're trying to do is get this reduced row echelon form so one's down the diagonal and a zero above and below and then over here we're just gonna have real numbers okay again this is reduced row echelon form and once more if you have a number here then it's row echelon okay so let's get started again by making this into a one very easy to do we're just going to multiply the top row by again the reciprocal of that number the reciprocal of 12 is 1 over 12. so again i'll write this as 1 12 times row 1 that's what we're going to replace row 1 with so this would be one eleven times one over twelve is eleven twelfths so eleven twelfths twenty divided by twelve let's write that over here twenty divided by twelve each is divisible by four so this would be 5 and this would be 3 okay so this would be 5 3. so 5 3. okay so now i want to get a 0 here let me kind of highlight that to show that it's done i want to get a 0 here so again if i have negative 5 i know that if i add 5 to that i get 0. so i'm going to multiply row 1 by 5 add the result to row 2. so 5 times row 1 plus row 2 that's what i'm going to replace row 2 with so 1 times 5 is 5 5 plus negative 5 is 0. okay and then 11 12 times 5 we know this would be what this would be 55 12. so 55 twelfths and then plus four they get a common denominator i'd say this is 48 twelfths so what's 55 plus 48 that's going to be 103 okay so this would be a hundred three twelfths 103 twelfths okay and then for this guy right here again we have five times five thirds five times five thirds so that's 25 thirds and then i'm going to add that to 26. again 26 times 3 is 78 so i could say this is 78 thirds and 25 plus 78 is 103. so you get 103 over 3. okay so let's erase this and i'll mark this one as done and now i want to get a 1 here okay i want to get a 1 here so again that's very easy to do we're going to multiply row 2 by the reciprocal of this number so i'm going to have 12 over 103 okay times row two again i'm going to replace row two with that so i know that zero stays unchanged because zero times anything is just zero and then this guy would be transformed into a one and then this guy i would have what i would have 103 over 3 times 12 over 103 so we know this would cancel with this and this divided by this would be 4 so i can erase this and just put a 4 here so again at this point i have my row echelon form so if i wanted to go back and substitute i could find my answer at this point and i'd be done but i like to take it to this form where this guy right here is a zero as well and i can just read my answer straight from the matrix so how can i get a 0 here again i want to think about what can i add to 11 12 to make it 0 what would be a negative 11 12 and so i would just multiply row 2 by a negative 11 12 and add the result to row 1. so i'm going to multiply negative 11 12 times row 2 add the result to row 1. okay that's how i'm going to get my row 1 okay that's what i'm going to replace it with okay so negative eleven twelfths times zero i know is zero zero plus one is still one no change there then negative eleven twelfths times one is negative eleven twelfths then plus eleven twelfths is zero so that's what we want there and then negative 11 12 times 4 this cancels with this and gives me a 3. so i'd have negative 11 thirds and then i'm going to add this to 5 thirds so negative 11 plus five is going to be negative six so this would be negative six thirds which is negative two so this is negative two and again we've found our answer so we know that at this point what x is equal to negative 2 x is negative 2 and y is 4. now let me show you what happens when you have a special case scenario again you know that you have systems that have no solution and you know you have systems with an infinite number of solutions again if you end up with kind of a false statement then you know you have no solution if you end up with just a true statement you know that you have an infinite number of solutions so let's go ahead and start this in the usual way i'm going to have a negative x i'm going to subtract 3y away from each side and i'm going to subtract 1 away from each side and then over here i'm just going to flip this and say this is 8x plus 24y is equal to 17. again that's perfectly legal so the numerical information from this i'm just going to take a negative 1 a negative 3 and a negative 1 and then an 8 a 24 and a 17. okay so we're good to go with this let me just copy this and i'll paste it in right here for us and again i'm trying to achieve this form one's down the diagonal a zero above and below again this is reduced row echelon form and then these will just be numbers over here and once more just so we don't forget if this is just a number over here and it's not a zero then this is row echelon form okay so that's the difference between the two is something you might get asked about okay on a test they might tell you to put it in row echelon form or they might say put it in reduced row echelon form and you need to know the difference between the two okay so i want to get a one as the first entry right here how can i do that again we've been doing this all day now so i'm just going to multiply row 1 by the reciprocal of that number okay so negative 1 is the reciprocal of negative 1 right because if i wrote this over 1 i would say that the reciprocal of it is one over negative one again it's just negative one so i would say that we have negative one times row one that's what i'm going to replace row one with so negative one times negative one is one negative 1 times negative 3 is 3 negative 1 times negative 1 is 1. okay so now what i want to do is get a 0 here so how can i do that again we've been doing this all day i know i need to add a negative 8 to 8 to get 0. so i'm going to multiply row 1 by negative 8 okay so row 1 times negative 8. i'm going to add the result to row 2. again this is going to give me my row 2. all right so negative eight times one is negative eight negative eight plus eight is zero then negative eight times three is negative twenty-four negative twenty-four plus twenty-four is zero should already see that there's a problem and then let's just do let's just keep going although you know there's no solution as a point just keep going negative eight times one is negative eight negative eight plus 17 is nine okay is not so let's stop for a minute and consider what we have we have 0x plus 0y equals 9. so 0x plus 0y equals 9. what is 0x it's 0. what is 0y it's 0. so basically you have 0 plus 0 which is 0 equals 9 which is always false so this is no solution okay and if we went back and we really observed this we would see that we have two lines that are parallel or two lines that have the same slope if i just look at this as three y equals you have negative x plus one again divide everything by three to put it in slope intercept form and the slope here is going to be negative one third okay for this one let me write this as 24y 24y is equal to negative 8x plus 17. i'm going to divide everything by 24. you can already see that you have the same slope right you can erase this negative 8 over 24 is negative one-third okay negative one-third and then 17 over 24 you can't simplify that but you don't need to because you can see that this slope and this slope those are the same when we have parallel lines again we know that they don't ever intersect so there's no point that lies on both lines so there's no solution for the system okay so this guy right here you put the symbol for the null or empty set or again just write no solution let's look at one more problem just to kind of wrap this up and then in the next lesson we'll look at some systems with three variables so now we have negative 1 equals 2x plus 5y so i just have to flip this so 2x plus 5y equals negative 1. and then for this one i have negative 6x minus 15y equals 3. so we're good to go at this point but before we even start you can immediately notice that 1. if i multiply equation 1 or this top equation by negative 3 i get equation 2. okay 2x times negative 3 is negative 6x 5y times negative 3 is negative 15y negative 1 times negative 3 is positive 3. so you know that this is going to be an infinite number of solutions for the answer okay because these two equations are the same so whatever works in the first equation also works in the second equation now let's show this with a matrix we have 2 5 and negative 1 and then we have negative 6 negative 15 and 3. so let's copy this so again i'm looking to get this kind of reduced row echelon form so one's down the diagonal a zero above and below and then these are just numbers okay so i'm going to get a one here in this first position we already know what we're doing multiply row one by one half which is the reciprocal of two so one half times row one that's what i'm going to replace row one with so this would be one this would be five halves okay this would be five halves and this would be negative one half so this would be negative one half now for this guy right here i want to make this into a zero and to do that i'm just going to multiply row one by six add the result to row two so i'm going to multiply again row one by six add the result to row 2. so that's what i'm going to replace row 2 with so 6 times 1 is 6 6 plus negative 6 is 0. okay then 5 halves times 6 this cancels it gives me 3. three times five is fifteen fifteen plus negative fifteen is zero negative one half times six the six cancels with the two and gives me three negative one times three would be negative three so i would have negative 3 plus 3 which is 0. so again i'm going to end up with 0 equals 0 which is always true so that's how i know i have an infinite number of solutions so let's just write that out we have an infinite number of solutions in this lesson we want to continue to talk about gaussian elimination and also gauss jordan elimination all right so in our last lesson we learned about the gaussian elimination and also the gauss-jordan elimination and we saw that we could take a linear system in two variables we could take the kind of numerical information and set up an augmented matrix and through these elementary row operations we could kind of pound our matrix into a specific form okay now with the gaussian elimination it's this one over here on the left this is referred to as the row echelon form where you have these ones down your diagonal okay so starting at the top left and then going down at a diagonal okay and then a zero below okay and what this tells you is that you can find out in the case of where you have your x variable the coefficients are on the kind of left column and the coefficients for the y variable are kind of next to it you could say that this bottom row here is 0x plus 1y is equal to whatever this is let's just call it d for this video so let's say this is d so really you know that y is d and you can go back up and plug in and find out what x is so that's our gaussian elimination with the gauss-jordan elimination it goes a step further okay so now the matrix is put into something known as reduced row echelon form so that's where we have these ones going down the diagonal a 0 above and below and so now directly from the matrix i can read my solution what we're going to do today is we're just going to kind of expand on this and we're going to look at some kind of linear systems with three variables and then in the next lesson we'll even look at some linear systems with four variables okay so i want you to see the kind of row echelon form for this so you're going to have ones down the diagonal zeros below and of course you have to back substitute to get your answer and then the reduced row echelon form you have ones down the diagonal zeros below and zero above so with the reduced row echelon form you get your answer directly from the matrix this guy on the right on the left the row echelon form you have to do some back substituting all right so let's go ahead and start this with kind of an easier example i am going to warn you that this process does get really tedious i'm going straight off of what your textbook would show you which is where you get a one in your column first and then you find your zeros okay there are faster methods to do this and you can use those if you want but i prefer to go with just what the textbook is going to show you so you don't get confused and then as we kind of progress we're going to find some easier ways to do this all right so let's go ahead and read this off we have x plus y plus 3z equals negative 1. we have 3x plus 3y plus 4z equals negative 3. and we have negative 5x minus 3y minus 6c equals 13. so again i want all my equations in the format of we have the ax plus b y plus c z is equal to d okay the reason i want that is i want all my x terms to line up my y terms my z terms and my constants because when i write this augmented matrix i want all the coefficients for x to be in one column all the coefficients for y to be in one column all the coefficients for z to be in one column and then all the constants to be in one column so let's go ahead and set that up i'm just taking the numerical information only we know that x has an implied coefficient of one y has an applied coefficient of one we have a three and we have a negative one then for this second equation i have a three a three a four and a negative three for my third equation i have a negative 5 a negative 3 a negative 6 and a 13. so let's just copy this and let me paste that in right here we know that we want to put a vertical bar here to separate the coefficients from the constants and then i'm going to wrap this whole thing in some brackets so the very first thing you want to do if you're doing gaussian elimination you want to get ones going down the diagonal and you want zeros below this is faster right because you don't have to go through and keep manipulating the matrix to get it into reduced row echelon form from gauss jordan elimination i'm going to do both with you i know it's going to be a lot of time i know it's going to be tedious but i want you to get a lot of practice with this so for the first one i'm just going to go through and do the gaussian elimination we're going to do row echelon form we're going to back substitute and then i'll go back into it okay and we'll we'll pound the matrix further and get it into reduced row echelon form so what i want is in the first column on the left i want the top entry to be a one it already is so that's going to save some time so then i want this to be a zero and this to be a zero okay so the reason your textbook always wants you to get a one in your column is because it's easy to use it to get a zero it's always easy to get a 1 hard to get a 0. so how do i get a 0 remember i can multiply any row by a non-zero number okay and add the result to another row so what's going to happen is i got to think about what i can add to 3 to get a result of 0 well 3 plus its additive inverse of negative 3 would be 0 and it's convenient to have a 1 there because 1 times anything is itself so if i multiply this by negative 3 i get negative 3 and then if i add negative 3 to 3 this is going to be 0. i don't care about any of this stuff over here i'm only worried about getting a 0 here so the way we're going to do this is we're going to say okay i'm going to multiply the additive inverse of 3 which is negative times this top row or row 1. let me just label these real quick so this is row 1 row 2 and row 3. so negative 3 is going to be multiplied by row 1 and then i'm going to add to row 2 and that's what i'm going to replace row 2 with okay so that's what that notation means and then let me do this one too real quick while we're kind of working on this we can do 2 at once so we have this negative 5 here so what would i need to add to negative 5 to get a 0 i would need to add positive 5. so i would multiply this top row here by positive 5 and it's really the same thing i'm just changing this from row 2 to row 3 and that's what i'm going to replace row 3 with okay so let's go through this it's a good idea to kind of write these things down on some scratch paper because one mistake here and it's going to kill you right you're going to get the wrong answer you've got to start the whole thing over so negative 3 times row 1. that's what we're going to do first so i'm just going to write these answers kind of down here so negative 3 times 1 is negative 3 negative 3 times 1 is negative 3 again negative 3 times 3 is negative 9 and then negative 3 times negative 1 is positive 3. so now what i'm going to do is i'm going to add these kind of results it's just this part right here i'm now adding it to row 2 okay the corresponding entries all right so negative 3 plus 3 is going to give me 0 and then i'll have negative 3 plus 3 again that's going to be 0. i'm going to have negative 9 plus 4 that's going to be negative 5 and then i'm going to have 3 plus negative 3 which is going to be 0. okay so i'm done with this and i'm done with this so i'm just going to erase this and now i'm going to move on to this one so if i do 5 times 1 that's five five times one again that's five five times three is fifteen and five times negative one is negative five so now i'm adding these results okay this part right here to row three okay if you ever get lost in what you're doing just look at your notation it's going to help you keep track of what's going on so 5 plus negative 5 is 0 and then 5 plus negative 3 is going to be 2 and then 15 plus negative 6 is going to be 9 and then negative 5 plus 13 is going to be positive eight okay so this part's done now i've got one up here i've got zeros below so as i move to my next column i want this to be a one again if i'm doing gaussian elimination okay gaussian elimination i want ones down the diagonal and zeros below i don't care about anything else in this particular case i can't just multiply by the reciprocal okay because that's what i normally do right if i had a 2 in that position i could multiply that whole row by half and a half times two would give me one but because i have zero and it's unique because zero times anything is zero what i'm going to do is i'm actually going to switch row two and row three remember we can always interchange two rows so i'm gonna say row 2 is going to switch with row 3. and so this is going to change in b let me kind of write this up here so we keep track so i'm going to have 0 0 negative 5 and 0. so i'm just going to erase this and i'm going to copy this so 0 2 9 and 8 and then i'll write this in here what i just did so 0 0 negative 5 and 0 and it's just a good idea to use a lot of scratch paper because again one silly mistake and the whole thing is blown all right so let's go through now and make this into a one i'm gonna multiply the reciprocal of two okay which is one half times that entire row so times row two that's going to give me my new row 2 and i want you to think about something i'm not affecting this to the left because 0 times anything is always 0. so there's no change on this this becomes what we want which is a 1. so i really only have to worry about this being a 9 halves and then 8 divided by 2 is going to be 4 okay so that takes care of that all right so now to complete this row echelon form i just need a 1 here and i'm done right i can go back and substitute and find my answer so i would want to multiply row 3 by the reciprocal of negative 5 which is negative 1 5 because negative 1 5 times negative 5 would give me 1 so that's what i'm going to replace row 3 with of course 0 times anything is 0 0 times anything is 0 0 times anything is 0. so i only need to change this into a 1 and at this point we've achieved our row echelon form so we can just go back substitute figure out our answer right so we can say that since this column here represents the coefficients for z we would have z or 1z is equal to zero these two guys are zero so those variables are eliminated you can think about this as saying you have zero x plus zero y plus one z equals zero well zero times anything is zero so this is zero plus zero plus one z which is just 1z or z equals 0. so now i can start back substituting if i know z equals 0 i can take these coefficients here i can just say this is 0x plus 1y plus 9 halves times zz is zero and this equals four well i know that this is gone right zero times anything is zero i basically just have this gone and y is just equal to four y is just equal to four so that's pretty easy and then in terms of x i know that 1x plus 1y okay y is 4 so just 4 plus 3z okay z is 0. so i can just say leave that off this would be equal to negative 1. well okay i can just subtract 4 away from each side of the equation and find that x is equal to negative 5. so x is going to be negative 5 y is going to be 4 and z is 0. so at this point if you're just told to find the answer and you want to use a matrix method you can just do it this way because this is a little bit quicker than going all the way through the process with the gauss jordan okay if you want to go through the remainder of this which we're going to do what you need to do is you need to get a 0 here you need to get a 0 here and you need to get a 0 here and what that does is it allows us to kind of read the answer directly from the matrix all right so let's go ahead and crank this out i'm going to start off with this guy right here in my middle column so again i'm going to use this one to my advantage i'm going to multiply row 2 i'm going to multiply row 2 by negative 1 which is the additive inverse of positive 1. i'm going to add the result to row 1. okay that's how i'm going to get my new row 1. so let's just go through and multiply negative one times everything in row two so negative one times zero is zero we have negative one times one which is negative one we have negative one times nine halves which is negative nine halves and then we have negative one times four which is negative four so what i'm gonna do now is i'm going to add these results here this part right here to the corresponding entries in row 1. so we know 0 plus anything is going to leave it unchanged so i need to worry about that negative 1 plus 1 is going to be 0. i know that negative nine halves plus three let's do that off to the side in a minute and then i would have negative four plus negative one which is negative five okay so let's work on this one right here we're trying to figure out what this is going to be so to get a common denominator i'd multiply this by two over two okay and i would have six let's just write this as six over two plus negative nine over two i know that negative nine plus positive six is going to be negative three so this would end up being negative three over two so this is done so let's erase this so now what i need is a zero here and here to finish this up it's actually going to work out pretty conveniently here because we have a 0 here a 0 here and a 0 here so what does that actually mean well remember if i'm trying to work on things in column 3 i'm going to use this 1 in row 3 to kind of work off of and what makes it easy is that when i multiply by zero i get zero if i add zero to something it leaves it unchanged so what happens is if i want to multiply three halves the additive inverse of negative three halves times my row three in every case it's going to be 0 0 and 0 over here so when i add those to the corresponding entries nothing's going to change so the only thing is going to change is this okay so if i add this to row 1 and i replace row 1 with this i only need to change this and i'm just going to change it into a zero it's the same thing here so if i multiply negative okay negative nine halves times row three okay for that one and then i add the result to row two that's what i'm going to replace row 2 with once again these zeros do not affect this okay because negative 9 halves times 0 is going to be 0. adding 0 to something does not change it so i'm only going to change this guy right here and it's going to be a 0. all right so we can see that we got the same answer by putting our kind of augmented matrix in this reduced row echelon form we have our ones going down the diagonal we have 0s above and below so we find that x is negative 5 we find that y is 4 and we find that z is equal to 0. all right so let's go ahead and take a look at another one again as we work more and more of these they get easier and easier so we have negative 6x plus y plus 7z equals negative 32. let me just write the negative 6 the 1 for the coefficient for y that's implied the 7 and then the negative 32 okay just taking the information from the first equation then from the second one you have negative 5x plus 5y plus 5z equals negative 10. again when you see something that can be simplified meaning everything here is divisible by 5 go ahead and take the opportunity to do that because it's going to mean that you're working with smaller numbers so if i divide everything by 5 this would be a negative one this would be a one this would be a one and this would be a negative two so you'd have negative one one one and negative two okay and i can erase that and then for the last one we have a negative three we have a two we have a one again that's implied to be one and then we have a one okay so let me put my vertical bar here to separate the coefficients from the constants and once again i just want to stress this this is already written in the format of ax plus b y plus c z equals d if it's not already written in that format you've got to do that first okay all right so let's set this up and let's copy this all right so let's paste that in so again i want this top left entry to be a one so the easiest way to do that is just to multiply this first row here by the reciprocal of this number so again i can label this as row one row two and row three you don't have to do this but it's just nice to kind of show what's going on so i'm going to multiply negative 1 6 which is the reciprocal of negative 6 by row 1. that's going to give me my new row 1. okay so if i do that negative 1 6 times negative 6 is going to be positive 1. negative 1 6 times 1 is going to be negative 1 6 then negative 1 6 times 7 is negative 7 6 and then negative 1 6 times negative 32 let's do that off to the side so we would have negative 1 6 times negative 32 we know that negative times negative is positive so you get rid of that sign there so we'll just erase these we don't need them and then 32 and 6 are each divisible by 2 right so 32 divided by 2 is 16 6 divided by 2 is 3. so this is going to be 16 thirds all right so the next thing i want to do is i want to get zeros below this one so i want this to be a zero and i want this to be a zero so what can i do again i just think about the fact that i have a one here already okay i have a one there the reason you get the one first is so that you can work with that so all i need to think about is the additive inverse of negative one which is positive one okay so i would just multiply row one by one or you just think of it as i'm just going to add row one to row two okay so what i'm going to say is i'm going to multiply row 1 by 1. i'm going to add the result to row 2 that's going to give me my new row 2. okay so all i'm going to do is just add so 1 plus negative 1 is 0. negative 1 6 plus 1 is what i'm going to write this as 6 over 6 and 6 minus 1 would be 5 so this would be 5 6. then for this one you would have negative 7 6 plus 1 i'm going to write as 6 over 6. negative 7 plus 6 is going to be negative 1 so this would be negative 1 6 and then lastly you would have 16 thirds plus negative 2. so 16 thirds plus you'd have negative i'll go ahead and say this is 6 over 3. so 16 minus 6 is 10 so we can say this is 10 thirds okay so that's taken care of so to get a 0 here again i use the same concept i'm going to use that 1 there and again the opposite of negative 3 is 3. so i'm going to multiply row 1 by 3 add the result to row 3. so i'm going to multiply row 1 by 3 add the result to row 3. that's going to give me my new row 3. so 1 times 3 is 3 negative 1 over 6 negative 1 over 6 times 3 is 1. this will cancel with this and give me a 2. so this is negative one-half this is negative one-half and then negative 7 6 negative 7 6 times 3 is 1. this cancels with this and gives me a 2 so this is negative 7 halves and then you have sixteen thirds times three the threes are going to cancel so this will cancel with this and i have 16. all right so let's add now three plus negative three is zero so let me erase this we'll have negative one half plus two so negative one half plus two so let me do plus four over two so this would be what four minus one is three so this would be three halves so three halves okay so this is gone then negative seven halves plus one let's add two over two negative seven plus two is negative five so this would be negative five halves and then lastly you would have 16 plus 1 which is 17. okay so this is 17. all right so now we have the first column done and we're going to move into the second column here and again i always want to start by getting a 1. i get the 1 and then i can get the 0 above and below pretty easily so to get the 1 i just multiply row 2 by the reciprocal of this 5 6 which is six fifths okay so six fifths times row two that's what i'm going to replace row two with all right so we know that this guy to the left isn't going to be affected because 0 times anything is just 0. okay but this guy is going to be a 1 so i don't really need to worry about that i really just need to do the calculations for these two so i would have 6 fifths times negative 1 6 okay so this would cancel with this and give me a negative one fifth so this would be negative one-fifth let me erase that and then lastly you would have your six-fifths times ten thirds we know that this would cancel with this and give me a 2 and this would cancel with this and give me a 2 2 times 2 is 4 so this would be 4. so now we want to get a 0 above and below this 1. so again i'm thinking about the additive inverse of negative 1 6 so that's going to be 1 6. so let me multiply row 2 let me multiply row 2 by 1 6. so 1 6 times row 2 then plus row 1. this is going to be my new row 1. okay so again to the left of this it's not going to matter because 0 times anything is 0 0 plus anything is just itself so i really need to just think about i know that this would turn into a 0. so i really just need to work on this negative 1 5 times 1 6 okay and then i'll add this result to this negative 7 6. so i know that this would be negative 1 over 30. so negative 1 over 30 and then plus negative 7 over 6. so i'll multiply this by 5 over 5. negative 7 times 5 is negative 35 so you would have negative 35 over 30 and negative 1 plus negative 35 is negative 36 so let's say this is negative 36 over 30 which is what each is divisible by 6 this would be negative 6 over 5. so i'll replace this with negative 6 fifths and one more okay we need to do one more so let me write that a little bit better now i also need to multiply four by one-sixth so four times one-sixth this would cancel with this this would be a three this would be a two so this would be two-thirds okay so two-thirds and then add that result to 16 thirds okay so sixteen plus two is eighteen eighteen divided by three is going to give you six and let's get this guy to be a zero now so i'm going to multiply row two i'm gonna multiply row two by what again i want the additive inverse of this so negative 3 halves then plus my row 3 that's going to give me my new row 3. i know that i don't need to really do anything this way because i'd be multiplying this by a zero and that's going to be zero and then adding it to this would leave it unchanged so really i just need to change this into a zero because i know it would be zero and i need to work on this one and this one so i would multiply negative three halves times negative one-fifth and that would give us three tenths okay and then i would add three-tenths to negative five-halves so three-tenths three-tenths plus negative 5 halves multiply this by 5 over 5. so this would be negative 25 negative 25 over 10 okay over 10. so negative 25 plus 3 is negative 22 so you would have negative 22 over 10. so this is negative 22 over 10. and then lastly i'm going to multiply negative three halves times this four okay that's going to cancel give me 2 negative 3 times 2 is negative 6. so negative 6 and then add the result to 17 that's going to be 11. what we want to do now is get a 1 as this entry here okay so how do we do that we multiply row 3 by the reciprocal of that guy so i'd have negative 10 over 22 times row 3. again that's going to give me my new row 3. so we know that this would be a 1. we know that this would be a one and we can just eyeball this and say that i know that 11 would cancel with 22 right 22 divided by 11 is 2. so let's just erase this this would become a 2 and negative 10 over 2 would be negative 5. okay so at this point you have row echelon form you have ones down the diagonal okay let me highlight that so one's down the diagonal you have zeros below okay so that's row echelon form this is what we get from gaussian elimination we're going to continue with the gauss jordan elimination and put this in reduced row echelon form so i want a 0 here and here again it's not that hard because we're we're almost there okay i know this is very tedious and that's just because we're taking our time so for this guy right here this negative one-fifth i can make it into a zero by multiplying row three okay row three by one-fifth add the result to row two so one-fifth times row three plus row two would give me my new row two okay so i know i don't have to work this way right i don't have to worry about that because if i'm multiplying a zero okay in each case here by this number i get zero and if i add zero to something it stays unchanged so i really don't even need to work on this one because i know it's going to be zero i just have to work on this one okay so that makes it kind of quicker so it's one-fifth one-fifth times this guy right here which is negative five that becomes negative one okay and that gets added to four so that is three okay so now the last one we're going to multiply row three by six fifths okay so by six fifths we're going to add the result to row one and that's going to give us our new row one okay now again when i think about this i don't need to do anything this way okay i know i'm going to replace this with a zero okay and i just need to work on this one so six fifths times negative five six fifths times negative five this would cancel give me negative one so it's negative six so negative six plus six would give me zero so now i have my solution finally okay so i can say that 1z equals negative 5 or z equals negative 5. i can say that 1y equals 3 or y equals 3 and i can say that 1x equals 0 or x equals 0. all right so let's continue with the next example so now we have negative 4x plus 7y minus 3z equals 33. so negative 4. we have 7 we have negative 3 and we have 33. then we have negative 7x plus 6y minus 5z equals 19. so negative 7 6 negative 5 and then 19. and then we have negative 5x plus 2y minus 4z equals 2. so negative 5 2 negative 4 and 2. okay so let me put this bar here and then let me put these brackets and we're ready to go so let me copy this let's go down here to a nice fresh sheet and again we want ones down the diagonal we want zeros above and below okay so i'm going to start by getting a one in this position here and to do that i'm just going to multiply the top row or row one let me label these real quick so row one row two and row 3. i'm just going to multiply row 1 by the reciprocal of negative 4 so i'm going to do negative 1 4 times row 1 okay that's what i'm going to replace row 1 with okay so negative 1 4 times negative 4 is going to be 1. then negative 1 4 times 7 is going to be negative 7 4. so negative 7 4. then negative 1 4 times negative 3 is going to be positive 3 4 and then negative 1 4 times 33 is going to be negative 33 4. so negative 33 4. okay so that part's done so now i want to get a 0 here and here so let me start with row 2 and getting a 0 here so again because i have a 1 here it makes it easy to get a 0 here because i can just say okay i want to add a 7 to this negative 7 to get a 0 so i can multiply this top row or row 1 by 7 so 7 times row 1 and then i can add that result to row 2 that's going to give me my new row 2 and this is going to be a 0. okay so this would be 0 and then for this guy right here i would have 7 times negative 7 4. so 7 times negative 7 4 and then i would add that to 6. so when that's done i'm going to add that to 6. so this would be negative 49 fourths and then if i add that to 6 i would make this 24 4. so this would end up giving me negative 25 fourths so i'm going to put negative 25 fourths okay so now i want to do 3 4 times 7 and add the result to this negative five so this would end up being what it would be twenty one fourths so twenty one fourths plus negative five so i could make this negative twenty over four and twenty one plus negative 20 would be one so this would be one-fourth okay so i'll erase this and put a 1 4 here so let's erase this and then lastly i have this negative 33 4 times 7 and then plus 19 when that's done so this would end up giving me negative 231 fourths so negative 231 fourths and then i'm adding to this 19. now what i can do to get a common denominator 19 times 4 is 76 so i can say this is 76 fourths so negative 231 plus 76 is going to be negative 155 and this is over 4. okay all right so that part's done now we're going to do something similar for this right here so i would multiply 5 by row 1 again think about that 1. 5 times 1 is 5. add that to negative 5 you get 0 and i would add the result to row 3. okay so that's going to give me my new row 3. so if i do this one would be 0 then i would take negative 7 4 and multiply it by 5 and add the result to 2. so this is going to end up giving me negative 35 fourths and then plus instead of 2 i'm going to write eight fourths so negative 35 plus 8 is negative 27. so this is going to be negative 27 4. that's going to go right here so negative 27 4. okay the next thing i want to do is take three-fourths so three-fourths and multiply it by five and then add the result to negative four so this is going to end up being 15 fourths okay and if i add this to negative 4 let's write this as negative 16 over 4. so 15 plus negative 16 is negative 1. so this will end up being negative negative 1 4. okay so negative 1 4 and then lastly what i want to do is multiply 5 by negative 33 4. so negative 33 4 times 5 that's going to give me negative 165 so negative 165 over 4 and then plus 2 again i'm going to write that as 8 over 4. so negative 165 plus 8 is negative 157 so this is negative 157 over four and let's move on now so the next thing i want to do is get a one here okay i always want to get a one first and then get my zero below and above in this case so i can do that by just multiplying row two by the reciprocal of that so i would have negative four twenty-fifths times my row two that's gonna give me my new row two okay so this is not going to be affected this will turn into a one okay this will be a 1. and then this guy right here you'd have 1 4 times negative 4 25. so what's going to happen is this will cancel with this i'll have a negative 1 over 25 okay and then let me erase this the last thing i want to do is i want to take negative 4 over 25 and multiply it by negative 155 over 4. so this is going to cancel and you basically have this negative times negative which would give you a positive between 155 and 25 you have a common factor of 5. so this would be 5 here and this would be 31 here okay so this right here is going to be 31 fifths so let's erase this so now we're done with that and now i want a 0 here and here so to get a 0 here on the bottom i would multiply row 2 by 27 4 okay so 27 4 times row 2 and then plus row 3 that's going to give me my new row 3. so again i don't need to worry about this stuff over here for this one i know this is going to end up being a 0. so i'm really going to only have to work with kind of this part right here in this part right here so i would have 27 4 times negative 1 over 25 i'm going to add that to negative 1 4. so what's going to happen is between 27 and 25 there's no common factors other than one so really i'm just going to say this is negative 27 over 100 so negative 27 over 100 for this i can multiply by 25 over 25 so i'd have negative 25 over 100 and that's going to give me what this would be negative 52 over 100 okay so let me erase this and we can simplify this before we write it 52 divided by 4 is 13 and 100 divided by 4 is 25 okay so i'm going to write this as negative 13 25 and then lastly i want to do 27 fourths times 31 fifths and then add that to negative 157 fourths so 27 times 31 is 837 so this would be 837 over 20. so 837 over 20. if i multiply this by 5 over 5 we get negative 785 over 20. so if i do this addition 837 plus negative 785 i'm gonna get 52 okay so i would get 52 over 20 52 over 20 which is going to simplify each is divisible by 4. 52 divided by 4 again is 13. 20 divided by 4 is 5. so this guy right here is going to be 13 fifths okay all right so let's work on this and getting a 0 here so again i want to multiply row 2 by seven fourths now and add the result to row one okay that's going to be my new row one and so i already know that this is going to be a zero okay then you would have seven fourths times negative one over 25 add that to three fourths let's scroll down a little bit i'll come back up so this right here would be negative seven over 100 plus if i multiply this by 25 over 25 i would have 75 over 100 negative 7 plus 75 is going to be 68 so this would be 68 over 100 which would simplify 68 divided by 4 is 17 so this would be 17 over 25 right because 100 divided by 4 is 25. so this would be 17 over 25. so 17 over 25 and then for this one right here again we're going to multiply 7 4 times 31 fifths and then plus negative 33 fourths okay so 7 times 31 is 217 so you'd have 217 over 20. so 217 over 20 and i'll multiply this by 5 over 5. so this would be negative 165 over 20. so 217 minus 165 is going to give me 52. so i'd end up with 52 over 20. so 52 over and we already know this is going to simplify to 13 over 5 right because 52 divided by 4 is 13 and 20 divided by 4 is 5. okay so i'm going to replace this with 13 fifths so now i want to move on to this column and i want to get a 1 here okay so how do i do that i multiply row 3 by the reciprocal of this so i would take 25 over 13 the negative of that multiply it by row 3 that's my new row 3. again this one and this one are not going to be affected this guy is going to be a 1 and then this guy right here i'd have negative 25 over 13 times 13 over 5. the 13s are going to cancel negative 25 over 5 is negative 5. so this is negative 5 and at this point again you could stop you have row echelon form okay so you could go back and substitute you know z is negative 5 at this point you could find out x and y but we're going to continue we're going to get a 0 here and here okay so what i want to do to make this into a 0 i want to use this one here the additive inverse would be 1 over 25 so times this row 3 and then plus row two okay so i know that i don't need to worry about this or this because multiplying by zero would leave a zero and then adding zero to that would leave it unchanged this guy i know is going to be a zero so i really only need to work with 1 over 25 times negative 5 and then plus 31 fifths okay so i know that this would cancel with this and give me a 5 here so this would be negative one-fifth so this would be negative one-fifth okay so plus thirty-one fifths would be thirty over five which is six so this is six so we know z is negative five and we know y is six let me just write that as we're doing this so we know z is negative 5 we know y is 6. okay now let's find out x so we're going to do is get a 0 here so again i'm going to use this row 3 with this 1 in it and i'm going to multiply row 3 by the additive inverse of this so negative 17 over 25 so then plus row 1 that's going to give me my new row 1. okay so i know that this guy and this guy are not going to be affected okay this guy's going to be a 0. all i need to do is figure out this guy so it is negative 5 times negative 17 over 25 and then plus 13 over 5. okay let's find out so this guy would cancel with this guy and give me a 5 negative negative is positive so let's just get rid of that and say this is 17 over 5 plus 13 over 5. that's 30 over 5 which is 6. okay so this is six so x is six y is six and z is negative five in this lesson we wanna finish talking about gaussian elimination and also the gauss jordan elimination all right so over the course of the last two lessons we learned about the gaussian elimination and the gauss-jordan elimination and we saw that we could use these two methods to kind of solve a linear system we looked at two variables and then also one with three variables and so now what we're going to do is just take the next step and look at a linear system with four variables this is very tedious i'll just tell you that before we start but i'm going to do some things that just kind of speed us up so i'm going to just take this guy real quick and set up our matrix so we can just get going right away so we know we want to take the numerical information from each equation here and you might notice that we've labeled our variables a little bit different in your book if you're taking precalculus or college algebra you might see x y z and then w okay so w would be first because it's in alphabetical order but i didn't do that because a lot of times you'll also see this with the notation of x sub 1 through x sub 4. so whatever you're comfortable with is fine if you don't like this you can replace x sub 1 with w x sub 2 with x x sub 3 with y and x sub 4 with z okay so i'm just going to follow the format of x of 1 coming first then x sub 2 then x sub 3 then x sub 4. okay that's what i'm going to do all right so we're going to start with this first equation we'll have a 3 a negative 2 a 5 and then a negative 1 and a negative 8. then in the second equation i'm going to have a negative 1 a 3 a negative 1 a 4 and then a 9. then in the third equation i have a negative 2 a negative 1 a 4 a 9 and a negative 9. let me make that a little better and let me scroll down a little bit then in this last equation i want you to notice that you don't have an x sub 2. whenever you're missing a variable you want to write a 0 as the coefficient for that variable to act as a placeholder so i'm going to put a 1 and then again for my x sub 2 i'm putting a 0 as the coefficient then for x sub 3 i put a 3 then my 2 then my negative 2. okay so this guy is now set up for us and we are ready to go so let me copy this real quick now the quickest way to solve this if you only know kind of gaussian elimination or gauss-jordan elimination if you just get this and this is where you are in your chapter i would suggest just doing the gaussian elimination putting this matrix in row echelon form meaning i'm just going to get ones down the diagonal and zeros below it's going to be really quick to do that and then kind of back substitute to get your answer if you go through to reduce row echelon form again that's from the gauss-jordan elimination it does take a little bit longer that is what we're going to do today just to get a lot of practice on these elementary row operations because i feel like you do need it it's just something that comes up it goes away and then later on if you have to go back to it it's something if you didn't practice enough you kind of forget it and you might struggle with it again all right so the very first thing i want to do is i want to get a 1 as this top position here and then i want to work below and get zeros okay every time i go to the right to a new column i want to get my one first and then my zeros so what i'm going to do i could multiply this first row which i'd label as row one okay let me label all these real quick i could multiply row one by one third but there's an easier solution remember for the elementary row operations i can swap two rows so i can say row one is going to swap with row four because i already have a one there and if i swap those rows this guy would come up here okay so let me just write this real quick i have 1 0 3 2 and negative 2. i'll just erase this from here i'll just copy this here 3 negative 2 5 negative 1 and negative 8. and then i'll just copy this right here so one zero three two and negative two okay so now that i have my one in this position here these ones going down are pretty easy to get and i'm gonna do multiple ones at once okay we're ready to kind of speed up this process so we know that if we want this to be a zero i've got to add the opposite of that number to it okay so in this case i've got to add a one in this case i've got to add a two and in this case i've got to add a negative three well this one is really convenient because whatever i multiply by 1 it's just itself so what i'm going to do in each case is i'm going to multiply row 1 by whatever i need to add to this to make it a 0 and then i'm going to add it to that row and then that's what i'm going to replace the row with okay so for the first one i'm going to say that i'm going to have 1 times row 1 plus row 2 this is what i'm going to replace row 2 with okay and when you multiply something by one it's just itself so really you could just say row 1 plus row 2 if you want then for the next one i'm going to end up saying that i'm going to have 2 times row 1 okay so the opposite of negative 2 is 2 so 2 times row 1 plus row 3 that's what i'm going to replace row 3 with and then for this last one here i'm going to have negative 3 okay because negative 3 plus 3 is 0. so negative 3 times row 1 plus my row 4 is what i'm going to replace row 4 with okay so let's go through and crank this out real quick so i know for this one i'm just adding row 1 and row 2 that's what i'm replacing row 2 with so 1 plus negative 1 is zero zero plus three is three three plus negative one is two two plus four is six and negative two plus nine is seven so this one is done then for this one 2 times row 1 plus row 3 that's what i'm going to replace row 3 with so 2 times 1 is 2 2 plus negative 2 is 0. 2 times 0 is 0 0 plus negative 1 is negative 1. then 2 times 3 is 6 six plus four is ten two times two is four four plus nine is thirteen two times negative two is negative four negative four plus negative nine is negative thirteen all right so this one is done so let's do this one so we have negative three times row one plus row four negative three times one is negative three plus three is zero we know that this would be still negative two negative three times three is negative nine negative nine plus five is negative 4 and then negative 3 times 2 is negative 6 negative 6 plus negative 1 is negative 7 then lastly we have negative 3 times negative 2 which is going to be positive 6 and then positive 6 plus negative 8 is negative 2. all right so that part's done i have a 1 here and 0's below now i want to move to my next column and i want to get a 1 here okay i already have a 0 above so i'm going to get that one first and then i'm going to get the zeros below okay so the easiest way to get a 1 in this position here i can multiply any row by a non-zero real number so i'm just going to multiply this whole row row 2 by the reciprocal of 3 because 3 times a third would give me 1. okay so i'm going to multiply one third times row two that's what i'm going to replace row two with so this would be one this would be two thirds this would be six thirds which is two and this would be seven thirds okay so that's pretty quick and now i want to get a 0 here and i want to get a 0 here so i know the additive inverse of negative 1 is 1. so again i could multiply 1 times row 2 and add that to row 3. that's what i could replace row 3 with if i want this to be a 0. and again if i'm multiplying by 1 it's just itself so you can just erase that then for this one the additive inverse of negative 2 is 2. so i want to multiply 2 times row 2 and then add that to row four that's going to give me my new row four okay so for this guy right here i'm just gonna add row two and row three and that's gonna replace row three so one plus negative one is zero then you would have two thirds plus ten well i know i could write this as over i could multiply this by three over three so this would be thirty-thirds and thirty plus two is thirty-two so this would be thirty-two thirds so this would be 32 thirds then i'd have 2 plus 13 which is 15 and then lastly i would have 7 3 plus negative 13. multiply this by 3 over 3 we get 39 there over 3 so negative 39 thirds so negative 39 plus 7 would be negative 32 so this would be negative 32 thirds okay so let's erase this and this this one's done so now let me work on this row here again i want that to be a zero so i'm going to multiply 2 times row 2 add the result to row 4 that's what i'm going to replace row 4 with so we know this is going to be a 0. 2 3 times 2 is 4 thirds then plus negative 4 multiply this by 3 over 3 you get negative 12 thirds 4 plus negative 12 is negative 8 so this is going to be negative 8 thirds okay then the next one is going to be two times two which is four four plus negative seven is negative three then the last one is two times seven thirds which is fourteen thirds then plus negative two which i'll write as negative six over three so fourteen minus six is going to be eight so this would be eight thirds so i'll put this as eight thirds okay so let me erase this so far so good so now i want a one here to get started okay these two columns to the left are done i want this to be a one so you already know the deal i'm going to multiply row 3 by 3 over 32 and this gets easier as we move on because i don't have to worry about this stuff over here this is zero and this is zero so multiplying that by the reciprocal doesn't do anything right it's still zero this is going to be one we already know that so really the only calculation i have to do is 15 times three over 32 which is going to be 45 over 32 and then you would have your negative 32 over 3 times 3 over 32 this cancels and this cancels you get negative 1. so this is negative 1 here okay so let's erase this okay so let's think about getting a zero here here and here okay so three different things we want to do again in each case because i've got a 1 there think about what the additive inverse is of what you're trying to make into a 0 multiply that by that row with the 1 in it and then add that to the current row that you're trying to make that into a 0. i know i said a lot there but basically if you think about this for row 1 i'm trying to make this 3 into a 0. so i'm going to multiply row 3 by negative 3 the additive inverse of that add the result to row 1. okay so that's going to take care of that that's what i'm going to replace row 1 with same concept as we move on so for row two i'm going to multiply negative two-thirds which is the additive inverse of two-thirds times row three okay that one and row three plus row two that's what i'm going to replace row two with and then lastly for row 4 i'm going to take positive 8 thirds multiply it by row 3 add the result to row 4 that's what i'm going to replace row 4 with all right so let's do these one at a time so let's start with this kind of row 1 here so i want negative 3 times this row 3 okay i'm going to add the result to row 1. now because these two guys here are zeros i don't need to worry about it zero times anything is zero adding zero to something doesn't change it i know this would be zero i don't even need to worry about that i just have to do the calculation for these two so i would start here with this kind of 45 over 32 and i would multiply by this negative three okay by this negative 3 and i would add to this this value of 2 okay so that's one of them that i have to do then the other one would be negative 1 so negative 1 times negative 3 and then i'm going to add to that negative 2. so let's kind of slide down here and do this and we'll come back up so 45 times negative 3 is negative 135 and this will be over 32 then plus for 2 to get a common denominator i would write it as 64 over 32. so if i sum these negative 135 plus 64 is going to be negative 71. so this would be negative 71 over 32. okay so for this one negative 1 times negative 3 is 3 right positive 3 and positive 3 plus negative 2 is positive 1. so i'm going to have negative 71 over 32 and i'm going to have positive 1. so let's erase this so we'll put this as negative 71 over 32. let me make that better so negative 71 over 32 and then again this was positive 1. okay so now i can erase this this part's done let me kind of slide this up so now let's work on this one so i know that this right here again i'm multiplying negative two-thirds times row three so this and this are not affected this is going to be a zero this right here i would have negative two-thirds times so negative two-thirds times 45 over 32 and then the result of that would be added to 2. so let's figure out what this is so this cancels with this and gives me a 16 down here this cancels with this and gives me a 15 here so this is negative 15 over 16. so negative 15 over 16. to get a common denominator going i'm going to say this is 32 over 16. and then 32 minus 15 is 17 so this would be 17 over 16. so that's what i'm going to put right here it's going to be again 17 over 16. all right and then one more to do so i'm going to multiply negative two-thirds times negative one so negative two-thirds times negative one which would just be two-thirds and then i'm gonna add the results to this seven-thirds here okay so two plus seven is nine so this would be nine over three which is three so i'm gonna replace this with a 3. okay so let's erase this one and now we're just going to work on this one so i've got 8 thirds times row 3 plus row 4. so if i go through again this one and this one it's not going to change these this guy is going to turn into a 0. so what i want to do is eight thirds times 45 over 32 so this cancels with this and gives me a four this cancels with this and gives me a 15. so you have 15 4 plus negative three so fifteen fourths you can go ahead and just say minus three three is twelve over four fifteen minus twelve is three so this is just three fourths three fourths all right then we want to do eight thirds times negative one which is just negative eight thirds and then we wanna go plus eight thirds which is obviously going to give us zero so this will be zero here all right so not much left so we want to get a one here now and to do that i can just multiply row four by four thirds obviously you can see that x sub four is going to be zero right because when i multiply this by four thirds this by four thirds this by four thirds and this by four thirds it's all going to stay zero this is going to change into a one right so four thirds times row four is what i'm going to replace row four with so this will just be one okay all right now i want a 0 here here and here so to get the 0 here i'm going to multiply row 4 by negative 45 over 32 add the result to row 3. again that's what i'm going to replace row 3 with and then i'm going to multiply negative 17 over 16 times row 4 add the result to row 2 that's what i'm going to replace row 2 with and then lastly i'm going to multiply 71 over 32 times row 4 add the result to row 1 that's what i'm going to replace row 1 with now before i do anything you might want to notice that this is a zero this is a zero this is a zero and this is a zero so in every case whatever this number is it doesn't matter when it multiplies these guys by zero and i add it to whatever row it is it doesn't change so none of these entries are going to change anywhere except for here here and here and in every case i'm adding exactly what i need to make them into a 0. so i can just go ahead and put 0 0 0 and say we have our solution so remember it's the coefficients for x sub 1 then x sub 2 then x sub 3 then x sub 4. so for x sub 1 it's going to be 1. for x sub 2 it's going to be 3. for x sub 3 it's going to be negative 1 and for x sub 4 it's going to be 0. so again in this particular case we just so happen to get x sub 1 through x sub 4 but if you had w x y and z remember the leftmost column would be w then x then y then z it goes in alphabetical order with subscripts you go 1 2 3 and 4 you go in numerical order in this lesson we want to start talking about the properties of matrices and specifically here we want to do an introduction to matrices all right so over the course of the last few lessons we learned how to solve linear systems with matrix methods specifically we looked at the gaussian elimination and the gauss-jordan elimination methods so at this point you had some exposure to matrices but over the course of the next few lessons what we want to do is just expand our limited knowledge and build a foundation for some other methods that can be used to solve linear systems so when we talk about a matrix in math normally we're speaking about an ordered array of numbers or you might say a rectangular array of numbers now when we talk about what's inside of a matrix these numbers here so for example in this guy right here you see if i read across i have 1 negative 7 5 and 13 and then here i have 3 negative 9 2 and negative 4 and then here i have 6 negative 2 1 and 5. so all of these numbers inside of these brackets represent the elements or the entries of the matrix okay so you might hear those two words interchangeably now when we talk about a matrix it's important to know when we're referring to a row or referring to a column so you see you have an image of a guy rolling a boat and that's meant to represent a row okay so a row goes across so you see you have a row here a row here and a row here so this matrix has three rows and you'll remember this from again our lesson on gaussian elimination and gauss-jordan elimination where we kind of labeled these rows we said this was row one this was row two and this was row 3 and i made that a b i don't know why so this was row 3. okay so three rows and then you see the image or the picture for a column okay columns are vertical so this is going up and down so this is vertical this is vertical this is vertical this is vertical so this is a column one this is a column two this is a column three and this is my column four okay so this guy has three rows and it has four columns so when we look at that information we can say it's a three by four matrix okay this is known as the order or the dimensions of the matrix so i'm just going to write here that this is a three by four matrix again that's called the order or the dimensions of the matrix now in this example what would the order be well again if i go through i can say i have a row here i can say i have a row here a row here a row here and finally let me make that better a row here so you have five rows right one two three four five again the rows are going across just picture the guy rowing the boat okay that's how you remember the columns are going up and down okay so we're going to say that we have one column here we have a column here a column here and a column here so you have five rows and four columns the rows always come first okay the rows always come first so it's a 5 by 4 okay when we're talking about the order or again the dimensions of the matrix what about this guy so here we have what we have a row here so this is row one we have another row here again the rows go across then the columns are up and down so this is a column here this is a column here this is a column here so this guy if i talk about the order it's a two by three right two rows three columns so it's a two by three all right so in a lot of cases we're going to use capital letters to name matrices and this has a lot of different uses mostly if you're going to kind of reuse a matrix over and over again you might be performing several operations with it so let's say you write capital letter a is equal to and again you have these elements you have 4 and 2 in the first row and negative 3 and negative 1 in the second row well i might have to do several things with a so it's just useful to name it that way i can keep talking about it without relisting the elements each time okay now another thing you might want to know is that if a matrix has the same number of rows as it has columns like we have here it's referred to as a square matrix right so this guy has two rows and it has two columns okay so this is a two by two otherwise known as a square matrix okay so if it's a three by three a four by four a 117 by 117 again same number of rows as columns it's a square matrix so again we have another example of a square matrix so this is matrix b and again we have three rows so one two and three again they're going across and you have three columns okay again the columns are going up and down this is a three by three matrix or again it's a square matrix all right so you also have a row matrix and a column matrix so a row matrix is a matrix with only one row and then a column matrix is a matrix with only one column so this d here is an example of a row matrix you only have one row here okay you've got several columns so column one column two column three so the order here is a one by three and again this is a row matrix because it only has a single row now e is going to be an example of a column matrix because it only has one column okay but it also has four rows so it has a row one let me make that a capital letter so a row one a row two a row three and a row four so this guy is a four by one if we're thinking about the order and again it's a column matrix because there's just one single column all right let's talk a little bit more about notation this is something you definitely need to know as we go deeper into this topic so generically speaking you'll probably see this in your book you have this capital letter a and it's equal to inside of the brackets you have all these lowercase a's and you have this subscript associated with each kind of entry or lowercase a so each guy here let's say i start with this one right here so this is a sub 1 1 okay so the lower case a is just a matter of having a capital a here okay if i had a capital letter b this would be a lowercase b if i had a capital letter c lowercase c so on and so forth the one one there is meant to say where i am in the matrix okay it's like a location so the first number okay the one here the first one tells me what row i'm in and the second one tells me which column i'm in well in this one at the top i'm in the first row in the first column if i move one to the right now i have a sub 1 2. so this guy right here is in the first row second column move to the right i have a sub 1 3. again i'm staying in the same row so i'm still having a row position of one but my column is just increasing as i move to the right so here it was one here it was two now it's three okay but the row always stayed the same and i'm going to continue out till i get to a sub 1 n so i'm still in the first row but i'm in the nth column okay so then this notation as we go down you see that in this case all the way to the left the column stays the same it's always a 1 right because i'm in the first column but the row is now increasing as i move down so we end up down here with the a sub m1 and if we go all the way here to the bottom right i have a sub m n so this tells me that what for this matrix it's an m by n right it's got m rows and it's got n columns so generally speaking that's what you're going to see but you need to know how to find specific entries they might ask you hey what's the value of something like a sub 3 3 okay so what that means is to go and find this entry in the third row in the third column so i would say okay this is the third row third column so that's this guy right here okay whatever that happens to be in this case it's generic but in a normal matrix you'd have some kind of number or symbol or you'd have something there that you could say this is equal to this so generically in your textbook what they're going to write is they're going to say a sub i j and again all this means is that it's the element that's going to be in the i throw or you could say row i and the column j right or the jth column if you want to say it that way it's always row first and column second now one thing i want to call your attention to because this does cause some confusion if you see it with commas it's really the same thing so i could say that a sub 32 this is really the same as if i said a sub 3 and then put a comma there for the two so some people get confused by that it's just a difference in notation if you have numbers involved that have two digits you need to use a comma so that you understand what it is okay let's say i had a sub 3 and then 12. well you don't know what this is is this a sub 31 2 is this a sub 312 what is it so that's why you put a comma in between them to say hey this location is on the third row twelfth column okay so that's when you definitely want to use a comma all right so to give a little example suppose we have uppercase b and it's equal to in our first row we have one negative five nine in our second row we have negative 7 4 and 12. so let's say i asked you to find lowercase b sub 2 3 okay sub 2 3. so again it's row first so this is the row and then this is the column okay so what is in the second row so second row is down here and what's in the third column that's here so it's going to be this guy right here so i could say this is equal to 12. okay and that's all you want to do if you get this as a question let's say i wanted to find b sub 1 2. again the first number which is this one is the row this guy is the column the second one so where's row 1 that's here it's on the top column 2 is here okay so that's going to be negative 5. very very easy suppose i gave you something like let me just kind of erase this one suppose i gave you b sub 3 2 b sub 3 2. what's the answer there well i know that i only have one two rows so i can't find a third row so they might give you this as a trick question right there is a second column but there's not a third row so this element doesn't exist okay so you could just write does not exist all right so another thing that you might see in this section it's pretty short and there's not a lot of questions but they're going to talk to you about how to determine if two matrices are equal so the rule is that two matrices are equal if and only if they have the same size okay so the same order in other words one is a three by three the other one's a three by three one is a three 3x4 the other one's a 3x4 they've got to have the exact same order or you could say have the same size and then each corresponding element has to be the same okay has to be the same so let's say for example we have a and it's equal to we have 2 and x in the first row z and 5 in the second row and we have b and it's equal to we have y and 3 in the first row and 4 and w in the second row so we might ask what are the values for the variables that make the matrix equation true if we said that a is equal to b okay and all you'd have to do is kind of say okay well i've got a 2 here and a y here so y has to be equal to 2 okay and then you'd say okay i have a z here and a 4 here so z has to be equal to 4. okay you get the ideas it's very simple then i have an x here and a 3 here so x has to be equal to 3 and then lastly i have a w here and a five here so w has to be equal to five because these guys are only going to be equal if we kind of set this up and say okay we have two x again was going to be three and then z was four and we had five as w so we have this and we'll say it's equal to this so two three four five okay kind of trivial and it seems like you know why would you go through this but it is something that is important to understand so they will give you kind of questions like this so every corresponding element has to be the same and the order has to be the same so you have a 2 and row 1 column 1 you have a 2 in row 1 column 1. you have a 3 and row 1 column 2 a 3 in row 1 column 2. right so on and so forth you have a 4 here and a 4 here in row 2 column 1 and then you have a 5 here 5 here in row 2 column 2. so every entry has to be exactly the same and the order has to be the same all right one more of these and it's a little bit more complex so again find the values for the variables that would make the equation we'd say a is equal to b make that true so here all we're going to do is end up setting up an equation in each case so we would say 3x is equal to 9 and so in this case we know that x is 3. let me kind of do this off to the side so we know x would be 3 because 3x equals nine we divide both sides by three x is equal to three that's very easy let's kind of grab a few more and let me kind of just erase that so we know we got this one done so let me just highlight the ones that we know and this one right here we'll have w minus one and that's going to be equal to six so let's solve that real quick so we have w minus one equals six add one to each side of the equation we get w is equal to seven so w equals 7 and again very easy just have to go through these it's more time consuming than anything so you have z plus 2 and that's going to be equal to 12. so z plus 2 equals 12. subtract 2 away from each side of the equation you're going to find that z is equal to 10. so we've got those done so now we just need to do these so i've got 19 here and i've got a 5y minus 1 there so 5y minus 1 equals 19. and again just add 1 to both sides you get 5y is equal to 20. divide both sides by 5 you get y equals 4. so we have y equals 4. and let's go back up all right so two more so i've got this 15 here and we're going to set that equal to 4k plus 3. so 4k plus 3 equals 15. 4k plus 3 equals 15. let's subtract 3 away from each side of the equation 4k is equal to 12. divide both sides by 4. we get that k is equal to 3. okay so just one more now again i know this is tedious but it's just something that you might get asked so it is important to cover it so then the last one here we have that q minus 9 is going to be equal to 11. so q minus 9 q minus 9 equals 11. again add 9 to both sides we get that q is equal to 20. okay so these are the values for the variables that are going to make that equation a equals b true you're just looking at the corresponding entries we know that in each case i have three rows and two columns okay so that's the first thing the matrices can only be equal if and only if they're the same order okay in this case they're each of three by two and the corresponding elements are the same okay so that means that nine and three x have to be the same or equal so we say three x equals nine we find that x equals three right you do that for each one and then you find out the values for the variables again we found that x is three z is ten w is seven y is four k is three and q is twenty so that's what you do if you get this as a question on your homework or on a test in this lesson we want to continue to talk about the properties of matrices and specifically here we're going to talk about adding and subtracting matrices all right so we're going to begin our lesson with the addition of matrices first and foremost we can only add two matrices together if they're the same size or again as we talked about in the last lesson the same order so these are the kind of dimensions of the matrix it's how many rows you have by how many columns you have so for example if we had a three by four matrix that matrix has three rows and four columns and if you had two three by four matrices you could add those two together okay that would be perfectly fine but if you had a three by four matrix and you had a two by three matrix you couldn't combine those two with addition that operation would be said to be undefined so the first thing you would do is verify that again they're the same size or order once that's done then you can simply add the two matrices together by adding the corresponding elements it's actually a very easy process overall so suppose i have matrix a and matrix a is equal to in the first row we have the elements 3 and negative one in the second row we have the elements five and two then suppose i also have this matrix b and in matrix b we say this is equal to in the first row negative four and three in the second row one and seven so in each case i have two rows and two columns so they're each going to be a two by two matrix so i can find the sum of a and b so to do that if i want to find the sum of a plus b it's just equal to what well i'm just going to sum the corresponding elements so all that means is i just take the element in the first row in the first column okay and the element in the first row in the first column in each case and i find the sum so what is 3 plus negative four okay we'll do that in a minute and then we just go through and do the same thing for kind of each case so here i would have negative one plus three so negative one plus three then down here i would have five plus over here i have a one then i would have a two plus a seven okay so that's all you're doing you're just adding the corresponding elements from the two matrices all right so let's crank this out real quick so we know that three plus negative four would give us negative one let me kind of slide down a little bit so i'm just going to say this is going to be negative 1 and then down here 5 plus 1 is going to give me 6 and then negative 1 plus 3 is going to give me 2 and then 2 plus seven is going to be nine okay so this is the matrix that is the result of adding a plus b in your top row you'd have negative one and two in the bottom row you'd have six and not all right let's take a look at another example so suppose now you have this as your matrix a and this as your matrix b so if i want to find the sum of a plus b okay first and foremost i've got to verify that the order is going to be the same in each case so we see that for matrix a i have three rows and three columns so it's a three by three and for matrix b it's three rows and three columns so again it's a three by three so because they're the same size or the same order or they have the same dimensions however you want to say that we can add them together right we can find the sum so again to do this it's very simple i just add the corresponding elements so if i start up here in row 1 column 1 i add those two guys together and i'm just going to do this right here i don't need to write it out so negative 2 plus negative 1 is going to be what that's going to be negative 3 okay now let me erase this and i'll highlight each one that i'm doing so let's do this one now so what is 1 plus 6 that's going to be 7. and then what is seven plus two that's going to be nine okay so that's my first row that's done so now let's move on here and i'll just do these a row at a time now to make it a little bit faster so what is five plus eight that's going to give me 13. what is 6 plus 3 that's going to give me 9 what is 9 plus 4 that's going to give me 13 and then let me erase this we'll go to this bottom row so this and this we'll say what is 1 plus 11 that's 12 what is 4 plus negative 6 that's going to be negative 2 and then lastly what is 8 plus 10 that's 18. okay so this matrix here is the sum of the matrix a plus the matrix b all right now you can also add more than two matrices together again they've all got to be the same order or the same size and in this case we have a matrix a a matrix b and a matrix c we're going to find the sum so in each case we have one two three four rows and one column so they're going to be a four by one matrix in each case and remember in the last lesson we said that these were a column matrix right because there's only one single column so to add these together we do a plus b plus c i want to just note here that the commutative property that we learned with addition of real numbers would hold here i can add these in any order also the associative property right i can group the addition however i want it won't change the sum either way because all i'm really doing is i'm adding things together and with addition we know that it's commutative we know that it's associative so nothing's really going to change the sum so let's go ahead and set this up so i'll just do this like this and again i'm just going to work here i'll do 3 plus negative 2 plus negative 8. so 3 plus negative 2 is 1 1 plus negative 8 is negative 7. let me move down to kind of the next row i'll do 5 plus 9 plus 14. 5 plus 9 is 14. 14 plus 14 is 28 and then let me move down to this row here and i'll erase this one up here so now i'm going to do negative 7 plus 3 which is going to give me negative 4 and then negative 4 plus 6 is going to give me 2. and then lastly i'm going to have this bottom row so i'm going to have 10 plus 1 plus 9. we know that 9 plus 1 is 10 and 10 plus 10 is 20. okay so pretty easy overall so a plus b plus c is going to give us this matrix that's still a column matrix right it's got the same dimensions as what we were adding and we've got negative 7 28 2 and 20 as that column there okay so four rows one column it's still a four by one all right so let's move on and talk about subtracting matrices now so again when you subtract matrices it's the same thing you have to have two matrices with the same order or again same size or dimensions whatever you want to say in this case we're going to start out with a which has the elements 5 1 in the first row and 9 negative 5 in the second row and then b which has the elements 7 2 in the first row and negative 9 12 in the second row or the bottom row so each of these is a 2 by 2 matrix right two rows and two columns now if we subtract it's not commutative right we have to know that a minus b is not going to be equal to b minus a you can kind of eyeball this and see that if i did let's say this guy right here so this 5 that's in row 1 column 1 and this 7 right here that's in row 1 column 1. if i did matrix a minus matrix b well 5 minus 7 would give me negative 2. if i did matrix b minus matrix a well 7 minus 5 would give me positive 2. those two elements wouldn't end up being the same so those two matrices would not be equal right so a minus b would not be equal to b minus a so for the purposes of what we're going to do here i'm just going to do a minus b again it's a very simple process i'm just going to match up corresponding elements and i'm going to do my subtraction in the correct order so if i have a minus b i'm going to take the element in matrix a and subtract away the corresponding element in matrix b so in this case i'm going to do 5 minus 7. again this is an a this is in b i'm going to do 1 minus 2 okay i'm going to do a 9 minus a negative 9. be careful there because you have a negative there okay so 9 minus negative 9 and then i'm going to do negative 5 minus twelve okay so five minus seven we know that's negative two okay so this is going to be negative two and then one minus two is negative one nine minus a negative nine is nine plus nine that's eighteen and then if we had negative 5 minus 12 that's negative 17. okay so this would be the result of a minus b now there's another way to do this and this is probably what you're going to see in your textbook i want you to remember that when we subtract integers right we learned that we could do a minus b and i'm just using a and b as any real numbers this might get confusing so let me use p and q so p minus q p is a real number q is a real number we can say that p plus negative q is the same thing right so if i had 5 minus 3 i could write this as 5 plus negative 3. okay so we're going to use that same concept here and we can say that if i have matrix a minus matrix b i can say this is a plus the negative of b okay and we haven't talked about multiplication yet with matrices we'll get to that in the next lesson but essentially if i have the negative of b i can just go into my matrix b and i can change the sign of every element okay it's just like multiplying every element by negative 1. so let's go ahead and set this up real quick let's say that we have the negative of b and i'll say this is equal to 1. instead of 7 i'll have negative 7. instead of 2 i'll have negative 2. instead of negative 9 i'll have positive 9 and instead of 12 i'll have negative 12. now what i can do is i can say let me make a little border here a minus b okay so a minus b is equal to a plus the negative of b so this will be equal to let me set up this matrix i'm going to say 5 plus this guy right here because i'm working with negative b so 5 plus negative 7 okay and then i'd have 9 plus 9 so 9 plus 9 i would have 1 plus negative 2 so 1 plus negative 2 and then i would have negative 5 negative 5 plus a negative 12. now because i switch this operation from subtraction to addition once i'm adding the opposite once i have this set up i can add in any order right if i do negative 7 plus 5 or if i do 5 plus negative 7 i get the same answer but you can only do that once you've changed it into its opposite and you've set up the addition okay so don't make that mistake you can't do that with subtraction all right so let's go ahead and crank this out so 5 plus negative 7 again we know this is going to be negative 2. 9 plus 9 we know that's 18 and then 1 plus negative 2 we know that's going to be negative 1 and then negative 5 plus negative 12 again we know that's negative 17. so it's the same answer just a different way to kind of look at it this might be easier for you it might be harder for you it doesn't matter you get the same answer either way so whatever you want to use all right let's take a look at one more example with subtraction so suppose we have a and we have b and we want to subtract let's do b minus a this time so b minus a okay so i'm not going to go through and say b plus negative a you can do that if you want again it's the same answer i'm just going to save time and do b minus a okay so what i'm going to do because these guys again are the same size this is a one two by one two three so it's a two row and three column matrix in each case so it's a two by three so we can subtract them i'm going to do each element of b minus each corresponding element from a so five minus a negative one five minus a negative one and let me change colors we would have one minus eight one minus eight we would have negative two minus two okay and then let me change colors again we'd have negative 4 minus 11 and then we'd have 9 minus 14 okay 9 minus 14 and then we would have 15 minus 6. all right so 5 minus a negative 1 is like 5 plus 1 so that's going to be 6. then 1 minus 8 is going to be negative 7 then negative 2 minus 2 is negative 4 and then negative 4 minus 11 is negative 15. so negative 15 and then 9 minus 14 is negative 5 okay so that's negative 5 then lastly 15 minus 6 is going to be 9. so this is the result of b minus a again if you wanted to you could do b plus the negative of a so to do that you would change each element inside of this matrix a into its opposite and then you could add and you would get the same result all right so now let's talk about the kind of zero matrix this is a matrix that is also referred to as the null matrix it's a matrix that contains only zeros as its elements okay so the zero matrix can be written as any size so whatever size you kind of need to work with you can make a zero matrix to kind of work with it so if you need a 2x3 a 30 by 30 715 by 220 you can do whatever you want is all going to be zeros now generally speaking we're going to notate this with a zero at least that's how i'm gonna notate it a lot of books do different things here you might see a zero with kind of the size so in this case i have one two three rows and i have two columns so you might see them write a zero and then say it's a three by two okay if it's known what the size is you typically just put a zero instead of writing out all these zeros but there's other notations that you might come across they might use a letter to represent this so just stick to what your book gives you now i want you to recall that when we worked with kind of real numbers we said that 0 was the additive identity and what that means is that if i add 0 to something it's unchanged so if i use p or q again let's just use q and i add 0 to that i'm going to get q back okay q is just some real number so for example if i had 111 and i added 0 i get 111 right if i had a trillion and i add 0 i get a trillion so 0 is the additive identity well when we do matrix addition because a zero matrix only has zeros as its kind of entries what happens is every time you add the zero matrix to another matrix you just get the original matrix back so the zero matrix is the additive identity in matrix algebra so for example here if i add if i add a plus this zero matrix i'm just going to put a zero there okay i'm just going to get a back because all i would be doing is saying that i would set this up and say let me kind of do this i would do 5 plus 0 which is 5. i would do 2 plus 0 which is 2. i would do 1 plus 0 which is 1. i would do 7 plus 0 which is 7. let me make this a little bit longer i would do 8 plus 0 which is 8 and negative 4 plus 0 which is negative 4. so i get a back right this is equal to a i haven't changed anything because in each case i just added a real number to 0 and so i got that real number back all right now additionally we saw that if you added a real number and its additive inverse together you got zero okay so it's going to be the same thing or the same concept when we work in matrix algebra if we have some matrix a okay and we add to it the opposite or the negative of a you're going to end up with a 0 matrix so in other words if i have this matrix a and i have these entries here 3 negative 5 8 11 and then 6 and 4 i could say the negative of a is what we talked about this earlier i'm just going to multiply each kind of entry here by negative 1 so this would be negative 3 this would be negative 8 this would be negative 6. this would be positive 5 this would be negative 11 this would be negative 4 okay so what happens is if i add a plus negative a i'm going to end up with a 0 matrix and in this particular case i'm going to have a zero matrix that's what it's one two three rows by two columns right so it's a three by two you could write that if you want or again it's going to be obvious that it's a three by two so you can just leave that zero as the notation again you can go through and kind of crank this out to prove it 3 plus negative 3 is 0. negative 5 plus 5 is 0. 8 plus negative 8 is 0 11 plus negative 11 is 0. 6 plus negative 6 is 0 and 4 plus negative 4 is 0. so again you end up with this 0 matrix instead of writing all this out you just put a zero okay to notate that you have a zero matrix all right so let's wrap things up with an introduction to matrix equations this is something we're going to be dealing a lot with kind of in this chapter and also as we progress in math in general so with a matrix equation we have an equation where variables are used to represent matrices so at this level it's very very simple let's say i have a matrix a and a matrix b they're both a two by two matrix so we know we can add them together and we have matrix a plus this matrix x which we don't know what that matrix is and we say it's equal to this matrix b so if i want to solve for x i can use the same strategies that we talked about when we solve linear equations here okay so i can subtract matrix a away from each side of this equation and i can say that the matrix x is going to be equal to the matrix b minus the matrix a okay so you see where we're going with this we can find out what x is by just doing b minus a okay so what is let me kind of make a little border here what is b minus a that would be equal to what well i would just take each element from b and subtract away the corresponding element in a so four minus two would be two negative one minus eight would be negative nine okay five minus a negative one is like 5 plus 1 so that's 6 and then 9 minus 3 would be 6. so this is b minus a and it's also equal to x because we said x was equal to b minus a so let me just write here that this is equal to 2 and negative 9 for the top row and then 6 and then 6 in the bottom row and you can check this that's the good thing about kind of working with equations you can always check well if it's true that the matrix x which is this guy right here is equal to the matrix b minus the matrix a well then it's also true that a this matrix here plus x this matrix here would give me b you can check that really quickly you can say okay if i had 2 plus 2 would i get 4 yeah that's a check if i had 8 plus negative 9 would i get negative 1 that's a check if i had negative 1 plus 6 would i get 5 that's a check if i had 3 plus 6 when i get 9 that's a check so our matrix x is this guy right here right with 2 and then negative 9 in the top row and that kind of looks like 2 minus 9. so let me make that a little bit cleaner and then 6 and 6 in the bottom row all right let's try one more of these and we'll just kind of call it a day on this and the next lesson we'll look at some matrix multiplication all right so we have our matrix a our matrix b again they're the same size in each case we have a three by three three rows three columns so we have that a this matrix here plus x which is our unknown matrix is equal to b which is this matrix here so again i'm just going to solve for x i'm going to subtract this matrix a away from each side of the equation and again we find that x is equal to b minus a and i don't need to check this this time we'll just crank this out real quick and call it a day let me kind of slide this over a little bit and so what is b minus a so again we have 12 minus the negative 1 which is going to be 12 plus 1 which is 13 okay and then we have 11 minus 5 which is 6 we have 8 minus 9 which is negative 1. we have 10 minus 2 which is 8. we have negative 6 minus 7 which is negative 13. we have 5 minus a negative 1 which is 5 plus 1 that's 6. we have 13 minus 15 which is negative 2 we have negative 9 minus 13 which is negative 22 and lastly we have 4 minus a negative 3 which is 4 plus 3 which is 7. so x our matrix that was unknown again is equal to the matrix b minus the matrix a which gives us this matrix here again in the top row you have 13 6 and negative 1. in the middle row you have 8 negative 13 and 6 and in the bottom row you have negative 2 negative 22 and 7. in this lesson we want to talk about multiplying a matrix by a scalar alright so when we talk about multiplying matrices there's going to be two scenarios that you're going to come across the first scenario which is much simpler involves multiplying some real number called a scalar by a given matrix the second scenario is where we're going to multiply two matrices together now the second case is much more involved and we're going to look at that process in the next lesson all right so how do we multiply a matrix by a scalar well again just to be clear here when we say a scalar this just means some real number or to be more specific a real number that is not inside of a matrix okay so when we see this in matrix algebra we call it a scalar so if we take that scalar and we multiply it by a matrix this is where we get the term scalar multiplication now what you're going to see in your textbook is something like this if matrix a is equal to again this just identifies the matrix by the individual elements that it's kind of made up of so you say this lowercase a sub ij again the i here is the row the j here is the column and again this is just generic notation so if this is an m by n matrix and k is a scalar so some real number that's not in a matrix the scalar multiple of a by k is the m by n matrix given by and again this notation is pretty simple overall you have k times the matrix a so the scalar k times the matrix a gives us a matrix that's made up of these elements here which is k multiplied by each individual element of a right this lowercase a sub i j and again when you see this kind of notation it may be a bit confusing but let's just jump into an example and you'll see that it's very very easy to do this process so suppose i have matrix a and it's made up of these elements here again in the first row we're going to have 3 7 and eight the second row will have negative two eleven and one in the third row we'll have five two and twenty two so this is a three by three matrix three rows three columns it's a square matrix so if i asked you to find negative five times matrix a what would you do again all you want to do is multiply negative 5 this scalar here by each and every element of matrix a so a very simple process so let me just kind of write this over here negative 5 a is going to be equal to i'm just going to multiply negative 5 by every element over here so 3 times negative 5 is negative 15 and then if we do 7 times negative 5 that's negative 35 if we do negative 5 times 8 that's negative 40. and then if i do negative 5 times negative 2 that's positive 10. if i do negative 5 times 11 that's negative 55 and then negative 5 times 1 is negative 5. for the last row we have negative 5 times 5 which is negative 25 then we have negative 5 times 2 which is negative 10 and then negative 5 times 22 which is negative 110. so this matrix right here negative 5 a is a scalar multiple of the original matrix a okay so that's all we're really saying and let's just look at another example it's pretty easy overall so for the second example we have matrix b and we have again in the first row 4 6 and 16 in the second row 2 10 12 and in the third and final row we have 10 4 and 14. so another three row and three column matrix so again it's a three by three a square matrix so if we want to find one half times b again all i would do is multiply every element in this matrix b by a half okay that's all we're doing very very simple process so one half times four is two one-half times six is three one-half times 16 is eight one half times two is one one half times ten is five one half times twelve is six one half times ten is five one half times four is two and one half times fourteen is seven okay so this would be one half times b so now let's kind of look at a combination of some things that we've learned already so in this section when you talk about scalar multiplication you get a few problems on scalar multiplication and then they kind of combine things together so what we'll see is that we're going to do some problems with addition and subtraction with scalar multiplication involved and then we're also going to look at some equations so suppose we have matrix a which is made up of with the first row 3 1 and five in the second row negative two zero and six and then for matrix b we have the first row as four seven and two and the second row is zero one and eight and each of these have two rows and three columns okay so it's a two by three in each case so we know that since they're the same size they can be added together now our problem is going to be to do 3 times a and then add the result of 2 times b okay so 3a plus 2b and so what i'm going to do first is find 3a okay i want to find 3a so let me start by just rewriting a here real quick for reference and then we'll delete it so we have 3 1 and 5 and we have negative 2 we have 0 and we have 6 okay so if i want 3 times a again a very very simple process i would multiply every element in matrix a by 3 okay that's all i'm doing so 3 times 3 is 9 3 times 1 is three three times five is fifteen and then three times negative two is negative six three times zero is zero and three times six is eighteen okay so let me erase this and let me just drag this out of the way and let me write down b here for reference so for b again we have 4 7 and 2 and then we have 0 1 and 8. and if i wanted 2 times b let me just write that here so 2 times b would be what i would multiply every element by 2. so this would be 8 this would be 14 this would be 4 0 would stay the same right because 0 times anything is still 0. this would be 2 and this would be 16. okay so now let me kind of arrange these in a way that we can see what's going on so i know 3a equals this and 2b equals this so i want the sum of these guys again you can only add two matrices if they're the same size or order and in each case we have a two row by three column matrix so we're good to go on that so if i want three a plus two b i would say it's what well i'm just going to take every element from 3a and add the corresponding element from 2b so 9 plus 8 would give me 17. 3 plus 14 would give me 17 again 15 plus 4 would give me 19 and then negative 6 plus 0 is negative 6 0 plus 2 is going to give me 2 and then 18 plus 16 is going to give me 34. okay all right so this is our result 3a plus 2b so it's a very easy process to not only do scalar multiplication but to combine it with some addition or subtraction operations all right so let's look at another problem that involves kind of combining the scalar multiplication with addition or subtraction so we have 4a minus 5b we're using the same two matrices so here's our a and here's our b if you want to copy those down real quick you can i'm just going to write them out real fast so i'm going to say that a is equal to again we have 3 1 and 5 in the first row and we have negative 2 0 and six in the second row then for matrix b we're going to have four seven and two in the first row and we're going to have zero one and eight in the second row all right so if we wanna do this four a minus 5b my suggestion to you when you have subtraction is to add the opposite so let's just do plus negative there so when i get to this scalar multiplication with matrix b i'm going to do negative 5 times b and then i can just add the matrices okay it makes it a little bit easier to not make a silly sign mistake because the worst thing with working with matrices is you go through all this work and then you get the wrong answer and you basically have to rip the page up and start over okay so if i do 4 times a 4 times a this is equal to 1. it's just 4 times every entry in a so 4 times 3 is 12 4 times 1 is 4 4 times 5 is 20 okay and then down here 4 times negative 2 is negative 8. 4 times 0 is 0 and then 4 times 6 is 24. so i can get rid of this i don't need this anymore and let me just kind of slide this down out of the way and then now i want negative 5 times b okay so negative 5 times b is going to give me 1. so it's negative 5 times every element so negative 5 times 4 is going to be negative 20. negative 5 times 7 is going to be negative 35 negative 5 times 2 is going to be negative 10 and then we have negative 5 times 0 which is 0 negative 5 times 1 which is negative 5 and lastly negative 5 times 8 which is negative 40. okay so let me erase this and i'll just drag this down here so it's out of our way and now what we want to do is this operation here so what is 4a plus negative 5b or 4a minus 5b as we originally saw it well again to add two matrices together or to subtract matrices you have to have two matrices that are the same size or the same order again in each case we have a two by three matrix so we're good to go right two rows three columns so we're just going to add corresponding entries so 12 plus negative 20 is going to be negative 8. then 4 plus negative 35 is going to be negative 31 then 20 plus negative 10 is going to be positive 10. then down here in my second row i'd have negative 8 plus 0 which is negative 8. i'd have 0 plus negative 5 which is negative 5 and i'd have 24 plus negative 40 which is going to give me negative 16. so this matrix here would be the result of doing 4a plus negative 5b or again 4a minus 5b all right so before we kind of move any further i want to go over some of the properties of scalar multiplication that you're going to see in your textbook just so that these things are clear and you don't get confused by them so let's define a and b to be these matrices of the order m by n and then k and h are going to be these scalars or again these real numbers that are not inside of a matrix so the first one is kind of the associative property we know about the associative property of multiplication with real numbers right that tells us we're multiplying three or more numbers together we can group that multiplication any way we'd like we always get the same result well the same thing is true here right so if i had k times h done first this is inside of parentheses these two scalars then the result of that is multiplied by the matrix a it's the same thing as if you did one of the scalars times the matrix a first and then took that result and multiplied it by the remaining scalar now the next two are extremely obvious if you have one as a scalar meaning you have one times your matrix a you get matrix a back and the reason for this is because one is the identity element in multiplication so if i'm going through and multiplying everything in matrix a by one i just get that element back so it's an exact copy of the matrix a the next one is if we have as a scalar so 0 times matrix a will give us a 0 matrix and again the notation for this varies but generally speaking you'll see a 0 with the kind of order we said these were m by n matrices so we'll say this is a zero matrix that is of the order m by n and let me make that m a little bit better okay and again because i'm multiplying zero by every element of a i'm getting a matrix with all zeros as entries and it's of the order m by n so that's why we define this as a zero matrix of the order m by n all right lastly we have a distributive property with scalar multiplication this one's pretty obvious as well so if something like k a scalar multiplied by these two matrices added together so a plus b we could say it's the same as doing k times a plus k times b all right then the second part of this would be if we had k plus h these two scalars inside of parentheses those guys are being added together first and then the result is multiplied by this matrix a this is the same as doing k the scalar times the matrix a plus h the scalar times the matrix a all right so let's look at a little matrix equation we're going to see more and more of these as we progress through the chapter and we're going to start getting into some scenarios where we'll actually be able to kind of solve a linear system with a method other than the gaussian elimination and the gauss-jordan elimination we'll see that we can kind of use matrices in a different way with the cramer's rule and then also with kind of solving by using the inverse of a matrix but we'll get to that later on so for right now we have this matrix a which has negative one and three in the first row and four and seven in the second row and matrix b which has six and five in the first row and zero and nine in the second row so they're each two by two square matrices so what we're saying for our equation is that 2 some scalar times some unknown matrix x plus our matrix a is equal to our matrix b so we would solve this the same way as if we had a linear equation kind of in one variable right so i want to isolate x my unknown matrix and to do that the first thing is to subtract matrix a away from each side of the equation so this would cancel and over here i'm just going to write this in line and say this is b minus a so now what i have is 2 times matrix x is equal to b minus a so how do i get rid of this 2 from over here well we can divide both sides by 2 or what i'm just going to say since we're working in terms of multiplication i'm going to multiply both sides of the equation by one half okay so i'm going to multiply this side by a half let me kind of move this down a little bit so it fits so i'm going to multiply this side by half and i'm going to multiply this side by half again i'm going to wrap this in some parentheses to make sure that multiplies each kind of matrix there so i know that this is going to cancel and i'm left with just my matrix x which is what i want so what we have here is one half on the right times the quantity matrix b minus matrix a and remember because of our distributive property we know that we could do one half times b minus one half times a or we could do it this way it doesn't really matter so let's go ahead kind of erase all this we know what we want to find let me kind of scooch this down a little bit it's gonna be hard to kind of do this on one sheet so let me just kind of copy this let's go down to a fresh sheet where we have lots of room to think and work so let's start out by just doing b minus a first okay and then we'll multiply the result by the scalar one-half i think that's a bit easier so if i do again matrix b this was we had six and five in our top row in the bottom row we had zero and nine and then for matrix a we had what we had negative one and 3 in the top row in the bottom row we had 4 and 7. so again if you want you could do plus negative or you could do minus so if i'm doing minus i would do 6 minus a negative 1. if i was doing plus negative i'd end up doing 6 plus the opposite of this which is positive so whatever you want to do it's fine i usually do plus a negative okay it makes it a little bit easier so i would change the sign of everything here so this would be plus this would be negative this would be negative this would be negative okay it just makes it easier to keep track of the signs so b plus negative a we're going to have what b plus negative a is going to be equal to 6 plus 1 is 7. 5 plus negative 3 is going to be 2 0 plus negative 4 is negative 4 9 plus negative 7 is going to give me positive 2 okay so this is b plus negative a and then to solve for x i just want one half of this matrix okay so let's just do this real quick we'll erase this kind of scooch this up a little bit and so what i'll say is this is equal to one half times seven is seven halves we'll have one half times two which is one we'll have a half times negative four which is negative two and a half times two which is one okay so our matrix x what we're trying to find is going to have a top row of seven halves and one and a bottom row of negative two and 1. now let me copy this real quick and let me just kind of erase this we don't need this anymore i'm just going to say that x is equal to again this matrix is going to be 7 halves and 1 and it's going to be negative 2 and 1. now if you want to check this and i advise you to check these when you first start think about making sure that this equation makes sense okay that it's true so 2 times x is what it's 2 times every element in x so 2x is equal to 2 times 7 halves is going to be 7 2 times 1 is 2 2 times negative 2 is negative 4 and 2 times 1 is 2. okay so this is 2 times x so if i then add this to a what i get b so again you'd add corresponding entries so 7 plus negative 1 gives me 6. 2 plus 3 gives me 5. negative 4 plus 4 gives me 0 and 2 plus 7 gives me 9 so we know here that our solution for x is correct in this lesson we want to learn about multiplying matrices all right so in our last lesson we learned about multiplying a matrix by a scalar again a scalar is just a real number that's not inside of a matrix and we found that that process was extremely straightforward right you just took the scalar and multiplied it by every entry of your matrix to get your result when you multiply two matrices together however the process is not so straightforward it's not what you would expect okay and so what we want to do is just kind of start out with a very basic very easy example and then just build up to the kind of tougher examples that you're going to get for your test or your homework or whatever you are encountering so let's say we have these two matrices here we have r which is a row matrix it's got one single row with three negative two and five as its elements again you could call this a row vector if you wanted to then c is going to be one single column with negative 1 0 and negative 3. so you could say this is a column matrix or a column vector so what we want to start out with is finding the product of r times c so what is r times c so i can write these two next to each other like this or i can put a dot in between them it doesn't really matter so what i want to do first is just write out the size of each okay this is going to be very important so i'm going to say that r is a one by three it's got one row and it's got three columns and i'm gonna say that c is what it's a three by one it's a three by one now this is very important this is something you want to write down you can only perform the multiplication if the number of columns in the first matrix meaning the leftmost matrix the r here matches the number of rows in the second matrix or the rightmost matrix the c here so you can just circle these two inside numbers if you write it in this manner and you can say are they the same and if they're the same you can put a check mark there and say okay i can proceed with my multiplication if those two guys are not the same you have to stop the multiplication is not going to be possible okay so you can say the product doesn't exist now the second thing you want to do here is figure out the size of the product and the way you're going to get that is you're going to look at these two outside numbers so you have this the number of rows from r and this the number of columns okay from c so this matrix will be a one by one okay so this will be a one by one so once i determine if i can perform the action and then i determine what the size of the product is i go ahead and write out my matrix i know it's a one by one so it's just a matrix with one single entry at this point i can erase all this i don't need it anymore and then i'm going to go through a process that might seem a little bit weird at first but it gets easier as you kind of do it so what i'm going to do is i'm going to think about the first element in this r this first one right here the first element in r or the leftmost one is a three okay what i'm going to do is i'm going to multiply by the top element or i could say the first element in c okay so this guy right here gets multiplied by this guy right here so 3 times negative 1. all right then you're going to add to this you're going to move down by 1 and you're going to move down by 1. so now what i'm going to do is i'm going to do negative 2 times 0. okay so negative 2 times 0. then i'm going to add to this i'm going to do this trick one more time i'm going to move down and i'm going to move down so 5 times negative 3 so 5 times negative 3 i'm going to find this kind of result and that's going to be what i'm going to put in there okay and this might not make any sense at this point that's okay as we practice we'll start to see what's going on here so 3 times negative 1 is negative 3 plus negative 2 times 0 is obviously 0 then plus 5 times negative 3 is negative 15. so negative 3 plus negative 15 would be negative 18. so that is going to be the result of r times c okay we're going to find that we're going to need to do a row vector times a column vector pretty often so we need to make sure we understand what we did there again we would start with this guy right here from this one times this guy right here from this one okay then we would add to that this guy right here times this guy right here then we add to that this guy right here times this guy right here okay so now that we have that down what we want to do is move on to kind of something else we know that when we talk about multiplication with real numbers it's commutative if i do 3 times 2 it's 6 if i do 2 times 3 it's also 6. 3 times 2 equals 6 and then 2 times 3 also equals 6. this is the commutative property of multiplication with kind of multiplying matrices it's not commutative okay so when you flip the order generally speaking you're not going to get the same answer and sometimes you're going to find that you can't even do the multiplication okay so what happens if we change the order here what if we did c times r well again let's kind of write this out and compare our size c is what it's a one two three row by one column matrix and r is a one row by three column matrix okay so now again we're comparing the inside numbers the number of columns from c the first one has to be equal to the number of rows from r the second one okay so those match these are the same again if they're the same you can put a check mark and say okay i can proceed then the size of this guy is going to be given by the outer numbers so it's a three by three three rows by 3 columns so let's erase this and let me kind of slide this down a little bit so we have c times r here and this is going to be a 3 by 3. so i always write this out first now here's where it gets a little bit tricky okay i want to write this out so you understand what's going on this is row one this is row two this is row three this is column one column two and column three okay so what i want to do here if i want to find this entry here i want to think about okay it's in row one column one all right row one column one so what i'm gonna do is i'm gonna multiply row one from the left one or c times column one from the right one or r or you could say row one from the first one or c times column one from the second one or r however you want to think about that so row one in c is this guy right here and column 1 in r is this guy right here again because we have such a simple example it's just going to be you know 3 times negative 1 in this case as we get to the tougher ones then you're going to have to go through what we did before where you're finding a couple of different products and summing them together but here it's very simple so it's just 3 times negative 1 which is negative 3. okay so that's this guy right here then we move down to this one okay so now i have my row one and column two okay so i'm basically going to stay with this one because again when i think about the row it comes from the leftmost one it comes from c it comes from the first one you've got to remember that so i stay here i'm still in my row one here but i'm now moving to column two here so i'm moving down so negative one times negative two is two okay and then again if i'm staying in row one here okay i'm staying in row one here i'm just moving down to column three here so i would have negative one times five which is negative five okay and if this is not kind of gelling right now it's okay as we work through more examples you'll start to pick this up now i'm moving into row 2 here in my answer so what i want to do is move to row 2 again in c because i'm always getting my row from the leftmost one so i'm in row 2 now here and i'm going to multiply to get this whole row by column one to get this one by column two to get this one by column three to get this one again because i have column one column two column three that's all i'm doing now because zero multiplied by anything is zero this whole row is going to be zero zero times three is zero zero times negative two is zero zero times five is zero then when i get to row three down here again i just change the row i'm in for this first one so now i'm here okay so row three column one negative three times this guy in column 1 so times 3 is going to be negative 9. the negative 3 times this guy in column 2 is going to be negative 2 so negative 3 times negative 2 is 6. the negative 3 times this guy in column 3 which is 5 is going to be negative 15 okay so that's how we get the entries as we get into a tougher example in a minute you'll see that you have more work involved but the process is basically the same okay so you see that r times c is negative 18 that one entry for that matrix and c times r is not the same right they're not equal to each other completely different size this guy has negative 3 2 negative 5 0 0 0 and negative 9 6 and negative 15 as its entries all right so let's go ahead and look at a tougher example now so suppose we have matrix a and matrix b so for matrix a we have two rows and one two three columns so a i'm gonna write my size down it's two by three and then for b it's what it's one two three rows okay by one two three four columns so can we find the product of a times b again you're looking at the inside numbers the columns from a the first one and the rows from b the second one so do these numbers match here yeah they're the same okay so we know we're good to go and let me make that a little bit better we know we're good to go on the multiplication here now if i flip the order of the multiplication if i said can i do b times a you're not going to be able to because if i kind of change this up and i said that let me just kind of drag this over here and let me drag them both over here so they fit so if i look at this now the two inside numbers do not match okay and this is very surprising again not being able to use this kind of commutative property that we're used to if we do a times b the product exists if we do b times a the product does not exist okay so we want to go back to a times b that's what we're going to be working with so let me just kind of slide this back over and let me just kind of slide this back over so we want to find a times b again we know that these numbers are the same so it works and we know that this guy is going to be a two by four matrix okay so the very first thing we're going to do is set up our matrix it's a two by four so a b is equal to okay and let me write this out so 1 two three four entries one two three four entries and another thing that i'm gonna do i did this before i'm just gonna put the columns the numbers in there just so we can keep track of what's going on this is something that really helps the students that i've tutored in the past just to keep track of where they are okay because a lot of times they go to a specific entry and they go i don't know what to do or they forget stuff or they kind of jumble it up this tells me immediately that what i'm in row one column one so where do i get that from again the row always comes from the first one the leftmost one just remember that the column comes from the second one or the rightmost one so i'm just going to start in row 1 in a and i'm going to highlight that or circle it or whatever you want to do just you're going to work with that for this entire row all the way across and all you're going to do is you're going to just keep going through the column so you're going to go this one times this one then this one times this one then times this one then times this one that's going to give your whole first row okay so for this first entry here i'm doing this top row times this first column row 1 column 1 row 1 column 1. that's all it is remember when you do this we have to go through the corresponding entries so you have the first one here and the first one here so 2 times 9 that's 18 then plus you have this guy here which is the second one times this guy here 3 times 10 is 30. then plus you have the third one here times the third one here zero times one is zero so 18 plus 30 is 48 plus zero is still 48 okay so it's not that bad overall but it does get you know a little bit tedious especially if you have something that's kind of bigger in nature so let's erase this and move down to this one so i'm still in the first row i'm just going to the second column now okay so let me highlight that we're going to do 2 times 2 which is 4 plus 3 times 5 which is 15 plus 0 times 0 is obviously 0. so you can just leave that off 4 plus 15 is going to be 90. okay so that's how we get that one there and again we just move down so now i'm going to be in row 1 column 3. so row 1 column 3. okay so two times zero is zero plus three times negative two is negative six and then zero times seven is zero so this is just negative six so pretty easy and let's erase this and let's go into this kind of last column here so we're going to say that we have 2 times negative 4 which is negative 8 plus 3 times 20 which is 60 and then plus we know 0 times anything is 0 so we don't need to worry about that and negative 8 plus 60 is going to be positive 52. so we've got the first row worked out okay so now what i'm going to do since i'm going to row 2 in the answer i'm going to move down to row 2 here again the row comes from the first matrix just remember that okay so my first matrix is a so i'm just going to highlight that and then i'm going to work column by column okay that's all i'm going to do so i'm going to say okay i've got this row times this first column to give me this guy right here so we do negative 1 times 9 that's negative 9 plus 5 times 10 that's 50 plus 11 times 1 that's 11. so 50 plus 11 is 61 and then minus 9 is going to give me 52. okay so this is 52 okay so let's erase this and again i'm just shifting 1 over going into column 2 and b because i'm in column 2 here okay so you've got negative 1 times 2 which is negative 2 plus 5 times 5 is 25 and 11 times 0 is 0. so negative 2 plus 25 is 23 okay and again i'm just shifting down so now i'm in this column and so negative 1 times 0 is 0 plus you've got 5 times negative 2 which is negative 10 plus you've got 11 times 7 which is 77. 77 plus negative 10 is 67 okay and then we have one more so we have this column here so we have negative one times negative four which is four five times 20 which is a hundred and then we have 11 times 11 which is 120 so if i sum all these together i know 4 plus 121 125 125 plus 100 is 225. so there we go we found our product a b okay not so bad again i recommend writing this stuff in so you can figure out where you are row one row two column one column two column three column four again if i'm wanting this entry here i find row one times column one okay i always find the row from the left one and the column from the right one okay or you could say the first and the second one however you want to define that once you do this a few times it becomes very very simple all right let's go ahead and do another one again this is one where you need a lot a lot of practice so we have matrix a and matrix b so first and foremost let's try a b so a b the size of a it's a one two by one two three so it's a two by three the size of b is a one two three okay by a one two so again if these numbers match up you are good to go you can put a check and say yes we can go ahead and do it and the size of the product comes from these outside numbers so it's going to be a 2 by 2 matrix okay so i'm going to go ahead and set that up i'm going to say a b is equal to again it's a 2 by 2. so just draw this out you've got 4 entries to fill out okay and again this should help you when you start this is row one row two this is column one and column two so how do i get this entry again again again i get my row from the first one the leftmost one however you want to think about that it's coming from a so i'm just going to highlight this row here okay that's where i'm going to be for here and here and then i'm just going to go through if i want row 1 column 1 i want row 1 times this column here column 1 in b so again we go through and figure this out negative 2 times 1 is negative 2 plus you've got 0 times 2 which is 0 plus you've got 1 times 0 which is 0 so this is just going to be negative 2 okay then i'm going to slide down because i'm now in row 1 okay stay in row 1 but i'm in column 2. so row 1 here times column 2 here so that's going to be 1. it's going to be negative 2 times 5 which is negative 10 plus 0 times negative 3 which is 0 plus 1 times 7 which is 7 this would give me negative 10 plus 7 is negative 3. okay so let's move on now so we're in row 2. so again now i'm going to shift from this row one down here to row two okay so since i'm starting here and it's in column one i'm going to multiply this row by this column okay so i'm going to have 5 times 1 which is 5 plus 9 times 2 which is 18 plus negative 5 times 0 which is 0. 5 plus 18 is going to be 23 okay 23. and then lastly what i want to do is this row times this column right here 5 times 5 is 25 plus 9 times negative 3 is negative 27 and then plus you've got negative 5 times 7 which is negative 35. so let's figure out what this is going to be well i know that 25 plus negative 27 is going to be negative 2 so basically i would have negative 2 plus negative 35 and that's going to be negative 37 okay so negative 37. so not that bad overall we found that our entries here are negative 2 negative 3 23 and negative 37. all right so let's erase this and then let's figure out if we can do b times a so b times a the size of b again is a three by two and the size of a is going to be a what it's a two by three so two here and two here the columns from b the first one match the rows from a the second one so we are good to go okay so we can do this multiplication the size of the product is going to be the outer numbers so 3 by 3. so ba b a is going to be this 3 by 3. so i'm going to have three numbers here three numbers here let me make this a little bigger so let me put my brackets in and again i'm just going to do this so row one row two row three column one column two and column three okay so for the first row first column so for this entry here again i want to think about getting the rows from the first one or b so this is my first row here okay and i'm going to start with this row okay times this column here because it's column one so this first column here so let me just kind of mark this one off so 1 times negative 2 is negative 2 plus you'd have 5 times 5 which is 25. if i sum those two amounts together i'm going to get 23. so that's my first entry so this is going to be 23 okay and i'm just going to work through this first row so i'm going to get this one and this one and so i'm staying with this first row with b okay so that's why it's going to be highlighted so i'm just going to go through this one first so 1 times 0 is 0 plus 5 times 9 is 45 so that's going to be 45 here okay and then i'm going to move to this one okay so i'm going to do 1 times 1 which is 1 plus 5 times negative 5 is negative 25 this is going to be negative 24. so negative 24. so that top row is now done okay so what i'm going to do now since i'm in row 2 i'm just going to move this down to row 2. okay so let me erase this and let me highlight this row 2 here and we'll start out with this one so we're going to do 2 times negative 2 which is negative 4 plus we're going to do negative 3 times 5 which is negative 15. so negative 4 plus negative 15 is going to be negative 19 and let me erase this and let me erase this and now we're going to move into this one so we have 2 times 0 which is 0 plus negative 3 times 9 which is negative 27 this is going to give me negative 27 here so let's erase this we've got one more for this second row so let me just circle this like this 2 times 1 is 2 plus negative 3 times negative 5 is 15. so this would be 17 here okay so we just have one more kind of row down here to fill in so let me just highlight this now and we'll go through them so you basically only have to this is a shortcut if you see a zero here you know that when you did kind of this one this one and this one in each case zero times this the top one okay is going to be zero so really all i have to do is figure out what is 7 times 5 that's 35 then what is 7 times 9 that's 63 and then what is 7 times negative 5 that's negative 35 okay that's a little shortcut if you see things like that always take that opportunity because again if i did this guy right here it would be 0 times negative 2 which is 0 plus you'd have 7 times 5 which is 35 okay so i just shortcutted that process to get these numbers all right so we've found ba and we've already found a b so we've got that result so now let's move on and let's talk about what happens when you multiply two matrices together that are square matrices okay of the same order so if you're multiplying a two by two and another two by two obviously you get a two by two if you multiply a three by three and another three by three you get a three by three if you multiply 117 by 117 and another 117 by 117 you get 117 by 117 okay so it's very obvious that you will always be able to kind of do the multiplication okay when you have two square matrices being multiplied together and they're of the same order so in this case a again it's a two by two and b is also a two by two so these numbers here are going to be the same and the outer numbers are gonna be the same so we know that the product here a b would just be a two by two matrix okay and so we'll just do this one more time so this is row one and row two this is column one and column two so again for row one to kind of find this guy right here i go to the row one from a the first one okay so this is my row one here and for the column i find it from here so this is my column one here so i take the product of this row times this column to find this guy right here so 10 times negative 2 is negative 20 plus we have 11 times 5 that's 55 55 minus 20 would give me 35 and then what would i do next well i want to move into this one so same row up here but different column over here okay so i would do 10 times 9 which is 90 plus 11 times 7 which is 77 and if i do 90 plus 77 i'm going to get 167 okay so again very very easy now i'm moving into row 2 and column 1. so for row 2 again it comes from the first one so this is my row 2 here and i want my column 1 here i want the product of those two so negative three times negative two is six plus i'd have one times five which is five this gives me eleven and then i just move down one so i'm gonna do this column here now so when we look at this we have negative 3 times 9 that's negative 27 plus 1 times 7 which is 7 negative 27 plus 7 is negative 20. okay so that's going to be my product a b if you want to flip this around just for practice and do ba okay again it's going to be the same size but the way that you calculate it's going to be different because b is now first so if i think about the row 1 i've got to think about the row 1 from b okay so this guy right here if i think about column 1 the column 1 is going to be for a okay so i want to find the product of this guy times this guy so this row vector times this column vector so negative 2 times 10 is negative 20 then plus 9 times negative 3 is negative 27. negative 20 plus negative 27 is going to be negative 47. again we just proceed so now i'm going to kind of move to this column here and i'm going to find the product of these two so negative 2 times 11 is negative 22 and then plus 9 times 1 which is 9. so negative 22 plus 9 is going to give me negative 13. and now what i'm doing is i'm moving into row 2 okay so moving into row 2 again that's from the first one so this guy right here i'm in row 2 and i'm going to start with column 1 here so row 2 column 1 5 times 10 is 50 plus 7 times negative 3 is negative 21. 50 plus negative 21 is going to be 29 and then we just have one more to do again i'm just going to shift this column here to this guy because i'm in column 2 now so i'd have 5 times 11 which is 55 plus 1 times 7 which is 7. 55 plus 7 is going to give me 62. okay so hopefully this is clear for you this is something that you do need to practice quite a few times to have it down okay you'll develop your own method your own strategy for doing this but typically i like to write out the row and the columns like i did in these examples here to keep track of where i need to be in my multiplication i've found that a lot of students that struggle with this if they just employ this kind of simple technique they end up really being able to improve their result quite substantially in this lesson we want to learn how to find the inverse of a matrix so in our last lesson we talked about how to multiply two matrices together we found that although the process is quite challenging at first because it's so foreign to us once we kind of nail down the steps it's not so bad overall if you're working with kind of larger matrices then the process is a bit tedious but again once you memorize the steps it's just some simple arithmetic it's going to be the same thing when we talk about finding the inverse of a matrix it's going to be a tedious process when we work with these kind of larger matrices but again we're just doing some simple arithmetic okay we'll find that when we work with a 2x2 matrix specifically there's a little shortcut and i'll get to that in a little while so the first thing we need to understand when we talk about this topic is the identity matrix so that concept so if i have a square matrix an n by n matrix that's an identity matrix what happens is i have ones going down the main diagonal okay so what that means is that in my row one column one i have an entry of a one and then in my row two column two i have an entry of a one if i have a bigger matrix let's say it was a three by three then in my row three column three i'd also have an entry of a 1. so those are your diagonal entries those are going to be ones you're going to have zeros above and below okay so you'll see this and say that looks very familiar when we did our gauss-jordan elimination remember we put the left side in reduced row echelon form which is this form here one's down that main diagonal zeros above and below and then the right side that was what we got our answer from okay but here we don't have a right side we just have this that we're working with so this is our identity matrix okay now specifically here when you look at this and you notate the size because it's an n by n same number of rows as columns i can put one single number down here to identify the size this is a two row by two column matrix so i can say that this is an identity matrix or i sub two to say it's a two by two identity matrix if i had a three by three identity matrix again i can just say it's i sub 3 and notice how you have a 1 in row 1 column 1 a 1 in row 2 column 2 and a 1 in row 3 column 3. again your diagonal entries are going down that main diagonal there zero's above and below so that's your identity matrix it's a similar process if you had a four by four a five by five a six by six so on and so forth now why is this important well this is kind of building up to something and essentially we need to understand that when we worked with kind of real numbers okay outside of matrices just real numbers going back to kind of grammar school we found that if we multiplied a number by one it was unchanged right one is the multiplicative identity in multiplication with real numbers well what happens is if we take an n by n square matrix like a okay so let's say if a is an n by n square matrix and we multiply it by an identity matrix of the same size right so that's why you have i sub n there you get a back now what's interesting here we learned in the last lesson that matrix multiplication was not commutative right if you change the order around you generally speaking will not get the same product and sometimes you could do it one way but there's no product that's going to exist the other way here you can go in two different ways so you can say a times i sub n would give me a and then also i sub n times a kind of reversing that would also give me a okay so it's not a problem to kind of switch that around so let's see a quick example of this so what we have here is a two by two identity matrix again we show this as i sub two the two tells me i have a two by two and then we have a matrix a which is also a two by two we have negative one and five in the first row two and seven in the second row so based on our rule above we know that a multiplied by i sub 2 or again a 2 by 2 identity matrix should give me a back and then additionally i sub 2 the 2 by 2 identity matrix times a should give me a back so i'm just going to do this first scenario i'll let you reverse it and do that scenario as well again you get a back either way so let's go ahead and crank this out real quick i'm just going to erase this for right now so first and foremost can we do the multiplication of course we can because we're multiplying two square matrices together of the same order if we think about it this is a two by two and this is a two by two so the columns from a the left one match the rows from i okay i sub 2 if you want to be specific the right one okay so you always circle these inside numbers and check to see if they're the same if they are you can proceed again the size of the matrix comes from the outside numbers so we know it's a two by two okay so let's go ahead and set this up and i'll put my entries in here and just to stay consistent with what we did in the last lesson i'm gonna put row one here and row two here and i'm going to put column 1 here in column 2 here again if you're really good at matrix multiplication you don't need this anymore when you first start out i find that it really really helps students figure out what's going on if i look at this kind of entry here i'm thinking about okay this is row 1 column 1 in my answer okay i always get the row from the leftmost matrix so in this case that's a so i want to be in row 1 of a so this is the first row i get my column from the rightmost matrix or you could say the second matrix if you want so in this one i have column one so i want to be in the first column here okay so to get this entry all i'm going to do is find the product of this row vector times this column vector that's all it is okay so we know how to do this we go through and we say okay the first entry times the first entry negative 1 times 1 is negative 1 then we're going to add to this we're going to have the second entry times the second entry so 5 times 0 is 0 so we know this is negative 1 so this is negative 1. and you can see already that your first entry in this matrix matches what it is in a okay so as we continue now if i move into this position here i'm still in row one okay so for a i'm still going to be in row one but now i'm shifting to column two so for this guy i'm gonna be in column two okay so that's all it is same process i'm gonna do this one times this one so negative one times zero is zero then plus i'm going to do this one times this one 5 times 1 is 5. we know 0 plus 5 is 5 and again we can see this matches exactly so now what i'm going to do is shift because i'm going to be in row 2 down here now so i'm going to start in row 2 here and i'll have column 1 here okay so row 2 column 1 this guy right here so 2 times 1 is 2 plus 7 times 0 is 0. this equals 2 so i get a 2 there and again this matches perfectly with a so let me erase this now we're gonna move on to kind of this one so for the last entry down here you're in row two column two so you've got the row from a and the column from i okay so you've got two times zero which is zero plus you've got 7 times 1 which is 7 which gives me 7 so i put a 7 there so you can see that we multiplied our matrix a by our matrix i sub 2 or the 2 by 2 identity matrix and we got a back again if you want to pause the video and reverse this if you wanted to do i sub 2 times a you would get a back as well okay so now let's move into talking about how to find the inverse of a kind of square matrix so first we're going to define a word here so we have this non-singular word okay and that just means that the matrix is going to have an inverse so when a matrix has an inverse we say it's non-singular you might also see the term invertible okay so those two words kind of mean the same thing you might see non-singular in one textbook you might see invertible another just depending on where you're kind of getting your information from if you have a matrix that doesn't have an inverse and we'll see an example of this in a little while it's known as a singular matrix okay so if a is a non-singular meaning it has an inverse n by n square matrix then this property is going to be true if i take a and i multiply it by the inverse of a okay then i'm going to get an identity matrix of the same order again i just put that in there to say it's an n by n so if a was a two by two the inverse of a would be a two by two and then this identity matrix would be a two by two so let's show an example of this real quick i'm giving you a and i'm giving you a inverse okay and what we're saying is that if you multiply a by its inverse you're going to get the identity matrix in this case you would get an i sub 2 because it would be a 2 by 2. okay so let's crank out a times a inverse and you could reverse this you would still get the identity matrix okay it wouldn't matter so what this is going to give me is what let me kind of write this out i know that it's a 2 by 2 times a 2 by 2 so the answer will be a 2 by 2. okay so this is row 1 this is row 2 column 1 and column 2. so i know for this guy right here row 1 column 1 i want this row here from the first one times this column here from the second one so this row vector times this column vector what is that so we have three times one which is three plus we have one times negative two which is negative two and we know this is going to give me one so i've got a one there already okay now let's move on so now i'm going to erase this and i'm going to move to this column here so this guy 3 times negative 1 is negative 3 then plus 1 times 3 is 3 and this is going to be equal to 0. so that's going to be this entry here okay so now we're moving into this row down here so let me kind of switch up my highlighting i know i'm going to be in this row here and this column here to start so 2 times 1 is 2 and then plus 1 times negative 2 is negative 2. this equals 0. so let me write that in here and then let me erase this one more to do and let's just highlight this now so we have 2 times negative 1 which is negative 2 plus 1 times 3 which is 3 and obviously this is 1. so we have our identity matrix we have our i sub 2 which is exactly what we were told from this rule here okay so now that we've proven that this is true with a simple little example let's talk about how we could find a inverse if we didn't know it okay so i'm going to give you the shortcut first okay and then i'm going to show you kind of the long way this shortcut that i'm going to give you is only going to work for a 2x2 matrix unfortunately when you get something larger you have a few different techniques that you can choose from the one i'm going to show you today is the one you're going to see in your textbook there's other methods that you can use they're not a big difference in time it just kind of depends on which kind of method you find less tedious i can tell you that this is not a process that is very enjoyable okay so for the 2x2 the shortcut goes as follows so i'm going to define a just generically to have these kind of entries of lowercase a lowercase b lowercase c and lowercase d okay now we haven't yet talked about the determinant of a matrix yet we'll get to that in two lessons from now but for now it's very easy for a two by two matrix it's just this guy times this guy so it is a times d then minus you're gonna go this times this so minus b times c okay so that's the determinant of a a lot of times you'll see d e t of a like this there's a lot of different notation for this and i'll talk about that when we get to that video for now you just need to know this is the determinant of a okay so let me erase this real quick and let me erase this i'm just going to drag this up here a little bit so that we know what this is okay so the way we find a inverse is that we're going to go 1 over the determinant of a and we're going to multiply this by we're going to do a little kind of swap root here so i'm going to take a and d and i'm going to swap them so i'm going to put d here and a here and then i'm going to make c and b into their opposite so i'm going to have the negative of b and the negative of c now the first time you kind of see this you're like what in the world is going on but after you do this a few times it's something you just memorize and i can tell you this comes up very often it's a very good shortcut it's going to save you a lot of time okay so let's say that i replace this with these numbers here and i know they're already copied but let's just say we have three one two and one and so i know that my a my lower case a is three my lowercase b is one my lowercase c is two and my lowercase d is one okay so let me start out by going 1 over the determinant again the determinant is 3 times 1 which is 3 minus 1 times 2 which is 2 so that's going to be 1. okay so 1 over 1 is 1. okay so that's pretty simple then times i'm gonna do this stuff here so these guys switch positions so the three and the one switch so one goes here and three goes here and then what's going to happen is this guy right here this guy right here i changed them into their opposite so this is negative 1 and this is negative 2. we already know that 1 times this in scalar multiplication just is itself so i know that a inverse is this and again we already know that that's the inverse right i have a 1 and a 1 a negative 1 a negative 1 a negative 2 and a negative 2 and a 3 and 3. so not too bad overall compared to what i'm about to show you so let's talk about the long way now and this is something you'll see in your textbook i'm going to show you later on at the very end of the video for those of you who want to see it where this kind of thought process comes from i'll start out with a 2x2 matrix and i'll go through the whole thing and i'll show you why this makes sense but for now if you don't want to look at that you can just take it as a given okay so if you want to find the inverse of an n by n non-singular matrix a so this works for a two by two a three by three four by four whatever you want to use it on okay you start out by basically setting up this in step one so you're going to put your augmented matrix where matrix a goes on the left and the identity matrix of the same size okay goes on the right okay then what you want to do is you want to use row operations to get this guy into the format of being the identity matrix and then this guy on the right which we'll call matrix b okay the result of doing the row operations to getting this guy into that format b is now going to be the inverse of a okay and again i'll show you exactly where this comes from at the end of the lesson the process takes a little while so i know some of you don't want to watch that all right so let's go back and i want to grab a real quick and i'll show you that you end up with the same thing so let's just take this real fast i'm going to copy it and i'm just going to paste this in here real quick and i'll show you that you get the same result either way so what i want to do is i want to take this matrix here this 3 1 2 and 1 and then i want to put a vertical bar and i want to write my identity matrix of the same size so one's going down the main diagonal a zero above and below okay so what i'm going to do is i'm going to use row operations to get this side over here to look like this over here i want this side to be an identity matrix and then this side would give me the inverse okay so what we're going to find is that we're going to end up with a right side that matches this okay so let's go ahead and do our kind of row operations we already know that this process kind of you know for lack of a better word is very tedious okay so the book usually tells you to get a one here first and then use that to get your zeros there's other methods you can use i'm going to stick with that method i know there's kind of quicker ways but i'm going to stick with that because i don't want a confusion okay so the first thing i'm going to do is i'm going to multiply row 1 by a third so that that entry becomes a 1. so i'm going to do 1 3 times row 1. that's what i'm going to replace row one with okay so this three would become a one this guy would become a one-third this would be a one-third and this would stay zero so let me erase all of these entries and let me put these in this is 1 1 3 this is 1 3 again and this is 0. okay so that's the first step now the second step is to get a 0 here and to do that i'm going to use my 1 that's above that okay so i think about the opposite of 2 which is negative 2 and i would say 1 times negative 2 would give me negative 2. if i added negative 2 to 2 that would be 0. okay so what i want to do is i want to take negative 2 and multiply it by row 1 add the result to row 2. that's what i'm going to replace row 2 with okay so let's go ahead and do that real quick so if i did negative 2 times 1 that would give me negative 2 and then if i added that to 2 i would get 0. okay then if i did 1 3 times negative 2 that would give me negative two-thirds so negative two thirds and then i'm going to add that to positive one or three thirds and if i add those two amounts together negative two plus three is one so i'd have one third okay and my screen is going to get too busy so let me kind of fill these in as i go so this is a zero and this is also a one-third so let me erase this we don't need this anymore and then for this one right here i'd multiply negative two times a third okay so that would give me negative two-thirds and then i'm gonna add the result to row two so i'm gonna add 0 we know that would still be negative 2 3. so this is negative 2 3 and this one's not going to move at all because negative 2 times 0 is 0 0 plus 1 is 1. okay so that part is done and just a little bit more to go so now i want to make this a 1 okay so i would just multiply that whole row by 3 right the reciprocal of 1 3. so 3 times row 2 would give me my new row 2. okay so 0 times 3 is 0. 1 3 times 3 is 1. if i did negative 2 3 times 3 i know that the 3s would cancel i'd have a 1 negative 2 times 1 is negative 2. so this is negative 2 and then i know that 1 times 3 is 3. okay so let me just put a 3 in there and you can already see that the kind of bottom row here works out to be this okay but we got to fix we've got to make this a 0 now okay so what i'm going to do is i'm going to multiply row 2 i'm going to multiply row 2 by the opposite of 1 3 which is negative 1 3 i'm going to add the result to row 1. okay that's what i'm going to replace row 1 with so i'm going to do 0 times negative 1 3 that's 0 zero plus one is still one one times negative one third is negative one third negative one third plus one third is zero so that's going to be zero and then negative one third times negative two would be positive two thirds and i can erase that sign i don't need it then plus one third well this would be one right two plus one is three three over three is one so this becomes a one here and then lastly okay i would do a negative one one-third a negative one-third times three and we know this cancels with this and gives me a one negative one times one is negative one okay and i would add that result to zero so that would be negative one now you can see here once the process is done this here matches this here so that's exactly what we said in our instructions right once you kind of start this way you use your row operations you get in the format of the identity matrix is on the left and on the right you're going to have the inverse of your matrix and we know that that's true already because we know that a inverse is the inverse of a because we already showed that okay so that's another way to go about doing it if you have a 2x2 you know it's not something you really want to do because it just takes too long you want to go ahead and use the shortcut so that's the only time i'm going to use that there i want to show you an example of a singular matrix okay so i'm just going to do one with a two by two and what you're going to find is that it's singular or meaning it doesn't have an inverse because the determinant is going to be zero okay the determinant is going to be zero so if you use row operations here you're not going to be able to get it into the format that you want but again we're going to use our shortcuts so i'm going to say that a inverse is equal to 1 over the determinant of a so 3 times 4 which is 12 minus 6 times 2 which is 12. so once you see that this is a 0 down here the determinant you can stop and say this is a singular matrix okay it's not going to have an inverse so you don't need to go any further i know the rest of it was to kind of multiply it by this these two guys would switch so you have 4 here and 3 here and these guys become their opposite so negative 6 here and negative 2 here but the whole thing is blown okay because you have a zero in the denominator and we know we can't divide by zero so this isn't going to work there's no inverse all right so let's take a look at an example with finding the inverse of a three by three matrix again as i said earlier when you get a three by three or four by four there are some tricks out there there are some different techniques but none of them are pleasant okay and you basically have to pick one of them and then grind through the work so we're going to look at the method that we learned today so basically i'm just going to copy this guy right here and i'm going to say i have a 1 a negative 1 and a 1 a 0 a 4 a negative 2 and then i have a 2 a 3 and a negative 1. okay so that's the first part remember you copy your matrix you're trying to find the inverse of on the left you're going to put your kind of vertical bar in here and then i'm going to write the identity matrix on the right so i'm going to put a 1 here a 0 here and a zero here then a zero a one and a zero and then a zero a zero and a one okay so again one's going down that main diagonal zeros everywhere else so you could say zero's above and below so let me go ahead and copy this because we won't be able to fit this on this page and let's go ahead and paste this in here and we'll just go through the work all right so we want to take the left side here and transform it into the identity matrix or you could say you're putting it in reduced row echelon form and again through that process the right side here is going to give us our inverse matrix of a okay so the first entry here we always want to get a 1 here if you go by what your book tells you right you get your one first in the column and then you use the one to get your zeros okay so that's already done for us so what i want to do is concentrate on getting a zero here and then getting a zero here okay so let's do those one at a time so i know that i can multiply row 1 by 1 which is just the same thing as having the row itself and then add the result to row 2 because 1 and negative 1 are opposites that's going to give me a 0 for that entry there so the way i'm going to write this is i'm going to say 1 times row 1 plus row 2 is going to give me my new row 2. you can get rid of the 1 if you want i'm just going to put it there for kind of emphasis okay so let's go through and basically just say that we're adding row 1 to row 2 and that's what we're replacing row 2 with so 1 plus negative 1 is 0. let me make that a little better and then 0 plus 4 is just 4 so that's going to stay unchanged 2 plus 3 is going to be 5. 1 plus 0 would be 1. 0 plus 1 would be 1 so that stays 1. 0 plus 0 is obviously 0 so that stays as a 0. all right so that's the first step then the second step here again we want to get a 0 here so to do that i want the opposite of 1 which is negative 1 so i'm going to multiply row 1 by that so negative 1 times row 1 then i'm going to add the result to row 3 down here that's what i'm going to replace row 3 with okay so let me just go through and multiply everything by negative 1. so 1 times negative 1 is negative 1 0 times negative 1 is 0. 2 times negative 1 is negative 2 1 times negative 1 is negative 1 0 and 0. if i multiply either of those by negative 1 i get 0. okay so those are going to be your entries that you add to this row so negative 1 plus 1 is 0. zero plus negative two is negative two negative two plus negative one is negative three and let me just erase these because we're done with them okay and then negative one plus zero is negative one and then obviously if i add zero to something it's unchanged so i don't need to worry about the rest of those okay so now let's move into the next column again if you follow what your book tells you and you don't have to you can do it different ways you want to get a 1 first okay so i'm going to put a 1 right there in row 2 column 2 okay considering the left part of the matrix and to do that i'm just going to multiply row 2 by 1 4 right the reciprocal of 4. so 1 4 times row 2 is what i'm going to replace row 2 with so we know that zero times a fourth would just be zero so that's not changed four times a fourth is one and then we know this would be what it would be five fourths so five fourths this would be 1 4 and this would be 1 4. so let me just change that in each case and then lastly we know 0 times 1 4 is 0. so that's done so now conveniently i already have a 0 above i just need a 0 below okay so to do that i'm going to use my 1 and i'm going to multiply row 2 by the opposite of this negative 2 which is 2. so 2 times row 2 and then plus my row 3 that guy right there is going to give me my new row three okay so let me just multiply everything by two so two times zero is zero then we have two times one that's going to be two then we have two times five fourths so two times five fourths so this would cancel with this and give me a two down here so that's going to be five halves so five halves and then two times one fourth we're doing that twice so we know that this cancels with this and gives me a two down there so in each case i'm going to have a half so i'm going to have a half here and i'm going to have a half here okay and then lastly 0 times 2 would be 0. so that's the result in each case of doing 2 by each one of these guys okay so now we want to take these results and add them to the entries in row three zero plus zero obviously that's not going to change two plus negative two is going to give me zero this one this one and this one is going to be a little bit of work so let's do those off to the side so we have 5 halves plus a negative 3. so negative 3. so i need a common denominator i'm going to multiply this by 2 over 2. so i could say this is negative 6 over 2. so negative six over two five plus negative six is going to be negative one so this is negative one-half okay so this is negative one-half so that's what i'm going to replace this guy with negative one half okay so let's move on now to this one so what we have is one half plus negative one so one half plus for negative one i'm just going to write it as negative two over two okay and so negative two plus one would be negative one so this is negative one half so this is negative one half and obviously one half plus zero is one half so that one's easy and then zero plus one is one so you don't have to do anything there okay so now we are getting close to the end okay so we wanna get a one here as our first thing to do in this kind of column here so to do that i'm just going to multiply row three by the reciprocal of negative one-half which is going to be negative two so negative two times row three that's going to give me my new row three and so let's go ahead and do that we know that this and this would be unchanged because they're zeros zero times anything is zero this guy would obviously be a one so i really only need to work on these and i know that this one right here this one would just be negative two so let's work on this one and this one okay so negative two times negative one half we know that this would cancel with this and be one so you basically have what you have negative one times negative one which is positive one so this is going to be positive one and then in the last case you have it negative two times one half and so this cancels with this and gives me a one negative one times one is negative one so this would be negative one okay so we have the bottom row here complete and now what we want to do is get a zero here and here and we'll be done okay so i'm gonna multiply row three by the opposite of five-fourths which is negative five-fourths add the result to row two okay you already know what's going on so for these first two it doesn't matter because i'm multiplying by zero so anything by zero is going to be zero and if i add zero to something it's not going to change so i don't need to worry about those entries i know that if i multiply negative 5 4 by 1 i get negative 5 4. if i add it to 5 4 i get 0. so i really don't need to worry about that either okay i only have to work on these over here so let's kind of do these entries real quick so negative 5 4 times 1 is negative 5 4. then we have negative 5 4 times negative 1 which would be positive 5 4. and then negative 5 4 times negative 2 what would that give me well i know it's going to be positive so i can just erase these signs and i can cancel this with this and put a 2 down here so that's going to be 5 halves okay and let me get rid of this sign here we really don't need it it's just for emphasis let me kind of drag these up a little bit more all right so let me just kind of put commas between them so we know what's going on all right so let's start with this guy right here we would have 1 4 plus this negative 5 4. we have a common denominator already so we have negative 5 plus 1 which is negative 4 so this gives me negative 4 over 4 which is negative 1. okay so this is negative 1. so i can erase this don't need that anymore and erase this and then now i'd have 5 4 plus 1 4 okay so 5 plus 1 is 6 so you'd have 6 over 4 which reduces to 3 halves right each is divisible by 2. so this is going to be 3 halves okay and then obviously 5 halves plus 0 is just 5 halves all right let me make this better because that doesn't look so good okay so we have that completed now and we just need to get a zero in this entry here and we will finally be done all right so how can we do that we multiply row three by negative two the opposite of two so row three gets multiplied by negative two add the result to row one okay so we don't have to worry about again these entries to the left of this because you have negative two times zero negative two times zero it's going to be zero add zero to something it's not going to change i know this is going to be zero because negative two times one is negative two negative two plus two is zero so that's taken care of so i really don't have to worry about these entries over here okay so what is negative two times one it's negative two what is negative two times negative one it's positive two what is negative two times negative two it's going to be four okay so let's add these negative two plus one is going to give me negative 1 then next you're going to have 2 plus 0 which is going to give me 2 and then lastly you're going to have 4 plus 0 which is going to give me 4. okay so we've completed our process all right so just to recap what we did we started out with a on the left and the identity matrix and i sub 3 because it was a 3 by 3 on the right we used row operations to basically transform the left side into an identity matrix we'll say i sub 3 because it's a 3x3 and then the right now let's just call this b or you could just say it's a inverse right because it's the inverse of a at this point okay so that's the process that's how you do it if you wanted to say we can just copy this real quick go back up and let's just paste this in we can say that a inverse okay is going to be this part right here okay on the right so let me just kind of set this up we'd say negative 1 negative one and one you would have two you would have three halves and negative one you would have four you would have five halves and negative two and so that would be the inverse of this matrix here and again this process is going to work if you you know have a four by four or five by five or something like that obviously those problems get very very tedious and there are some tricks out there but usually in your course you're only going to be given a three by three and occasionally maybe for like a bonus question they'll give you a 4x4 these guys who are teaching these classes they understand how tedious this process is all right so let's wrap up the lesson by kind of returning to this generic formula i want to show you where this comes from some textbooks show you where it comes from and you might get lost if you're trying to read it other textbooks just skip it okay so if we find the inverse again of an n by n non-singular matrix a again the first step is to put this kind of augmented matrix there where a what you're trying to find the inverse of is on the left and your identity matrix is on the right then through your row operations the left side you change that into the identity matrix one's going down the diagonal 0s above and below and this matrix that's formed on the right let's just call it b is the inverse of a so where does this come from let's go ahead and start with a fresh example okay so we have a and it's equal to this we have 5 and negative 1 and we have 3 and 6. so let's start out by using the little shortcut to figure out what we should get so again a inverse if we use our shortcut is equal to 1 over the determinant of a which is 5 times 6 which is 30 okay minus negative 1 times 3 which is negative 3 minus a negative is plus a positive so this should be plus 3. so this would be plus 3 here so this would be 1 over 33 so it would be 1 over 33 times okay remember the little trick you have to do over here you switch these two so this becomes a 6 and this becomes a 5. these two become their negatives or you could say they're opposites so this would be negative 3 and this would be positive 1. okay so if i did let me kind of do this down here i'll continue and i need a bigger bracket so 1 over 33 times 6 would be 6 over 33 of course each is going to be divisible by 3 there 6 divided by 3 is going to give me 2 and 33 divided by 3 is going to give me 11. so that's my first entry then over here if i did negative 3 over 33 each is divisible by 3 this would be a 1 and this would be an 11 okay and then for this entry here 1 over 33 times 1 would just be 1 over 33 and then for this last entry i would have 1 over 33 times 5 which is 5 over 33 okay so that would be a inverse and let me just kind of write this over here just kind of scoops this up now let's say that you didn't know that shortcut and you didn't know the rule that i just gave you okay so you didn't know about kind of going through the process how could you figure this out how could you derive the formula well if i just started out with the notion that a times a inverse gives me an identity matrix in this case it's going to be a 2 by 2 well i can come up with a little formula so let's kind of go down a little bit let me just copy a real fast so it's 5 3 negative 1 and 6. let me copy this real fast and i'm going to go to a fresh sheet we'll come back there and we'll check our answer eventually so we have this guy here and i know that if i multiply it by its inverse let's just say i don't know what that is we'll say it's x y w and c these are just kind of stand-ins for those entries okay we know that if we multiply these two together if they're inverses i get an identity matrix that's a two by two so one's down the diagonal and i've got zeros above and below so we know this so what i would do is i would say okay i can multiply these two together so the way i would do that i know i would get a 2x2 matrix and i need a lot of room here so let me just not put my brackets in yet i know for this entry here i would do this row times this column so 5 times x is 5x and then i would have plus negative 1 times w so minus w then for this entry i would do this row times this column so 5 times y is 5y and then negative 1 times z we'd have minus z okay then for this entry i would do this row times this column so i would have 3x plus you'd have 6w okay and then lastly i would do this row times this column 3y okay plus six z all right so so far so good now i know that this guy if i set it equal to this what has to be true let's think about what we know so far or what we've learned in kind of previous lessons we know that two matrices can only be equal if and only if their size is the same in each case here we have a 2x2 and their corresponding entries are exactly the same so what that means is that this guy right here this 5x minus w must be equal to 1 if this is true okay we also know that five y minus z must be equal to zero again corresponding entries so this guy right here this three x plus six w must be equal to zero and this guy right here this 3y plus 6z is equal to this one here okay so so far so good now let's go down a little bit and let's continue to think about what's going on i have four variables that are unknown and i can solve it here because i can split it into kind of two different systems with the same two variables each so on the left i'd have a system with x and w on the right i'd have a system with y and z okay now you can go through and solve this pretty quickly but what i'm going to do is solve it it's going to take me a little bit longer but i want to show you what your book's going to do so i would solve this using a kind of augmented matrix so in the first one i would have kind of the way this is lined up i have my x on the left and my w on the right okay of this left side of the equation so 5 and negative 1 and 3 and 6. and notice the order is important i've got to have the x lined up and the w lined up so that this column is the x's this column is the w so this is for the x's this is for the w's right then this guy is going to give me my constants my 1 and my 0. let's stop and think about this for a second i know that if i put this in reduced row echelon form this guy would be a one this would be a one this would be a zero and this would be a zero so x one x plus zero w would be this so this guy is going to give me what x is okay and similarly this guy is going to give me what w is okay so let's erase this and let's just keep that information so we keep track of what's going on now similarly i can set up this system over here 5 negative 1 3 and 6. and through a similar thought process i can say that this guy right here is going to be my answer for y when it's done and this guy right here is going to be my answer for z okay and i'm going to erase this kind of circling we don't need that and it would just get in our way okay now what do you notice here this is where people get lost in this process the left side here matches the left side here right 5 negative 1 3 6 5 negative 1 3 6. okay so what happens is if i'm going to solve this using row operations what i do on this would match what i do on this this they're the same numbers right so what your book's going to do they're just going to erase this and they're going to squeeze this down and say hey i'm going to combine these two operations because i can't and what's going to happen is i already know i've already kept track of what's going on so when i put this guy into reduced row echelon form where i make it into its identity matrix okay then i can get my answer by saying this is what x is this is what w is this is what y is and this is what c is okay so let's copy this and go to a fresh sheet and we'll paste this in and we'll finally crank this out i know this takes a while but nothing kind of worth seeing is quick okay so let's go through and use our kind of row operations we're trying to get this left side here into reduced row echelon form or again you could say the identity matrix so the first thing i would want to do is get a 1 here so how would i do that i would multiply row 1 by 1 5th that's what i'm going to replace row 1 with okay so i know that this would be a one i know that this would be a negative one-fifth i know this would be a one-fifth and this would stay a zero so that's pretty easy then the second thing i would do is i would make this into a 0. so i would multiply row 1 by negative 3 i would add the result to row 2 that's what i'm going to replace row 2 with so negative 3 times 1 is negative 3 negative 3 plus 3 is 0. then negative 3 times negative 1 5 gives me 1. negative 3 times negative 1 is 3 and then over 5. so this is three-fifths and then i'm going to add to this 6. so i'm going to have 3 fifths plus might as well say 6 is 30 over 5. okay so this would be 33 fifths this is 33 fifths all right and then we have negative 3 times a fifth that's negative three-fifths if i add that to zero it's just negative three-fifths so that's pretty easy and then lastly we have negative three times zero that's zero so we don't have to worry about that so that part is done now let me get this to be a one so i'm going to multiply row 2 by 5 over 33 that's what i'm going to replace row 2 with and so we know this is a 1. we don't even have to worry about this this right here if i did 5 over 33 times negative three-fifths the fives are going to cancel the three over the 33 we know they're each divisible by three right so we say this is a one and this is 11. so this right here is going to be negative 1 over 11. and we know that lastly i'm going to have 5 over 33 here because 1 times anything is itself all right so one more thing to do here i just have to make this into a zero and to do that i'm going to multiply row two by one fifth and i'm going to add that result to row one that's what i'm going to replace row one with let me kind of slide that down because it took a little bit too much room slide that over here all right so one-fifth times zero is zero don't have to do anything with that one-fifth times one is one-fifth one-fifth plus negative one-fifth is zero all right so now i'm gonna do one-fifth times this negative 1 over 11 okay so that's going to give me negative 1 over 55 okay so negative 1 over 55 and i'm going to add to this 1 over 5. so to get a common denominator going i'm going to multiply this by 11 over 11 okay so that would give me 11 over 55 and if i add these two amounts together i'm going to get 10 over 55. now i can simplify this because each one of these guys is going to be divisible by 5. if i go ahead and cancel a 10 with a 5 10 divided by 5 is 2 and 55 divided by 5 is 11 okay so that's what i'm going to replace this with this is going to be 2 11 and then lastly i'm going to have 1 5 five over 33 and so this cancels with this and gives me a one i'm going to have one over 33 so 1 over 33 plus 0 is just 1 over 33 okay so let's erase all of this didn't mean to erase that bracket and let's kind of copy this real fast so let me paste this in here real quick and i'm saying that this guy right here should be 2 over 11. so this is my x it should be 2 over 11. this guy right here should be 1 over 33 and this guy right here should be negative 1 over 11 and this guy right here should be 5 over 33. okay so we already know that this is the case because we found it using the shortcut so let me copy this real quick so let me just paste this in here real quick and let me erase this so what you see is that you have a inverse right this is exactly what we found from the shortcut you have two elevenths and two elevenths one over thirty three and one over thirty three negative one over eleven and negative one over eleven five over thirty three and five over thirty three okay so it's the same and what's cool about this is that we ended up deriving kind of this formula that we see in our textbook right so we ended up saying that a times a inverse gives me the identity matrix we had to set up some kind of variables to represent our unknowns for the a inverse but what we ended up with was solving kind of a system that's set up this way we had a on the left and we had the identity matrix on the right we tried to solve this system so we try to put this guy in reduced row echelon form where we have ones down the diagonal zeros above and below so that's the identity matrix right and then once we got in this format we found that this guy right here ends up being the inverse okay of what we started with which was a so really cool to know where that comes from if you were reading that in your textbook and you got lost now you know the steps now you know how it works and you could really do this with a 3x3 a 4x4 whatever you want you can extend what we just did out to as large of a system as you wanted to work with again as long as you're working with a matrix that's invertible or again non-singular in this lesson we want to talk about solving linear systems using inverse matrices all right in the last lesson we talked about how to find the inverse of a matrix we found that only square matrices had an inverse and not all of them would have an inverse right so when a square matrix has an inverse it's called non-singular or invertible if it doesn't have a matrix it's called a singular matrix now we know that for a two by two matrix there's a very good shortcut that we can use to find the inverse when you get into a three by three or higher there are some different options you can kind of employ but all of them are tedious they all involve a good amount of work okay so what we want to do here is we want to take the next step and we want to show how we can solve a linear system using an inverse matrix but it's not necessarily going to be a good method to use okay in a lot of cases this is going to be way more work okay but it's something that you do need to learn because it's taught in every textbook and you're probably going to get tested on it now let's go ahead and start out with this kind of easy system to get started we have 8x minus 7y equals negative 17 and we have negative 6x minus 8y equals 26. so the very first thing you want to do is make sure that your equations in your system are in standard form okay just like when we try to set up an augmented matrix you want the ax plus b y equals c so everything has to line up your x's your y's your constants make sure all that lines up now the next thing you want to do in this particular case you don't have to do this this is just something that i like to do to make things simpler notice how everything there is divisible by 2 so i can really rewrite this and say this is negative 3x and then minus 4y and this is equal to 13. so i'm just going to rewrite my system real quick and say that it's 8x minus 7y is equal to negative 17. and negative 3x minus 4y is equal to 13. okay so let's copy this real quick i'm just going to go to a fresh sheet because things are going to get busy very quickly here and let me just paste that in so now what i'm going to do is going to seem a bit strange at first but i'm going to set up three different matrices okay so one's going to be for the coefficients one's going to be for the variables and one's going to be for the constants so the way your book is going to do this they're going to say that a is going to be the matrix for the coefficients so if i just grab them in order i have 8 and negative 7 and then i have negative 3 and i have negative 4. okay so that's going to be my first matrix of this kind of equation i'm going to be setting up which we'll get to in a minute then the second matrix is going to be a column vector so a matrix with one single column it's just going to have the two variables of the system so i have x and i have y so i'm going to put x and y okay so this is for the variables of the system and then i'm going to have a matrix b which is going to contain these constants okay so let me write that it's going to be another column vector so i'm going to say we have negative 17 and then we have 13. okay so the way we're going to set this up is we're going to say that a this matrix here multiplied by x this matrix here should be equal to b this matrix here so you might be saying why does that work out well what happens is if i go through the process of multiplying this matrix by this matrix i'm going to get entries that exactly match this kind of left side okay and it's going to be set equal to this right side here now let me show you that real quick let me just kind of scooch this out of the way so we have a little bit of room and i'll just move this up it's going to be very tight to do this but first and foremost can we multiply a times x we know a is a 2 by 2 and we know that x is what it's a two by one so the columns from a match the rows from x so we know we're good to go there and we know the output of this is going to be a two by one right the outer dimension so two by one so if i do ax i would get a two row okay by one column matrix so i would take this row times this column a times x is 8x and then negative 7 times y would be minus 7y and then for this one right here i take this row times this column negative 3 times x is negative 3x and then negative 4 times y is minus 4y now again this exactly matches what we have on the left side here so let me kind of slide this down now and let me put some brackets around this and if we say this is equal to b b is down here this is negative 17 and this is 13. this is exactly what we see with our system remember two matrices can only be equal if they're the same size in each case it's a two by one okay and they have exactly the same elements okay so we're saying that this is going to be equal to this this is equal to this okay we're saying this is equal to this this is equal to this so that's where this equation comes from i just wanted to show you that so that you're not confused about why this is mathematically legal okay so now let's talk a little bit about how we could solve this equation right now we know what the kind of entries are for a and the answers are for b but we don't know the answers for x that's what we'd like to find so we can figure out what x and y are so with normal equations right linear equations in one variable it's very easy we divide both sides okay by a and we can get our solution right we would say x is equal to b over a well it's not going to work out that way for us because we can't divide a matrix by another matrix when we work with kind of matrices that operation is undefined so we have to have a different way and what we're going to do is use some of the knowledge from the kind of previous lessons and we're going to say that we know if we take a inverse and we multiply it by a i'm going to get an identity matrix of the same order okay so if i took a inverse multiplied by a i would get i sub 2 right a 2 by 2 identity matrix so let's just go ahead and do that real quick i'm going to slide this down here i'm going to say i have a inverse times this on this side and i've got to make sure the order because it's going to be very important to make sure the order is correct so i'm going to slide this in on the left and what happens is on this guy a inverse times a gives me an i sub 2 and this is times x and this equals a inverse times b now stay with me for a second we have x which is two rows by one column and we have i sub two which would look like this right it would be a one's going down the diagonal zeros above and below so this is my i sub 2. now if i multiply i sub 2 by x first off is that legal this is a 2 by 2 this is a 2 by 1 so yes it is and i would get x back okay and you can pause the video and do that real quick you can multiply this by this you will get x back so really what i can do at this point is i can just erase this because it doesn't really do anything it's just like if i had 1 times 5 and i said ok well this is just equal to 5. that's all i'm doing there so at this point i can say that x this unknown right here is going to be equal to a inverse times b okay so that's where we're kind of going with this and really quickly we know that we can do the operation this way because a inverse is going to be a two by two and b is going to be or is i would say a two by one so mathematically we can multiply these together if you flip the order though okay if i did it like this if i said i had b times a inverse you're going to run into a problem right because this is a 2 by 1 and this is a 2 by 2 you cannot do that operation okay so you've got to make sure that when you're setting these equations up you pay attention to kind of which order you're multiplying everything in all right so what i'm going to do now is just kind of copy these we're going to go to a fresh sheet because we're going to run out of room we'll paste this in here and i'm going to get rid of this for a second what i need to find right now i'm just going to get rid of this too i need to find a inverse to start this off so i need to find a inverse and then i'll multiply it by b and i'll know what x is so how do i find a inverse again if you didn't watch the last lesson let me give you the shortcut real quick so if we have a and let's define the entries to be lowercase a lowercase b lowercase c and lowercase d what happens is if i want a inverse all i need to do is put 1 over the determinant of a and we haven't talked about this yet but it's very easy to find this it's basically just a times d so these two multiplied together so a times d minus these two multiplied together b times c okay that's very easy then multiplied by what i'm going to do here might not make a whole lot of sense but we'll kind of get to this later on so a and d would swap so d and a notice how those positions swap and then b and c these guys are going to become the opposite so negative b and negative c okay so this is something you probably want to write down for your notes because we're going to use it a few times today and the main thing here is that don't be frustrated if you can't memorize this with kind of one go i've been using this forever so i already know it but it's one of those things that after you practice enough times it just gets committed to memory so write it down use it when you're solving your problems and you'll memorize it very quickly all right so let me just erase this i can do this from memory and so i'll say a inverse is equal to but it's going to be 1 over the determinant so this guy times this guy 8 times negative 4 is going to be negative 32 and then minus negative 7 times negative 3 is going to be positive 21 if i did negative 32 minus 21 that's going to give me negative 53 okay so that's the first part of the formula then we have to multiply it by remember we have all this stuff that goes on so these two switch so the position of this one and this one so this becomes up here and this one becomes down here and then these guys get changed into the opposite so instead of a negative seven i get a seven instead of a negative three i get a three so from this point i'm just multiplying a scalar by a matrix so very very simple let's just do this down here real quick and then we'll replace everything so negative 1 over 53 times negative 4 would just be 4 over 53. okay so that's going to be your first entry so it's going to be 4 over 53. so let me erase this and i'll put this over here so 4 over 53. we kind of slide that down because we need some problem so a inverse is going to be equal to this then the next one is negative 1 over 53 times 3 this is just going to be negative 3 over 53 and then the next one is going to be times 7 so obviously that's just going to be negative 7 over 53. and then the last one is going to be times 8 so that's just going to be negative 8 over 53 okay so let me erase this from here and let me try to make this a little bit more compact because we're going to run our rump so we're going to have 4 over 53 we're going to have negative 3 over 53 we're going to have negative 7 over 53 and we're going to have negative 8 over 53. all right so that's about as good as i can do now what we said and i have to go to kind of the next one i can erase a i don't even need it so what we need to do let's go back up we know from this equation and i can just erase all this we don't even need this anymore that x is equal to a inverse times b here's b let me grab this real quick and again you got to do this in the right order so a inverse is first b is second so let me put this in here okay so we're going to say that we have a inverse times b again this is a two by two and this is a two by one so we can do this these numbers match and it's going to be a two by one so a two by one so you're gonna have one there and one there so let's go ahead and crank this out real quick we want to start out with this kind of entry here and that's going to come from this row times this column so you're going to have 4 over 53 times negative 17 that would be negative 68 over 53 so negative 68 over 53 and then you'd have plus you have this guy times this guy so negative 7 over 53 times 13 so that's going to give me negative 91 okay negative 91 over 53. so if i add negative 68 and negative 91 i get negative 159 so this would be negative 159 over 53 and if i do that kind of calculation i'm going to get negative 3. okay so this first entry here is going to be negative 3. all right so now i just need this entry down here and all i'm going to do is get kind of this row times this call okay so negative 3 over 53 times negative 17. we know negative 3 times negative 17 would be positive 51. so this is positive 51 over 53. and then you're going to do this guy which is negative 8 over 53 okay times this guy which is 13. so that's going to give me negative 104 okay so negative 104 over 53 can add these two together so you're basically doing negative 104 plus 51 which gives me negative 53. so you know what this is going to be it's negative 53 over 53 which is negative one okay so at this point we've found our answer a inverse times b is x so let me just copy this real quick or i can copy the whole thing okay so let me paste this in here real quick and we know that a inverse times b is x so let me just kind of replace this and i can replace this and i'll just drag this up here so we have our solution right this guy represented x this guy represented y so we can say that x is negative three and y is negative one okay so let me erase all of this and we can just check this real fast i know this process is very tedious and again it's not a recommended method if you used any other method you'd be done a long time ago right but it's just something they teach so we do want to cover all right let's check this real quick if i plug in a negative 3 for x i get 8 times negative 3 which is negative 24 and then negative 7 times negative 1 would be plus 7. does this equal negative 17 yes it does and then for this one you get negative 3 times negative 3 which is 9. negative 4 times negative 1 would be 4 so plus 4 equals 13 that works as well all right let's take a look at another example so we have 3x minus 9 equals 3y we have negative 8y equals 48 plus 16x so in this case we have to do some cleaning up right we can't just start because everything we get messed up you want every equation in a x plus b y equals c and here you have some opportunities to kind of work with smaller numbers so you want to take advantage of that i'm going to for the first equation i'm going to subtract 3y away from each side and i'm going to add 9 to both sides and then i'm going to divide everything by 3. so this is going to give me x minus y equals 3. so this is going to be my first equation let me just kind of drag this up here and we're going to bring this to another sheet so don't even worry about this right now so i'm going to subtract 16x away from each side here so minus 8y and this equals 48. for this one everything is divisible by 8 right because you divide this by 8 you get negative 2 then times x you divide this by 8 you get negative y and this equals you divide this by 8 and you get 6. so you can say this is negative 2x minus y is equal to 6. okay so you can look at that and say that hey i could solve that with substitution in probably like less than a minute but we're going to do this using the kind of inverse of a matrix method which will probably take us about 10 minutes all right let's go ahead and paste this in and let's set everything up so the first thing i want is matrix a which is going to be the coefficients so you have a 1 and a negative 1. you have a negative 2 and you have a negative 1. okay then i want my matrix x which is going to be again a column vector with x and y then i want my matrix b okay let me write that here b is going to contain a 3 and a 6. okay again another column vector so we already know that this sets up to be a times x is equal to b we've already shown that that's legal and then we also know that if i multiply this side over here by a inverse and again slide this down as long as i do it on the left side i am good to go right a inverse times b because again this is a two by two this is a two by one so this way is legal if i flip it around it's going to be illegal okay it's not going to be defined so a inverse times a is the identity matrix and again it would be a two by two and this is a two by one so that's legal so i would get x back so x would be equal to this so i need to find the inverse of a multiply by b and again i'll have my solution for x so let's copy a real quick and go to the next sheet and we're going to quickly find the inverse now that we know the method so again it's 1 over the determinant multiply this by this 1 times negative 1 is negative 1 then minus this times this negative 1 times negative 2 is positive 2. if i have negative 1 minus 2 that's going to be negative 3. so this is negative 1 3 times interchange these two so what i want to do again this one and this one they change positions so this becomes negative one and this becomes positive one and then what i'm going to do for these for this one and this one i change the signs so this one becomes positive one and this one becomes positive 2. okay so again when you see that it might not make sense the first few times but then after you practice you're going to very quickly memorize this kind of formula it's a very good shortcut all right so if i multiply everything by negative 1 3 i'm going to get basically positive 1 3 here i'm going to get negative 1 3 here i'm going to get negative 2 3 here and i'm going to get negative 1 3 here so this is the inverse of a so this is a inverse and let's just move this over here and i'm going to erase a at this point i don't need it anymore it serves me no purpose i need to go back up and get b so let's copy this and let's go back down and paste this in and so now i want a inverse times b again in that order because it's only defined in that order so a inverse times b again this is a two by two and this is a two by one so the dimensions will be two by one okay so let's go ahead and set this up we do this row times this column so you have one third times three and then plus you're gonna have negative one third times six so what's that going to give me one third times three is one and then you have negative one third times six this cancels with this and gives me 2. negative 1 times 2 will be negative 2 so this is negative 2. so what's 1 plus negative 2 that's negative 1. so this is negative 1 and then for this one and i shouldn't put that bracket in there for this one i'm going to go this row times this column so negative two thirds times three plus negative one third times six and of course this is going to cancel you'll have negative two and this cancels with this and gives you two negative one times 2 is negative 2 and so this is going to give me negative 4 right negative 2 plus negative 2 is negative 4. so we found our solution we know x is negative 1 and y is negative 4. so let's go ahead and copy this real quick just go back up and prove that to you real fast so let me just paste this in here real quick so we know that a inverse times b is x so let me erase this and i'll just slide this down okay so we can record our solution we know this one is x so x is negative one we know this one is y y is negative four just going to erase everything at this point we don't need it anymore it's going to list our solution and be done so we'll say that this is negative 1 comma negative 4 as an ordered pair again if you want to check i highly recommend it when you start negative 1 minus a negative 4 so plus 4 equals 3. yeah that works out and then this one is negative 2 times negative 1 which is 2 and then minus a negative 4 is plus 4 does that equal 6 yes it does so that's the correct solution all right so let's wrap up our lesson and look at a more challenging problem or you could say a more tedious problem we're going to look at a linear system with three variables and again we're going to solve it by kind of finding the inverse of one of the matrices that we set up now again i just want to emphasize this this is definitely not the best method to do it it's just something we're showing you because you will probably be tested on it all right so the first thing you want to do if you get a three variable system again again again you've got to make sure these equations are set up correctly first if you don't it's going to give you the wrong answer right so you've got to make sure that the x's are lined up the y's are lined up the z's are lined up the constants are lined up so that when you load the information into your kind of matrices that you're setting up everything corresponds correctly okay so you're going to have a column with the coefficients for x's then a column with the coefficients for y's you know so on and so forth so in this format we have everything as ax plus by plus cz equals d so we're good to go but again if it wasn't like that you'd have to set it up correctly now what i'm going to do is i'm going to make a coefficient matrix just as i always have i'm going to call it a and so i'm grabbing all the coefficients so this has an implied coefficient of 1. this is going to be negative 3 this is going to be positive 2. so that's from the first equation that's this one that's this one and that's this one okay from the second equation i would have a two i would have a negative five and i would have a positive three okay so again just to highlight that's from here here and here okay and then from the last one i've got a three a negative 6 and a 4 and again that's from this one from this one and this one okay so all i did was i set up a coefficient matrix and i'm just calling that matrix a now next i'm going to set up a matrix that's going to represent the variables of the system we're going to call that capital letter x and so that's going to have three variables x y and z so x y and z so this is just a column vector and then lastly i'm going to have my matrix b okay so i'm going to write that up here so b is going to be a column vector with my constants so again it's going to be negative 24 it's going to be negative 39 and it's going to be negative 40 not okay so we know from the previous two examples that if i do a the coefficient matrix times x the variable matrix i can set this equal to b right this kind of matrix with my constants in it and that would represent kind of our system right so you can prove that to yourself you can stop the video and you can multiply a times x and set this equal to b and you'll see that you get the same thing all right so what we want to do now is kind of set up the solution for this and we already know how to do it so let me kind of slide this down just a little bit so we have a little room to work it's going to be pretty tight so we already know this is going to set up as a inverse times a times x and you don't need those dots there but i'm just putting them there for emphasis is equal to a inverse times b okay now we know that a inverse times a is going to give me an identity matrix and in this case it's going to be a 3 by 3. so this would be an i sub 3 times x is equal to a inverse times b now we already know that if i multiply a 3 by 3 identity matrix by a 3 by 1 matrix x i'm just going to get x back okay so i can really erase this i don't need it anymore and i can just declare that x is equal to a inverse times b of course we already knew this from kind of previous examples but again it's just good to go over it again so really all i need to do here is find a inverse multiply it by b and then i'll have my values for x y and z okay so let's copy a so let's find the inverse first that's going to be the hardest part all right so what we want to do right now is find the inverse of a again we talked about this in the last lesson for a two by two matrix if it's got an inverse it's very easy to find there's a shortcut for a three by three there's a variety of methods that you can use we'll talk about another method kind of later on after we talk about determinants for now it's a little bit beyond us but the method that we know the method that we learned in the last kind of lesson that works for any kind of invertible square matrix what we do is we set up an augmented matrix and it's going to look something like this okay so let's say a is an n-by-n matrix in this case it's a 3x3 and then you have i which is the identity matrix in this case it would be a 3x3 so you're setting up an augmented matrix where a is on the left the identity matrix is on the right you're going to use row operations to put it in this form so you want the identity matrix on the left and your a inverse is going to be on the right and again we talked about why this works in the last lesson so let's go ahead and set this up real quick and we'll get our solution so i'm going to start with just this guy right here i'm just going to erase the a equals i'm going to put a vertical bar here and i'm going to put a 1 0 0 a 0 1 0 and a 0 0 1 okay so this is just the identity matrix on the right and this guy on the left is our a the matrix we're trying to find the inverse of so we're going to use our row operations to put this side you could say in reduced row echelon form or we could say it's going to become an identity matrix it's just going to match this over here so the very first thing we want to do if we follow the method that's given in every textbook is we want to get a 1 there and we already have that so we don't really have to do that and then based on that 1 we're going to get our zeros so we want a 0 here and we want a 0 here okay so what i'm going to do is i'm going to multiply top row here by negative 2 and i'm going to add the result to row 2. so i'm going to notate that by saying i have negative 2 times row 1 and i'm going to add the result to row 2 that's what i'm going to replace row 2 with so let me just go through and multiply negative two by everything in row one so negative two times one is negative two negative two times negative three is positive six negative two times two is negative four negative two times one is negative two and then negative two times 0 is going to be 0 in each case so for both of those okay now what i'd want to do is add each of these to the kind of entries in row 2 that's what i'm going to replace row 2 with so negative 2 plus 2 is 0. then 6 plus negative 5 is 1 then negative 4 plus 3 is negative 1 then negative 2 plus 0 is obviously negative 2 and then in each case i'm adding 0 to something so it's not going to change so that part's done so now i move to this row right here and i want this to be a 0. so what i do is i multiply row 1 okay row 1 by negative 3 okay and then i add the result to row 3 that's what i'm going to change row 3 with okay so let me just multiply everything by negative three first get that out of the way negative three times one is negative three negative three times negative three is positive nine negative three times two is negative six negative three times one is negative three and of course negative three times 0 in each case is 0. okay so now we're going to add each of these to the entries in row 3. so this is going to be a 0 9 plus negative 6 would be 3 this is going to be a negative 2 right 4 plus negative 6 would be negative two and then in each case here and here i'm adding to zero so it's just going to be negative three and zero and then here zero plus one is just going to be one right so this is going to be my row third so now we're going to move to this column here and conveniently we already have a 1 again so we don't have to do anything there we just need to use that 1 to get our 0 here and our 0 here okay so i'm going to multiply row 2 by i'm going to get rid of this one first i'm going to make this a 0. so i'm going to multiply row 2 by positive 3 and i'm going to add the result to row 1. that's what i'm going to replace row 1 with okay all right so let me multiply everything by 3. so i'm going to go ahead and say 0 times 3 is 0 and then 1 times 3 is 3 and then 3 times negative 1 is negative 3 3 times negative 2 is negative 6 3 times 1 is 3 and then 3 times 0 is 0. so let's add everything so zero plus one is still one three plus negative three is zero negative three plus two is negative one negative six plus one is negative five and then we have three plus zero which is three and then obviously zero plus zero is zero so that takes care of that now we move into kind of making this into a zero so multiply row two by negative three add the result to row three that's what i'm going to change row three with so negative three times zero is zero negative three times one is negative three negative three times negative one is three negative three times negative two is six negative three times one is negative three and negative three times zero is zero so let's add now zero plus zero is zero so i don't have to do anything there negative three plus three is zero we have three plus negative two which is positive one now we just have six plus negative three which is positive three we have negative three plus zero which is negative three and of course 0 plus 1 is 1. all right so looking pretty good so far as we move to the next column again conveniently we already have a 1 so we just need to get a 0 here and here now we have the same number in each case we have a negative 1 and a negative 1. so whatever i do to make this into a zero we'll also make this into a zero so we can kind of do this as one step i'm going to break it up into two just so you don't get lost but you could do it as one step so really what i need to do is just multiply row three by one okay or you could just say you have row three and add the result to row two that's what i'm going to replace row two with and again it's going to be the same thing i'm gonna multiply row three by one and add the result to row 1 that's what i'm going to change row 1 with again because in each case 1 plus negative 1 would give me 0 there now if you want to write a 1 there just for emphasis that's fine but it's the same thing either way so let's start with kind of this one i'm just going to add the entries in row 3 to row 2 that's what i'm going to replace row 2 with so 0 plus 0 is 0 no change there 0 plus 1 is 1 no change there 1 plus negative 1 is 0 okay and then 3 plus negative two is going to be one negative three plus one is going to be negative two and then one plus zero is one okay so that one's done and then lastly we're going to have zero plus one which is one zero plus zero which is zero one plus negative one which is zero and then we have three plus negative five which is negative two we have negative three plus three which is zero and we have one plus zero which is one okay so now we're finished and essentially you see that you have the identity matrix you have an i sub 3 here on the left and you have your a inverse on the right okay so we can just copy that real quick a inverse is going to be equal to you're going to have this negative 2 0 and 1 then you're going to have 1 negative 2 and 1 and then to make my bracket big enough and then 3 negative 3 and 1. so let me copy this real quick and we'll go to a fresh sheet and let's paste this in here real fast all right so let's get rid of all this now we need to let me kind of go back up we need to multiply a inverse which we just found times b and that's going to give me x these three entries here that are going to give me the solutions for my system so let me copy b real quick and let me paste this in and remember we've got to do this in the correct order if i do a inverse times b it's defined if i do b times a inverse it's not defined okay it's not going to work so a inverse times b is going to be equal to what well again i know that this is a three by three and this is a three by one so it's going to end up being a three by one so you can have one two three entries here let me make that a little better so for the first entry let me kind of slide this up just a little bit for the first entry what i'm going to have is this first row times this column so in each case let me just kind of set this up so we can blow through this in each case when i'm multiplying a row by a column this isn't going to change so in other words i'm going to have negative 24 i'm going to have negative 39 and i'm going to have negative 49 okay so let me just kind of wrap these and i'm going to copy this real quick and i'm going to paste it twice right there and i'm going to paste that again just to save us a little bit of time okay so what's going to happen is i'm going to multiply again this row by this column so you'd have negative 2 times negative 24. so let's put negative 2 here and plus you're going to have negative 39 times 0 and then negative 49 is going to be multiplied by 1 okay so we can do that real quick negative 2 times negative 24 is 48 this is 0 okay so you have 48 here this is going to be negative 49 so this is gonna be negative one so that is your first entry so this is negative one then for this one again you're doing this row times this column so one times negative 24 is negative 24 then plus you have negative 2 times negative 39. we know that's positive and 2 times 39 is 78. so let's just say this is 78 and then you have this 1 times this negative 49. so that's just negative 49. so what does this come out to if i do negative 24 plus negative 49 i get negative 73. if i add that to 78 i get positive 5. so this will be 5. and kind of get rid of these bars at this point i just have this one left so this row times this column so you have 3 times negative 24 that's negative 72 okay so negative 72 then plus you'd have negative 3 times negative 39 which is positive 117 so let's put 117 and then lastly you have 1 times negative 49 so let's put plus negative 49. okay so what is negative 72 plus negative 49 that's going to give me negative 121. so i'm going to put plus negative 121. and if i do 117 plus negative 121 i get negative 4. okay all right so again it takes a lot longer to do it this way i think it's kind of cool if you do this just a few times just to kind of get the concept down but then again it's not an efficient way if somebody gave you the inverse to start then maybe it'd be pretty quick to use this because all i'm doing is some very simple matrix multiplication to get my answer but if you don't have the inverse if you have to calculate it even with the shortcut that we're going to learn kind of in a few lessons it's still going to be a very tedious process and it's going to be easier to use other methods like cramer's rule which we'll be talking about here shortly all right so let me kind of copy this and let's go back up and i don't know where i'm going to fit this let's just kind of paste this in here for a second and let me just erase some stuff here we know that x is equal to this so i'm just going to erase this and just kind of slide this down we know that this first entry was x this was y and this was z okay so x is negative 1 y is 5 and z is negative 4. so let's erase everything now we need anything anymore we've found our solution so as an order triple i'm going to say it's negative 1 it's 5 and it's negative 4. okay and we can just check this real quick just to make sure we didn't make a mistake so in this first one you'd have negative 1 you would have minus 3 times 5 then plus 2 times negative 4 should be equal to negative 24. well if i do my multiplication first this is 15 and this is negative 8 so you can say that you have negative 1 minus 15 which is negative 16 minus 8 which is negative 24. so this one's good all right for the next one we have what we have 2 times negative 1 which is negative 2 minus you have 5 times 5 which is 25 plus you have 3 times negative 4 which is negative 12. this should equal negative 39 negative 2 minus 25 is negative 27 and if you add negative 27 and negative 12 you do get negative 39. this one's good to go so for the last one you have 3 times negative 1 which is negative 3 and then minus 6 times 5 which is 30 and then plus 4 times negative 4 which is negative 16. this should equal negative 49 and of course it does right if you do negative 3 plus negative 30 you get negative 33 and negative 33 plus negative 16 is negative 49. so it works in all of them again not the fastest way to kind of go about this but it's good to have the concept down it's good to understand how you can kind of solve something using this method again if somebody gave you an inverse to start it would be a decent kind of technique to use but if they didn't give it to you it would probably be a much slower method to use in this lesson we want to talk about finding the determinant of an n by n matrix so when we work with a square matrix or again an n by n matrix a matrix with the same number of rows as columns we're going to have a real number that's associated with that matrix that's known as its determinant so this is a very special number for us we use it all the time when we're working with matrices and we already saw that when we work with the inverse of a two by two matrix there's a little shortcut that we can use that saves us a lot of time it involves taking the determinant of the given 2 by 2 matrix in order to kind of find the inverse okay so that's part of the formula so what we're going to do today is just go very deep into this topic and we're going to show you how to find the determinant of any n by n matrix i'll show you the shortcut that you can use for a three by three matrix and after this in our next video i'll show you the shortcut that you can use for the four by four matrix okay it would be too much to put in one video so we're going to kind of break it up all right so we're going to start out with the 2 by 2 matrix which is the easiest scenario so let's say we start out with this matrix a and it's equal to so in the first row you have this lowercase a and this lowercase b as your entries in the second row you have the lowercase c and the lowercase d as your entries basically if i want to ask for the determinant of this matrix a i can put these vertical bars around a okay that's one way you can ask for the determinant or you could also say d e t of a like this so depending on which book you're using you'll see different notation just follow whatever your book is doing or whatever you're doing in your class okay but finding the determinant for a two by two matrix is very simple you'll see i already have the formula written it's a times d minus b times c so essentially you just think about multiplying down like this and then you can multiply up like this or you can multiply down like this it doesn't really matter most books will go like this okay but realize the fact that c times b and b times c those are the same right because we're working with real numbers so multiplication is commutative so it doesn't really matter how you list that if you want to say it's a times d minus c times b or if you want to say it's a times d minus b times c it's the same result in the end so let's look at an example real quick so we have a is equal to we have 2 and negative 1 in the first row and 5 and 6 in the second row so the determinant of a okay would be equal to and another way you might see this you could put vertical bars around the entries so 2 negative 1 5 and 6 and this is how you would ask for the determinant of something if you didn't have a name for the matrix okay so that's one way you would see that so the determinant of this is what again i'm just going to multiply down so what is 2 times 6 that is 12 and i'm going to subtract away again you can multiply down or up it doesn't really matter i'm just going to go ahead and multiply down so negative 1 times 5 is negative 5. and again notice if you did 5 times negative 1 it's also negative 5 right so it doesn't matter 12 minus a negative 5 is 12 plus 5 so 12 plus 5 and that's 17. so the determinant of a is 17. all right so let's look at one more of these and then we'll move on to a 3x3 so for a it's equal to we have 3 and 8 in the first row and 7 and negative 4 in the second row so the determinant of a is equal to again multiply down 3 times negative 4 is negative 12 then minus i'm going to just multiply up this time so 7 times 8 is 56 so what is negative 12 minus 56 that's going to be negative 68 okay so that's my determinant of a all right so obviously that's a very easy process when you get into a three by three or higher if you use the kind of book method it can be a little bit tedious but it's really not that bad it's definitely not as bad as finding the inverse but luckily for us again we have a shortcut for a 3x3 which is really easy to use it takes under a minute to calculate the determinant of a 3x3 matrix so it's a really good shortcut and again in the next lesson i'll talk about the shortcut for a four by four all right so first and foremost before we get into this kind of official method we got to talk about something known as a minor and a cofactor okay so these are terms that you're going to use even outside of talking about finding the determinant so it's important to understand what they are so i'm going to highlight this word right here so if a is a square matrix the minor which is notated with this capital m sub i j remember this sub i j this is the ith row j column that's just generic notation to say i'm at a given position in the matrix so then of the entry you have this lowercase a sub i j again that just tells me i'm at some given entry in this matrix a what we're saying is the determinant of the matrix obtained by deleting the ith row and jth column of a okay so remember this generic notation sub i j here just tells you you're at a given position some given row and column so all this is telling you to do if you want to find the minor of some given entry delete the row it's in delete the column it's in and find the determinant of the matrix that is kind of left after you've done that so to see this with a generic example this is something you probably want to write down because it's very important so if you have capital letter a to name this matrix and it's equal to you've got all lower case letters here it's entries a b c d e f and then g h i if i wanted to find the minor of this kind of entry here this lowercase a what i would do is delete this column and i would delete this row okay so the matrix that's left is made up of these elements you have e you have f you have h and you have i okay so if i wanted to find the determinant of that again i can put these vertical bars around it okay and then i can just say it's equal to what e times i you multiply down so e times i minus you'd have f times h so f times h so this right here is the minor okay the minor of this a so i can say m capital m sub 1 1 okay some people put a comma between them and say it's row 1 column 1 is going to be equal to this right here this e times i and then minus f times h and i know this generic kind of stuff will trip some students up but i just want to give you a generic example first and then we'll get into some ones with some numbers involved so let me just do one more here okay let me do the minor of this element here so this lowercase b so what's the minor again i'm going to delete this column in this row what's left it's this d this g this f and this i okay so i'm just going to put vertical bars around this to say i want the determinant that's d times i d times i minus f times g so f times g so that's my determinant so the minor capital letter m sub again i'm in row one but now i'm in column two so you can put a comma there if you want or you can leave it as one two because it's clear this is equal to this so it's d times i minus f times g okay so very generic and now what i want to do is just talk about a cofactor real quick and then we can move on and we can talk about some examples with some numbers and i'll show you how to use this to calculate a determinant so what you're going to see in your book is that the cofactor now you have a capital c sub ij of the entry again this is just some generic entry in our matrix a so lowercase a sub ij is this negative 1 raised to this power so i plus j remember i is the row okay generically and j is the column okay so i'm just going to put col to abbreviate column so negative 1 to the power of i plus j and then it's multiplied by the minor remember this is the capital letter m sub ij okay so what you need to understand here if i go back to this example here let me just kind of erase this real quick and we already have our minor so we have our minor for what this is the minor for again this element here which is in row one column two and let me slide this down i want to make this crystal clear okay this is row 1 this is row 2 this is row 3. this is column 1 column 2 column 3. so i'm in row 1 column 2. okay so that's what that 1 comma 2 tells me so if i find the sum of those two numbers 1 plus 2 that's 3. so the cofactor capital c sub 1 2 is equal to negative 1 raised to the power of this 1 plus 2 the row number plus the column number i'm in row 1 column 2 1 plus 2 is 3. so it's raised to the power of 3 okay so we know that if we raise this to an odd power it's going to be negative 1 so it's going to change the sign of this guy right here this d times i minus f times g okay so if i get a result okay for the exponent that's odd it's going to stay negative if i get a result that's even it's going to be positive okay so for example let me erase this real quick if i go back to the earlier example where i found the kind of minor of a this lowercase a that m sub 1 1 well this was equal to what it was equal to we delete this and this it was e times i right minus f times h okay it's just the determinant of what's left so this guy right here so e times i minus f times h so it's this now the cofactor okay would be equal to what it's negative 1 raised to the power of what you're in row 1 column 1. so you can just look at these numbers down here 1 plus 1 is 2. so raised to the second power i know this is even so this is going to be positive 1 so it's just going to be the same as the minor in this case okay so to make this really easy you can make yourself a little matrix with the signs involved so that you don't get confused okay this is going to come up pretty often so when you talk about cofactors you have to have the signs down so again if you just look at this i know this is confusing at first but if you really think about it it kind of starts to make some sense so this is row one column one right here one plus one is two right so this is an even number then this is row one column two so one plus two is three that's an odd number so that's why this is negative and this is one 3 right row 1 column 3. 1 plus 3 is 4. so that's why this is positive these signs are going to alternate right if this is negative this is positive this is negative this is positive this negative is positive okay so they alternate going across they also alternate going up and down right positive negative positive negative positive negative positive negative positive so really if you just remember the first one is positive you can go through and make a little matrix of signs so that you can figure out what the cofactor is all right now let's look at an example real quick and this is going to lead to something so i'm going to ask you to find the cofactor okay for every number in the kind of first row up here okay so let's start by calculating the minor first of each one and then we'll think about the cofactor okay remember that's just thinking about negative 1 raised to the power of the row plus the column okay times the minor that's all it is so what is let's start with the minor what is the minor of this element here i'm going to delete this and i'm going to delete this the matrix that's left is a 2 and a 4 in the first row and a 0 and a 1 in the second row so again the vertical bars tell me i want the determinant so 2 times 1 is 2 minus 4 times 0 is 0. so the minor here is going to be 2 minus 0 which is 2 okay so let me just write this down here real quick we're going to say that m sub 1 1 is equal to 2. so the next thing is what's the cofactor so what is c sub 1 1 that's equal to what again it's negative 1 raised to the power of in this case this is row 1 row 2 row 3 this is column 1 okay column 2 and column 3. you really don't need that because you could just sum these numbers down here okay but it's in row 1 column 1. 1 plus 1 is 2. so again this is raised to an even power so we know it's going to be positive one so the sine of the minor is not going to change when it becomes the cofactor in this case so this is going to be equal to 2 as well okay and we don't need the minor we're just going to keep the cofactor so let's move this up okay and let's talk about the cofactor of this one so again i'm going to find the minor first so delete the column delete the row what's left i'm left with 7 6 4 and 1. so 7 times 1 is 7 minus 4 times 6 is 24. if i do 7 minus 24 i get negative 17 okay so that's going to be my minor my minor is negative 17. so m in this case i'm looking for the minor of this guy right here it's going to be in row one and column two okay so row one column two so this guy is going to be a negative seventeen but if i want the cofactor of this so c sub 1 2 this is equal to what it's negative 1 raised to the power of again use these numbers down here it's row 1 column 2. row 1 column 2. so 1 plus 2 is 3. this is an odd number so this is going to stay negative 1 times the minor okay which is negative 17 so this becomes positive 17. okay so this is positive 17. all right so let's erase this and let's drag this up all right so the last thing we want to do is find the cofactor for this entry right here so again i'm going to delete this column and this row what's left i have 7 i have 6 i have 2 and i have 0. so what's the determinant of this guy well i'm going to have 7 times 0 which is 0 and then minus i'm going to have 2 times 6 which is 12. so it's going to be 0 minus 12 which is negative 12. so let's erase this so my minor again we're in row 1 and now we're in column 3. this is going to be a negative 12. so the cofactor c sub 1 3 is what we know that we're in row 1 column 3 okay so 1 plus 3 is 4 negative 1 to an even power would be positive one so the sign of this is not going to change okay so let's drag this up and again you can always reference the sign chart that we have here so this one's positive negative positive okay and if we look back we see that's exactly what happened this one was positive this one was negative and this one was positive right so this sign didn't change this one did and this one didn't okay so now that we have these cofactors let's talk about the method to find the determinant of any n by n matrix this would work no matter what you're working with okay so if it's a three by three a four by four a two hundred by two hundred whatever you're working with you can use this method so what you want to do is you want to multiply each element in any row or column so it doesn't matter which row you use doesn't matter which column you use but you're going to just use one of them okay so you're going to multiply each element in any row or column of the matrix by its kind of cofactor then the sum of these products would give us the value of the determinant so in this particular case i already have my cofactors for this whole row okay so what i'd want to do is i'd want to multiply 5 by its cofactor so 5 times 2 then plus negative 1 times its cofactor which is 17. let me change this notation up a little bit i want to be consistent so if i use a parenthesis there i want to use a parenthesis here just so you don't get confused then plus i'm going to have 3 times its cofactor which is negative 12. okay so if we go through and do this operation we're going to end up with our determinant 5 times 2 is 10 and then plus you'd have negative 1 times 17 which is negative 17 and then plus 3 times negative 12 is negative 36. okay so what is this going to give me 10 plus negative 17 is obviously negative 7 and then negative 7 plus negative 36 is going to give me negative 43. so this is my determinant of a okay so if i erase this i can say that my determinate of a is going to be negative 43. okay let me slide it down just a little bit just right here now before we go any further let me just show you the shortcut okay and we'll do some mixed practice we'll kind of practice it each way and i'll give you a generic formula that you can memorize in case your teacher tells you that you have to do it the long way okay so for the shortcut let me kind of move this out of the way for now for the shortcut all you need to do is copy the first two columns okay so column one and column two just copy them to the right of this so five seven and six and then negative one two and 0. okay so now what i want to do is i want to multiply the numbers starting at this top left okay i want to multiply down this diagonal so i'm going to do 5 times 2 which is 10 times 1 which is still 10. okay so that's 10. then i'm going to keep going so i'm going to add to this i'm going to multiply these numbers going down this diagonal so negative 1 times 4 is negative 4 times 6 is negative 24. and then i'm going to add to this i've got one more to do 3 times 7 times 0 we know that's 0. so you can just erase this okay now you have to be very careful what you're going to do next put brackets around this so you don't make a silly sign mistake you're going to now subtract away the whole thing so put brackets you're going to go up now so starting at the bottom left you're going to go up so 6 times 2 is 12 times 3 is 36 and then plus you're going to go up again we know there's a zero involved in that multiplication so that would be zero right zero times four is zero zero times five is zero and then here you're going to go up one last time so 1 times 7 is going to be 7 times negative 1 is going to be negative 7. okay all right so 10 plus negative 24 is negative 14. and then 36 plus negative 7 is going to be 29 okay so what you're going to have here is negative 14 minus 29 which is also negative 43 okay so either way you find that the determinate is negative 43. so you can see how much faster that method is it does take a few tries to kind of memorize it okay and we'll practice this much more throughout this lesson let's go ahead and look at another example so looking at this example i want to do another example where we talk about the cofactors again and then we'll practice the shortcut so let's go ahead and do this row again so i'm going to take this row right here so i'm going to start with this guy and i'm going to find its cofactor so i'm going to eliminate this and this and so i'm going to have the determinant of 0 8 3 and 7. what's that going to be 0 times 7 is 0 minus 8 times 3 is 24 so this is negative 24. so if i look at that one remember this guy is in row 1 column 1. so my sign there 1 plus 1 is 2 that's an even number sine won't change right because negative 1 raised to the second power is going to be positive 1. so i would say c sub 1 1 is equal to a negative 24 okay and then i'm going to find this one so i'm going to delete this and this so i want c sub i want again row 1 column 2 it's equal to the determinant of this guy you would have 6 8 negative 2 and 7. so 6 times 7 is 42 and then minus you have 8 times negative 2 which is going to be negative 16. this is plus a positive right because you have minus a negative so this is going to end up being 42 plus 16 which is 58 okay so this is 58. so i'm going to put 58 here but remember i've got to multiply this minor that i just calculated by negative 1 raised to the power of 1 plus 2. 1 plus 2 is 3. 3 is an odd number so this is going to be negative right negative 1 raised to the third power is negative 1. so this would be negative 58 there then lastly let's look at this guy right here so i'm going to delete this column and this row so c sub 1 3 is going to be equal to we would take this determinant so we have 6 0 negative 2 and 3. so 6 times 3 is 18 and then minus you'd have 0 times negative 2 which is obviously 0. so this is going to be 18 and again the sign isn't gonna change here because if i look at my numbers down here i'm in row one column three one plus three is four four is an even number right negative one to the power of four is positive one so this will stay as 18. so now that we have our cofactors we can multiply them by their given entries okay and then we can sum those amounts so negative 24 times negative 1 and then plus we're going to have negative 58 times 2 and then plus we're going to have 18 times 4 okay so what's that going to give us so this is positive 24. let's just erase that and put that negative 58 times 2 is negative 116. so negative 116 18 times 4 is going to be 72 okay so 24 plus 72 is 96 so you'd have 96 okay minus 116 and that would be negative 20. okay so we can say our determinant is going to be negative 20. all right so the determinant of a is negative 1. okay let's look at our shortcut again again it's a very good shortcut so copy these columns so negative 1 6 and negative 2 your first one and then copy your second one so 2 0 and 3. again i'm just going to multiply start at the top left that's easy to remember and go down so negative 1 times 0 times 7 we know that's 0. so i don't have to do anything there then 2 times 8 times negative 2 we know that 2 times 8 is 16 16 times negative 2 is negative 32 and then plus multiply down this last diagonal 4 times 6 is going to be 24 and 24 times 3 is okay so let's sum this real quick so we don't make some silly sign mistake if i do 72 plus negative 32 that's going to give me 40. so i have 40 over here remember i'm subtracting away this guy when i go up so again you're subtracting the whole thing away so if you're not doing it all in one go make sure you use brackets okay i'm just going to put a bracket there so i don't make a silly sign mistake all right so let's go up now so i'm going to start here at the bottom left and i'm going to go up i've got a 0 involved so i know any i can just skip that because 0 times anything is 0. then here going up 3 times 8 is 24 times negative 1 is negative 24 and then going up one last time 7 times 6 is 42 42 times 2 is 84. so plus 84. so what is negative 24 plus 84 that's going to give me positive 60 okay so what i have now is 40 minus 60 which is obviously negative 20. so either way you do this you get a determinant that is negative 20. okay so again a very very good shortcut overall something you definitely want to use okay so i want to do one more example but before i do it i want to return to my generic example so in case you're in a class that doesn't want you to use the shortcut i've seen teachers do this i want to give you a generic formula that is very easy to remember okay and this is based on what we've done so far and i'll explain it as i go so you want to take if you want the determinant of a you want to take this guy right here your a and multiply it by the determinant of this guy right here okay so we know the determinant of this guy right here would end up being what it's e times i minus f times h but this is probably a little easier to remember so i'm just going to leave it like this then you're going to put a minus here okay and i'll explain why in a second you're going to put a b there and then times the determinant of again what would be left if i marked this out it would be d it would be f g and i okay so that's that and then lastly you would put a plus here okay times your c and let me write this down here because i won't be able to fit that all the way then times your determinant of again what would be left your d your e your g and your h okay so this is a very easy way to remember this a lot of books will actually give you this as the shortcut and they will give you different notation based on what they're using as entries but essentially you just take this guy and multiply it by the determinant of this guy okay then minus this guy times the determinant of again if i mark this out this that's left and then plus you have this guy the c times the determinant of this guy okay so very easy to remember the reason this is a minus sign here is remember you're in row one column two one plus two is three so that's where your cofactor would end up taking the minor and making it negative okay but in this case here it's one and one one plus one is two that's even in this case it's one and three one plus three is four so it's even remember your signs alternate this one's plus this one's minus this is plus if you're talking about getting the sign for the co factor right taking the minor and applying that kind of sign change if it's needed all right so let's go back and i want to just do one more example again it's just it never hurts to do a lot of examples of these things just to make sure that you have everything covered all right so let's do it both ways okay so i'm not going to use the cofactor method i'm going to use the thing that i just told you that method so i'm going to start by taking this and multiplying it by the determinant of this okay so i'm going to say that the determinant of a is equal to negative 2 times the determinant of this so let me just write that out so 1 3 0 and 6. then minus okay minus you have this so 5 times the determinant of put the 0 and the 4 the 3 and the 6 okay and again that just comes from marking this out and this out that's what's left and then lastly you have plus this guy times the determinant of this so let me fit this down here negative 1 times the determinant you have 0 1 4 and 0. okay all right so let's calculate the determinant here 1 times 6 is 6 and then minus 3 times 0 3 times 0 is 0. so this would just be 6. so negative two times six is negative twelve so that's your first one then here zero times six is zero okay you'd have zero minus three times four is twelve so this is negative twelve so you'd have negative five times negative twelve which is going to be positive 60. so plus 60. so plus 60 okay and erase this and then what do you have here you have negative 1 times 0 times 0 is 0 then minus 1 times 4 is 4. so basically what you have is negative 1 times negative 4 which is plus 4. okay so when you use that kind of technique that i just showed you even if you don't use the shortcut it's not that bad okay if you just kind of memorize that it doesn't take that long to do it's just a few kind of practice problems away for you and you'll have that one down as well so negative 12 plus 60 is going to be 48 right positive 48 and then plus 4 would be 52. alright so i want to show you one more time the shortcut just so you have that down as well again it never hurts to practice so copy the first two columns so negative 2 0 and 4 and then 5 1 and 0. again multiply down and let me make this a little bit better because it's not going to line up so again 0 and 4 and then 1 and 0. so multiply down so start here and go down what is negative 2 times 1 that's negative 2 times 6 is going to be negative 12 okay and then plus you're going to multiply down 5 times 3 is 15 times 4 is 60 then plus we already know that's going to be a zero so you don't even need to write that negative 12 plus 60 is going to be 48 okay then minus again i'm going to put a bracket here so i don't make a sign mistake it's so easy to make a sign mistake so i always recommend putting a bracket start at the bottom left and go up so 4 times 1 is 4 times negative 1 is negative 4 and then go up you have a 0 there so don't even worry about it and then go up you have a 0 there so don't worry about it so what is 48 minus the negative 4 that's 48 plus 4 which is 52 so again either way you do this you find that the determinant is 52. so again this is a very good shortcut and we're going to use it throughout our course in the upcoming lessons we're going to talk about cramer's rule okay we're going to be using this shortcut to find our determinant very quickly and we'll see that we can finally solve a system of linear equations using these kind of matrix methods that really save us a lot of time in this lesson we want to talk about finding the determinant of a by four matrix so in the last lesson we talked about the laplace expansion method which is also known as the cofactor expansion method for finding the determinant now using this method we know the determinant for any n by n matrix any square matrix can be found as the sum of the entries in any row or you could do a column multiplied by their respective cofactors okay so we saw that for kind of working with a 3x3 this process really wasn't that bad it's a little bit tedious but it's not that bad and then we also found a shortcut for the 3x3 that made the process very very simple and quite a bit less tedious well when we start working with a 4x4 or higher we might need to kind of look at an alternative strategy because we don't have such a shortcut for that so i want to introduce a strategy that you can use in some cases it's going to be a little bit better than the kind of laplace expansion method in other cases it won't be okay but at least you have another tool in your tool kit to kind of find determinants for something that's a four by four five by five a six by six if you had to do it by hand so the first thing we need to do today is introduce kind of a new term force in our matrix algebra section and this is known as a triangular matrix so first and foremost you have a lower triangular matrix which occurs when a square matrix has all entries above the main diagonal as zeros remember when we talk about these kind of diagonal entries the row number and the column number are going to be the same so row 1 column 1 that's that entry there row two column two that's that entry there row three column three that's that entry there okay so going down those kind of diagonal entries you could say this main diagonal if i look above that the entries should be all zeros okay which is what we have here for the lower triangular matrix or you can say this matrix is in lower triangular form now similarly when we look at the upper triangular matrix this is going to occur when all the entries below the main diagonal are zeros so again here's your main diagonal and all your entries below are zeros okay so the reason we're kind of showing you this is because there's a very special property when you work with a triangular matrix okay so this could be the lower triangular form or the upper triangular form basically your determinant can be found by finding the product of those diagonal entries okay so i can just go down the main diagonal and say that my determinant for this matrix u is equal to a times b times c okay that's all it's going to be similarly if i come up here to matrix l okay i can say that the determinant let me make that a little better okay the determinant for l is going to be again a times b times c so i think you see where we're going with this we're going to give you a four by four kind of matrix and then essentially you're going to use some row operations to put it into kind of this upper triangular matrix format okay where you have zeros below this kind of main diagonal and then you can find the determinant by just finding the product of those entries along the main diagonal now before we can kind of jump into this process we have to talk about some of these properties of determinants there's a few of these i'm just going to focus on three of them the ones that involve what happens when you perform row operations on a matrix so let's talk about the first one the first one is where we swap two rows or you can say interchange two rows or you could also say this happens when you interchange two columns okay so first and foremost what's the determinant of this two by two matrix we know that we start by multiplying down one times four is four then we subtract the weight you can multiply up or you can multiply down it doesn't matter two times three is six so this gives me negative two okay so what happens if we swap two rows here well what's going to happen is the sign of the determinant will change the absolute value would be the same but the sign will change so instead of negative 2 you'll have positive 2. so if i put 3 and 4 in the first row and one and two in the second row again the sign will change it will be positive two if i multiply down three times two is six and then minus if i multiply up one times four is four so this is positive two so it's just a sign change okay so that's all it is the second one that you need to understand if we multiply a row okay a row by some non-zero real number then essentially you can just multiply the determinant by that same non-zero real number so let me give you an example of that let's say that i multiply to row one here by positive two so instead of one i have two instead of two i have four okay and i'm just doing this to row one row two will be unchanged so that's gonna be a 3 and a 4. so the effect it's going to have is is going to change this by a factor of 2. so i can just take negative 2 the old determinant multiply by 2 to get the new determinant so this guy will be equal to negative 4 and again you can prove this if i multiply down 2 times 4 is 8 and then minus if i multiply up 3 times 4 is 12. 8 minus 12 is negative 4. now this isn't going to come up today okay but you do need to know this for future kind of lessons especially if you get into a linear algebra course now let me show you the one that we're going to use the most today and this is very important we know that when we kind of perform row operations to kind of change an entry let's say we want this to be a 0 for example we can multiply one of the rows by some non-zero real number and add the result to a given row okay we can replace the row with that and the way your book will normally say this is it'll say if you replace a row with the sum of that row and a non-zero constant multiple of some other row when this occurs you're going to have no change to the determinant okay so that's very important to understand so let's say for example i went through when i multiplied the top row by 2. so 2 times row 1 okay who would that be so we would have 2 and then we would have 4 and then i added the result to row 2 and that's what i replaced row 2 with so let me kind of move this over because i'm going to run out of room it's going to put this up here for now and i forgot my plus sign and so i know that multiplying 2 times row 1 would give me a 2 and it would give me a 4. i'm going to add these two results to this bottom row and the top row will be unchanged so let me just kind of scooch this over for a second so i'm going to put 1 and 2 those are unchanged i'm not doing anything 2 plus 3 is going to be 5 and 4 plus 4 is going to be 8 okay so everyone should understand what i did there this is something we did when we did our gaussian elimination our gauss-jordan elimination again i'm just multiplying a given row which was row one by a non-zero real number which was two and then i added that result to the entries that were in this row two that's what i replaced row 2 with okay so 2 times 1 was 2 2 plus 3 was 5. that's how i got that 2 times 2 is 4 4 plus 4 is 8 that's how i got that so what we're saying is if you perform this operation there will be no change to the determinant right so the determinant should still be negative 2 and it is if we multiply down 1 times 8 is 8 and then minus if we multiply up 5 times 2 is 10 8 minus 10 is negative 2. okay so with that being said let's take a look at an example so we have this four by four matrix again the vertical bars tell us to take the determinant so what i'm going to do is i'm just going to copy these kind of numbers we'll go to a fresh sheet so we have 3 1 4 and 10. we have two negative one six and three we have zero negative five three and negative two we have one zero one and five okay so let me copy everything here and let's go to this fresh sheet and we'll have plenty of room i'm going to put my vertical bars here again to tell me that i want to find the determinant again i'm thinking about kind of this main diagonal here all the entries below i want those to be zeros okay if i achieve that through these kind of row operations then what happens is the determinant will be found as the product of those kind of entries going down that main diagonal but remember if we end up swapping two rows which we're going to do here you've got to take and make it the negative of what you find as the determinant there and again we'll see that as we go on so let's start out by just kind of working on this first column what i'm going to do is i'm going to use this 1 okay to get a 0 here and here and then i'm going to swap rows okay so let's go through that i'm going to multiply my row 4 by negative 2 and i'm going to add the result to row 2. okay that's what i'm going to change row 2 with so let me just go through and do the multiplication first so negative 2 times 1 is negative 2. negative 2 times 0 is 0 negative 2 times 1 is negative 2 and negative 2 times 5 is negative 10. okay so now if i add these to the entries in row 2 negative 2 plus 2 is obviously 0. 0 plus negative 1 is still negative 1. negative 2 plus 6 is going to be positive 4 and then negative 10 plus 3 is negative 7. all right so let's erase this and this so i'm going to use my 1 again to get this into a 0. so i'm going to multiply row 4 okay row 4 by negative 3 i'm going to add the result to row 1 that's what i'm going to change row 1 with so negative 3 times 1 is negative 3 negative 3 times 0 is 0. negative 3 times 1 again is negative 3 and the negative 3 times 5 is negative 15 okay so adding negative 3 plus 3 is going to be zero and we know that zero plus one is still one negative three plus four is one and then if we do negative fifteen plus ten that's negative five okay so good to go there let me erase this and now what i'm going to do is i'm going to swap row 4 and row 1. so row 1 will swap with row 4 okay and again if we do that we've got to change the sign of the determinant so i'm going to put a negative 1 out in front to remind me to multiply the determinant of this new kind of matrix that i'm forming by this kind of negative 1 to find the determinant of that original matrix which is up here okay so if we go back let me kind of copy this 0 1 1 and negative 5. so i can erase that and let me grab this so 1 0 1 and 5. and then i'm going to write this in here so 0 1 1 and my negative 5. okay so we're good to go and again i put that negative 1 out in front if you want to keep track of it a different way you can you can put a negative on top as long as you remember that you know when the end process we multiply down that diagonal you've got to take that and then multiply by negative 1 to find the determinant of again that original matrix you started with all right so let's look at our main diagonal now so that's this guy right here and i want to make that negative 5 into a 0 and that 1 into a 0. so let's start with that negative 5 because it's a little bit harder to do so what i'm going to do is i'm going to multiply row 4 i'm going to multiply row 4 which has a 1 in it okay by positive 5. i'm going to add the result to my row 3 that's what i'm going to change row 3 with so 5 times 0 is 0. 5 times 1 is 5. 5 times 1 is 5 again 5 times negative 5 is negative 25 so let's do our addition now zero plus zero is zero five plus negative five is zero we know that five plus three is eight we know that negative twenty-five plus negative two is negative twenty-seven okay and let's erase this and now i want to make this guy into a zero i can easily do that by just adding this row two to row four okay that's just like me saying that i have one times row 2 plus row 4 and that's what i'm going to replace row 4 with because again if i had negative 1 and i added it to 1 that would give me 0. so 0 plus 0 is obviously 0. negative 1 plus 1 is 0. we have 4 plus 1 which is 5 okay and then we have negative 7 plus negative 5 which is negative 12. so we're good to go there and now all i need to do again if i look down these kind of diagonal entries the last thing i need is to change the five into a zero but you might be tempted to use this kind of one here right because it's easy to say okay if i multiply row one by negative five add the result to row four this guy would be a zero but you have a problem there because this is a one okay so that's not going to work because you're going to end up changing this it won't be a 0 anymore so what you're going to have to do is use row 3 here so i'm going to figure out what i need to multiply row 3 by that when i add row 3 to row 4 okay this guy ends up being a 0. so in other words 8 this entry right here times what let's just say it's x is equal to negative 5 which is the opposite of that number there okay so if i divide both sides by 8 i find that x is negative 5 8. so i would say negative 5 8 times my row 3 this row right here plus my row 4 that's what i'm going to replace row 4 with so i know negative 5 8 times 0 n times 0 again would be 0 and 0. negative 5 8 times 8 would give me negative 5 right we already know that and then negative 5 8 times negative 27 we know that would be positive what is 5 times 27 well that's 135 okay so this would be 135 over 8 okay so now let's go through and add so obviously zero plus zero in each case is going to be 0. you can get rid of that negative 5 plus 5 is 0. you can get rid of that so really the only thing we have to kind of work on is this so we're going to add negative 12 and so to get a common denominator i would say this is negative 96 over eight so if i do 135 minus 96 i'm going to get 39 so this would be 39 8. okay so that's going to be that entry there so 39 8. okay so nice and easy not too bad so now finding the determinant is pretty easy because we just go down this kind of main diagonal because everything below is a zero okay we can just multiply those entries okay again this is when you're in upper triangular form which is what we have here now don't forget we swapped rows so that's why we have this negative 1 out here so i'm going to lead with that and say that my determinant would be negative 1 times 1 times you've got negative 1 times you've got 8 times you've got 39 8. okay so what's going to happen is you have a negative 1 here and a negative 1 here those will basically cancel that's positive so you can cancel this 8 with this 8 and you have 39 as your answer so the determinant is positive 39. so if we go back we could say this is equal to positive 39. in this lesson we want to talk about how to find the transpose of a matrix alright so finding the transpose of a matrix is a very very simple concept we're just going to jump right into an example so suppose i have matrix a and it's a two by two matrix two rows and two columns with the first row being negative nine and eight and the second row being 4 and 2. so if i want to find the transpose of this matrix a i would take that name a and i would put a superscript t up here and it's going to be equal to all we need to do is take the first row which again consists of negative 9 and 8 and make that the first column in the transpose so this will be negative 9 and 8 like this and then similarly the second row will become the second column so 4 and then 2. so basically when you see the definition for this in your textbook what they're trying to tell you is that the transpose of a matrix is formed by kind of swapping or you could say interchanging the rows and columns so in other words if i kind of go back through this this first row became this first column this second row became this second column okay or you could also say this first column became this first row this second column became the second row so it's very very easy to do this let's take a look at another example so here we have our matrix a and we want to find the transpose now this is a good point to stop in the last example we saw a two by two matrix which is a square matrix so when you look at the order of the transpose and you look at the order of kind of the matrix you started with it's going to be the same but if you look at kind of the order of the transpose of a non-square matrix it's going to be reversed so what do i mean by that well this is a two row and three column matrix it's a two by three well what happens is when i find the transpose the a with the superscript t it's going to be a three by two because this row will now become a column and this row will now become a column so it's going to have three rows and two columns so let's see this real quick so i'm going to take this guy right here and make it a column so negative 1 5 and negative 7. so the first row became the first column and then i'm going to take this second row and make it my second column so it's going to be 0 8 and 9. okay it's just that simple and now let me kind of move this over here so this is a 2 by 3 and this guy again the order is reversed so it's a 3 by 2. so again just to go through this this first row became this first column this second row became the second column or you could also say that this first column became this first row this second column became this second row and this third column became this third row let's look at one more for the final one again we have our matrix a and this guy is going to be a three by four right it's a three by four we've got three rows and four columns so again when i find the transpose that order is going to be flipped it's going to be a 4 by 3. so the transpose of a again we're going to put a with a superscript t here is going to be equal to again the way i always do it is i just take the row the first row and make it the first call that's how i go about it i take the second row make it the second column third row make it the third column but again it's just as valid we'll just kind of do this in the opposite way now let's take the first column and make it the first row so it would be 5 11 and then five okay and then take the second column make it the second row so nine three and six and then take the third column make it the third row so zero negative two and zero and let me kind of slide this up just a little bit and then for the last kind of column we have here the fourth column it'll be the fourth row so you're going to have negative 4 you're going to have 44 and you're going to have 19. okay so that would be your transpose of your matrix a okay so remember the first row became the first column the second row became the second column and the third row became the third column or again you could say the first column became the first row the second column became the second row the third column became the third row and the fourth column became the fourth row in this lesson we want to talk about how to find the adjoint of a matrix so when we talk about the adjoint of a matrix which is also referred to as the classical adjoint if you're in a linear algebra course or the adject it's found using kind of two easy but i would say tedious steps so the very first thing you want to do is make a matrix of cofactors now we talked about cofactors when we did our lesson on determinants and specifically when we looked at our method known as the kind of laplace expansion method or the kind of expansion by cofactors method now after you've kind of got your matrix of cofactors the next thing you want to do or the second part of the process is to find the transpose of that cofactor matrix so like i said it's two easy steps because you're basically just doing some arithmetic but it's quite tedious especially if you get into something like a three by three a four by four five by five these are kind of matrices that when you find the adjoint for them it takes a little bit of time so without further ado let's just look at an example real quick and we're just going to start out with something that's very generic and i'm going to show you that you've seen the adjoint of a matrix already you just didn't know it so first we have our matrix a and it has a and b in the first row and c and d in the second row now we already said that the inverse of a the little shortcut method was 1 over the determinant of a okay so that's why i have those vertical bars surrounding that a there times it's going to be the adjoint of a okay so this is the adjoint of our matrix a what we're going to do is we're going to swap these two so i'm going to swap a and d so d comes up here a goes down here and then these two the c and the b i'm going to make them into their opposite so i'm going to put a negative b and a negative c just to say hey whatever we have in those positions we're going to take the negative of it or just change the sign now this is the a d j of a or again the adjoint of a or the adjective of a however you want to think about that so where did this kind of come from well again what we could do is we could start out with this guy right here we could make a matrix of cofactors and then we can take those cofactors that matrix and we can find the transpose of that guy and it will give us this exactly so to see that real quick let me just kind of copy this let's go to a fresh sheet and i'm just going to paste this in and let me just kind of clean this up a little bit so what we're going to have here i'm going to put a d j of a is equal to this okay so our original matrix a was what it's a b c and d and the reason i'm going through this generically is that once you kind of look at this you'll know the shortcut for two by two to find the kind of adjective or again the adjoint and you can quickly do it with a 2x2 just from that shortcut when you get to a 3x3 or higher you're going to have to do more work and i'll explain why in a little while so the very first thing with this guy is to make a matrix of cofactors now when we talk about a co-factor remember there's kind of two parts to it the first part is you have to calculate a minor okay so how do we find the minor well let's say i wanted the minor of this element here i would delete the row and i would delete the column and i would look at the determinant of the matrix that's left well if you do this with a two by two the determinant of the matrix that's left is just the entry that's left because the determinant of a one by one matrix if i wanted the determinant of this it would just be this itself so when you work with these you're just grabbing the entry that's left that's all you have to do but if you go into a three by three or higher you're going to end up having to actually go through the work of finding a determinant because it's not just going to be the single entry that's left okay so what we want to do is say that the minor and i'm going to notate this as m sub 1 1 because again this is row 1 this is row 2 this is column 1 this is column 2. this is telling me i have the minor element for row one column one so it's this entry right here and what is it equal to again if i mark out this row mark out this column it's the determinant of what's left in this case it's just d okay so that is how i find the minor but that is not the cofactor there is an extra step and there's a few different ways you can do this let me show you the long way and then i'll show you how to make like a little sign table so you can speed this up so the cofactor c sub 1 1 okay so of this entry right here you take the minor and you multiply it by negative 1 raised to the power of the row plus the column so in this case you can get that information from here so 1 plus 1 is 2. now notice how if you're raising negative 1 to an even power you're going to get 1 and the sine of the minor is not going to change so this guy is equal to positive 1 times d which is just d so the cofactor for this kind of entry in row 1 column 1 is just going to be d so let's erase all of this and let's put that c is equal to okay we're just going to put a d here to start now for the rest of these if you want to kind of make it a little bit quicker for yourself when you set up the kind of cofactor matrix instead of having to go through and multiply negative 1 raised to the power of the row plus the column each time you can predetermine the signs you'll see a lot of different sign charts out there and basically they alternate so again if i look at the kind of row and the column i'm in in each case this is a row number of one and a column number of one right here 1 plus 1 is 2 2 is an even number so we know negative 1 raised to an even number okay is going to be positive 1 so the sign is going to be positive or just stay what it is okay that's what that means then if i go here it's going to be row one column two one plus two is three it's negative see how it alternates then if i come down here it's row two and column one so it's two plus one which is three which is odd okay so again that's going to be negative and this last one is row two column two so two plus 2 is 4 that's even so it's positive so if you have this as a little reference you can find your minor and just attach the sign that way so you have to go through multiply by negative 1 raised to kind of the power of the row plus the column each time so let's just keep this here for reference and let's just burn through the rest of this so if i want the minor for this guy again all i do is i delete the row and the column it's in and i take the determinant of what's left in this case it's going to be c and for this one let's delete the row in the column and i'm left with b and for this one i'm going to delete the column and the row it's in and i'm going to be left with a now we already did the sign change for this one there was no sign change but for this one notice how again there's a sign change so i'm going to put a negative here for this one there's a sign change so i'm going to put a negative here and for this one right here there is no sign change so when you have a plus you don't change the sign because the minor and the cofactor are the same but when you have a minus you do change the sign so you want the negative of whatever is there so this is the cofactor matrix now we've taken all the minors and applied the sign changes when they were necessary now to get the adjoint you'll notice that this is still not the same the entries along the diagonal are same but these two are not so we have to do is take the transpose of this guy so the transpose of c of c and we put that capital letter t the superscript above our kind of letter c there to show this all we need to do this is a very simple process we take the first row and make it the first column so we'll take d and negative c then we take the second row and make it the second column so negative b and a so basically if you've never taken a transpose before you're just interchanging the rows and columns so i took the first row and made it the first column the second row became the second column now another way you could do this you could say that the first column became the first row and the second column became the second row so however you want to do that so i'm done with this matrix here and basically at this point i can say that the adjective or the adjoint of our matrix a is equal to the transpose of our matrix c which is this guy right here and you'll notice that these two match up perfectly you've got a d and a d you've got a negative b and a negative b a negative c and a negative c and an a in a so what happens is when you work with the two by two matrix you can always use this little shortcut by just remembering the generic entries here and remembering the adjective or again the adjoint turns into this one so these two switch and these two on the kind of opposite diagonal are going to be their opposite so let's look at an example real quick so we have our matrix a and it's equal to we have negative 1 and 3 in the first row and 5 and 2 in the second row so what we can do here again if it's a two by two it's very easy we can just use our shortcut so i can switch these two so i can say this is two and this is negative one and i can say that these two are going to be their opposite so this is negative three and this is negative 5. so look how fast and easy that is using that shortcut unfortunately in a moment when we start looking at three by threes we've got a lot more work to do but let's just do this the slow way real quick just so we get a little bit of practice before we get into the 3x3 and we have something more difficult so if i want the adjective of this guy the slow way i would find a cofactor matrix first and again let me just set up the signs so in this kind of first row first column here again it's row one column one so one plus one is two that's even so there'll be no sign change and then it alternates so this guy is going to be negative and then down here will be negative and here will be positive again just look at the sum of the row number and the column number row one column one one plus one is two two is even so that's why it's positive if it's row one column two one plus two is three it's odd so that's why it's negative okay so you can just look at that sum the row number and the column number if it's odd it's going to be negative okay if it's even it's going to be positive meaning there's no sign change to your minor all right so let's go through this real quick if i want the minor of this guy delete this row delete this column we want the determinant of this which is just this number itself so it's going to be 2. so let's write our 2 in there and i'll put it in a different color so let's move on to this one now so i'm going to delete the row and the column i'm left with just 5. again you take the determinant of that it's just that but notice how i put the negative 5 there i want the opposite of this because there's a negative sign there okay i'm basically just multiplying it by negative 1. then as i move down here i'm going to delete the row and the column i'm left with just 3. the determinant of that would obviously be 3 but again there's a negative there so i want negative 3 the opposite of that number then lastly for this entry here i want to delete the row and the column i'm left with negative 1. so obviously the determinant of that is negative one okay so we had a plus there so there's no sign change again again again i can't stress this enough because this is what confuses people if you end up with kind of a negative sign there if you're looking at a sine table take the opposite or you could say multiply by negative one by what you get for your minor if it's a plus there leave the minor alone there's no sign change needed all right so now that we have this we're not quite there yet because we have to take the transpose so the transpose of c is equal to the adjective or again the adjoint of a which is equal to what well let me kind of move this out of the way because we're going to run out of room so i'm going to define this down here so what we would do is take this first row and make it our first column so 2 and negative 5 the first row became the first column and then the second row will become the second column so negative 3 and negative 1. so you'll notice this matches our shortcut exactly 2 and negative 3 in the first row negative 5 and negative 1 in the second row okay let's look at an example that's a lot more tedious but again the concept is the same so we have a 3x3 here and obviously what we want to do is start out by making a matrix of cofactors then find the transpose of that and that's going to give us our adjoint okay so c is going to be equal to that's my cofactor matrix i'm just going to go through and put some signs in so we don't have to deal with them so this guy is row one column one so it's plus and then all the signs are going to alternate so it's going to be minus and then plus okay and you can go through and see that for yourself but then going down here this is going to be minus plus and minus and then plus minus plus so the signs always alternate okay so now what i want to do is go through and find those entries so i'm going to start with this one right here i'm going to mark out the row and the column and take the determinant of what's left okay so here's where it's more work because before with the 2x2 you really didn't have to you do take the determinate but the determinant is just the number that's presented because you only have one number here i actually have a 2 by 2 matrix that i have to find the determinant of so it's much more work so i'm going to do 2 times 0 which is 0 minus you have 3 times negative 1 which is negative 3 0 minus a negative 3 is 0 plus 3 which is 3. okay so no sign change here so this will just be 3. so let's go through the next one so we want this one so i'm going to mark out this column and this row so i would have 5 and negative 4 and negative 1 and 0. and i'm just writing this out because it's kind of hard to see this with kind of this highlighting in between so the determinant here would be 5 times 0 which is 0 minus you have negative 1 times negative 4 which is 4. so this is negative 4 but you have a sign change there so this is positive 4. okay all right so let's go to the next one so now we're here we're going to mark out this column in this row so what we want to do is find the determinant of this 5 times 3 is 15 then minus you have 2 times negative 4 which is negative 8. 15 minus a negative 8 is going to give me positive 23 and there's no sign change here so we'll put just 23. okay all right let's erase this and let's do this one so i'm going to delete this row and i'm going to delete this column so what we're going to have here is 1 and negative 6 and then 3 and 0. so we want the determinant of this 1 times 0 0 minus 3 times negative 6 is negative 18 again minus a negative is plus a positive so this would be positive 18. so we're doing a sign change here because it's right here where there's a negative so this is going to end up being negative 18. so let me write that in negative 18. all right let's go to the next one now so we're going to do this one so i'm going to delete this column in this row so what do we have we have 3 and negative 6 then negative 4 and 0. so the determinant of that 3 times 0 is 0 minus negative 6 times negative 4 is 24. so we have 0 minus 24 which is negative 24 and there's no sign change there so it's just negative 24. all right let's go to the next one so now we have this one right here so delete the column delete the row what's left we have three and 1 negative 4 and 3. so the determinant of this guy 3 times 3 is 9 minus 1 times negative 4 is negative 4. 9 minus a negative 4 is the same as 9 plus 4 which is 13 but again you've got that negative there so you've got to do the sign change so this is negative 13. all right let's look at this final row so we have this guy right here so i'm going to delete the column i'm going to delete the row so what i have left is this guy right here so the determinant of that 1 times negative 1 is negative 1 then minus you have negative 6 times 2 which is negative 12 again minus a negative is plus a positive so it's negative 1 plus 12 which equals 11. no sign change so it's just going to be 11. and let me change colors there so now we're here so we're going to delete this column in this row so we would have 3 and 5 negative 6 and negative 1. so the determinant of this guy 3 times negative 1 is negative 3 minus negative 6 times 5 is negative 30 and this equals we would have negative 3 plus 30 again minus a negative plus a positive so this is going to be positive 27. now when we look at this we have to do a sign change so this will be negative 27. all right last one so we want to find the cofactor of this guy so delete this row and this column so what do i have i would take the determinant of this matrix 3 times 2 is 6 minus 1 times 5 is 5. so this is 1 and there's no sign change here so this is just going to be 1. all right so now i have my cofactor matrix or my matrix of cofactors at that point it's pretty easy i'm going to say that the transpose of this cofactor matrix is equal to the adjoint of a which is going to be equal to what we're going to transpose this guy so i'm taking this row and making it this column so let me use a different color we're going to have 3 4 and 23. again first row becomes first column then the second row becomes the second column so negative 18 negative 24 and negative 13 then the third row becomes the third column so 11 negative 27 and then one okay now if you want you can delete this notation because you're done with it if you're asked for the adjoint just give them this you don't need to tell them it's the kind of transpose of the cofactors because they already know that okay so this would be your answer your adjoint or your adjective of a all right let's go ahead and take a look at another one it never hurts to practice a little bit more if this is something you're already comfortable with you can already start doing your practice problems otherwise let's go ahead and knock this guy out so we want to find the adjective or again the adjoint of a so the first thing i want to do is find the kind of cofactor matrix and the first thing i'm going to do again is just make a matrix of sines because i know that that's going to help me right i don't have to go through and do negative 1 raised to the power of the row plus the column each time so again if you can just remember that this first one's positive because it's in the first row and first column 1 plus 1 is 2 raising negative 1 to that even power gives you positive 1 you just alternate from there so this is negative this is positive and then you would come down here and you alternate so this is negative positive negative come down here and alternate positive negative positive very easy to remember that this first one's positive and that they alternate so if you remember that then you're basically good to go make your first row then drop down here and say okay if i'm alternating it was positive now it's going to be negative and you just alternate that's how you can keep going so once we have this kind of set up we just go through and get our minors and then apply the sign change if we need to find the transpose of that matrix and we're good to go all right so the first one here to find the cofactor delete the row delete the column what's the determinant of what's left negative 2 times 9 is negative 18 and then minus zero times one is zero so this is just negative eighteen there's no sign change so let's just put negative eighteen in there and then let's move on to the next one so now i'm going to be here i'm gonna delete the row and the column and let's just do this from where it is we would take the determinant of this matrix here it would be 6 times 9 which is 54 minus you have 0 times 0 which is 0. so this is 54 but again you have that sign change so this is going to be a negative 54. all right let's move on to this one so i'm going to delete the column and the row and the determinant of this guy 6 times 1 is 6 then minus negative 2 times 0 is 0. so this is just 6 no sign change because it's positive so let's just put a 6 there so let's start right here now we'll be in the second row so i'm going to delete this row and i'm going to delete this column so what do i have left i have this matrix with negative 1 and 3 and 1 and 9. so the determinant negative 1 times 9 is negative 9 then minus 3 times 1 is 3. negative 9 minus 3 is negative 12. you've got a sign change there so this would be positive 12. so now i'm going to be here so i'm going to delete this column and this row so the determinant of this 1 times 9 would be 9 then minus 3 times 0 would be 0. so this is 9 no sign change so it's just going to be 9. then let's move on to the next one so i'm going to be here delete this row delete this column and take the determinant 1 times 1 is 1 minus you have negative 1 times 0 which is 0. so this is 1 but again you have a sign change there so it's going to be negative 1. all right let's erase this and let's highlight this guy right here so we're going to delete this row and this column okay so now what we're going to have is the determinant of this so negative 1 times 0 is 0 then minus 3 times negative 2 is negative 6 0 minus a negative 6 is 0 plus 6 which is 6. so this is 6. all right now let's look at this one right here so i'm going to delete the row it's in and the column it's in and i'm going to take the determinant of this guy so it'd be 1 times 0 which is 0 then minus you have 3 times 6 which is 18 0 minus 18 is negative 18 but you have your sign change there so it's going to be minus a negative 18 or you could say negative 1 times a negative 18 however you want to think about that it's going to be positive 18. let me just change the color of this guy real quick i'd like for my rows to have different colors so let's do the last one we're almost there now so this guy right here i'm going to delete the row and the column and so the matrix that i have left the determinant of it 1 times negative 2 is negative 2 then minus negative 1 times 6 is negative 6 minus a negative is plus a positive so negative 2 plus 6 which is 4. so this will just be 4 here so let's erase this so now we have a matrix of cofactors and what we want to do is find the transpose of this guy and that's going to give us our adjoint so you can say c transpose is equal to the adjoint of your matrix a and this is equal to again you can just take the first row and make it the first column or you can take the first column and make it the first row let's do it a little differently so let's take the first column and make it the first row so negative 18 12 and six again all i did was i took the first column made the first row now i'm going to take the second column and make it the second row so negative 54 9 and 18 and then the last thing i'm going to do is i'm going to take the third column and make it the third row so i'm going to have 6 negative 1 and 4 okay again it's equally valid to take the first row and make it the first column it's the same you can take the second row make it the second column take the third row make it the third column you are interchanging the rows and the columns when you find the transpose so either way we found our adjoint for a and i can erase this notation we really need it this guy right here is the adjoint of our matrix a in this lesson we want to talk about how to find the inverse of a matrix using the determinant and its adjoint so at this point we're kind of finally ready to talk about an alternative approach that can be used to find the inverse of a non-singular or you can say invertible square matrix so let's suppose we have our matrix a and it's an n by n again that just tells you of a square matrix same number of rows as columns invertible matrix again invertible or non-singular just means that you can find an inverse okay if you can't find an inverse that's a singular matrix so if this is true then a inverse can be found using this formula so we say that it's 1 over the determinant of a times the adjoint of a now we already know how to find the determinant if you didn't watch kind of the last two lessons okay you probably don't know what the adjoint of a is but i'm going to cover that again here okay i'm not going to cover it in as much detail but we're going to go through it really quickly so let's go down and let's just start off with an example and the very first thing i want to do we know for the 2x2 we have a nice little shortcut and you're going to see where this shortcut comes from in a minute but for now let's just use the shortcut so we know that if i generically define a to be equal to let's just say it's a b c and d then a inverse is found as what it's equal to 1 over the determinant of a times what you do is you take a and d these guys along the main diagonal and you swap them so d goes up here let me change colors real quick d goes up here and a goes down here and then these other guys the b and the c you just change the sign so this becomes negative b and negative c okay so following this format if i want a inverse really quickly what i can do is i can say it's 1 over the determinant of a now let's just stop and get that for a minute it would be what it would be zero times three which is zero minus you'd have one times negative one which is negative one be careful there it's minus a negative one so that's plus one so the determinant is one so one over one is just one so you can just get rid of that you don't even need it so really all i need to work on is this part of the formula and again what do i want to do i want to swap a and d so i want to swap this one and this one so 3 goes up here 0 goes down here and then these two i want to make them into their opposite so negative 1 and positive 1. so this is the inverse of a we already know how to do that it's really easy and cool but where did it come from well let's talk about that for a minute if you go back to that formula let me just go back up here this is something you probably want to write down in your notes because we're going to use it all day again a inverse is 1 over the determinant times the adjoint of a now we did this but we really didn't know we were doing it okay so let's go back and i'm just going to copy this i'm going to need a fresh sheet to go through all this so 0 1 negative 1 and 3 and let me just copy this real quick and let's go to a fresh sheet so we have some room to work and we're going to see that we're going to get the same answer so let's paste this in okay so the first thing is we know the determinant is one so let me just kind of write that off here the determinant of a is one so we don't really need to work on that how do we find the adjoint of a well we talked about this over the course of the kind of last two lessons first you need to find a matrix of cofactors okay and then you need to take the transpose of that it sounds kind of fancy but it's really not that hard it's just more tedious than it is hard we know that finding a cofactor is kind of a two-step process the first thing is you have to find the minor and then you have to apply a sign change in some cases so really what we would say is a given entry in our cofactor matrix which i'll call c let's just say that c sub i j this is just some generic entry is equal to negative 1 raised to the power of the row plus the column so i plus j okay and then times your minor element so m sub i j and if this doesn't make any sense if this doesn't ring a bell from our lesson on determinants or the last two lessons it's okay it's very easy to do when you see generic notation sometimes it's a bit confusing what this means is let's first start out by calculating a minor okay to calculate a minor let's say i want the minor of this guy right here well all i do is i mark out the row and the column what's left i take the determinant in this case i'm just left with a kind of one by one matrix just the number three is involved so the determinant of that is just three okay so i can say that if i'm building a minor my m sub one one because in the first row first column remember i'm taking the minor of that guy right there well that's going to be equal to 3. now if i want the cofactor if i want c sub 1 1 well then i just multiply by negative 1 raised to the power of the row plus the column so we know this is row one row two this is column one and column two so if it's row one column one one plus one is two if i take negative one and raise it to an even power it's positive one so there's no sign change so in this case the minor and the cofactor are the same let me erase this we don't really need this anymore and let's just say the cofactor c sub 1 1 is 3. and i'm just going to put a capital letter c to be my matrix of cofactors and the first one up here is going to be a 3. okay so let's put that in and now i want to find c sub 1 2 okay so the first thing i look at when i look at this i say okay this is row 1 column 2 okay so 1 plus 2 is 3. i know that 3 is an odd number so whatever i get for the minor i want to make it negative and you can go through and do that and apply your signs before you start a lot of people like to have a table of signs just to refer to so that they don't have to go through this each time okay so we have c sub 1 2 and again this is equal to negative of the minor okay and the minor will be found by what let me put that 0 back in we would look at this guy right here in row 1 column 2 i would mark out the row i would mark out the column what's left it's a negative 1. so the negative of negative 1 is what you're going to have there and i put a negative out in front but i can just erase that now because i have the negative of negative 1 which would be 1. all right so let's look at the other two entries so moving into this bottom row you're going to have your row number as 2 your column number as 1. so c sub 2 1 okay and so 2 plus 1 is 3 that's an odd number so we know this is going to be negative and again if i mark out the column in the row what's left it's the number 1 okay so it's just negative 1 there so c sub 2 1 would be negative 1 and then lastly i want c sub 2 2. so let me erase all this and let me highlight this guy so that's what i want 2 plus 2 is 4 again negative 1 to the power of 4 negative 1 to the to an even power is positive 1. so we don't need to change the sign whatever the minor is that's what we're going to keep mark out the row mark out the column you get 0. so that's what we have there okay and i can just erase this i don't even need it so that's my matrix of cofactors now i'm not done i don't have my adjoint yet because what i need to do so c and then i'm going to put a t up here is going to be equal to remember to find the transpose of a matrix i've got to take the kind of rows and interchange them with the columns and all that means is that i would take this first row here it will become my first column here so 3 1 would go like this you would have 3 1 that would now be my column it was a row now it's a column and then for this guy negative 1 0 it would now be negative 1 0 it would be a column there so this is my transpose of c which is my adjoint of a so we found the adjoint of a and that's equal to this guy right here which is 3 negative 1 1 and 0. now remember the formula let me just go back up it's 1 over the determinant of a in this case the determinant of a is 1 times the adjoint of a if we go back the adjoint of a is this so if i just copy this real quick i'm just going to copy it and go back up and i'm just going to paste this in and you'll see that we got the same thing the adjoint of a here is 3 negative 1 1 0 3 negative 1 1 0. so because i don't have to multiply by anything because the determinant is 1 so 1 over 1 is 1 it just becomes this guy by itself but normally it doesn't work out that way in this case we just happen to have an easy example so again we could say a inverse is this guy right here so let's go ahead and look at a harder example now so we're going to have this matrix a it's going to be a 3x3 matrix we're going to use the same procedure unfortunately there's no shortcut for this you just have to kind of grind through the work so we know we need to find the determinant of a and we need to find the adjoint of a and then we're going to multiply kind of 1 over the determinant of a by the adjoint of it so let's just start out by finding the kind of cofactor matrix i think that's the one that takes the longest so let's go through and figure this out so if i start with this one first off let me just go through and get the signs so we can just take care of that right away you already know that these alternate in signs so this is row one column one so it's a plus then as you move it's row one column two so this is a minus right the signs are going to alternate so this is plus this is minus plus minus and this is plus minus plus okay so this is what we're going to have when i get the minor i'm just going to apply the sign and i'll erase this okay that's one thing you can do to just kind of speed things up there's a lot of kind of tricks out there but that's a good one that you can use so if i want the kind of cofactor for this one right here i would delete the row and the column and i would take the determinant of what's left so this is what's left so we would multiply down here negative 1 times negative 1 is positive 1 and then minus i would do 4 times negative 2 which is going to be negative 8. so 1 minus a negative 8 is going to be 1 plus 8 which is 9. so again it's positive so i'm not going to change the sign so it's just not all right let's move on now so for the next one let's just go to this one if i again highlight this row and this column what's left i'm going to have 4 negative 2 negative 2 and negative 1 and i'm going to write this out here because it's kind of hard to see so again if you want the determinant of that 4 times negative 1 is going to be negative 4 and then minus negative 2 times negative 2 is positive 4. negative 4 minus 4 is going to be negative 8. now when you put this up here remember you've got to change the sign so this is minus a negative 8 which is going to be positive 8. all right let's move on now so the next one is this guy right here again delete the column and the row what am i left with i want the determinant of this 4 times 4 is 16 minus negative 1 times negative 2 is positive 2 16 minus 2 is 14. again i don't need to change the sign here because that's already a plus all right let's erase this and let's move on to this one right here so i want to delete this row and i want to delete this column so what's left and let me write this one out because it's a little hard to see so negative 3 1 4 and negative 1. so i want the determinant of that negative 3 times negative 1 is 3. 4 times 1 is 4. so 3 minus four is negative one and of course in this particular case we're changing the sign right we have minus the negative one so this will be plus one so now let's move on to the next one so we have this one now and so i'm going to delete this column and this row so what's left again i'm going to write this out because it's kind of hard to see so you have 1 and you have 1 you have negative 2 and negative 1. so if i take the determinant of this 1 times negative 1 is negative 1 and then minus if you do negative two times one it's going to be negative two so negative one minus a negative two is the same thing as negative one plus two which is positive one so again there's no sign change here so i'm just going to put this as positive 1. so now we want this guy right here so i'm going to delete this column and this row so what i have is 1 negative 3 negative 2 and 4. again if i want the determinant of this 1 times 4 is 4 minus negative 3 times negative 2 is 6 4 minus 6 is negative 2 but i've got a negative here so i've got to change the sign so this becomes positive 2. all right let's move to the kind of final row and so what we want to do here is we want to delete this row and this column so what am i left with i'm left with this matrix here so negative 3 times negative 2 is 6 and then minus 1 times negative 1 is negative one six minus a negative one is the same thing as six plus one which is seven there's no sign change there so let's just put a seven all right just two more and i know this gets very tedious but sometimes it's faster sometimes it's not so let's put this guy and we'll kind of highlight this column in this row so the matrix that's left is a 1 and a 4 in the first column and a 1 and a negative 2 in the second one so let's go ahead and say 1 times negative 2 is negative 2 minus 1 times 4 is 4. negative 2 minus 4 is negative 6. you've got a negative there so you've got to change the sign so this would be positive 6. all right one more of these not too bad so i want this one so again i'm going to highlight this column and this row if i deleted that this is the matrix that's left so 1 times negative 1 is negative 1 then minus negative 3 times 4 is negative 12. again negative 1 minus a negative 12 is negative 1 plus 12 which is 11. so there's no sign change here so i'm just going to put an 11 in so now i have my cofactor matrix okay and when i take the transpose of this guy that's the adjoint of a so c and then i'm going to put a t up here a superscript t okay and that's equal to again you take the row and you make it into the column so row 1 becomes column 1. so i'm going to write that we have 9 8 and 14 for the first column because it was the first row the second row becomes the second column so 1 1 and 2 1 1 and 2 and then the third row becomes the third column so 7 6 and 11. so now we have this guy so i can erase this i don't even need it anymore and i can say that the adjoint of a is going to be equal to this guy right here so we have our 9 our 8 and our 14 we have our 1 our 1 and our 2 we have our 7 our 6 and our 11 okay so that's my adjoint of a now the other thing i need is the determinant of a because i have to take this guy and multiply it by 1 over the determinant or you could say i could take this guy and divided by the determinant however you want to think about that so the fast way to get the determinant let me kind of scooch this out of the way we want to copy the first two columns so 1 4 and negative 2 and then negative 3 negative 1 and 4. so let me multiply down 1 times negative 1 times negative 1 would be 1. then plus you'd multiply down again negative 3 times negative 2 times negative 2. we know that's going to be negative okay 3 times 2 is 6 times 2 again is 12. and then you're going to multiply down one more time 1 times 4 times 4 is going to be 16. so let me just find the sum real quick we know 16 plus negative 12 is 4 4 plus 1 is 5. so the first part of this formula is 5 and then you're subtracting away the second part of this formula so you use brackets or parentheses something to let you know that you need to kind of make sure you respect that sign so now i'm going to start at the bottom left and go up so negative 2 times negative 1 times 1 is going to be positive 2 then plus we have 4 times negative two times one that's going to be negative eight then lastly plus negative one times four times negative three we know that's positive and it's basically just four times three which is twelve so inside here two plus negative 8 is negative 6 negative 6 plus 12 is positive 6. so you would have 5 minus 6 which is negative 1. so that's going to be your determinant okay that's going to be negative 1. so the determinant of a is equal to negative 1. so now we're ready to find the inverse again the formula let me just show you one more time a inverse is equal to 1 over the determinant of a times the adjoint of a so if i know this is negative 1 1 over negative 1 is just negative 1 right so really all i have to do is say i'm going to take this guy right here and i'm going to multiply everything by negative 1 because that's all you're really doing so i can say that a inverse is equal to i'm just multiplying this guy by negative one multiplying by a scalar so every entry i'm just going to put a negative in front of it they're all positive so they'll all become negative so negative nine negative one and negative seven you have negative eight you have negative 1 and negative 6 and then you have negative 14 you have negative 2 and negative 11. so that is your inverse of a again is it quicker than if we had used kind of row operations maybe maybe not depending on how fast you are at that i'm kind of quicker at using row operations the times i go through it with you i go slow when i use row operations i can usually go pretty quickly so for me this method is a little bit more time consuming but not by a great deal so just another method you can use all right so suppose we have matrix a and again we want to find the inverse of this matrix kind of using this new method so again you've got to find the determinant of a and you've got to find the adjoint of a whichever one you'd want to do first is fine last time we found the adjoint first i think that's a little bit harder to do so let's just do that first again so i'm going to start by finding a matrix of cofactors so again the signs if you think about it they start out in this one as positive right because it would be negative 1 raised to the power of you have row 1 column 1 1 plus 1 is 2 that's even then it alternates so this would be minus this would be plus this would be minus this would be plus this would be minus this would be plus minus plus if you could just remember that this one's plus and that they alternate going this way and this way you're good to go right because you basically go plus minus plus then when you come down here you know it's minus plus minus and when you come down here you know it's plus minus plus again it's not hard to do because you're just looking at the row number and the column number you sum those numbers okay and if that result is even it's going to be a plus because negative 1 to an even power is going to give you a positive 1. if that result is kind of odd then we know it's going to be negative or negative 1 because negative 1 to an odd power is negative 1. all right so with that being said let's find the minors apply the sign changes that's going to give us the kind of cofactors and then we'll find the transpose and from there we'll have our adjoint so the first thing is we're going to start right here i'm going to mark out that and that right the row and the column i'm left with this guy right here so i find the determinant 3 times negative 3 is negative 9 minus 1 times negative 1 is negative 1. negative 9 minus a negative 1 is negative 9 plus 1 which is negative 8. there's no sign change applied so this is just negative 8. all right now we move to this one so again i'm going to mark out this column and this row and what's left and you can write this one out because these are kind of hard to see so 0 4 1 and negative 3. i want the determinant of this 0 times negative 3 is 0 the minus 1 times 4 is 4 0 minus 4 is negative 4. there's a sign change there because there's a negative so let's just say this is positive four okay so let's go ahead and go to the next one now so again if i go on this one i'm marking out this column i'm marking out this row so what's left is this guy right here zero times negative 1 is obviously 0 the minus 3 times 4 is 12. so this is negative 12 there's no sign change so negative 12. all right let's move down to the second row now so i'm going to mark out this row and this column and what's left you'll have negative 4 negative 3 negative 1 and negative 3. let's find the determinant of that guy negative 4 times negative 3 is 12 then minus negative 3 times negative 1 is 3. 12 minus 3 is 9. we're going to put a negative on that because again there's a sign change there so this would be a negative 9. and so i'm here now so i'm going to delete this row and this column let me write this one out again so you have 4 and 4 negative 3 and negative 3. so we want the determinant of that 4 times negative 3 is negative 12 minus negative 3 times 4 is negative 12. we know that minus a negative is plus a positive so negative 12 plus 12 is 0. so this guy is going to be 0 and of course you don't put a plus or minus with 0 it's just 0. so now i'm going to move to this guy so i'm going to highlight this row and this column so i want the determinant of this matrix so we have 4 and negative 4 and then 4 and negative 1. so 4 times negative 1 is negative 4 and then minus negative 4 times 4 is going to be negative 16. we know minus a negative again again again is plus a positive so this is negative 4 plus 16 which is going to give me 12. now we have a sign change here so i'm going to put this as negative 12. so now we're going to start with this one so i'm going to highlight this row and this column so the matrix that's formed is this guy right here so negative four times one is negative four and then minus negative three times three is negative nine again minus the negative is plus a positive negative four plus nine is going to give me five so no sign change here this is just going to be 5. all right let's erase this two more to do so now i have this guy right here so i'm going to highlight this column in this row and what do we see we see that we would have 4 and 0 and then negative 3 and 1. so if i take the determinant of this guy 4 times 1 is 4 then minus negative 3 times 0 is 0. so this is just 4 but then i want to apply that sign change so this is going to be negative 4. all right then lastly let me kind of erase this we have this guy right here so it's this column and this row that i'm going to block out and then determine it here 4 times 3 is 12 and then minus negative 4 times 0 is 0. so this is 12 and there's no sign change here so this is just 12. all right so we have our matrix of cofactors now and what we want to do is find the transpose of this so c and then we're going to put that superscript t up here and this equals again this is very easy to do take the first row make it the first column so you have negative 8 4 and negative 12. take the second row make it the second call so you have negative 9 0 and negative 12. take the third row make it the third column so 5 negative 4 and 12. so this is your transpose of the cofactor matrix which is the adjoint of a so let's put that the kind of adjoint of a is equal to this guy right here so let me kind of slide this down so i have some room and let me just rewrite this so it's negative 8 negative 9 and 5. we have 4 0 and negative 4 and we have negative 12 negative 12 again and positive 12. okay put my brackets around this and i'm just going to erase this at this point i don't need it anymore and i'll just kind of drag this down here so now let's find the determinant so we have 4 0 and 4. again i'm copying the first two columns negative 4 3 and negative 1. i'm going to start by multiplying down this diagonal 4 times 3 times negative 3 is the same thing as 4 times negative 9 which is negative 36. so that's the first part here then plus we're going to go down this diagonal negative 4 times 1 times 4 is going to be negative 16 then plus we're going to go down this diagonal notice there's a zero in the multiplication so you don't even need to do that because it's going to be zero so what's negative 36 plus negative 16 that's going to give me negative 52. so we'll have negative 52 minus again i'm going to put some brackets there so i don't make a sign mistake so what i want to do now is just start here and multiply up so 4 times 3 times negative 3 is again 4 times negative 9 which is negative 36 and then plus you would do negative 1 times 1 times 4 that's just going to change the sign of 4 so it would be negative 4 and then this one you know you have a 0 involved so you can just leave that off 0 times anything is always 0. so negative 36 plus negative 4 is negative 40 okay so what we see here is we have minus a negative so this is going to be negative 52 plus 40 which is negative 12. so that's my determinant for a so let's erase all of this and we'll say that the determinant of a is equal to negative 12. now the formula tells us that a inverse is equal to 1 over the determinant of a which in this case is negative 12 times the adjoint of a okay which we have down there so let's go back so basically what i'm going to do is just take 1 over negative 12 and multiply by each term here that's how i'm going to get my inverse so let me erase this and say a inverse is equal to and let me just erase this notation we don't really need it we'll say this is negative 1 over 12. again this is just multiplying by a scalar so everything in here is going to get multiplied by negative 1 over 12. so let's start out with negative 8 over negative 12. and we know that this would be positive and 8 divided by 4 would be 2 12 divided by 4 would be 3 so the first entry is going to be 2 3. let me erase that so then moving over here i'd have negative 9 over negative 12 and so everything there is divisible by three so this divided by three would be three this divided by three would be four of course negative over negative is positive so the second entry would be three fourths then when i look at five there's nothing really to simplify it would just be negative 5 12. so negative 5 12. so let's move on to this next row so you have 4 0 and negative 4. so let's figure out 4 first so 4 over negative we know each is divisible by four four divided by four is one twelve divided by four is three so this would be negative one third let me erase this a little bit so again negative one third we know zero would be unchanged because zero times negative one twelfth is still zero and then this guy is just going to be different by the sign right it's the same thing as this it'd just be positive one-third so looking at this final row here we have negative 12 negative 12 and 12. we know that negative 1 12 times negative 12 would just be 1. and you have that again here and then for this one it would just be negative 1 right because basically if you had 12 times negative 1 12 of course these cancel and give you one but you still have the negative so it's negative one so this is going to be the inverse of a again it can be faster to do it this way or it might not just depending on your kind of speed of doing kind of row operations but i will say that this is another tool that you can use to find the inverse of a matrix and this lesson we want to talk about cramer's rule for a 2x2 linear system all right so up to this point in our course we've learned kind of a variety of methods that can be used to solve linear systems with kind of these matrices that we've set up but probably the best or the quickest method that we can use is known as kind of cramer's rule okay so not in every scenario but in most scenarios this will be true so what i want to do today is just show you kramer's rule i'll show you how easy and effective it is and then at the end of the lesson i'll show you where this comes from a lot of students will kind of read how it's derived in their book and get a little bit lost because you use a lot of generic notation so i'll take the time to show you where it comes from at the end of the lesson for those of you who want to see that so i'm just going to start out today with a simple system we have two variables and we have two equations right so we have this negative six x plus y equals nineteen this negative three x minus two y equals seven so again the very first thing you wanna do is put the equations in standard form okay so that's something that we've been doing for a while now so the ax plus b y equals c and a b and c could just be any real number that you want them to be okay so after that you're going to start out by making a kind of matrix so the matrix is just going to contain the coefficients for the system your book's going to call this capital letter d okay so this is going to be equal to and we're going to be looking for the determinant of this matrix so we're going to be putting our kind of vertical bars here so i'm going to take my negative 6 which is the coefficient for x this has an implied coefficient of 1. this has a coefficient of negative 3 and this has a coefficient of negative 2. okay so notice how everything is in order so i have my negative 6 and my negative 3 on the left right those are the coefficients for x i have my 1 and my negative 2 on the right okay those are my coefficients for y so you've got to make sure that you set this up correctly and that's why it's important to put this in standard form all i need to do let me just kind of drag this over here we're going to say d okay and then sub x and this is going to be a determinant okay of this matrix we're going to form so again i'm going to use my vertical bars what you do is you think about the columns you set up so this is for the x's right the coefficients and this is for the y's right the coefficients so i'm going to replace if this is an x i'm going to replace the column with the x's those coefficients with these constants okay so i'm going to put a 19 here and a 7 here okay and then the y's those coefficients stay the same so a 1 and a negative 2. and if the first time you see that you're like how am i going to remember that just trust me after you do this a few times it's really really quick and easy then similarly if i do this kind of d sub y okay it's going to be the determinant of this kind of matrix we're going to set up it's the same thing so the x's will now stay so negative 6 and negative 3. so you're going to take the coefficients from that y variable okay so the 1 and the negative 2 replace it with the constants so 19 and then 7. okay so what we want to do at this point is kind of calculate these three so let me copy these i'm going to go to a fresh sheet and then i'll give you the actual formula which is really easy so i'll paste this in and let's just do this one at a time so for my capital letter d what's what's the determinant what is d equal to so to calculate d which again is the determinant of this guy d is going to be equal to you're going to multiply down so negative 6 times negative 2 is 12. then you're going to subtract the way you can multiply up or you can go down it doesn't matter negative 3 times 1 is negative 3. so we all know at this point that minus a negative 3 is plus a positive 3. so this is 12 plus 3 which is 15. okay so that's my value for d let me go back up and i'm going to erase this now and i'm just going to put that the value is 15 okay because we're going to use that in a moment now let me calculate this d sub y in this d sub x so let's go back and i'm just going to erase this because i don't really need it anymore okay i'm going to erase this i don't need it anymore so for this guy let me just kind of drag this over here my d sub x again i'm going to multiply down 19 times negative 2 is negative 38 and then minus 7 times 1 if i go up that's 7. so negative 38 minus 7 would be negative 45. okay so let me go back up and my d sub x would be negative 45 okay and let me just kind of drag this over here so that everything's kind of in line and now i need my d sub y okay so let me go down here we'll erase this we don't need it anymore my d sub y again i'm just going to multiply down negative 6 times 7 is negative 42 and then minus negative 3 times 19 is negative 57 so remember minus a negative is plus a positive so this is negative 42 plus 57 which is going to give me 15. okay so that's the value of d sub y all right so let's go back up and i'll show you how easy this is now so d sub y is 15. okay so you can get your solution once you have these kind of values calculated these determinants so your x will be equal to you're going to take your d sub x so this always matches this here okay over your d okay and then if you want your y it's equal to your d sub y again this always matches this right here over your d okay so it's very simple very easy we already know that d sub x is negative 45 okay and we already know that d is 15. okay and then for d sub y we already know that that is 15 and we know that d is 15. so that tells me that x is negative 3. negative 45 divided by 15 is negative 3 and that tells me that y is 1. so really quick overall and let's check this so negative 3 comma 1 would be our ordered pair solution so negative 3 comma 1. we don't need this anymore let's get rid of it and let's plug in so negative 6 times negative 3 would be 18 then plus we'd have one times one which is just one and this equals nineteen of course it does that works then the next one we have negative three times negative three which is nine then minus you have two times one which is just two this equals seven and of course it does so really really easy to kind of calculate this especially with the 2x2 when you get into a 3x3 obviously it's harder to find a determinant but we have the shortcut for that so it's not that bad but if you start getting into a 4x4 or 5x5 no matter what you do those are kind of hard systems to solve let's look at another example all right so for this one we have five x minus three y equals negative five and we have two x minus six y equals negative twenty six so again it takes you a few times to kind of remember what's going on the first thing is to make sure the equations are set up in standard form okay again if you don't do that you're going to not set up the correct matrices and you're going to end up with the wrong answer okay so once that's done which it's already done for us here okay we want to set up our capital letter d okay and that's the determinant of the kind of coefficients from your system setting up that matrix so you would have 5 and you would have 2. again these kind of coefficients correspond to that x variable right the x's and then for the y's again you're grabbing these so you have negative 3 and you have negative 6 okay so you take the determinant of that so we can go ahead and kind of set this up over here get out of the way and then you want to have your d sub x again how do we find that well we look at the column where the coefficients from the x variable came from so that's the 5 and the 2 and we replace it with the constants in the system so this guy and this guy okay so you're going to say that you have your negative 5 and your negative 26 there and then this part stays the same so the negative 3 and the negative 6 those don't change and then for your d sub y again it's the same thought process but now you're taking the column with the y coefficients right the negative 3 and the negative 6 you're replacing them with the constants so the x part stays the same the 5 and the 2 and then these get switched so it's going to be a negative 5 and a negative 26 okay and again the first few times you do this it is a little bit confusing takes a little while to kind of remember what's going on but after you do it a few times it becomes something you just commit to memory okay so let's copy these just as we did before and let's go to another sheet and i'm just going to paste these in we'll calculate the values real quick and then we'll use them to get our solution so for this one for capital letter d this is equal to what 5 times negative 6 is negative 30 and then minus you have negative 3 times 2 which is negative 6 minus a negative 6 is plus 6 so this would be negative 24 okay negative 24. let's erase this and now let's do this one so multiply going down negative 5 times negative 6 is positive 30 and then minus negative 26 times negative 3 is positive 78 30 minus 78 is negative 48. so this is negative 48. so let's erase this and just say this is negative 48 and let's erase this so now 5 times negative 26 is negative 130 and then minus if we go up 2 times negative 5 is negative 10 again minus a negative 10 would be plus 10 so this would be negative 120 okay so negative 120 so let me drag this over here and let me kind of drag this down and we'll just copy these and we'll bring them to the next sheet okay so we don't need any of these things anymore we'll just erase all of it again once you've calculated your determinants you've set everything up and calculated you don't need that stuff anymore we just need the values from them because again if i want the solution for x it's what it's d sub x it's always this guy matches this notation here so d sub x over your d okay this guy's always in the denominator okay and then for y it's what it's d sub y over d okay so again you do this a few times it becomes very easy d sub x is negative 48 and d okay is going to be negative 24 obviously that's positive two and then d sub y is negative 120 and then d is negative 24 and that's going to give us positive 5 okay so the solution here is 2 comma 5. so let's erase this now and we'll check it real quick it's always good to check things when you're first starting out because again you might make some silly mistake and then you turn in your test and you got it wrong and it's something you could prevent right just by checking so 5 times 2 is 10 and then minus 3 times 5 is 15 does this equal negative 5 yes it does 2 times 2 is 4 then minus 6 times 5 is 30. is this equal to negative 26 yes it is okay so 2 comma 5 is the correct kind of ordered pair solution again x is 2 y is 5. all right let's look at one more of these and then i want to talk to you about what happens for like a special case scenario what if you have no solution or an infinite number of solutions what do you do there and then i'll go through the process of kind of deriving the formula for you okay so what we have here is x plus 2y equals 7 and negative 2x minus 7y equals negative 35. again everything's in standard form already so we don't need to do anything there we just want to set up our again capital letter d it's the determinant of again take the coefficients so you have 1 you have 2 you have negative 2 and you have negative 7. again this is for the x's this is for the y's okay you've got to know where kind of everything is because you're going to be replacing stuff so when i do my d sub x again what i do is i replace the column with the kind of coefficients for x with the constants so 7 and then negative 35. okay just grab that from there and then i'm going to keep my y kind of coefficients the same so my 2 and my negative 7 there's no change there okay and then for d sub y it's the same thought process but now i take these y coefficients and i swap them with the kind of constants in the system so you're going to have your 1 and your negative 2 you know from the x's that's going to stay the same and then this part right here is going to change it's going to be the 7 and the negative 35 okay so now that we have this again let's just calculate these determinants and i'll just paste this in right here real quick and this was a d it didn't copy but we can just write it all right so for our kind of determinant for this d we have 1 times negative 7 which is negative 7 minus you have negative 2 times 2 which is negative 4 again minus the negative 4 is plus 4 so this is negative 7 plus 4 which is negative 3 okay so this is negative 3 and then the next we want to do let's do d sub x so 7 times negative 7 is negative 49 then minus you've got 2 times negative 35 which is negative 70. again again again minus the negative is plus a positive so negative 49 plus 70 which is going to give us 21. okay so this is going to be 21 okay and then let's find d sub y so d sub y we're going to go 1 times negative 35 which is negative 35 minus you have negative 2 times 7 which is negative 14 right minus a negative is plus positive so you have negative 35 plus 14 which equals negative 21. okay so let's erase all of this and we'll put negative 21 here and let me just kind of line these up and then we'll go back we'll kick our formula off and we'll have our solution really quickly so let's go up and we'll erase all this we don't need any of it anymore again once you've calculated all those determinants you're done with that information okay so let's just paste this in we have the values and so we can say that again x is what it's d sub x okay that part matches over d d is always in the denominator then for y it's what it's d sub y over d so d sub x we know is 21 and d we know is negative three okay so this is going to be negative seven right 21 divided by negative three is negative seven then for d sub y it's negative 21 okay and this is over d which is negative three so this is going to be positive seven so the solution here is that x is negative seven and y is positive seven or the ordered pair negative 7 comma 7. all right so let's go ahead and check this it's always good to check stuff just in case again you made a silly mistake so for the first one you'd have negative 7 plus 2 times 7 which is 14 equals 7 which is true right so this one works for the second one you'd have negative 2 times negative 7 which is 14 minus 7 times 7 which is 49 equals negative 35 that's true as well all right so let's take a little bit of time and talk about the scenario where you have kind of a special case scenario right so you have an infinite number of solutions or you have an inconsistent system meaning you don't have a solution at all so you won't be able to use cramer's rule to kind of identify which scenario you have okay unfortunately what's going to happen is you're going to have that kind of capital letter d that determinate for that matrix you set up with the coefficients is going to end up being 0 okay so let's go ahead and see that real quick everything's already in standard form if i set up my d okay again this is the determinant of we have our four and our negative two just grabbing the coefficients and then our six and our negative three if we multiply down four times negative three is negative 12 and then minus if you do negative 2 times 6 that's negative 12 as well minus a negative is plus a positive so you might as well just say this is plus 12 which is 0. okay so why is that a problem let's think about the formula one more time if this is zero well when i try to find x it's d sub x whatever that is over d well d is zero i can't divide by zero that's undefined okay so once you see that you've got to stop and say okay i'm dealing with a special case scenario and i can't use cramer's rule to kind of identify which one it is do we have an inconsistent system meaning there's no solution or do i have an infinite number of solutions i can't figure that out with kramer's rule so i've got to use some other method right i can use kind of gaussian elimination or gauss-jordan elimination if i want to do kind of matrix methods or i can go through and just kind of use simple substitution or elimination or whatever you want to do in this particular case i would probably just use elimination right i would multiply this first equation by let's say positive 3 so that would give me 12 x minus 6 y is equal to negative 60 and then the second equation i would multiply it by negative 2. so that would give me negative 12x that would give me positive or plus 6y and that would be equal to negative 36. now looking at this we don't need to go any further we can just say there's no solution right that's very obvious that there's no solution because again if you go through this this is going to cancel become zero it's equal to something that's not going to be zero okay so you know it's not going to have a solution negative 60 minus 36 would be negative 96 but again because this ends up being a false statement okay you can say there's no solution if you go through this and it ends up being a true statement you know there's an infinite number of solutions all right so what i want to do now is just kind of wrap up the lesson and show you where this kind of cramer's rule comes from it's very easy to do i'm not going to go through the whole thing because although it's easy to do it is a little bit tedious and time consuming so i'm just going to show you where we get that x is equal to that d sub x over d okay where that comes from and to do that i want to start out by just talking about the setup in general so you have two equations let's just label this top one as equation one and this bottom one as equation two you'll notice that this system just contains x and y and each equation is already written in standard form for us okay now the coefficients look a little bit weird you have a sub 1 and a sub 2 as the coefficients for x what we're doing here is we're saying hey this is just something generic we don't have numbers involved so we're using a as a stand in and to kind of tell the difference between this coefficient here in the first equation and this coefficient here in the second equation we have the a sub 1 and the a sub 2. okay so that's what that notation is for similarly you have the b sub 1 and b sub 2 as the coefficients for y and then c sub 1 and c sub 2 those are your constants okay so let's start out before we even get into anything let's just write what we know and we'll compare it to how we could get a general solution so we know already that d okay is equal to the determinant of if i set up a matrix of just the coefficients so in this case everything's already in standard form so i'd want my a sub 1 my b sub 1 my a sub 2 and my b sub 2. okay so that's my capital letter d that's this matrix here and again i'm taking the determinant of that and if you wanted to make this super clear for you remember this leftmost column corresponds to the kind of coefficients for the x variable so that's why i'm putting x's and then this is going to be for the y's these are the coefficients for the y variable and so you want to make sure you know what's going on with that because when you do your d sub x remember what you want to do since this is sub x i take this column that has the coefficients for the x's and i replace it with the constants so what i want here is c sub 1 and c sub 2 and then this part would stay the same so b sub 1 and b sub 2 and then similarly if we do d sub y then what i'm going to replace the column with the coefficients for y with the constants so c sub 1 and c sub 2 goes there and then on the left it's the same as over here so a sub 1 and a sub 2. so generally speaking if you are setting these things up again if you have d sub x you just take d and you take the column with this guy in it so the coefficients for that variable and you replace it with the constants okay and so the same thing for d sub y i take the column with the coefficients for that kind of variable in this case again it's y so i take that and replace it with the constants so very easy to remember once you do this a few times let me copy this real fast i'm going to bring it to another sheet and let me put this over here i'm just going to paste this in and really quickly i'm just going to drag this over here and generically we want to say that this is equal to what this is a sub 1 times b sub 2. again i'm just multiplying down and then minus if i multiply up i'll do a sub 2 times my b sub 1 okay and this is equal to what it's equal to c sub 1 times b sub 2 and then minus you're going to do c sub 2 times b sub 1. and then lastly this one's equal to a sub 1 times c sub 2 and then minus you're going to have your a sub 2 times your c sub 1. so this is our value for d this is our value for d sub x and this is our value for d sub y okay so we have that let me go to another sheet so i just want to write really quickly that x is equal to d sub x over d and y is equal to d sub y over d now i'm not going to show the y one i'm just gonna do the x one okay but you could do the y one on your own so you can kind of continue this process so if we go back and i'm just going to copy this real fast and trust me once we get to the other parts is going to go really quickly so let me paste this in so this is d in each case let me put equals here and let me put equals here let me kind of paste this in and all i'm doing is i'm just pasting in a value for d we already figured that out that it was a sub 1 times b sub 2 minus a sub 2 times b sub 1. okay then for d sub x i can grab that from here let me copy that and i will paste that in and just get rid of that extra equal sign that's not necessary and then let's go back up and we'll grab this d sub y okay copy that and we'll come back down and paste this in okay so now we're going to come back to this in a moment i'm going to again show you how you can get this result so let's go up let's say i wanted to solve this using elimination and what i'm going to do is i'm going to eliminate the variable y again you could use a similar thought process to eliminate x and kind of go through the other scenario i'm only going to do one okay so what i'm going to do is i'm going to first start out by multiplying equation 1 by b sub 2. so that would be a sub 1 times b sub 2 then times my x then plus i would have b sub 1 times b sub 2 then times my y and this equals c sub 1 times b sub 2. so all i did was i multiplied equation 1 by b sub 2. so every term has a factor of b sub 2. okay so now what i'd want to do again if i'm trying to eliminate the y variable i want this to be the opposite of this so negative b sub 1 times b sub 2 times y so how could i accomplish that well i would multiply equation 2 by negative b sub 1. so this one right here would be negative a sub 2 times b sub 1 times x okay so again i'm just multiplying by negative b sub 1. that's what i did there this is already done so you can skip that and then it's equal to it's going to be negative c sub 2 times the b sub 1. okay so now if i use my elimination process we know on the left sides if i add those together this is going to be gone okay so all i'm going to have let me kind of scroll down so we have enough room on the left i have my a sub 1 times b sub 2 times x then minus my a sub 2 times my b sub 1 times x and this equals you're going to have your c sub 1 times b sub 2 minus your c sub 2 times b sub 1 okay now at this point it probably looks like a lot of nonsense but i want you to realize that you can solve for x you have x here and you have x here so to solve for x we want to factor that out okay so if i factor that out inside the parentheses that have what's left so i'd have that so a sub 1 times b sub 2 then minus my a sub 2 times b sub 1 okay that guy right there and then let's just close the parentheses and say this is equal to c sub 1 times b sub 2 minus c sub 2 times b sub 1. okay so how can we solve for x well all i'm going to do is i'm just going to divide both sides of the equation by what's multiplying x and in this case it's this a sub 1 times b sub 2 minus a sub 2 times b sub 1 okay i'm going to do that on both sides so a sub 1 times b sub 2 minus your a sub 2 times b sub 1 okay so notice that this would cancel with this it's just a complicated form of 1 and i'm left with a solution to say that it's x is equal to your c sub 1 times b sub 2 minus your c sub 2 times b sub 1 over your a sub 1 times b sub 2 minus your a sub 2 times b sub 1 okay so let's copy this real quick and let's go back to this page here and let me paste this in right down here so i want you to notice something and what you're going to notice is that this formula right here for x and this is the same okay so you have c sub 1 times b sub 2 c sub 1 times b sub 2 minus c sub 2 times b sub 1 minus c sub 2 times b sub 1. then down here it's the same as well it's the exact same thing so that's where this comes from you just realize that all you really need to do is find kind of this determinant and this determinant and basically you've found your solution for x okay now if you wanted to prove this for y you could do it in the same way you can go back to the original kind of set up here and now you would want to make kind of these two terms here opposites you could eliminate the x variable and you could solve this guy for y and all you're going to find is that your solution is this guy right here okay it's just going to turn out that way so y is equal to your d sub y over d in this lesson we want to talk about cramer's rule for a three by three linear system so in the last lesson we talked about using kramer's rule to solve a two by two linear system now we're just going to take the next step and we're going to look at some examples with kind of a three by three linear system this is no more difficult it's just a little bit more tedious okay because it takes more time to kind of calculate things so let's just start out with this example here we have our system and notice how every equation is already written in standard form force again this is something that's very important to check okay if you have a three variable system okay and you have three equations you want to make sure every equation is ax plus b y plus cz equals t okay and you might get different notation you might have x sub 1 through x sub 3 but you want to make sure that everything lines up right so the x's line up the y's line up the z's line up the constants line up because when you load information into your kind of matrices it's got to make sense okay all the columns have to correspond to kind of the same thing whether it's you know the coefficients for x or y or z or the constants it's all got to line up and make sense okay so with that being said let's go ahead and get started with this problem the very first thing we want to do as we saw in kind of the last lesson we want to set up our kind of d okay and this is the determinant of the coefficient matrix so if i take my coefficients from the system in the first equation i'd have negative 3 i'd have 2 and i'd have a negative 1. okay so that's an implied coefficient of negative 1. so let's just write that in the top so negative 3 we'd have 2 and then negative 1. okay and then for my second kind of row here i'd have a negative 1 and you could write that in if you want okay let me make that a little bit better so again a negative 1 we would have a 3 and a negative 3. so a negative 1 a 3 and a negative 3. okay then we're going to have a negative 1 again let me write that in so negative 1 you're going to have a 2 and a negative 3. so negative 1 2 and negative 3. okay so this is finding the determinant of the coefficient matrix now we need three other kind of setups like this we need d sub x d sub y and d sub z so let me kind of write d sub x first and each time i write one of these i'm going to go ahead and just move it to another page because we're going to very quickly run out of room okay so d sub x would be equal to what well essentially what i do is i take this guy right here and i take the column where the kind of coefficients correspond to this variable that i'm looking at so in this case it's x you know if it was y if it was z you've got to look at that kind of column that corresponds to the coefficients for that variable in this case i'm looking at this guy because this column represents the coefficients for the kind of x variable what i want to do is i want to replace that column and that column only with the constants of the system okay so what i'm going to do is i'm going to take and put negative 15 negative 9 and negative 13. so i just took these values put them here in this place here okay and this the rest of it's going to stay the same so you have 2 3 2 negative 1 negative 3 and negative 3. so that part's the same and i'm just going to cut this away i'm going to go to a fresh sheet and i'm going to paste this in and i'm going to put that right there okay so that's going to stay there and we'll come back to this in a little while let's get the rest of it going so i'm going to erase this highlighting here and now what i want to do is i want to find d sub y well again it's the same thought process except now the kind of column that corresponds to the coefficients for the y variable that's going to be this middle column here so everything else would be the same so negative 3 negative 1 and negative 1. again this guy's the coefficients for the x variable this is what's going to change in the middle so i'm taking the constants so negative 15 negative 9 and negative 13 write that in and then this this guy right here the coefficients for the kind of z variable that's going to stay the same so negative 1 negative 3 and negative 3. so let me cut this away and then grab all of it so let me cut this away okay so let me go back up so now we just need to find d sub z okay so that's the last one so again it's the same thought process so these first two columns will be the same so negative 3 negative 1 negative 1 2 3 and 2. and again i'm looking at the kind of column here that corresponds to the coefficients for this variable okay so i'm going to replace this with the constants of the system it's just that easy so negative 15 negative 9 and negative 13. okay so let me copy this and get rid of it from here and let me paste that in so each one has its own page okay so let me go back up and the very first thing we want to do is copy this and figure out what this kind of d is going to be equal to so let's calculate the determinant of this first because it's going to be used in every formula and again if you don't remember that i'll talk about that in a second but the first thing is we need to calculate the value of d okay so i'm going to use the shortcut here i'm going to copy the first two columns so negative 3 negative 1 negative 1. we have 2 we have 3 and we have 2 okay so what i want to do is multiply going down first so i want to start here and i want to multiply down so you'd have negative 3 times 3 times negative 3. so you know that would be positive and 3 cubed is 27. so i would say this is 27 then plus i'm going to multiply down this diagonal here so 2 times negative 3 times negative 1 that's going to be 6 then plus i'm going to multiply down this diagonal negative 1 times negative 1 times 2 is 2. okay so i'm going to put some brackets around that or you can do the operation now if you want but basically you're going to subtract away put some brackets around this now i'm going to multiply up so this is getting kind of busy to where we can't see stuff so let me erase these arrows and we'll get rid of that so we can see what's going on so negative 1 times 3 times negative 1 is going to be positive 3 okay then plus we're going to go up 2 times negative 3 times negative 3 i know that would be 9 times 2 which is 18 then plus if i do negative 3 times negative 1 times 2 that's going to be positive 3 times 2 is 6. okay so what's this going to be equal to well 27 plus 6 is going to be 33 and then plus 2 is 35 and then minus 3 plus 18 is 21 plus 6 is 27. if i do 35 minus 27 i get 8 okay so d here is equal to 8. so let's just erase all of this okay i'm going to erase this and i'm just going to say that the value here is going to be 8 and again this is going to be used in every formula so let me go back up to this part right here i'm just going to erase all this highlighting we don't need it anymore and i'm just going to put that d is equal to 8 okay now i also need to know what d sub x is i need to know what d sub y is and i need to know what d sub z is because in each case we're going to be using this to get our answers so let's go down and let's calculate these guys so again i'm going to use my shortcut i'm just going to write this as negative 15 negative 9 and negative 13. just copy the first two columns so 2 3 and 2. so i'm going to start by multiplying down okay so i'm going to multiply down so negative 15 times 3 times negative 3 which is going to be 135 okay then plus i'm going to put some brackets around this i'm going to multiply down again so 2 times negative 3 times negative 13 is going to be 78 okay then plus i'm going to multiply down again so negative 1 times negative 9 times 2 is going to give me positive 18. okay so then we're going to subtract away i'm just going to erase this normally i leave it but it's a little bit heavy so it might block us from seeing stuff i'm going to multiply up now so negative 13 times 3 times negative 1 which is going to be 39 okay and then plus we're going to multiply up so you've got 2 times negative 3 times negative 15. and so that's going to be 90 okay and then plus we have negative 3 times negative 9 times 2 so that's going to be 27 times 2 which is 54. so let's go ahead and crank this out so 135 plus 78 is 213 plus 18 is 231 okay then minus if we do 39 plus 90 we get 129 plus 54 is 183 okay so what we want is 231 minus 183 which is 48 okay so this is 48. so let's erase all this and just put that this is equal to 48. okay so let's go back up and let's say this is 48. so now let's find d sub y so again copy the first two columns so negative 3 negative 1 negative 1. you've got negative 15 negative 9 and negative 13. again i'm going to multiply down to start so i'm going to multiply this way so negative 3 times negative 9 times negative 3 is negative 81 okay then plus i'm going to multiply down here so you've got three negatives so you know it's going to be a negative and then 15 times 3 is going to be 45 then plus if i go down this way negative 1 times negative 1 is positive 1 then times negative 13 is negative 13. okay so let's just go ahead and find this sum real quick so negative 81 plus negative 45 is negative 126. then if i add negative 13 i get negative 139 so let me just put that in there so negative 139 and then minus i'm going to multiply up now let me get rid of these kind of arrows so i'm going to start here and go up so you've got three negatives so you know it's negative and it's just going to be 9 there then for this guy you've got negative 13 times negative 3 times negative 3 so you know that's negative and 13 times 9 is basically what you have and that's 117. so this would be minus 117 or you put plus negative 117 if you want and then lastly you want to do this one so you've got negative 3 times negative 1 which is 3 times negative 15 which is minus 45 or plus negative 45. so negative 9 minus 117 is negative 126. if i subtract away another 45 i get negative 171. so this is negative 171. now you have minus a negative here which becomes plus a positive you get negative 139 plus 171 which is equal to 32 okay so this is 32. so let's erase all of this and say that this is equal to 32 okay and let's go back up and let's say this is 32. okay let's go back down okay so this is the last one to do let me erase that highlighting we don't need that and i'm just going to copy again the first two columns and so i would have negative 3 negative 1 negative 1 again 2 3 and then 2. so i'm going to multiply down so negative 3 times 3 times negative 13 is 117 okay then plus you've got 2 times negative 9 times negative 1 that's 18. then plus you've got negative 15 times negative 1 times 2 that's going to give me positive 3 okay so 117 plus 18 is 135 plus 30 is 165. so the first part is 165 then minus let me kind of erase these because they just get in our way so let me go up now so this way negative 1 times 3 is negative 3 times negative 15 is positive 45. then plus if i go up here 2 times negative 9 is negative 18 then times negative 3 is positive 54 okay and then plus let me go up here negative 13 times negative 1 is 13 then times 2 is 26 okay so 45 plus 54 is 99 plus 26 is 125 okay so this is minus 125 and this would be equal to what 165 minus 125 would be 40. okay so pretty simple so let's erase all of this and erase that and just say this is equal to 40. so let's go back up and say this is 40. it does take a while to kind of go through and get your determinants but once you have that set up you're basically at a point where you have your solutions so your x is equal to your d sub x over d your y is equal to your d sub y over d and your z is equal to your d sub z over d okay so in each case the denominator is the same it's going to be 8. so let me just erase this or i can actually just slide this down so i'll just say this is equal to equal to equal to i'm going to put an 8 in each denominator okay and the only thing that's going to change is the numerator so d sub x is 48 d sub z is 40 okay and d sub y is 32 okay so if you go through and do the calculations let me slide this down just a little bit more so we can fit everything 48 over 8 is obviously 6. 32 over 8 is 4 and 40 over 8 is 5. so x is 6 y is 4 and z is 5 okay so as an ordered triple this is 6 comma 4 comma 5. remember this is your x this is your y this is your z all right let's go ahead and take a look at another example again it's the same thing it's just a tedious process but we're going to get through it so in this example we already have all the equations written again in standard form that ax plus b y plus cz equals d so let's go right into it we want to start out with d again this is the determinant of the coefficient matrix so for the coefficients here i've got negative five i've got negative five again i've got a negative one i can go ahead and write that in so negative one okay and then for this one i've got a negative five a negative 4 and again a negative 1. and then for this one i've got a negative 1 i've got a negative 3 and then a 2. okay so let's recap i got negative 5 negative 5 and negative 1 negative 5 negative 5 negative 1. negative 5 negative 4 negative 1 negative 5 negative 4 and negative 1 and then negative 1 negative 3 and 2 negative 1 negative 3 and 2. so you always want to check stuff to make sure you got it right because the simplest error for getting a negative or something like that you're going to get the wrong answer okay so the next thing i want to do is find my d sub x so what's that equal to again i just take the column with the kind of coefficients for x in this case it's going to be this one and i replace it with the constants of the system so i'm going to take and put a negative 21 a negative 22 and a negative 11. and then the rest of this is the same so it's negative 5 negative 4 and negative 3 and then it's negative 1 negative 1 and positive 2. okay so let me cut this one away and i'll drag this over and now let me find my d sub y okay so what is d sub y again the same thing i'm going to kind of look at my column that is the coefficients for the y variable so that's this middle one here and i'm going to replace that with the constants so everything else is the same so you have your negative 5 your negative 5 and your negative 1. this is going to be my constants so negative 21 negative 22 and negative 11 and then for the last column again it stays the same so negative 1 negative 1 and positive 2. okay so let's cut this one away i'll just kind of slide this down and let's get the last one so we want d sub z as well okay so that's going to be equal to what so now i'm going to look at this kind of last column these are the coefficients for z i'm going to replace that with the constants so everything else is the same so you've got your negative 5 your negative 5 again you're negative 1 you got your negative 5 your negative 4 and your negative 3. again the last column just going to replace it so negative 21 negative 22 and negative 11. okay so let's go ahead and get rid of this okay so let's go back up and set everything up now so i'm going to grab this and move it to another sheet and i'm just going to erase all of my highlighting and let me go to that other sheet and paste this in for us and then we'll be ready to go okay so i can erase all this highlighting i don't need any of that and let me go back up here and we're just going to write some things down so we know that we want to find d we want to find d sub x we want to find d sub y and we want to find d sub z okay so we're going to have these values we're going to use them to calculate our x y and z so we already know that x is equal to d sub x over d okay then y is equal to what it's d sub y over d let me give myself some space i'll come over here and say z is equal to d sub z over d okay so we have all this set up and let's go down and let's start out with the value for d because this is going to be used in everything so let's just knock this one out first so again the quick way to do this is just to copy the first two columns so you have your negative 5 negative 5 and negative 1 negative 5 negative 4 and negative 3. again you're going to multiply down on that diagonal so i'm going to go down this way so negative 5 times negative 4 times 2 negative 5 times negative 4 is 20 20 times 2 is 40. then plus you're going to go down this diagonal so you have negative 5 times negative 1 times negative 1 which is negative 5. and then you're going to go down this diagonal you've got negative 1 times negative 5 which is 5 that's negative 3 which is negative 15. okay so negative 5 plus negative 15 is negative 20 plus 40 is 20. so this first part of this is 20 then minus let me kind of erase these and let me go up now so if we go up you have negative 1 times negative 4 times negative 1 which is going to be negative 4. so this is negative 4 then plus if i go up again negative 3 times negative 1 times negative 5 is going to be what was basically just negative 3 times 5 which is negative 15. okay so negative 15 again because you have three negatives there and then going up here 2 times negative 5 times negative 5. i know that's positive and basically 5 times 5 is 25 then times 2 is 50 okay so what is this going to give me negative 4 plus negative 15 is negative 19 plus 50 is 31 okay so basically what you have here is 20 minus 31 which is negative 11 okay so we can erase this stuff we have our first one we'll say this is negative 11 we'll go back up and just put that this is negative 11. so in each case i can go ahead and write negative 11 down here okay it's going to be in every formula and now we can go back and start getting some solutions okay so i find the determinant here i can go and plug it into my formula for x and then i'll find the determinant that's d sub y and then d sub z plug those in for their respective formulas and we'll be done so to calculate the determinant the fast way again i'm going to copy the first two columns so negative 21 negative 22 and negative 11 and then negative 5 negative 4 and negative 3. again i'm going to multiply going down first so i'm going to go this way negative 21 times negative 4 times 2 is going to be 168 then plus i'm going to do negative 5 times negative 1 times negative 11. so i know that's negative and 5 times 11 is 55 so this would be negative 55 here then plus if i go down this way you have negative times negative times negative which is negative okay and you basically have 22 times 3 which is 66. so what's this sum here 168 plus negative 55 is 113 plus negative 66 is 47 okay so the first part of this is 47 then minus let me erase those okay and then i want to go up now so negative 11 times negative 4 times negative 1 well we know that's negative so it's going to be negative 44 right 11 times 4 is 44. so negative 44 goes there and then going up here negative 3 times negative 1 is 3 and then times negative 21 is negative 63. so let's put plus negative 63 and then lastly we have 2 times negative 22 times negative 5. if you do 2 times negative 2 you get negative 44 and then times negative 5 that's going to be 220. so plus 220. so what is negative 44 plus negative 63 that's going to be negative 107 okay so if i did negative 107 plus 220 i get 113 okay so i'm going to put minus 113 here so this is going to be negative and basically what is 113 minus 47 that's going to be 66 so this would be negative 66 okay so let's erase all of this and we'll say this has a value of negative 66 and of course we can just go back up and get our solution real quick so for this one it's negative 66 so we plug it into this formula negative 66 over negative 11 is positive 6 so we know that x is going to be 6 right that's my x value okay so let's go down and now we'll find y okay so let's go ahead and expand this by copying the first two columns so negative 5 negative 5 and negative 1 negative 21 negative 22 and negative 11 okay so let's go ahead and just multiply down negative 5 times negative 22 is 110 and if i multiply by 2 i get 220 okay so that's the first one and then plus if i multiply going down here so we would have negative 21 times negative 1 times negative 1 that's just negative 21 okay and then plus if i multiply down here i know that's negative i'll go ahead and put that in and 5 times 11 is 55. so what is the sum of 220 plus negative 21 well that's 199 and then plus negative 55 that's 144. so the first part of this is going to be 144 and then minus remember we're going to go up now let me erase these when i go up i'm going to start here and go up so i know that's negative 22. if i go up here that's going to be negative 55 right because again it's 3 negatives 11 times 5 is 55 so it's negative 55 okay then this one 2 times negative 5 is negative 10 then times a negative okay so i know that's going to be positive 21. well 10 times 21 is 210. okay so what is negative 22 minus 55 that's negative 77 and if i add 210 i get 133 okay so what is 144 minus 133 that's going to give me positive 11. so let's erase this and let's put an 11 here and let's go back up and we'll say that this is 11. and if i put an 11 here 11 over negative 11 is negative 1. so my y value here is negative 1. okay so now i just need to find z and i just need to find the determinant of this guy first so i'm going to copy negative 5 negative 5 and negative 1 negative 5 negative 4 and negative 3. okay so we're going to multiply going down negative 5 times negative 4 times negative 11 is going to be negative 220 okay then plus negative five times negative 22 times negative one well we know that's negative what's 5 times 22 that's 110 so this would be negative 110 okay then plus you've got negative 21 times negative 5 times negative 3 which is negative 315 okay all right so now let's just go ahead and sum these so negative 220 plus negative 110 is negative 330. then if i add negative 315 i get negative 645. so the first part is negative 645. again then we subtract away this kind of second part where we're going to multiply up so if i go up here negative 1 times negative 4 is 4 then times negative 21 is negative 84 okay so that's negative 84 then plus negative 3 times negative 22 times negative 5 well negative 3 times negative 22 is 66 then times negative 5 would be negative 330 okay so then lastly i'm going to do this one so you have a negative there because it's 3 negative so let's just put that in first we know that 5 times 5 is 25 then times 11 would be 275 so this would be negative 275 here okay so what we want to do is sum these amounts of negative 84 plus negative 330 that's negative 414 plus negative 275 is negative 689. so this would be minus a negative 689 which is negative 645 plus a positive 689 which is equal to what it's just 689 minus 645 which is nothing more than 89 minus 45 which is 44 okay so this is going to be 44. so let's write that in we'll just erase this okay we'll say this value is 44. you can erase this we don't need it anymore and we can go back up now we have our solution so this is 44 so this is 44 so my solution here would be negative 4 right so 44 over negative 11 is negative 4. so i'll just write a negative 4 in here for z okay and so my order triple here is 6 comma negative 1 comma negative 4. again x is 6 y is negative 1 z is negative 4. in this lesson we want to talk about finding the area of a triangle using determinants all right so an application of kind of matrices and determinants is being able to find the area of a triangle whose vertices are given as points in our coordinate plane so you have these three vertices that you're going to be given and the first thing you want to do is you want to label one as x sub one y sub one one is x sub two y sub two and then kind of the last one is x sub three y sub three and the first thing you're going to ask me is does it matter which point gets labeled as which and it doesn't okay and i'll explain why in a moment so then you're going to plug into this kind of formula here for the area of a triangle so you see it's equal to you have plus or minus this one half times the determinant of this guy remember if you see these vertical bars it means to take the determinant okay so first and foremost before i explain anything just notice where you're plugging things in at you've labeled stuff as x sub 1 y sub 1 through x sub 3 y sub 3 you're plugging in x sub 1 y sub 1 in this first row so you have that there and there okay and then you always have a 1 in the third column then you're going to plug in x sub 2 y sub 2 always have a 1. x sub 3 y sub 3 always have a 1. okay so that takes care of that and then you're going to take the determinant now this is where i need to start explaining things so you have that plus or minus there that might be super confusing for you and basically what this is telling us is that we want to make sure that we get a positive area okay so if the determinant evaluates to be negative you multiply by negative one half to get a positive area if the determinant evaluates to be positive well then i can just multiply by positive one-half and again i get a positive area now the other scenario is you end up with an area that's zero and we're going to talk about this specifically in the next lesson if your area evaluates to zero the three points that you had are actually on the same line okay so that's going to be our test for collinearity and we're going to talk about that specifically in the next lesson for today we're just going to look at two problems and neither is going to have an area of zero so let's talk about the final question and that's why am i able to label you know any of these as x sub 1 y sub 1 or x of 2 y sub 2 or x sub 3 y sub 3 why don't have to go in a certain order well it deals with the properties of determinants that we've already talked about if i had something like 2 3 4 5 like this the determinant is what 2 times 5 is 10 minus 3 times 4 is 12 this is negative 2 okay what happens if we swap rows so let's say i put the 4 and the 5 into row 1 and the 2 and the 3 in row 2. well what happens is the sign of the determinant changes right the absolute value is the same the absolute value will be 2 in each case but now instead of getting a negative 2 i'm going to have a positive 2. so 4 times 3 is 12 and then minus 5 times 2 is 10. so this is going to be positive 2 versus the negative 2 there okay so when we swap rows okay we just change the sign now think about this here if i had a certain configuration here i've kind of labeled everything if one yields a positive and then i switch what got labeled as kind of x sub 1 y sub 1 with what got labeled as x sub 2 y sub 2 i'm basically just swapping rows so it's just going to change the sign of the determinant the absolute value will still be the same and i have this plus or minus here to take care of any situation where i get a negative so i'm good to go all right so with that being said let's jump into the first example and so we have this triangle with the following vertices we have this one comma three which is going to be this guy right here okay one units to write three units up five comma two that's this guy right here five units to the right two units up and then seven comma nine so seven units to the right 9 units up so right there so those are our three vertices let's just copy them real quick so 1 comma 3 and then we have 5 comma 2 and then we have 7 comma 9. okay so let's copy this and let's go back up and i'm just going to paste this in here real quick and we'll just kind of fill this out we'll go to a fresh sheet because i don't think i can fit everything so the area is equal to plus or minus one half times the determinant of this guy so i'm just going to label these in order i'll just say this is x sub 1 y sub 1 i'll say this is x sub 2 y sub 2 and i'll say this is x sub 3 y sub 3 and again it does not matter what gets labeled as which and just to prove that to you i'll swap it after i do this okay so i'm going to plug in a 1 a 3 and then i'll always have a 1. then i'm going to plug in a 5 a 2 and again i'll always have a 1 and then a 7 a 9 and i'll always have a 1. okay so again i'm just plugging in for x sub 1 y sub 1 that's all i got a 1 and a 3. plugging in for x sub 2 y sub 2 that's how i got a 5 and a 2 plugging in for x of 3 y sub 3 that's how i got a 7 and a 9. so let's copy this real quick so how do i find the determinant quickly let's just copy this one three one five two one seven nine one again the shortcut is to copy the first two columns so you have 1 5 7 3 2 9 and then you're going to start at this top left and multiply down this diagonal 1 times 2 times 1 is 2 and then plus you're going to multiply down this diagonal 3 times 1 times 7 is 21 and then plus you're going to multiply down this diagonal 1 times 5 is 5 times 9 is 45. if i sum that amount 2 plus 21 is 23 23 plus 45 is 68 okay so the first part of this formula is 68 and then for the second part we're going to multiply up now so starting at the bottom left i'm going to go up 7 times 2 is 14 times 1 is still 14 and then i'm going to multiply up 9 times 1 times 1 is 9 then plus i'm going to multiply up 1 times 5 times 3 is 15. so 14 plus 9 is 23 plus 15 is 38. so you have 68 minus 38 which is 30. okay so the first part of this is that the determinant became 30. so the area okay the area is equal to in this case because we got a positive determinant i'm just going to use the plus one-half so i'm just going to say one-half times 30 which is 15. now when you work with triangles remember your formula from basic geometry one-half times the base times the height remember you have units involved with the base and the height so you've got to think about the fact that's going to be square units so if you were working with inches it would be you know square inches or inches squared or whatever that is so you would really want to give a precise answer and say it's 15 square units so i'll just say the area and let me erase this is going to be 15 square okay square units now really quickly just to prove this to you let's say i swap these two rows so let's say i put 5 2 and 1 on the top i'm just going to erase that and i'll drag this up here so really this would correspond to what it would correspond to me choosing this point to be x sub 1 and y sub 1 okay and this point to be x sub 2 y sub 2 okay and because i swapped two rows the only effect this is going to have is going to change the sign of the determinant right so now the determinant is going to be negative 30 and so to get the area i'm going to multiply by negative 1 half and so i'm still going to get 15 it's going to be 15 square units but again let me prove this to you real fast because i know a lot of people say nope you have to go in a certain order you do not all right so let me erase this and let me paste this in really quickly so five one seven two three nine one one one and again i'm copying the first two columns so five one seven two three nine all right so you already know the deal we're gonna start at the top left and we're gonna multiply down five times three times one is fifteen and plus two times one times seven is fourteen then plus one times one times nine is 9. so you'll notice that this is 38 now okay and you're going to be subtracting away 68 which is going to give you negative 30. but let's go through it 15 plus 14 is 29 and then 29 plus 9 again is 38 okay so this is 38 and then minus again if we go up 7 times 3 times 1 is 21 9 times 1 times 5 is 45 and then 1 times 1 times 2 is going to be 2. okay 21 plus 45 is 66 plus 2 is 68 again 38 minus 68 is negative 30. so again i've just proved this to you that swapping these two rows is basically like we've chosen different points to be x sub 1 y sub 1 and x sub 2 y sub 2. all i did was change the sign and i can now just multiply this by negative one-half to get a positive 15 and so again my answer for the area is 15 square units okay let's just look at one more of these it's a very very easy concept so the vertices we're given now is 6 comma 2 negative 4 comma negative 5 and 2 comma 7. so here's your 6 comma 2 here's your 2 comma 7 here's your negative 4 comma negative 5. so let's copy those you have 6 comma 2 you have negative 4 common negative 5 and you have 2 comma 7. okay so let's copy this and let's just go to a fresh sheet we should know the formula now it's pretty easy so the area is equal to one-half and again you've got that plus or minus out in front times the determinant of let's go ahead and label this so this i'm going to call x sub 1 y sub 1 this i'm going to call x sub 2 y sub 2 and this i'm going to call x sub 3 y sub 3. so we know that the first row is coming from this right the x sub 1 y sub 1. so 6 and then 2 the last column the third column is always a 1. so i'm going to just fill that in then this is x sub 2 y sub 2 so negative 4 and then negative 5 then this is x sub 3 y sub 3 so 2 and 7. so very easy formula to remember we don't need these points anymore you just get rid of them they're just going to take up space and let me get rid of that and slide this up out of our way and let me just copy our kind of matrix so we have six two and one negative four negative five and one two seven and one so copy the first two columns six negative four and two and then 2 negative 5 and 7. so let's multiply again starting from here and going down 6 times negative 5 is negative 30 times 1 is still negative 30. then we're going to multiply down this way 2 times 1 times 2 is 4 then this way 1 times negative 4 times 7 is negative 28. so we know that negative 30 plus negative 28 would be negative 58 and negative 58 plus 4 would be negative 54. okay so the first part of this is negative then minus now we're going to go up so we're going to start here and we're going to go 2 times negative 5 times 1 2 times negative 5 is negative 10 and then times 1 is obviously still negative 10. 7 times 1 times 6 is going to be 42 and then we have 1 times negative 4 times 2 which would be negative 8. okay so negative 10 plus negative 8 is negative 18. so what is negative 18 plus 42 that's going to give me positive 24 okay so what you have here is negative 54 minus 24 which is negative 78 okay so negative 78. so let me erase this the value of the determinant of this guy is negative 78. so our area will be equal to because it's negative you're going to use the negative one-half so negative one-half times your negative 78 okay so what is that going to give us well forget about the sign because we know it's going to be positive 78 divided by 2 is 39 so this equals 39 and again when you write your answer you want to put square units so let's just erase this and we'll say that the area is 39 square units in this lesson we want to talk about testing for collinear points using determinants so in our last lesson we talked about how we could use determinants to calculate the area of a triangle given three vertices of the triangle and really quickly i just want to review the formula because we're going to use it here today so the area is equal to you have plus or minus this one-half times the determinant of this guy so you would label one of those kind of given points as x sub 1 y sub 1 another is x sub 2 y sub 2 and then the last one is x sub 3 y sub 3. we went through and proved that it didn't matter what got labeled as which okay because we have this plus or minus here so if the determinant becomes negative okay just multiply by negative one-half if it's positive multiply by positive one-half and you're good to go now what we didn't talk about in the last lesson specifically was what we do when this formula gives us a result of zero meaning there is absolutely no area well it turns out that if you have these kind of three points that you are given they are supposed to be vertices of a triangle and you plug it into this formula and you get zero those three points absolutely lie on the same line right so those three points are collinear again they lie on the same line so we can kind of simplify this formula and say that okay if this part right here evaluates to zero then we know those three points that you gave us are collinear so let's go ahead and use this real quick and look at an example so we're given these three points here we just want to test for collinearity we have 4 comma 1 negative 1 comma negative 2 and negative 6 comma negative 5. so do they lie on the same line so let's just say this is x sub 1 y sub 1 let's say this is x sub 2 y sub 2 and let's say this is x sub 3 y sub 3. it does not matter which gets labeled as which so let me go back up i'll give you a chance to just copy this formula real quick it's very easy to remember the third column of this matrix is all ones and everything else goes in order so you have x upon y sub 1 then you have x sub 2 y sub 2 and x sub 3 y sub 3. so the notation here the sub 1 matches that you're in row 1. the sub two matches that you're in row two the sub three matches that you're in row three right so it's very easy to remember so i'm just gonna plug into that so i would have my x sub one y sub one so i would have a four and then a one this third column's always a one okay so that's easy to remember then you'd have your x sub 2 y sub 2 so negative negative 1 and then negative 2 and then this is always a 1 and then your x of 3 y sub 3 so negative 6 and then negative 5 and this is a 1. so we need to find the value of this if this is 0 those three points are on the same line or we could say they are co-linear what's the quick way to find the determinant again what we want to do is copy the first two columns so 4 negative 1 and negative 6 we have 1 negative 2 and negative 5 and we want to multiply down okay so 4 times negative 2 times 1 would be negative 8 and then plus we want to multiply down 1 times 1 times negative 6 is negative 6. and then we want to multiply down 1 times negative 1 times negative 5 is going to give me positive 5. so negative 8 plus negative 6 is negative 14 then plus 5 would be negative 9. okay so this is negative 9 that's the first part remember you're going to subtract away you're going to go up now so negative 6 times negative 2 is 12 times 1 is still 12. negative 5 times 1 times 4 would be negative 20. and then you would have 1 times negative 1 times 1 which is going to be negative 1. okay so if i do 12 plus negative 20 that's negative 8 okay and then negative 8 plus negative 1 is negative 9. so you see that what you have here remember you have minus a negative so that's plus positive you have negative nine minus a negative nine so this is really negative nine plus nine which equals zero so because this is zero because those kind of three vertices that you were given okay that were supposed to represent a triangle give you an area that's zero when you punch them into the formula you know that those three points lie on the same line they are co-linear points let's look at another example so now we have 3 comma 7 5 comma 10 and 6 comma 6. so again i'm just going to label these in order x sub 1 y sub 1 this is going to be x sub 2 y sub 2 this is going to be x sub 3 y sub 3. so again just plug into the formula so you already know this is going to be what x sub 1 y sub 1 so 3 7 and then there's always a 1. then x sub 2 y sub 2 5 10 always a 1. x sub 3 y sub 3 6 6 always a 1. okay copy the first two columns 3 5 and 6 7 10 and 6. we're going to multiply down 3 times 10 times 1 is 30 and plus 7 times 1 times 6 is 42 and plus 1 times 5 times 6 is going to be 30. 30 plus 30 is 60. 60 plus 42 is going to give me 102. then minus for this one i'm going up so 6 times 10 times 1 is 60 and then plus we have 6 times 1 times 3 which is 18 and then plus we have 1 times 5 times 7 which is 35 so we have 60 plus 18 which is 78 plus 35 which is 113 okay so it's 113. so at this point we can stop we don't even need to do this calculation even though the result will be negative 11 it's not zero so these points are not collinear all right let's look at one more of these i think it's a very easy concept and you just need a few practice problems you basically have it down this is a much better formula to use much faster formula to use versus kind of using the distance formula that we talked about earlier in the course so this is my x sub 1 y sub 1 again i'm just going in order this is my x sub 2 y sub 2 this is my x sub 3 y sub 3. okay so just plug in so again i'm going to say i have negative 12 and negative 4 x sub 1 y sub 1 and then a 1. then i'm going to have my negative 6 my 0 and my 1 again x sub 2 y sub 2 and then a 1 and then i'm going to have 6 8 and 1 again x of 3 y sub 3 and then a 1. so find the determinant of this i'm going to put negative 12 negative 6 and 6 copy the first column then i'm going to copy the second column so negative 4 0 and 8. and let me make this negative 4 over here a little better make it line up a little better so i'm going to multiply down starting so negative 12 times 0 times 1 there's a 0 in there so you know that's 0. forget about it then let me kind of move this down so it lines up we're going to have negative 4 times 1 times 6 that's going to be negative 24 then plus you have 1 times negative 6 times 8 that's going to be negative 48 okay so negative 24 plus negative 48 is negative 72. so that's the first part so let me change colors it's negative 72 and then minus we're gonna go up six times zero times one is obviously 0. 8 times 1 times negative 12 is negative 96 and then 1 times negative 6 times negative 4 we know that negative 6 times negative 4 would be 24 so plus 24. so what is negative 96 plus 24 well that gives us negative 72 okay so again if you have this minus a negative okay if you have minus a negative it's plus a positive so you have negative 72 plus 72 which is zero so these three points lie on the same line they are co-linear in this lesson we want to talk about finding the equation of a line using determinants all right so over the course of kind of the last two lessons we've been kind of working with this formula which uses determinants to find the area of a triangle given three vertices and then we also kind of expanded on that and we looked at a formula to determine if three points were collinear so let's just recap that real quick we know that for the area of a triangle we can basically get three points or three vertices for that triangle okay and we can label one as x sub one y sub one the other is x sub two y sub two and kind of the final one is x sub 3 y sub 3. doesn't matter what gets labeled as what okay so you'd plug into this guy right here you take the determinant and if that result was negative you'd multiply by negative one-half so that you got a positive area if that result was positive you'd multiply it by a positive one-half so that you got again a positive area so we found that if the area ended up being zero well the only way that can be true is if those three points that you kind of started with were on the same line okay so that becomes the test for collinearity so if this part right here okay evaluates to zero so i take the determinant i get zero i know those three points that you gave me which were x sub 1 y sub 1 x sub 2 y sub 2 and x sub 3 y sub 3 are collinear again that just means they lie on the same line now we can expand on this even further and use this kind of formula to find the equation of a lot so we already talked about how to do this kind of earlier in the course we use kind of the typical algebra one method which is if you're given two points that lie on the same line what you do is you first calculate the slope okay and you do that with your slope formula and once you have that you can plug into the point slope form of the line okay that y minus y sub 1 equals m the slope times the quantity x minus x sub 1 and from there you can get the equation of the line you can solve for y and put it in slope-intercept form or you can put it in standard form if you want you can do whatever you need to do okay but we can do the same thing kind of using this formula here so you'll notice instead of kind of three given points now we just have two okay so we have x sub one y sub one that's one point and then x sub two y sub two that's another point so the only thing that's changing here is we don't have three points that are kind of known we only have two the first point this x and y okay are unknown so we just leave them as those two variables those are going to be the two variables in our equation okay when we set it up so all you need to do here is just kind of go through your process for getting a determinant and then you're going to set that equal to 0. so let's just look at an example real quick this is not a hard concept so we're given these two points so we have 3 comma 1 and 2 comma 7 and we want to find the kind of equation of the line that goes through those points so i'm going to i'm going to do the equation in slope intercept form the y equals mx plus b form okay but you can do it in standard form if you want as well so what i'm going to do is just kind of plug into my formula okay so i know this equals zero and again it's x y and then a one it's x sub one y sub one and then a one and then it's x sub two y sub two and then a one so let's just label these in order it doesn't matter you get the same result so let's say this is x sub 1 y sub 1 let's say this is x sub 2 y sub 2. and we're just going to plug in so i'm going to erase x sub 1 y sub 1 and i'm going to put a 3 and a 1. i'm going to erase x sub 2 y sub 2 and i'm going to put a 2 and then a 7. okay so from this point let me kind of copy this i want to get some room going and let me paste this in and let me just get rid of this stuff so how do we find the determinant let me kind of move this down how do we find the determinant we copy the first two columns that's the quick way to do it so we'd have x you would have 3 you would have 2 you would have y 1 and 7. and just go through your normal procedure okay and we're going to set that equal to zero so the first thing i would want to do is multiply going down okay and this isn't really written too well so let me kind of scooch this down so everything kind of lines up a little better so i would start at this kind of top left and i would multiply down this diagonal so x times 1 times 1 is x then plus you'd multiply down here so you'd have y times 1 times 2 and i could make that a little better so let me try to angle that a little better so y times 1 times 2 is 2y and then plus you're going to multiply down this diagonal 1 times 3 times 7 which is 21. so that's the first part let's just put that in brackets and then we're going to go minus i'm going to erase this so we can see what we're doing so now i'm going to go up so i'm starting at the bottom left and i'm going to go up 2 times 1 times 1 is 2 and then plus we're going to go up 7 times 1 times x is 7x and then plus we're going to go up one more time 1 times 3 times y is 3y okay so let me erase all of this we don't need this information anymore i'm just going to scooch this back down okay and we'll do that like this and what i'm going to do is i'm going to say that this guy right here since this is the determinant i'm going to say it's equal to zero okay so we're just setting up an equation it's the same thing as this i'm saying this determinant is equal to zero this is the determinant just written out okay as the steps we need to take to find it it's still equal to zero so let's go down a little bit and get some room to work so i'm going to drop the brackets from kind of the first part so just x plus 2y plus 21. from the second part i'm going to be distributing that negative to each part so it would be negative two minus seven x and then minus three y okay so this equals zero that's why it's so important to use brackets so you don't make a sign mistake so now i'm just going to combine like terms we have x minus seven x which is going to give me negative 6x and then we have 2y and negative 3y which is minus y and then we have 21 minus 2 which is plus 19 and this equals 0. so from this point we can again we can put it in standard form if we want we just subtract 19 away from each side but what i want to do is i want to put it in slope intercept form so i am going to add 6x to both sides of the equation and we know that this part would cancel i would have negative y and then plus 19 is equal to 6x let me scroll down a little bit get a little bit more room i want to subtract 19 away from each side and so this cancels obviously so we have negative y is equal to 6x minus 19. so to finish this up i'm just going to multiply everything by negative 1. so this would become positive this would become negative and this is positive so i can just erase this and say that my equation in slope-intercept form for this line is y equals negative 6x plus 19. now let me copy this real quick and let me go back up and let me paste this in here real fast so we're going to use that as our reference i'm just going to go through this really quickly with the kind of old-fashioned algebra 1 method just so that you see that this does work so again if this is x sub 1 y sub 1 and this is x sub 2 y sub 2 okay the slope formula says what m is equal to y sub 2 minus y sub 1 so 7 minus 1 over x sub 2 minus x sub 1 so 2 minus 3. 7 minus 1 is 6 2 minus 3 is negative 1 so the slope is negative 6 and we already know that because it's right there so what we want to do next now that we have our slope and let me just label that i'll say m m my slope is equal to negative 6 i would use one of the points again it doesn't matter which one and plug into that point slope form of the line so that's the y minus y sub 1 for y sub 1 i'm going to use 1 is equal to m my slope which is negative 6 times the quantity x minus x sub 1 in this case x sub 1 is 3. okay so what i would do to solve for y i would distribute this so negative 6 times x is negative 6x and then negative 6 times negative 3 is plus 18 okay and this equals over here i have minus 1 and then my y so to solve for y what do i need to do just add 1 to both sides of the equation and i would wrap this up and say that y is equal to this cancels so you have your negative 6x plus 19 which is exactly okay exactly what we have here those are the same so it's the same thing i would argue that kind of using this method with determinants is probably a little bit faster just depending on you know your speed with kind of doing things definitely if you use the shortcut for determinants it's going to be a little faster because you don't have to go through and calculate the slope first and then plug in and kind of get your equation let's take a look at another one so now we have negative 1 comma 5 and 7 comma negative 3. again i'm just going to label this as x sub 1 y sub 1 and this as x sub 2 y sub 2. it does not matter the way that you label the points you get the same answer either way so what i want to do is plug in to my formula again the first one here the first row is just x y and 1. then the second row is x sub 1 y sub 1 and 1. and that is x sub 2 y sub 2 and 1. so it's very easy to remember again this gets set equal to 0. for x sub 1 i have negative 1 okay negative 1 and then for y sub 1 i have 5 then for x sub 2 i have 7 and for y sub 2 i have a negative 3. all right let's copy this again we're just going to go a fresh sheet so that we don't run out of room and let me just paste this in here and i'm just going to slide this down so i'll just put this out here so this equals 0. again i'm just going to copy the first two columns so x negative 1 and 7 y 5 and negative 3. so what we want to do is multiply down to start x times 5 times 1 is 5x then plus multiply down y times 1 times 7 is going to be 7y then plus you've got 1 times negative 1 times negative 3. you've got two negatives there which makes a positive so it's basically just 3 times 1 times 1 which is 3. so let's put this in some brackets again it's not going to change anything but we want to do that just to be consistent now this is where the brackets kind of help you because it reminds you to kind of distribute that negative to everything so let me erase all of this and now we're going to go up so 7 times 5 times 1 is 35 and then plus negative 3 times 1 times x is going to be negative 3x so let me just put minus 3x and then lastly you have 1 times negative 1 times y which is going to be negative y i'm just going to say this is equal to 0 and the reason that works is because again if i go through and kind of calculate the determinant this is what i would do and that's supposed to be equal to 0 so that's why this is legal so let's just go through and erase this and we can just slide this down if you want or just erase it and kind of redo it and let's just kind of scroll down and get some room and go through and set this up so i can drop the brackets from the left side i don't need to worry about that so 5x plus 7y plus 3 no change when you drop the brackets the negative here has to be applied to each term okay so that's why the brackets are very important there so you have negative 35 you'd have plus 3x and you'd have plus y and of course this equals 0. so all i really need to do now is just combine like terms i see that i have a 7y and a y that's basically 1y so that would be 8y so let's write that first then you have this 5x and 3x so you combine those and you're going to get plus 8x and then lastly you have this 3 minus 35. so what is 3 minus 35 that is negative 32 and of course this equals zero okay so what i want to do now if i want to solve this for y you can divide everything by 8 now or you can wait it doesn't really matter if i divide everything by 8 what i'm going to get is let me kind of scroll down a little bit i'm going to get y plus x minus 4 equals 0. 8 divided by 8 is 1 8 divided by 8 is 1. negative 32 divided by 8 is negative 4 and 0 divided by 8 is 0. so at this point all i really need to do is subtract away x and add 4 to both sides so in other words i would do minus x and plus 4 to both sides of the equation so minus x and plus 4. and so that's going to give us a final answer of y is equal to again this is all going to cancel negative x plus 4 okay so that's the equation of our line all right let's just look at one more it's a very easy concept once you kind of look at two or three of these you basically have the concept down so we have one point that's negative two comma negative two again let's just say this is x sub one y sub one and the other point is three comma eight this will be x sub two y sub two you already know what to do we're plugging into the formula so you have x y and 1 then you have x sub 1 y sub 1 and 1 so x sub 1 is negative 2 y sub 1 is negative 2 and then you have a 1 and then lastly you have x sub 2 which is 3 y sub 2 which is 8 and then 1 again and this is equal to 0. okay always set this equal to 0. and then let's just copy this real quick and come to this sheet here that way we have lots of room to work let me kind of scooch this down a little bit so we have some room again i'm just going to copy the first two columns so x negative 2 and 3. you have y negative 2 and 8. and again i'm just going to multiply down x times negative 2 times 1 is going to be negative 2x then plus let me put some brackets around this we have y times 1 times 3 which is 3y then plus 1 times negative 2 times 8 is negative 16. so let me just put minus 16 there close the brackets and this is minus again we're going to use some brackets so we don't make a sign mistake so let me erase this and now we're going to go up okay so we're going to go 3 times negative 2 times 1 which is negative 6 then plus we're going to do 8 times 1 times x which is 8x and then we're going to go up 1 times negative 2 times y is going to be minus 2 y okay and of course this equals 0. so let's scroll down a little bit and let's just apply this negative to everything on the left side i can just drop those brackets they don't do anything so negative 2x plus 3y minus 16. distribute the negative to each term over here so this would be plus 6 minus 8x and then plus 2y again i'm just changing the sign of each term there so this equals 0. now i can just combine some like terms so i have negative 2x and negative 8x that's going to give me negative 10x i have 3y and 2y that's going to give me plus 5y i have negative 16 plus 6 which is going to give me negative 10 so minus 10 and this equals 0. so everything is divisible by 5. so i can just go through and do that now divide this by 5 this by 5 this by 5 and this by 5. so negative 10 divided by 5 is negative 2 so that would be negative 2x and then plus 5 over 5 is 1 so this is just y negative 10 over 5 is again negative 2 and this equals we know 0 divided by anything that's not 0 is 0. right so 0 divided by 5 is 0. so at this point let me kind of scroll down i can just solve for y by adding 2x to both sides and adding 2 to both sides okay so this is going to cancel and this is going to cancel so you'll have that y is equal to 2 x plus 2. in this lesson we want to talk about partial fraction decomposition with linear factors all right so when you get to this section on partial fraction decomposition you might get a little bit confused it is a bit of a dense topic it's just one of these things that you have to work a lot of examples on and so what i'm going to do is split this up we're going to have this lesson today where we talk about linear factors this is an easier scenario just to kind of get our feet wet and then in the next lesson we'll talk about the quadratic factors which is a little bit more complex all right so to get things started we already know how to add fractions or how to add rational expressions we know we need a common denominator so if i started out with something like 7 over x minus 4 plus negative 3 over x plus 3 we know that we have to get a common denominator and the least common denominator here would be the product of these two denominators so x minus 4 times x plus 3. so to add these two what i would do is i would multiply each fraction by what's needed to have the lcd as its denominator so in this particular case i would multiply 7 over x minus 4 by what's it missing it's missing the x plus 3. so x plus 3 over x plus 3. right you got to make it legal then plus the negative three over x plus three so what's this guy missing it's missing the x minus four so times the x minus four over the x minus four okay so at this point we already know what to do and i'm just going to kind of write this up here so if we did 7 times x that's 7x and then plus 7 times 3 is 21 and then you have negative 3 times x which is minus 3x and you have negative 3 times negative 4 which is plus 12. now if you simplify 7x minus 3x is 4x and if we did 21 plus 12 you get 33 okay so that would be my numerator and my denominator would be either written as x minus 4 times x plus 3 if you wanted to leave it in factored form okay or you could go through and write it as x squared you would have minus x and then you'd have minus 12 okay so you could write it either way now i'm going to erase this factored form we'll need it in a second but i just want to present the problem as you will be given it by your textbook or your teacher or however you're getting these problems now we know how to go from this form to this form but what you want to do now is go from this form to this form okay so you want to take it from this and go to this this is going to be very important when you get to calculus and start working with integration now what we want to do is first determine if we have a proper fraction okay so in this case what i'm saying here is that the degree in the numerator needs to be smaller or less than the degree in the denominator so this polynomial up here has a degree of one this guy has a degree of two so it's a proper fraction we're good to go so once you've determined that you have a proper fraction you're going to factor your denominator okay so we know what this factors into it's pretty easy in this case so 4x plus 33 again i'm going to factor the denominator so it's x minus 4 times x plus 3. okay so here's where it gets a bit tricky for each distinct linear factor okay in this particular case we factor this this is a linear factor this is a linear factor those are each different okay so for each distinct linear factor you're going to set up a fraction where that distinct linear factor is in the denominator so i'm going to have one fraction where this is the denominator i'm going to have another fraction where that guy's the denominator okay so depending on how many factors you have will determine how many fractions you end up with okay so in this case i have two of these guys so i have two here okay now what's unknown up here those are going to be the constants so what's going to be over x minus 4 well we already know that that's going to be 7. but we don't know that just yet because again we're pretending we haven't worked the problem so we're going to use something like a to represent that unknown okay then for x plus 3 we know what's on top of that it's going to be negative 3 but again we don't know that so we're going to say that it's going to be b okay so your book uses a and b capital letters you could use a sub 1 a sub 2 things like that if you want it really doesn't matter you're just saying that there's some unknown constant up there that we're going to solve for now there's different ways to solve these and the way i'm going to show you right here is from your textbook it does take a little bit longer but it works every time okay so i'm going to clear the denominators and the way i'm going to do that is multiply both sides of the equation by the lcd so if i multiplied 4x plus 33 over this x minus 4 times x plus 3 which would be the lcd so i'm going to multiply by that and this would cancel and i'm just left with this okay so i'm just going to erase this as i do this okay so this is what i have now now if i multiply this a times this x minus 4 times x plus 3 and again this is over x minus 4. well what's going to happen is this will cancel with this and i'll have a times this x plus 3. so let me just kind of erase this and i'll erase that and i'll say this is a times this quantity x plus 3. okay and one more to do so let me erase this we'll do this one more time so we have b over x plus 3 times again this quantity is x plus 3 times x minus 4. so this is going to get erased and we'll put plus b times this cancels with this you have your quantity x minus four okay so we've cleared all the denominators so at this point what you want to do is you want to match your form from this side to this side so what do i mean by that well you have some number times x and then plus some constant okay so that's what i want i want some number times x plus some constant okay and this is going to make more sense in a second so let me go through and just distribute things so this is a times x plus 3 times a and then plus b times x is bx and then minus b times 4 is 4b okay so over here this is the same 4x plus 33. now what i want to do is get all the terms with x's together so i can kind of factor okay so let me show you that so we're going to have 4x plus 33 is equal to let's say that we have a times x plus b times x i'm going to put this as a group and then plus i'm going to make another group with just 3a minus 4b okay and this all makes sense in a minute so let's scroll down a little bit more and get some room so what i want to do now again i have this 4x plus 33. so again it's some number times x okay and then plus some constant so how can i do that well i have this x that i can factor out right it's common to here and here so pull it out so you have x and then you're going to have this a plus b and if you want it to look like this you can put the x behind it it doesn't matter okay so let's scooch this down a little bit and then plus you have this 3a minus 4b okay so what do we have here let me just erase this so we can kind of match things up we have this guy right here has to be equal to this guy right here okay because again this equality here says that 4x plus 33 is equal to a plus b this quantity times x plus this quantity 3a minus 4b okay so i could just match up this coefficient with this coefficient okay and then i can also match up this guy with this guy okay that's all i need to do first time you see this it does get a little bit confusing but you work enough examples and again it's very very simple all right so let's scroll down and get some room going and what i want to do now is just say that what this 4 is going to be equal to a plus b and i'm also going to say that 33 is equal to this 3a minus 4b okay so basically what i have here is a system of equations a very very simple one so at this point you can solve this with substitution elimination whatever you really want to use i'm just going to use substitution because i think it's the easiest so i'm going to actually just copy this because we're going to run out of room and come over here to a fresh sheet and i'm just going to say that a is equal to if i subtract b away from each side i'm going to have 4 minus b okay so now i can plug in here so i'll have 33 is equal to 3 times i'm going to plug in 4a so i'm going to plug 4 minus bn then minus 4b okay so we have 33 is equal to 3 times 4 is 12 and the minus 3 times b is 3b and then minus 4b okay so negative 3b minus 4b is negative 7b so let's just write that if i subtract 12 away from each side of the equation this is going to give me 21 over here and over here i'm going to have negative 7b and if i divide both sides by negative seven okay i'm going to find that b is equal to negative three and we already knew that because we kind of started out with that information so b is negative three so let me erase this and come back up here we'll just say b is equal to negative 3. and let's find a now so if i plugged into this simpler one 4 is equal to a and then you have minus 3. add 3 to both sides of the equation and of course we find that a is equal to seven okay but again we already knew that so b is negative three and a is seven so if i come back up here and just write my answer so a is again seven and b is negative three again the form that we had we said that we had a over x minus four and then plus you had b over x plus three okay so again this is equal to this so we can say this is equal to 4x plus 33. over x squared minus x minus 12. so again a is 7 and b is negative 3. and again we already knew this because we started out with this problem in this format we went this way and now we've gone in reverse so that's a simple example i realize it's tedious as time consuming but you got to work a lot of these to kind of get used to what's going on let's look at another example and this one is going to be a little bit more challenging so we have negative 15x plus 10 over 9x squared minus 6x plus 1. okay so again the first thing is make sure that you have a proper fraction so this guy has a degree of one this guy has a degree of two so we're good to go okay the next thing you want to do is factor the denominator now you'll notice that this guy is a perfect square trinomial okay because 9x squared is a perfect square 1 is a perfect square so you think about this would factor into 3x and then you would do minus 1 that quantity squared right so i'm going to write this as negative 15x plus 10 over you have your 3x minus 1 quantity squared now i'm going to tell you whenever you get something like this you have to be very careful okay if you have a linear factor raised to a power okay that's not 1. so if it's 2 or 3 or 4 you have to do something called build up the power okay so let me explain what that is basically what you want to do when you're setting this up you're going to take this guy raised to the first power so i'm going to have my a over this 3x minus 1. then plus now i've got to do this again so b over now i'm going to do 3x minus 1 but it's going to be squared okay now here's where students get confused they see this down here and they just include this and they leave this off you don't want to do that if this was let's say cubed you'd need to do another one you'd need to go plus c over 3x minus 1 cubed so you need the first power the second power the third power you're going all the way up to whatever this power is if this was a four you'd put plus you'd go d over you'd have three x minus one to the fourth power you know so on and so forth so you've got to be careful when you're working with these guys this is called a repeated linear factor so you've got to build up that power so if it's 3x minus 1 that quantity squared you've got to have 3x minus 1 to the first power and also you've got to have 3x minus 1 squared so make sure you include both of those very important all right so i'm going to set this equal to this negative 15x plus 10 over and you can leave this in factored form it doesn't really matter so 3x minus 1 quantity squared okay so let's go through and multiply everything by the lcd we should be able to do this mentally because it's pretty easy if i multiply this guy right here by this 3x minus 1 that quantity squared i know i'm just going to be left with the numerator so i'm just going to have negative 15x plus 10. then over here this is going to cancel i'm just left with b then over here what's going to happen let me do this one kind of manually here because it's hard to visualize so a over 3x minus 1 times you'd be multiplying by 3x minus 1. you could say squared or you could just do this 3x minus 1 again one of these is gonna cancel so i'll have a times the quantity 3x minus one okay so at this point we've cleared the denominators so what i'm going to do is again just stick with the textbook method so nobody gets confused and i'm just going to distribute the a so i'm going to have 3 a x and then minus a and then plus b this equals negative 15 x and then plus 10. okay so at this point i want to really just group this i'm going to put this as plus negative here and i'm going to say that this guy right here is what it's the coefficient of x in each case so that means that 3a is going to be equal to negative 15 divide both sides by 3 and we'll say that a is equal to negative 5. so we already know what a is now how do we find b well you could plug in for a in the equation or again you can just take this negative a plus b and say it has to be equal to 10. we already know what a is so we're pretty much there so we would say that negative a plus b is equal to 10. again a is negative 5. so the negative of negative 5. got to be careful there because you have a negative out in front plugging in a negative so this would be plus 5 plus b is equal to 10 subtract 5 away from each side you get b is equal to 5. let me make that a little better so a is negative five and b is positive five so coming back here we know that a is negative five so let me just put negative five here we know b is positive five okay so if you went through and kind of did this addition you would end up with this exact okay so this is the point we can go from this form to this form and from this form back to this form let's look at a different type of example so now we're not going to have a proper fraction so what do you do in this scenario so we have negative 2x squared plus 3x minus 4 and x squared plus x minus 2. so what do we do when this happens right the degree here is 2 the degree here is 2. for a proper fraction the degree in the numerator has to be less than the degree in the denominator so we're just going to do some long division and work with the remainder okay so it's pretty easy in this case so you have negative 2x squared plus 3x minus 4. this is only going to take a second i know everyone hates long division and then you have this x squared plus x minus 2. well really all you're going to be able to do is you do x squared into negative 2x squared that's going to give you negative 2. okay and you're basically done you should be left with a remainder so negative two times x squared is negative two x squared negative two times x is negative two negative two times negative two is going to be plus four but you're subtracting this away okay so remember you change the sign of everything i could put a minus out in front but really i'm just going to drop those parentheses and change all my signs and just add so that's plus that's plus and this is minus so what you end up with is this canceling this being 5x and then minus 8. so that's your remainder okay so really i could write this as negative 2 plus 5 x minus 8 over that denominator which is x squared plus x minus two okay so we got that part done and once you're finished with that you have a proper fraction here so i'm going to work with this don't forget the negative two part it's very important when you report your answer but for right now we're just going to work with that proper fraction so 5x minus 8 over x squared plus x minus 2. so we know it's proper so factor the denominator this guy is going to factor into what so it's going to be x here and x here i know that it's going to be plus 2 n minus 1. okay let's just erase that and move this up here and again what do we do well for each distinct linear factor i've got to have a fraction with a denominator with that guy so i'm going to go a over you've got x plus 2 plus b over you've got x minus 1. okay so if i go through and clear the denominators again the lcd is going to be this guy right here so i know in this case it would clear it so i would just have five x minus eight in this case the x plus two would cancel with the x plus two so i would have a times the quantity x minus one okay and then in this case i would have the x minus 1 cancel with the x minus 1 and b would multiply by x plus 2. so plus b times x plus 2. all right so we got that worked out so let's just distribute now so this would be ax minus a plus bx plus 2b and again this equals you have your 5x minus 8. okay so let me get everything with x together so i would say this is a times x plus b times x and then plus i'm going to group these together so negative a plus 2b okay so let's write this over here and again all i'm trying to do is relate the two sides so this coefficient here is going to relate to this guy over here that i'm going to set up i'm going to say this is a plus b out in front times x okay all i did was factor the x out let me just kind of make this a little better so we have 5 x minus 8 equals so this is equal to that okay then plus this is going to be your constant so your negative a plus 2b that's your constant so this is going to relate to that okay so once you've set this up and again you just need to work a lot of examples so you get used to this all you need to do is say that okay i know that a plus b is equal to 5 and i know that negative a plus 2 b is equal to negative 8. now this is a system you do have to spend a a few seconds to solve it it's not too bad i know a plus b equals 5 so a if i wanted to solve for that is what it's 5 minus b so if i plugged in for a here i would have the negative of 5 minus b and then plus 2b is equal to negative 8. so let's go ahead and solve this really quick so i'd have negative 5 plus b and then plus 2b is equal to negative 8. so this would end up being 3b and if i add 5 to both sides i would get negative 3 over here so this is negative 3. if i divide both sides by 3 i'll find that b is negative 1. okay so let's erase everything here we don't need any of this so if b is negative one so if this is negative one or minus one then what i add one to both sides and i find that a is equal to six so b is negative one and a is six okay so here's where you got to be careful with this when you write your answer remember we did a division first so you got another part to this so the a is going to be 6 and the b is negative 1. i'm just going to erase this real quick remember this is only part of your answer so if you come back up here you had this okay so what i'm going to do is i'm going to say this is negative 2 plus this right here okay so this right here and you can erase this part is equal to this okay so your negative 2x squared plus 3x minus 4 over x squared plus x minus 2 would be equal to negative 2 plus 6 over x plus 2 plus negative 1 over x minus 1. again don't forget this part here it's very important to have a full answer i know a lot of students they run through these problems they forget the negative two and then they get points off okay let's run through one more problem this one is not not harder but it's more time consuming you'll get these sometimes so you have negative x squared minus 30x minus 13 over x cubed minus x squared minus 5x minus 3. so it is a proper fraction so we don't have to do any division to start but when you go to try to factor the denominator it's not something you can factor with grouping this is something where you have to do the rational roots test okay and then start using your synthetic division to go through it i don't want to make this video any longer than it needs to be so i'm just going to tell you in advance that this guy is going to factor into x plus 1 that quantity squared let me make that better and then times your x minus 3. okay i'm just going to put that out there again if you're working these in a workbook or something like that you're going to have to take the time to use your rational roots test and then go through the possibilities to see if you can factor this guy all right so once you've figured this out and you have this factored again all i'm going to do is i'm going to say let me just kind of move this over here and let me put this up here so negative x squared minus 30 x minus 13. okay so again what i was saying is all you want to do is set this up where each one of these guys each distinct linear factor has a fraction where it's in the denominator okay but again if you have something with a power here it's raised to the second power so i need x plus 1 to the first power and i also need x plus 1 squared again this if this was to the third power i would need x plus one to the first power x plus one to the second power and then x plus one to the third power right so on and so forth so what i'm gonna do is i'm gonna say that i have a over i've got my x plus one and then plus b over i've got my x plus 1 that quantity squared and then plus my c over you've got your x minus 3. okay so this is the setup for this one it's going to be a little bit more time consuming so let me just kind of scroll down just a little bit and try to fit everything on the screen so we know that we would clear the denominator so this would be equal to this i'm not going to be able to fit that side by side but if we multiply both sides by the lcd we know that this is going to cancel completely so i'm just left with this numerator so i'm going to squeeze this in and say we have negative x squared minus 30x and then minus 13. okay so then equals for this guy when i go through and multiply each one of these guys by the lcd you can think about what's going to cancel well in this case i have one factor of x plus one here have two of them so i would have a times one factor of x plus one times this x minus three and i'm not going to fit all this so i need to go through and simplify each time so this would be what this would be x squared and then this would be minus 3x and then plus x and then minus 3. so this would be minus 2x right here and then minus 3. okay so from this point let me just kind of erase this we know that a would multiply that so minus 2x minus 3. and if you wanted to you could go through and distribute it now but i'm going to do it in a minute so then you move on you have plus b over this quantity x plus 1 squared again times this lcd so this cancels with this and you have the x minus 3. so b times the quantity x minus 3. okay and i'm not going to be able to fit this in one row so i'm just going to continue this down here so i'm just going to draw a little arrow say we've got to go down there so now what i'm going to have is the x minus 3 cancel and i'm going to have c times i'm going to have this x plus 1 this quantity squared okay so let me scroll down a little bit now that we have everything we'll get some more room going all right so i'm going to have negative x squared minus 30x minus 13 is equal to i'm just going to distribute this to everything so a x squared minus 2 a x minus 3 a and then plus you've got b x minus 3 b and then plus i'm going to continue this down here now you should expand this to not make a mistake a lot of students will just say this is cx plus c okay that's wrong remember this guy means you have to foil it out so really i could say this is x squared x squared plus 2x plus 1. what this would turn into down here is c x squared plus 2 c x plus c what you want to do here is just think about the fact that you have a negative 1 for the coefficient of x squared a negative 30 is the coefficient for x and a negative 13. so you want to just match the format of what you have on the left side okay in some cases you might not have a quadratic over here but you might have a quadratic over here and in that case what you're going to do is you're going to treat it like if you had 0 on this side okay so if you didn't have an x squared over here you would just say that you had 0x squared okay and then 0 would be matched up okay but that's not what we have here this is a little bit easier and basically what i want to do is say i have my negative x squared minus 30x minus 13 is equal to so everything with an x squared so you have ax squared and you have cx squared so i'm just going to have a plus c as that coefficient and then times x squared and then i'm just going to check these off as i go so i can keep track of what i've used then i'm looking for everything with an x so you've got an x here and here and also here okay so we would do plus i'd have negative 2a and then plus b and then plus 2 c and then times x again all i'm doing is i'm taking these terms you can imagine i'm just writing them next to each other i'm just factoring the x out so the x is just being pulled out in each case and it's outside of a set of parentheses that's all i'm doing okay then i'm going to continue this down here so plus now we're going to have everything that's kind of a constant so this is gone this is gone and this is gone so now i have negative 3a negative 3b and c so negative 3a i have negative 3b and i have c okay so now we have something where we can go through and set up a system so we know that again we're just kind of equating the two sides so this is negative one here and you can say that this has to be equal to this so i know that a plus c has to be equal to negative 1. i know that this right here the coefficient for x has to be equal to this negative 30 okay so i can say that negative 2a plus b plus 2c is equal to negative 30. okay and then i know that this guy right here okay which represents the constant is going to be equal to negative 13. so negative 3a minus 3b plus c equals negative 13. okay so let me copy this because we're definitely going to run out of room and let me just go over here to refresh sheet and we can solve this system pretty easily what i'd want to do is take these two right here and eliminate b okay that would give me a linear system with just a and c right so i'm going to add these two together and what i'm going to do is just multiply this top equation of these two by positive 3. okay so if i do that i would have negative 6a and then plus 3b and then plus 6c and this equals negative 90. okay so then i have this negative 3a minus 3b plus c equals negative 13. okay so if i add these two equations together what i'm going to have is this would be negative 9a this would cancel i would have plus 7c and this equals this would be negative 103. so now let me just kind of write this over here i have negative 9a plus i have 7c and this equals negative 103. okay so that's going to allow me to find a and c so what i'm going to do let me just kind of drag this down here put it off to the side i'm going to solve this for a and i'll say that a is equal to negative 1 minus c then i'm just going to plug in over here so negative 9 times this negative 1 minus c and then plus 7c equals negative 103. okay well this would be positive 9 plus 9c and then plus 7c equals negative 103. 9 plus 7 is 16 so this would be 16c and then if i subtract 9 away from each side of the equation i'm going to have that 16c is equal to negative 112 divide both sides by 16 and we're going to find that c is equal to negative seven okay so we know what c is so c is negative seven okay so what's a well if i plugged in a negative seven there we'd have a minus seven equals negative 1 add 7 to both sides you get a is equal to 6. so a is 6. okay so at this point you can kind of get rid of everything except for this equation here you can really use either equation with b but i'm just going to use this one so i'm going to plug in now so i'd have negative 2 times 6 plus b plus 2 times negative 7 equals negative 30. let's just get rid of that we don't need it so we'd have negative 12 and then 2 times negative 7 is negative 14 negative 14 plus negative 12 is negative 26. so let's put negative 26 plus b equals negative 30. let's add 26 to both sides of the equation so this is gone and we find that b is equal to negative four okay so b is negative four all right so now we found a b and c again some of these are going to take a very long time okay so we come back up here and again c was negative seven so you can either put minus seven like this or you put plus negative it doesn't matter whatever you wanna do a is six and then b is negative four so again you just put minus four like this so when we write this out we could say that our negative x squared minus 30 x minus 13 over x cubed minus x squared minus 5 x minus 3 is going to give us this 6 over x plus 1 and then minus 4 over x plus 1 squared and then minus 7 over x minus 3. in this lesson we want to talk about partial fraction decomposition with quadratic factors all right in the last lesson we introduced the concept of partial fraction decomposition and we looked at the easier scenario where you just have these linear factors so this is really a simple process and i went through the textbook method so that when we got to the harder examples you would already know what we were doing okay i know there's a lot of shortcuts that are involved with this but if you stick to the textbook method with the easy ones when you get to the harder ones it's very easy to follow the process all right so i'm going to start out with something that's going to end up giving me a quadratic factor and we'll explain as we go what to do so we have negative 5x squared plus 4x plus 37 and then over x cubed plus 3x squared minus 5x minus 15. okay so the very first thing you do when you get one of these problems is you ask yourself do you have a proper fraction do you have a numerator that has a degree that is less than the denominator well yeah we do the degree of this polynomial up here in the numerator is 2 the degree down here is 3. remember if you don't have a proper fraction you have to stop and do long division okay and then move on from there and we talked about that in the last lesson all right so now once we've established that we have a proper fraction we need to factor our denominator for some of these problems when they get more difficult they'll usually give you a factored denominator because they just take so long okay so for this one it's pretty easy to factor we can use grouping from the first group i could pull out an x squared and that would give me x plus three from the second group i could pull out a negative five that would give me x plus three so we can see that this would factor into what x squared minus five so x squared minus five that's one factor and then the other factor would be x plus three okay so i can go ahead and erase this and i'm just going to rewrite this problem with a factored denominator so negative 5 x squared plus 4 x plus 37 okay so now when we go to break this up let me scroll down and get some room going when we go to break this up remember if we have a linear factor something like x plus 3 that's linear because again the x here the variable is raised to the first power okay so that's linear we set up a fraction where this guy's in the denominator and in the numerator you're going to have an unknown constant so we call it a or b or whatever you want to use it doesn't really matter as long as you're picking different letters each time okay so the thing to notice here is that this is to the first degree right this is a first degree polynomial or it's a linear factor that came from over here okay so this guy up here is going to be a constant what's in the numerator when you set these up is going to be one degree less than what's in the denominator this has a degree of one this has a degree of zero okay because i can say this is a times x to the power of zero if you wanted to be super technical all right so now what do we do with this factor over here this is a quadratic factor and notice how i say it's a quadratic factor a lot of times you get a quadratic factor but you could factor that and make linear factors that's not the case here okay so this is irreducible something you can't factor any further using rational numbers so x squared minus 5 what do i do with that well i'm going to set up a fraction in the same way and let me make that a little better but what's going to change is what goes up here now i'm not going to have just a constant i've got to have a linear expression up here okay so what i want is something that looks like this so bx plus c okay so this is a second degree and this is a first degree so the numerator if you just remember this is always one degree less this guy is one so this guy is zero this guy is two so this guy is one okay so that's a good way to remember it now remember if you have a repeated linear factor or repeated quadratic factor you've got to build up the power we'll see an example of that in a minute so once i have this set up what i want to do is i want to clear the denominators and the way i would do that is i would multiply both sides by the lcd so this is going to be the lcd right here and if i multiply this left side by that i would just get the numerator so i would just have negative 5 x squared plus my 4x plus my 37. now if i multiply this guy by this the x plus 3s would cancel and i would have a times the quantity x squared minus 5 then plus as i move on to this if this was multiplied by the lcd then the x squared minus 5 would cancel with the x squared minus 5 and i would just have the x plus 3. remember you have to foil this though because this is a quantity times this quantity so bx plus c and then times you've got your x plus three okay i'm just going to warn you in advance when you work these problems with these quadratic factors it is way more tedious than when you just have linear factors okay so what i want to do again is on the right side i want to match the form of the left side so i want something times x squared plus something times x plus some constant okay so the way we do that is we just simplify everything so i'll say this is ax squared minus and this is going to be a times 5 or 5a if you want to put it that way and then plus over here i've got to do foil so bx times x is bx squared my outer would be 3bx so plus 3bx my inner would be plus cx so plus cx and my last would be plus 3c okay so let's copy this left side so negative 5 x squared plus 4x plus 37 all right so at this point again if you get confused just think about you need it to match this form here so i've got to get everything with x squared together so i can factor okay i've got to get everything with an x together so i can factor so this has an x squared and this has an x squared so put them together i would have a plus b times x squared okay if i wrote those next to each other i could factor out the x squared and that would give me that so a plus b is the coefficient for x squared then we would have plus now i'm going to work with the x so that's going to be here and here okay so i'm going to factor that out i'm going to say that i have 3b plus c and then i'm going to factor out the x that's going to be right there so 3b plus c that's going to be your coefficient for x then your constant term look for stuff without the variable x you have negative 5a and 3c so plus you've got negative 5a and then plus 3c okay so let's make this match up over here we're going to set up a little system of equations so let's write this out okay so we know that for these two to be equal right for this to hold that a plus b has to be equal to negative 5 right this is the coefficient for x squared this is the coefficient for x squared again we have an equality here so it must be true that a plus b is equal to negative five then it must also be true that three b plus c is equal to so 3b plus c is equal to 4 okay and then lastly it must be true that negative 5a plus 3c is equal to 37. so negative 5a plus 3c is equal to 37. so let's copy this we're going to go to a fresh sheet because we would definitely run out of room here okay let's paste this guy in here and basically what we do here we have three equations and three unknowns very easy system to solve what i'm gonna do is take these first two and i'm gonna eliminate the variable b and to do that i would multiply the top equation by negative three so i would have negative three a minus three b is equal to 15. and then for this guy i would have 3b 3b and then plus c equals 4. i'm just going to leave it as it is so negative 3a if i add that there's nothing to really add it to you could say plus 0a if you wanted to but it's just negative 3a there and then negative 3b plus 3b that's 0 right that's gone and then you just have the c so plus c and this equals 15 plus 4 is 19. okay so this gives me this new equation here which contains all the information from these two so now i can take these two okay and i can solve for a and c and then i can go back and plug in and find out what b is so if i have negative 3a plus c equals 19 i can solve that and say c is equal to 19 plus 3a and i can plug that in here so let's say that negative 5a plus i'm going to have 3 times this 19 plus 3a and this equals 37 okay so we have negative 5a and then plus 3 times 19 is 57 and then plus 3 times 3 is 9. so 9a and then this equals 37. so negative 5a plus 9a would be 4a so let's put 4a here and let's go ahead and subtract 57 away from each side of the equation so this is going to cancel and we'll say that we have 4a is equal to negative 20. divide both sides by 4 we'll find out that a is equal to negative 5. okay so a is negative 5. okay so let me erase this and once you find out the first one you're basically good to go all right so b now is very easy to find if i plugged in a negative 5 for a well b has to be 0 because basically if i plugged in a negative 5 there negative 5 plus what gives me negative 5 well only 0. now for c i can plug into this middle one here 3 times 0 again b is 0. so 3 times 0 is 0 you just get c is equal to 4. okay so we get a is negative five b is zero and c is four so let's copy this and let's just paste this in right here so for a i'm gonna put negative five and then for b it's zero right so zero times x is zero right so this term is gone then you have c which is four so we're just going to put a four here so our answer here is we have negative five over x plus three plus you have four over x squared minus 5. and again you can always check these by getting a common denominator and performing the addition all right let's move on and look at one that's really really tedious so you're going to get problems like these and i think it's important to work through some examples so that when you get to the test you're not overwhelmed so we have something like three x minus one over four x to the fifth power plus four x cubed plus x okay so the first thing is again is it a proper fraction yes the numerator has a degree of one the denominator has a degree of five again again again if it's not a proper fraction you have to do your long division first okay so keep that in mind but now it is a proper fraction so we move on to just factoring the denominator again in a lot of cases if you get a tougher problem the denominator will be factored for you but here it's not so bad i can pull out an x immediately i would have four x to the fourth power plus four x squared plus one if you notice when you factor something that this is a perfect square and this is a perfect square you can immediately think okay is this a perfect square trinomial if you check it you'll find out that it is okay so this is going to end up factoring into x times the quantity 2x squared plus 1 and this is squared okay so let's erase this and put this 3x minus 1 on top all right so now we're kind of getting somewhere so we have the denominator factored but we have to be careful here because this is a linear factor okay so let's just go ahead and start with that so i'm going to put my let me kind of move this over a little bit let me put my a over my linear factor of just x and then plus i have a quadratic factor here and it's irreducible right it's a prime quadratic polynomial right it's 2x squared plus 1 i can't factor that with rational numbers the other thing to keep in mind here is that this is raised to the second power so again you've got to build up the power don't just write you know bx plus c over 2x squared plus one and then you know quantity squared this is only one of them okay and i'm going to start with the first power you have to go plus you have to go and i'm just going to use d so d x plus e okay over you would have 2x squared plus 1 and this quantity would be squared okay now this is going to get really long there's no way i can fit all of this on you know my screen so this is something where you want to just write down notes so this is going to be equal to this okay now i already know that because i have my notes in front of me but you have to have that written down so you can follow so i'm going to scroll down and what we're going to be doing is multiplying both sides by the lcd okay and let me come back up for a minute i know that this is the lcd the x times the quantity 2x squared plus 1 squared okay so if i multiply this right here by the lcd i'm just left with the numerator so i'm just left with 3x minus 1. i got to try to write as small as i can this is a lot to fit on the screen so let's come back down and again if i multiplied this x times this quantity 2x squared plus 1 squared by this the x will cancel with the x and i'm left with this so this would be a times this two x squared plus one quantity squared now what i would tell you is when you get these make sure you four them out so we know that this guy would be what the first guy squared would be four x to the fourth power plus two times the first guy times the second guy so 2 times 1 i already know is 2 and then 2 times 2x squared is 4x squared and then plus the last guy squared which is 1. okay so let me erase this and let me put this in here as four x to the fourth power plus four x squared and then plus one you can even distribute the a at this point if you want so we would say this is four a x to the fourth power plus four a x squared plus a okay so now this is all taken care of so let's scooch this up here so now i'm working with this one again if i'm multiplying this guy by x times 2x squared plus 1 quantity squared well one factor of this would cancel with one factor of this so basically what i would have is bx plus c okay times you'd have times x and then times one more factor so 2x squared plus one but only one factor because again one of these canceled with one of those so to make this a little bit easier i'd multiply x by this first so bx squared plus cx okay just to get things going and then you could use foil to further get this going okay i'm just going to write this directly up here as i get an answer so bx squared times 2x squared would be what it would be two times bx to the fourth power then we would have our outer so bx squared times one would be plus bx squared our inner we'd have plus cx times 2x squared would be 2 times we'll go ahead and just say it's cx cubed and then i'm going to run out of room so i'm just going to keep this going down here so i'm just going to draw an arrow to continue so the last thing would be cx times 1 which would be cx so now the last one we're working with here is a little bit easier because this is going to cancel okay and i'm just going to be left with the x times this okay so x times dx let me write this out the long way so no one gets confused so x times dx is dx squared and x times e is plus e x okay let's get rid of that and let's drag this up here all right and again i know this is long so it's really important that you follow along with your own notes all right so let's go down a little bit and again what i'm trying to do this side right here has to match this side well you might be thinking yourself i've got x to the fourth power i've got x cubed i've got x squared i've got x to the first power and i've got a constant so how are we going to set this up well what happens is you can always write this side over here i can always say this is zero x to the fourth power plus zero x cubed plus zero x squared plus three x minus one i can always do that because zero times anything is just zero okay so if i kind of move this up i can say i'm going to be matching things so this is equal to where's my x to the fourth power well i have it here and i have it here right nowhere else so what i'm going to do is i'm going to factor out the x to the fourth power and i would have a 2b okay plus 4a and then the x to the fourth power is factored out again if i had these two terms next to each other i could factor out the x to the fourth power that's how i got that so i know that 0 is going to be equal to 2b plus 4a we'll get to that in a second when we set up our system but just a little foreshadowing so then we'll put plus we're going to come down here so let me just draw a little arrow so now i'm looking for what's cubed so i have 2c x cubed and nothing else so this guy right here so let me write that so we have 2 c x cubed okay then plus so what's squared so we have this guy and this guy and this guy okay so i'm going to factor out the x squared so what i'd have is 4a plus i'd have b plus i'd have d okay and then times x squared which was factored out all right almost done so we have for x we have what we have ex and cx and nothing else right let's make sure okay good so we'll put plus and i'm just going to factor it out so i'm going to say c plus e and then times x okay and then lastly you have your constant okay and we just have a as our constant here so plus our a so if we look at everything what we're going to see is what again if i match stuff up this is the coefficient for x to the fourth power this is the coefficient for x to the fourth power again it doesn't matter that's zero right i'm going to set that as 2 b plus 4 a is equal to 0. okay then i want to move on to x cubed so i see that this is 0 and this is 2c so 2c is equal to 0. so at this point i can already solve and say c is equal to 0 but we'll do that in a moment then i have that zero is going to be equal to this guy right here so we can say that four a plus b plus d is equal to zero okay moving on so now we know that this three here the coefficient for x has to be equal to c plus e so c plus e is equal to 3 okay and we also know that a is negative 1 right so we know a is negative one so let's copy this and let's go to a fresh sheet and once you get it to this point in a lot of cases it's gonna be a real easy system like this it's going to be five equations and five unknowns but a lot of these are given to you pretty much immediately we already know that a equals negative one and we know that c is zero okay because again if you just divide both sides here by two you would find that c is zero very easy so now what else can i find out well i know that if i plugged in a 0 here for c e is 3 okay so very easy so far and i already know what a is so if i plug in a negative 1 there 2 times b plus 4 times negative 1 is negative 4. so let me just write minus 4 is equal to 0. if i add 4 to both sides i'll say this is equal to 4 divide both sides by 2 and b is 2. okay so b is equal to 2 okay and so what are we missing we have a b c we're missing d okay so d is here and basically what we need to do is just plug in so four times negative one is negative four and then b is two so plus two and then plus d equals zero so negative four plus two is negative 2 so we'll go ahead and say we have negative 2 plus d equals 0. if i add 2 to both sides we would find that d is equal to 2. [Music] okay so let's copy this and let's go ahead and paste this in and so i know a is negative one so let's put that in and we can erase these as we go and c is zero so you can just get rid of it okay and b is two so b is two and let me erase that and so d is 2 so d is 2 and e is 3. okay so e is 3. okay so now we have our answer i know this is a very tedious problem but you will get examples like this that take 15 minutes where you just have to work your way through it right you just have to go through it so we started with 3x minus 1 over 4x to the fifth power plus 4x cubed plus x and when we break it down we end up with negative one over x plus two x over two x squared plus one plus two x plus three over the quantity two x squared plus one squared in this lesson we wanna talk about graphing horizontal parabolas all right so we've already talked about how to graph a vertical parabola now we just want to follow up and talk about how to graph a horizontal parabola so of course with a vertical parabola it opens up or down and with a horizontal parabola it's going to open to the left or to the right so let's start with something that we know we know this is the vertex form for a parabola that opens up or down so we've already seen this the f of x equals a times the quantity x minus h squared plus k so most books call this the vertex form some will call it the standard form i'm going to refer to it as the vertex form now in a lot of cases you need to be able to go from this standard form the ax squared plus bx plus c to this form here the reason you want to do that is because in this form here the vertex form the vertex is given as h comma k so you can immediately see that you can also see a which you can see here as well but a tells you whether this guy opens up or down if a is positive it opens up if it's negative it opens down additionally you're going to be using a with that little step pattern that we talked about when we discussed graphing parabolas we'll go over that in a moment if you never learned it all right so we know that to go from this form to this form we can use the completing the square method or we can use the vertex formula so this h is found as negative b over 2a so i can just take again this is b the coefficient for x to the first power this is a the coefficient for x squared so that gives me h or my x coordinate for the vertex and then what happens is once i have that i can plug in whatever that value is in for x in my function okay and i can find out what y is so you could really just say that for k for k it's f of h okay because all i'm doing is i'm plugging this in whatever h is in for x in that function to find out the y value for the vertex now what i want to do is talk a little bit about going into this horizontal parabola or the sideways parabola so all i did here for this top one is i wrote y instead of f of x and i subtracted k away from each side of the equation okay for this one this is the sideways parabola and i want to just show you that the roles of x minus h have kind of been swapped right so here it's inside of parentheses now it's outside and then y minus k it's been swapped right it's been coming inside of parentheses from outside of parentheses so you can say that the x minus h and the y minus k those guys have really interchanged roles so we're going to have to think about things a little bit differently here okay so i can solve this for x and say this is a times the quantity y minus k squared and then plus h now what i need to know here is how to go from this form to this form okay because this is going to come up immediately again we want to put it in this form because when it's in this form the vertex is still going to be given as h comma k okay so it's still those h is here now and k is here now okay so they're in different positions but it's still the same letters so that we understand what h is it's the x coordinate for the vertex what k is it's the y coordinate for the vertex now another thing you need to understand is that a now is going to tell you if it opens to the right or to the left if a is positive it's going to open to the right if a is negative it's going to open to the left so the vertex formula that we talked about a minute ago where we said h was equal to negative b over 2a well now since the x and the y's have swapped rolls this isn't going to be h anymore now it's going to be k so that's how you find k and once you have that you can plug in for y okay and then you can find your x coordinate now let's go ahead and just look at an example this isn't too tough so we have x equals 2y squared plus 4y minus 3. so i want to start by putting this in the format of x equals a times the quantity y minus k squared plus h again you can complete the square if you want if you prefer that that's fine for myself i like the quick method so now i know that k is going to be found as negative b over 2a so what is b in this case again this is a it's the coefficient for y squared this is b is the coefficient for y to the first power you could label this as c if you wanted to you don't really need it here so i'm just not going to do it so what i'd have is negative 4 so negative 4 over what's a a is 2. so 2 times 2 is 4. so this is negative 1. okay so i could write this as x equals i know what a is that's 2 times the quantity y minus a negative 1 and you could put plus 1 but for the sake of matching this up i'm going to put minus a negative 1 here and then plus we still need our h okay so how do we get that again you just plug in all right so i'm going to plug in a negative 1 there and there so what would that give me x is equal to negative 1 squared is 1 2 times 1 is 2. and then negative 1 times 4 is minus 4 and then minus three so two minus four is negative two negative two minus three is negative five so i'm gonna put plus negative five here okay so now that we have this we know the vertex occurs where again it's h comma k so h is negative 5 and k is negative 1. so if you write this as y plus 1 like this you've got to know that it's in the format of minus whatever k is you have to do that little transformation once you do this a bunch of times it's something you can do mentally but i want to write it like this so that it's very clear for you what h is and what k is so we would plot negative five comma negative one so that's our first step so we would go to negative 5 and then negative 1. so let me do this in a different color so that's right there and we know that a was positive right if i go back a was 2 right so since a is 2 let me just write that here a is equal to 2 we know that this guy opens to the right again if it was negative it would open to the left additionally we know about this axis here which kind of splits the parabola into two equal halves right so in this case now it's going to be y equals negative one right so y equals negative one so the reason you know that is now it's going to be a horizontal line okay whereas before was a vertical line and that guy is always going to intersect the vertex so if it's a horizontal line it's going to have y equals something and it's going to be your y-coordinate from the vertex so in this case it's negative 1. let me just erase this we don't really need that so now we've plotted our vertex again that's negative 5 comma negative 1. what we want to do now is use that step pattern so remember it's that 1 3 5 7 9 you know so on and so forth so what we do is we multiply our a value by each one of these so 1 times 2 is 2 then 2 times 3 is 6 then 2 times 5 is 10 and we don't need to keep going because we're not going to be able to get all these points in but basically before what you would do is you would say okay i go one unit to the right and then i go this many units up okay but now since things are changed now what i'm going to do is i'm going to go one unit up and this many units to the right so i'm going to go one unit up and two units to the right so that puts me here then i'm going to go one unit up and i'm going to go 6 units to the right so that would put me here then if i wanted to kind of expand this i'm not going to be able to fit it here but you could on your paper you could go one unit up and you could go 10 units to the right so that puts you at 13 for the x value and 2 for the y value okay so once you have these points you can just reflect across that axis okay you know that from here to here is a distance of one unit so you just flip it over here okay and then from here to here is two units so you go one two this way and again once you have these points it's very easy to sketch your graph all right let's just look at one more of these i think this is something that you definitely will understand especially if you already know how to graph a vertical parabola is basically the same process so we have x equals negative 3y squared minus 18y minus 22. okay again if i want the format of x equals a times the quantity y minus k squared plus h all i need to do to get my k value is say it's negative b over 2a okay so what is b in this case it's going to be negative 18. so be careful there it's the negative of negative 18. so that's 18 over what is 2a a is negative 3. so 2 times negative 3 is negative 6. so this is going to be negative 3 okay so that's k or the y coordinate for the vertex so let me just put here that this is going to be negative 3. now to find h again i'm just going to plug in a negative 3 for y so negative 3 squared is going to give me 9. negative 3 times 9 would be negative 27. so let's say that x equals negative 27 and then negative 18 times negative 3 is 54. so let's go ahead and put plus 54 here and then minus 22. so if i did negative 27 minus 22 i'd get negative 49 and if i add 54 to that i'm going to get positive 5. so this h here is going to be 5 okay so we can write this as x is equal to we know a is negative 3 times the quantity let me make that better y minus a negative 3 so you put minus a negative three or you could just put plus three and then plus five but you would know that the vertex would occur at five okay and then comma you'd have your negative three so you take that again plot that point and then use your step pattern in this case it's negative 3 for a so you know it's going to open to the left and let me just come to the sheet so again we have 5 comma negative 3. you can plot that really quickly so here's 5 comma negative 3. and again your axis is going to be y equals negative 3. so it's y equals whatever the y coordinate is for that vertex now again once you have that you use your little step pattern your a here was negative 3. so that's what you're going to multiply by that pattern so the 1 3 5 7 9 you know so on and so forth so negative 3 times 1 is negative 3 and then negative 3 times 3 is negative 9 and we're not going to go any further than that right but you can keep going if you wanted to kind of make your paper a little bit longer but basically what i'm going to do is i'm going to go up 1 and i'm going to go to the left 3 right negative 3 tells me i'm going to the left 3. so up 1 to the left 1 2 3. so here's a point and then the next one is negative 9. so up 1 to the left 1 2 3 4 5 6 7 8 9 arrive at that point and again you can just reflect these points across this axis to get this point here and this point here so that's how you could sketch the graph of a horizontal parabola in this lesson we want to talk about conic sections and here we're going to be focusing on parabolas all right so we've already spent a lot of time working with parabolas in the course now we're just going to go a bit deeper and explore a geometric definition we'll also talk about two new terms the focus and the directrix all right so let's start with what we know we know that a vertical parabola looks like this we have y equals a times the quantity x minus h squared plus k so we know it's vertical because the x is involved in the squaring operation and again when it's in this form we can say that h comma k is the vertex now if a is greater than zero it opens up if a is less than zero it opens down we also have the horizontal parabola so these guys open left and right so x equals a times the quantity y minus k squared plus h vertex still occurs at h comma k even though things are moved around and again we know it's a horizontal parabola because y is involved in the squaring operation now here if a is greater than zero it opens to the right if a is less than 0 it opens to the left so we know this already but what we want to do is develop this geometric definition for a parabola this is something again you see in your textbook when you get to conic sections so i've already graphed this guy is y equals 1 8 x squared so this is the example we're going to work with okay so we already know how to sketch the graph of this that's nothing new for you but what's probably new is this point here okay which is called a focus let me put an f there for focus and this line down here which is called the directrix let me put a d here okay so when we have the official definition for parabola it says that it's the set of all points where in this plane they're equidistant from a fixed point f which is the focus and a fixed line d which is your directrix so you can pick let's say this point here to make it really simple and say okay what's the distance from here to here and what's the distance from here to let's say here okay so this is zero comma two this is zero comma two and this right here is four comma two okay and we can see that let me kind of draw this line not perfect but you can see this is a horizontal line here and the distance from if you think about this x coordinate being 0 and this x coordinate being 4 4 minus 0 is 4. so the distance between those two points would be 4 units well what we're saying is that the distance between this right here and this right here would also be four units okay not a perfect line but you get the idea so this point right here again is four comma two and this point here would be what it would be four comma negative two okay so four comma negative two so the distance here this is a vertical line so you think about two minus a negative two which is two plus two that's four so the distance from here to here is four units and the distance from here here is four units so that's exactly what we're saying now there's a lot of things to go through in this lesson it's a really long lesson it's a really detailed lesson but what i want to start with would be just a few points and then i'm going to derive an equation that we're going to use and then we'll build on that the first thing is the focus is always going to be inside of the parabola okay so if this focus is above the vertex then you know that this guy opens up if it's below the vertex you know it opens down okay if it's to the right of the vertex it opens to the right if it's to the left of the vertex it opens to the left okay we're going to need that later on additionally the directrix is always going to be outside of the parabola okay so what you can think about here is that the distance from here to here and the distance from here to here is always going to be the same okay so the distance from the focus to the vertex and from the vertex to the directrix is the same in fact this guy right here is the midpoint for the line segment that connects the focus to the directrix okay so you can see that from the focus to the vertex is two units from the vertex to the directrix is two units as well okay so a lot to throw at you now what i want to do is use this kind of definition to derive a little equation and this is something you're going to see in your book it doesn't take that long so i thought we could go through it so let's let me come and pop back up here real quick i'm going to call this let me just erase all this i'm going to call this right here 0 comma p okay so we know the x location is 0 the y location i'm just going to call it p just generically i'm going to call this point x comma y okay generically and i'm going to call this right here we know the x location is the same and the y location is the negative of p because again this went up two units this went down two units so it's the negative of p so if this is p this is negative p okay so now that we have these three points we can say that f our focus is zero comma p we can say that our point p is x comma y and we can say that our directrix okay the point that we're working with is going to be x comma negative b okay so that's just a point on the directrix the line itself is y equals negative p but let's grab this real quick cut it away let's come over here and you'll see this in your book so basically what we're saying is that the distance between the focus and the point so between the focus and the point is equal to the distance between the point and your directrix okay so we know this but we can show an equation by just doing this generically okay we're going to use this equation throughout our lesson so what we'll do is we'll do our distance formula so square root of let's say this is the second point and this is the first point so it's x sub 2 minus x sub 1. so x minus 0 is x and that's squared then plus it's y sub 2 minus y sub 1 quantity squared so y minus p quantity squared okay so we've got that part set up then this is equal to do the square root over here so i'm going to call this 2 and this 1 now and i'm just doing this for convenience so i'm going to say x minus x is 0 and i'm going to say y minus the negative p is y plus p so y plus p squared is all you'd have there okay again i don't have anything squared with the x because x minus x is 0 0 squared is 0. okay so to kind of get rid of the square roots i'm going to square both sides and i'm going to be left with the radicand in each case so i'm going to have that x squared plus just go ahead and foil this out this is going to be y squared and then minus 2 and then you're going to have p times y and then plus p squared equals this guy would be y squared plus you'd have two p y plus p squared okay you can immediately get rid of this and this and this and this okay because again same thing on each side of the equation subtract the way it's gone okay so that's a piece of cake so let's scroll down a little bit and now what i'd have is x squared is equal to if i add 2py to both sides of the equation we would get 4py over here so this is the equation you'll see in your book you might see 4a y just depends on what book you're using but generically you want to think about p let me kind of just bring this back up here you want to think about p as this coordinate here this y coordinate in this specific case okay now this is going to change when we look at a horizontal parabola in this case we're looking at a vertical parabola and we're looking at one specifically that has a vertex at zero comma zero so these formulas are going to change we'll get to that later for right now we can use this to derive this equation so let's say i told you that the focus the focus occurred at zero comma two okay we would know that let's say we were given that you said hey find the equation find the directrix okay well the way you do this is you say that you have x squared is equal to you've got your 4 the p is the only thing you get you get that from right here this is p okay so it's 0 comma p well in this case p is 2. so 4 times 2 and then times y so x squared equals 8y or you can solve for y and let me just erase all of this because it's in the way you could solve for y by dividing both sides by 8 and you would get y equals 1 8 x squared which is exactly what you have there okay so not a magic trick nothing fancy there just plugging in and simplifying okay let's talk about a few things that we're going to find useful here let me just kind of erase all of this and start with something that's fresh so when we talk about these formulas again in this case it's x squared is equal to four p y okay so the focus the focus is at zero comma p okay the directrix is going to be y equals negative p again you can think about this as the example this is zero comma two and this line here is y equals negative two okay so this is always going to set up this way again again again this is going to be the case where you have a vertical parabola and the vertex is at zero comma zero if your p ends up being negative this guy's going to be upside down okay so your directrix is now going to be above the x-axis okay and your focus is going to be below okay so you just got to keep that in mind all right let's look at the first problem this is a very easy problem something that you're going to see so they're going to ask you to find the equation okay with a focus of 0 comma negative 1 12 and a vertex of 0 comma 0. so again what you're looking at here is the focus is your focus is 0 comma p okay in this case i know i'm using this formula okay later on we'll see if the focus is p comma zero well i'm using the other formula for a horizontal parabola with a vertex of zero comma zero we're not there yet we'll get there in a second for right now we know this is the focus we know the directrix okay is going to be y equals the negative of p well p is this negative 1 over 12 or negative 1 12. so it's the negative of that so it's just 1 12 okay y equals 112. that's your directrix so you're done there for your equation you just plug in very simple plug in for p and i know it's a lot to remember but x squared is going to equal 4 times 1 12 and it's negative so 4 times negative 1 12 times y okay let me do this up here we know that this cancels with this and gives me a 3 down there so you could say x squared is equal to you have your negative 1 3 y of course most of the time we want to solve for y so we multiply both sides by negative 3 and you'd have that y is equal to negative 3x squared so there's your equation and again you see that p is negative and you know that you're going to open down right you know that if this right here which is traditionally our a value if that is negative we know we open down if it's positive we know we'd open up so again for your focus if it's zero comma p and p is negative you're opening down if it's zero comma p and p is positive you're opening up okay so another type of problem you're going to be given is you'll get an equation and they'll say what's the focus what's the directrix okay so you've got to match things up to the form again y equals one-fifth x squared so i know that the vertex is zero comma zero here okay additionally what else do i know i know this is a vertical parabola so i'm gonna use this formula x squared equals four p y all i need to do is match things up so right now this is y equals something i want x squared equals something so multiply both sides by 5 you get 5y equals x squared and then just match stuff up i know that this is equal to this okay so what i'd say is 4p is equal to 5. again i have x squared i have x squared i have y i have y it's the 5 and the 4p that get matched up and you solve for p divide both sides by 4 you get that p is equal to 5 4. so again the focus is 0 comma p okay it's 0 comma p so the focus 0 comma 5 4. the directrix what is it y equals the negative of this negative five fourths it is a lot to take in and remember but you write your notes you do your practice it will come okay let's look at another scenario so now what you'll notice is that your focus changed a bit now you have some value here for x y is zero okay so this is the case where you're going to open horizontally right you're going to be to the right or to the left well if this p-value now is positive you're open to the right if it's negative you open to the left so your focus in this case is some p-value comma zero okay your directrix in this case is x okay because it's opening again horizontally so you're going to have a vertical line now that's your directrix right because it's behind the parabola so x equals the negative of p okay so if we wanted to find the equation and the directrix well for the directrix it's x equals the negative of this so negative 1 over 80. that's it you're done for the equation plug in for p again this is 1 over 80. okay so times 4 this cancels it gives me 20 down here so really i could just say this is y squared equals you're going to have 1 over 20 times x if you want to solve this for x multiply both sides by 20 and you would say that x is equal to 20 y squared okay so one more of these and then we'll go to the more challenging scenario where the vertex is not zero comma zero so you're given a point one comma three you're given a vertex again zero comma zero you're told that it has a horizontal axis so you know this opens sideways right it's opening to the right or it's opening to the left okay so what i want to think about is my formula so it's y squared is equal to you have your 4 p x okay just plug in the point so we're plugging in the point so i plug in a 3 for y i plug in a 1 for x so you just get it right 3 squared is 9. so 9 equals 4 p divide both sides by 4. and you get that p is equal to 9 4. okay so once you have that okay we already know what p is so what is the focus the focus again is p comma zero right if it's horizontal so it's going to be nine fourths comma zero okay the directrix is what it's x equals the negative of this so negative 9 4. okay so that's done what about the equation itself so we have y squared is equal to you've got your 4 times p p is 9 4 and then times your x so what we'll do here is cancel this with this you'll say you have y squared is equal to 9x you can solve for x again sometimes you want to do that just divide both sides by 9 or multiply by 1 9 so you would have that x is equal to 1 9 times y squared okay now let's talk about the more challenging scenario and this is where the vertex is not at 0 0. so this involves a lot more thought okay it's really not that bad once you do a few of these but the first time you get to them it's not obvious what's going on because your p-value you have to calculate it now okay it's not given to you and the same goes with your directrix so you have to kind of use your knowledge of what we talked about earlier in the lesson so let's just go through this real quick we know when x is involved in the squaring operation it's a vertical parabola we know when y is involved in the squaring operation it's a horizontal parabola so this is x minus h quantity squared equals 4p times the quantity y minus k you have y minus k squared equals 4p times the quantity x minus h so these are the two equations we're going to use okay let's look at an example so we're given the focus and we're given the vertex and what we're going to be doing here is finding the directrix and we're going to find our equation okay that matches the given information now this is a lot more difficult okay so what do we do to start we figure out does it open up does it open down does it open left does it open right first time you do this it's a little bit challenging but it gets a lot easier so we know that the x value is the same in each case if you see that then you immediately know that it's vertical okay how do you know that well just think about the coordinate plane real quick if you do 2 on the x-axis so that's right there and that's the same well the only thing that's changing are the y-coordinates well i'm changing vertically okay so that means that my vertex is at two let's say negative one is there okay and then my focus is at two and then it's going to be negative one half let's say that's right there okay well this guy's going to open up because the focus is above the vertex okay so that's all you have to remember you don't have to draw a picture each time you know if your focus is above the vertex it opens up if your focus is below the vertex it opens down okay it's just that simple if your focus is to the right of the vertex it opens to the right if your focus is to the left of the vertex it opens to the left so now that we know that okay you know immediately that that p-value is greater than zero okay if it opens up okay and it's less than zero if it opens down so you've got to remember that so i'm just gonna write here a note that p is greater than zero okay now the next thing i'm going to do is write out my equation that i was given so i know that i'm going to use the vertical one so x minus h quantity squared and then this equals you've got your 4p times the quantity y minus k now i plug in for h and k the vertex this is h this is k so don't get confused by that h is the x coordinate k is the y coordinate okay so plug in here plug in a two plug in here minus a negative one put plus one okay so from this point you might be thinking well how do i get a p because last time we would just take zero comma p right so a lot of you will say okay well p is equal to negative one half but that's not how you get that in this particular case okay what you want to do is think about that geometric definition that we talked about earlier i know that the distance between the focus and the vertex is my p if i go back to this example here the distance from here to here okay this was zero comma p okay so this ended up being two so it's zero comma two but this was my p my p was two okay because that's the distance so coming back down what i want to think about is the distance between this and this okay just think about that don't worry about anything else so what is the distance between negative one-half and negative one okay well that's a half okay and i know that it's convenient that it works out here in that particular case but it doesn't always work that way okay so you have to think about that so the distance between this and this is one-half okay so my p-value okay i know it's positive because it opens up is going to be equal to one-half okay and again don't just take it from here you've got to do the calculation it's not always going to work out that way i plug that in and i'm basically good to go i say x minus 2 quantity squared is equal to 4 times a half is 2 times the quantity y plus 1. so you can leave it like this some teachers will accept it or you might want to simplify and solve for y in that case you just want to distribute this and say this is 2y plus 2. and let me erase that and just kind of drag this down and then to solve for y i'll say that we have 2y is equal to i'll subtract 2 away from each side so x minus 2 quantity squared and then minus 2 and let me just erase this and let me divide both sides by 2 now and so what would they give me this gets erased you could really write this as y equals one half times the quantity x minus two squared minus one okay so this would be the answer that i would give if i was in class it's solved for y and it's nice and neat for the directrix you have to think about the fact that the distance from here to here is going to be the same as the distance from there the vertex to the directrix so this is a distance of one-half units okay you got to think about the direction so from the vertex to the focus i traveled up right i went from negative one to negative one half so for my directrix i've got to travel in the opposite direction remember the focus is inside okay the focus is inside and the directrix is underneath okay so i'm going to go this direction now so i'm going to subtract a half away from negative 1. so negative 1 negative 1 minus 1 half would be negative three-halves right if i got a common denominator this would be negative two over two negative two over two minus one-half would be negative three-halves so the directrix would be y equals negative three-halves okay so i know that's a lot to take in but let me just go over this one more time you've got to find p first it's not just grabbing this guy right here okay you've got to think about the difference between this and this it just happened to work out that way for this example okay so it's the difference between this and this okay so that gives you your p use that to plug in and get your equation again the directrix all you have to do is go in the opposite direction by that number of units okay let's look at another example i know this is super confusing this is one of the ones that you just have to work a lot of examples so the focus 17 4 comma negative 7 the vertex 4 comma negative 7 okay so in this particular case how do we open up down left to right well again this is negative 7 and negative 7. so the y is the same now okay the x's are changing so that means i'm moving horizontally so that means i'm opening horizontally okay so i know that now think about the vertex the vertex is 4 the focus is 17 4 the focus is larger right it's to the right so that means we're opening to the right so immediately write down that p is greater than zero okay so you remember that if we're opening to the left you've got to remind yourself that p is going to be negative if you're opening down p is going to be negative you've got to account for that okay very important so what i want to do now is think about my p let's just go ahead and get it so it would be the distance between this and this so again don't worry about this you're just subtracting 17 fourths minus 4 16 fourths so that's going to be 1 4 okay so your p value is positive 1 4. again notice how it doesn't work out to be this in this case so don't grab that like we did when we had a vertex at zero comma zero that is a very very common mistake when you get on these problems so we have p equal to one fourth so now what's our equation well we have that y minus k quantity squared is equal to we've got this four p times the quantity x minus h p is a fourth if i plugged in a fourth there four times a fourth is one get rid of that if i plugged in four again this is h this is k so h goes over here now so this is four and this is negative seven so minus the negative seven is plus seven and if you wanted to drop the parentheses you can and you could just add four to both sides and solve this for x so x is equal to the quantity y plus seven squared plus four okay now your directrix let me erase this your directrix how do we find that again we know that this guy is now going to be a vertical line so x equals something because it opens horizontally opens to the right okay so we know that the distance between this and this is the same as the distance between the vertex and the directrix again this was the distance of one-fourth we went to the right so now we need to go to the left because again if you think about this this is opening to the right okay so your focus is here your directrix needs to be over here so you need to go this way now okay so i would subtract 1 4 this value here away from 4. so what is 16 4 minus 1 4 that's 15 4. okay so x equals 15 4 is going to be your directrix here so i know these problems are quite challenging when you first see them they can seem kind of overwhelming just finding the value of p but once you get used to this it becomes very very easy to eyeball this and say okay i know it opens to the right i know p is going to be you know the distance between this and this and then once i have that i can find the directrix i can find my my equation by substituting in it's really not that bad now i want to leave you with this this is something that might be in your textbook it might not be in your textbook i find it super helpful when i'm working these problems just as a reference so if you open to the right your focus is h plus p comma k okay so if you open to the right your focus is h plus p comma k and we can go back up we know p is a fourth so the focus the focus is what it's h which is four plus 1 4 which is 17 4 comma k which is negative 7. does that match that yes it does okay and i probably should have used the highlighter there because i kind of covered it up so let me highlight that but basically you see what your focus is it matches exactly that formula now your directrix is x equals h minus p so we saw that h in this case was 4 and we subtracted away 1 4 and we got 15 4. okay so that's that your equation we already know this now when you go to opening to the left you have to change what p is okay it's going to be negative that's another common mistake all right it's going to be negative so your focus is h minus p comma k your directrix is now x equals h plus p okay remember you're going in the opposite direction and this changes right you have a negative here because the p is negative you've got to remember to change that to negative okay now when you go up or down it's the same kind of scenarios but now your h is constant here that's not going to change but you get k plus p if you're going up and then k minus p if you're going down we saw this earlier your directrix if you're going up y equals k minus p if you're going down y equals k plus p your equations here are the same except that again if you're opening down if you're going down then you get a negative here because the p is negative in this lesson we want to talk about conic sections and here specifically we want to focus on the ellipse all right in the last lesson we talked about the parabola and specifically we learned a new geometric definition for a parabola we learned that no matter where we were on our parabola that given point that given x comma y ordered pair is the same distance away from a fixed point called the focus as it is from a fixed line called the directrix now when we get into talking about an ellipse we see the concept of a focus again but now we're going to have two of them so we're going to call it the foci okay so that's for two and basically an ellipse is the set of all points in a plane the sum of whose distances from two fixed points these foci is going to be constant when we work with an ellipse when we start looking at these equations you're generally going to start out with ones that have a center at the origin this is the easiest case scenario and then you're going to look at shifting this for harder problems so we're going to start out with x squared over 25 plus y squared over 9 equals 1. and what i'm going to do is i'm going to go immediately to the coordinate plane where i've already graphed this and i'm just going to walk you through some of the terminology that you're going to see in this section this is not something that's very difficult to kind of work with but there's a lot of terms that you have to remember so it's more of a kind of tough vocabulary lesson where you just have to memorize a lot of stuff so the first thing is let's write our equation again we have x squared over 25 plus y squared over 9 and this equals 1. okay so the first thing i want to talk about would be this point here and this point here so where do those guys come from what do they mean and basically if you think about this these are the x-intercepts right so in this case because our center is at the origin these are the x-intercepts how do i get the x-intercepts i plug in a 0 for y so if y was 0 this whole thing would be gone i would have x squared over 25 equals one now when you work with the standard form of an ellipse you're always going to have one on the right side so this denominator here if i multiply both sides by it it's always going to set up to x squared equals 25 okay now i solve this with the square root property by saying i have the positive square root of 25 and the negative square root of 25 so that's 5 and negative 5. so let me erase this real quick i can just say and i'll just write this real fast so y squared over 9 i can just say that for my x intercepts that i basically take the square root of 25 which is 5 and then comma y is 0. and then i take the negative square root of 25 which is negative 5 and then comma 0. okay so the quick way to do that is just to look at this number right here this denominator under your x squared term and just say what's the principal square root what's the negative square root so if y is 0 you've got the principal square root if y is 0 you've got your negative square root so that's how you can get that quickly you can use a similar thought process to get your y-intercepts so here and here so now x is 0 in each case okay x is 0 in each case and now what's the principle of negative square root of 9. so 3 and negative 3 okay so 3 and negative 3. okay so now we have those values we know where those points came from now let's talk a little bit more about the terminology because this ellipse is what we say a horizontal ellipse it's shortened fat right so it's stretched more horizontally than it is vertically what happens is we call this line segment from right here from negative five comma zero to five comma zero the major axis okay and the endpoints of the major axis are known as the vertices so i'm just gonna say this is a v here for vertices and then the shorter axis okay so from this guy right here which is zero comma three down to this guy right here which is zero comma negative three that line segment there you can call that the minor axis okay and those endpoints are known as the co-vertices so c and then v okay co-vertices so again a lot to remember just something you want to take for your notes so the longer one that's the major axis the shorter one that's the minor axis when we work with this guy generically what's going to happen is we're going to label the distance from the center to the vertex okay this vertex over here we'll call this distance a okay we'll call the distance from the center to one of the co-vertices here we'll call that b okay so we can show mathematically and i'm not going to go into this that c squared okay and c is going to be the distance from this guy right here to one of your foci okay i'm going to call that c c squared is equal to a squared minus b squared okay so what happens is i can call a 5 right because the distance from here to here is 5 units right if i go from here to here or from here to here it's 5 units right so a squared i know is 25 and b squared again b is 3 this is 3 units so b squared is 9. so 25 minus 9 is 16. so c squared is equal to 16. if i take the principal square root of this guy because i'm thinking about a distance it's going to be 4. okay so the focus here would occur let me just kind of erase this because everything's just kind of getting kind of jammed up here the focus would occur four units to the right of this guy and four units to the left of this guy so your focus or your foci in this case because you have two you would have four comma zero and then you would have negative four comma zero okay so that's where those two points came from now your foci are always going to occur on the major axis okay always going to occur on the major axis so now we know where all these points came from we know where these two came from okay again these are the vertices or the endpoints for the major axis we know where these two points came from again these are the co-vertices or the endpoints for the minor axis and then these guys right here are the foci again we know that c squared equals a squared minus b squared now generically let me give you a formula so if we see x squared over a squared plus y squared over b squared equals one and a is greater than b this is going to be a horizontal ellipse okay think about what we just looked at here 25 is greater than 9 okay think about the fact that we took the principal and negative square root of this 9. just just think about the principal square root the principal squared of this guy is bigger than the principal squared of this guy so if we're thinking about from the center going this many units to the right and from the center going this many units up well if this kind of square root gives me a distance if this amount as a distance is greater than this amount as a distance it's horizontal because if this number is bigger it's stretched further horizontally than it is vertically okay so in this particular case your vertices are given as what all i have to do is take the principal and negative square root of that guy so i would say a comma 0 and then negative a comma 0 and i'm good to go the co-vertices all i have to do is take the principal and negative square root of that guy so zero comma b and then zero comma negative b and then the foci okay remember it's c squared equals a squared minus b squared and then what happens is once you find that value of c you're going to say you have c comma 0 and then negative c comma 0. okay so you can plug in and figure that out so now let's talk about a vertical ellipse because things are going to change so most books are going to keep a as the larger number okay so a is greater than b so they're going to switch this up and say this is a squared and this is b squared now okay so things are going to change because in this particular case the vertices again because this is larger it's going to involve taking the principal and negative square root of this number okay so the vertices i'm going to change this up and now 0 here and 0 here and i'm going to put a here and negative a here okay so this is just something you got to get used to the co-vertices now i'm going to erase this and this and i'm going to say that i have b and negative b 0 and 0. and for my foci this is going to change and i'm going to say 0 and 0 c and negative c let me make that better okay so again it's always going to be the case that c squared equals a squared minus b squared make sure that you assign a properly so a is going to be greater than b if you want to use these formulas in this setup all right so let's look at an example of a vertical ellipse so we see x squared over 16 plus y squared over 36 equals 1. so this is larger than this so that's how i know we have a vertical ellipse because when i take the square root of 36 at 6 when i take the square root of 16 it's 4. so think about i'm traveling from the center okay from the center i'm going to go up 6 units okay up 6 units i'm going to go down 6 units i'm going to go down 6 units from the center i'm going to go to the right 4 units and i'm going to go to the left 4 units so this guy is stretched more vertically than it is horizontally so that's why it's a vertical ellipse okay now let's think about let's think about some points here and we'll look at the graph in a minute so the vertices again these are the end points for the major axis now your major axis is going to be involved on the y-axis okay and so what we're going to say is that it's going to be 0 comma square root of this is 6 and then 0 comma negative square root of that is negative 6. okay so again if i start at the center i go 6 units up so to the center i go 6 units down okay then the co-vertices again now that's going to be on the x-axis okay so i'm going to have square root of this guy is 4 comma 0 and then the negative square root of that is negative 4 comma zero okay so again start at the origin and you go four units to the right start the origin you go four units to the left then again for your foci you want a squared and then you want b squared remember c squared equals a squared minus b squared assign these appropriately to where a is greater than b if you want to use this formula so i know that 36 minus 16 is 20 okay so i would have c squared equals 20. okay now technically c is a distance if we want to think about it this way so if i take the square root i really only want to do the principal square root but to fill out this kind of information here i want to do the principle in the negative square root just to make things faster so i'm going to say that c is equal to plus or minus the square root of 20 okay just to get these coordinates so it's going to be zero comma whatever that is and then zero comma whatever that is okay so this is going to be if you simplify it c equals plus or minus this is four times five and 4 is a perfect square so that's 2 times 2 so you can pull that out so plus or minus 2 times square root of 5. so you have 2 times square root of 5 and negative 2 times square root of 5. again c is technically a distance okay but you're traveling from the center to two times square root of five right you're going up and then you're traveling from the center down by two times square root of five so that's how you get those coordinates okay so let's copy this and let me erase this and i'll just go through and label some points real quick so we have the vertices again zero comma six and zero comma negative six so you see zero comma six is here and zero comma negative six is here so again these are the vertices because this line that's purple here this is your line segment that's the major axis okay from zero comma six down to zero comma negative six this is longer than this one okay so that's why this is the major axis that's why these are the vertices then the co-vertices i don't know if this is readable here let me just kind of erase this and i'll move this up you've got 4 comma 0 and negative 4 comma 0. so 4 comma 0 and negative 4 comma 0. okay so those are your co-vertices again that's the label we give for the endpoints of the minor axis again this one's shorter than this one so let's get rid of that we've already plotted that and then your foci again these are always going to be on the major axis 2 times square root of 5 is about 4.472 so you can just go from zero up to about 4.5 and say that'll be your point and then from zero down to again negative 4.5 okay it's not going to be that but i'm just giving you an approximation that you can look at on the graph okay we're not going to be able to actually plot that using kind of a scale with integers okay so we've got everything plotted once you've done that and you've labeled everything you just sketch your curve right so you're done and basically you want to just to reiterate this think about the center here as the midpoint between these two co-vertices okay and it's also the midpoint between these vertices and it's also the midpoint between these two foci okay so if you're ever asked what's the center and you're given the vertices or the co-vertices you can figure out what the midpoint formula okay so let's talk a little bit about shifting things around now we're just going to look at an example we have the quantity x minus 1 squared over 25 plus the quantity y plus 1 squared over 16 equals 1. now the first thing i'm going to tell you is just think about the formula i just worked out with you and then just shift things okay so how does this shift well because everything's wrapped in parentheses here you think about things as the opposite as you would okay so this is minus one with x okay with x you're thinking horizontally and again because it's wrapped in parentheses you think about the opposite effect if it's minus one i want to think about it shifting to the right one so i'm going to say this shifts right one unit okay now inside of parentheses here we've got plus one and now we're thinking about y y is vertical okay so do the opposite of what you think it's plus one so that's going to shift me down one unit okay so shifts shifts down one unit okay so think about this just as you would if it was centered at the origin okay what would we do well we know this is horizontal because this is larger than this okay so my formula is x squared over a squared plus y squared over b squared is equal to one okay and i know that my vertices my vertices will occur where well again i'm just going to take the positive and negative square root of this guy so it's a comma 0 and then negative a comma 0. okay well in this case this is 25 square of 25 is 5. so it's 5 and negative 5. okay so we'll shift this stuff in a minute this is not going to be the final thing we'll shift in a minute just so you know then we have the co-vertices okay so from our formula again we turn to this b squared and we say it's zero comma square root of this guy square root of 16 is four and then zero comma negative square root of that is negative four okay and then for the foci okay for the foci let's put f here we have what well we think about the fact that we have a squared minus b squared that's what's going to give us c squared so c squared equals 25 minus 16 is 9. so again for practical purposes we know c is the distance so technically c is three but we need negative three and positive three so just say that it's plus or minus the square root of nine so we'll say that we need three comma zero and we need negative three comma zero okay so let's erase that and let's think about how we can shift things now so one unit right one unit down so all the x coordinates if i'm going one units to the right i just add one okay so that's all i need to do so i'm going to erase this and put six erase this and put negative four i'm going to erase this and put four i'm going to erase this and put negative two i'm going to erase this and put one and i'm going to erase this and put one okay then i'm going to shift all the y coordinates down one okay so i'm just going to subtract one everywhere so this is minus 1 and minus 1. and then this is minus 1 and minus 1. and then this is going to be 3 and this is going to be negative 5. okay so let's copy this so i've got everything pasted here we just go through the points again this is how you would graph this guy so your vertices you've got 6 comma negative 1 so 6 comma negative 1 and then negative 4 comma negative 1 so negative 4 comma negative 1. and then for your covertices you've got 1 comma 3 so 1 comma 3. then you've got 1 comma negative 5. so 1 comma negative 5. and then your foci you've got 4 comma negative 1 and you've got negative 2 comma negative 1. okay so your center is right here and this is going to be at 1 comma negative 1. okay so everything has just been shifted one unit to the right and one unit down okay let's talk a little bit about something you might see you may need to go from general form to standard form before you can graph something so i'm not going to graph anything else i think you get the idea that it's pretty simple but i do want to cover how to go from general form to standard form and i also want to cover how you can get the equation if you're just given some general information how can you kind of go through the thought process so with this guy we've already seen this when we work with circles a long time ago you should know that a circle is a special case ellipse where your a and your b are basically the same value okay but essentially what you want to do here is just do completing the square twice so group your x terms together group your y terms together okay create two perfect square trinomials factor those into binomials squared and then set up your little equation okay so what i'm going to do first is say that i have 36 x squared minus 72x okay and plus i'm going to do y squared plus 20y and this will be equal to 8. i'm just adding 8 to both sides okay so i didn't do anything fancy you need the variable that's raised to the second power your squared term to have a coefficient of 1. when you work with this with the quadratics you generally will divide everything right if you're trying to solve a quadratic equation with completing the square but here you can't really do that so what we're going to do is we're going to factor okay so we're going to pull this 36 out and what i'm going to have is x squared minus 2x and i'm going to leave a little space here because i've got to add something and then plus i don't have to do that here so y squared plus 20y a little space because i'm going to add something this equals 8. how do i complete the square again look at the coefficient for the variable raised to the first power cut in half and squared negative 2 times a half is negative 1 squared is 1. so you add 1 and subtract 1. here half of 20 is 10 10 squared is 100. so add 100 and if i can fit this in subtract 100. okay so what do i do about this what's inside the parentheses so to close this down i've got to distribute the 36 so what this would look like okay is 36 times i'm going to say x squared minus 2x plus 1 i'm going to distribute this to the negative 1 so minus 36 then moving on we'll put plus there's nothing out here so i can just say y squared plus 20y and then plus 100 again i can just close this down because nothing's out here then minus 100 equals 8. so negative 36 minus 100 is negative 136. so let me scroll down a little bit and we'll rewrite this and say we have 36 times the quantity x squared or actually let me just go ahead and factor this this is x minus 1 quantity squared and then i'm going to skip this for now i'll get to in a second i'm going to say plus you've got y plus 10 quantity squared and then negative 36 again minus 100 is minus 136 this equals eight and let me scroll down a little bit so now add this to both sides of the equation okay so plus 136 plus 136 you're going to have 36 times the quantity x minus 1 squared plus you're going to have y plus 10 quantity squared equals 8 plus 136 is 144. to finish this up you want this right side to be one okay so whatever this is over here just divide everything by that okay that's all you need to do and 36 over 144 144 divided by 36 is 4. so just put a 4 down here and then this will stay and this just becomes 1. okay so that's your equation the quantity x minus 1 squared over 4 plus the quantity y plus 10 squared over 144 equals 1. okay let's talk about one more type of problem that you might encounter you might be told to get the equation given the vertices and the foci now how could we do this we want something like this x minus h quantity squared over and this guy would say would be a squared plus y minus k quantity squared over b squared equals 1. okay how am i going to get the center the h comma k well again if i think about the vertices here you've got 11 comma negative 8 and you've got negative 9 comma negative 8. well here's 11 okay i'm just going to put this as 11. so 11 comma negative 8 would be something like right here and then negative 9 comma negative 8 would be like right here okay so we know that if these are the vertices okay we know that this is going to be a horizontal ellipse okay we know that because the vertices are the endpoints of the major axis and the major axis here is running horizontally okay so additionally we know that the midpoint okay of this line segment is the center so if you wanted to kind of count this out and cut in half you could but i mean really the easy way is just to eyeball this and say okay well 11 minus 9 is going to give me 2 and then 2 divided by 2 is 1. so it's 1 comma negative 8. that's going to be the center so i can erase this and put a 1 and then i can erase this and go minus the negative 8 is plus 8. now additionally we need a squared and we need b squared so how do we get that well a is the distance from the center which is one comma negative eight okay so the distance from the center to one of your vertices okay so it can go you can go either direction so it's what one two three four five six seven eight nine ten okay so a is 10. and again you can just kind of count this you don't need a graph so if a is 10 i know a squared is a hundred and to get b squared i use my little formula that c squared okay is equal to a squared which is 100 minus b squared okay so what's c squared again c is the distance from the foci to the center so just take either one of these so i'm going to say from nine to one is going to be eight units because again if i plotted nine comma negative eight that's going to be right there okay so the distance from here to here it's one two three four five six seven eight right eight units so you can count it like that or you can just eyeball and do it it's pretty easy so we know if it's eight and i square eight i get 64. okay so this is 64. and you can easily see that this is going to be 36 right so if you subtract 100 away from each side this would be what it would be negative 36 is equal to let me make that better you'd have your negative b squared divide both sides by negative 1 and you have b squared equals 36 right we don't need b we just need b squared so this is 36 and then we're done right we have our equation and basically it's the quantity x minus 1 squared over 100 plus the quantity y plus 8 squared over 36 equals 1. so we know that this guy is going to be a horizontal ellipse and we know how to go through and get what's missing which is the co-vertices okay we can kind of label everything and then sketch a nice graph in this lesson we want to talk about conic sections and here specifically we want to focus on the hyperbola alright so now that we just got finished talking about graphing an ellipse and the properties of an ellipse we move on to the hyperbola so the hyperbola is going to be very similar in terms of its equation but very different in terms of its shape and how we go about graphing it so the hyperbola is the set of all points in a plane such that the absolute value of the difference of the distances from two fixed points is constant again these two fixed points are called the foci of the hyperbola now when we start looking at the equation of a hyperbola again it's going to look very similar to the equation for an ellipse so this guy is centered at the origin and we're going to find out why in a minute but it's going to open to the right and to the left so we're going to have one what we call branch opening to the right and one branch that opens to the left so just to show you a little preview of this we're going to graph this in a minute but one branch opens to the right and the other opens to the left so let's talk a little bit about why that happens and to do that i want to just stop and talk about an ellipse for a moment so we know if we saw something like x squared over 9 plus y squared over 16 equals one this is an ellipse okay so the plus here gives me an ellipse the minus here is what gives me a hyperbola okay so why are these two going to be such a different shape we saw the shape of the hyperbola if we look at the ellipse it's an oval shape so what's creating the disparity between the two shapes what's that minus sign x squared over 9 plus y squared over 16 equals 1. forget about the formulas and all that stuff i just want you to think back to x and y-intercepts okay this is how this is going to make sense how do we get this point this point this point and this point let's think about this if i plug in a 0 for y i'm getting my intercepts plug in a 0 for y this is gone so you have x squared over 9 is equal to 1. okay multiply both sides by 9 i've got x squared is equal to 9 which gives me a solution of x is equal to 3 and then also x is equal to negative 3. so you could say x equals plus or minus 3 if you want okay so that's where this came from because again y is 0. so we would say 3 comma 0 there that's where this came from again y is 0. so negative 3 comma zero okay so that's where those two points came from now additionally we're going to have points here and here so where do those come from again think about plugging in a 0 for x to get your y-intercepts so here you'd have y squared over 16 equals 1 and again y squared equals 16. if i multiply both sides by 16 and y is going to equal plus or minus 4. okay so that's how i get 0 comma 4 and 0 comma negative 4. we already know about the equation of an ellipse you know so on and so forth i just want to show you that you end up with these kind of intercepts in this case so you're able to sketch this graph through these four points so that's what's creating this shape now when we go back to the hyperbola you're not going to get these four points to connect okay and the reason is because of this minus sign let's think about this for a second if i have x squared over 9 minus y squared over 16 equals 1 plug in a 0 for y it's a similar thought process x squared over 9 equals 1 x is going to equal plus or minus 3. okay so that means that 3 comma 0 is on the graph and negative 3 comma 0 is on the graph we can see those points right here right here's 3 comma 0 here's negative 3 comma 0. okay so from the center i'm going three inches to the right i'm going three units to the left we're going to find out in a minute that that distance again is called a okay in this particular case we know that when we work with these equations we have these generic forms so right now i'm just going to write that a is equal to 3. okay we'll revisit that in a moment how do we get the y intercepts here i'm just going to erase that well i could plug in a 0 for x this would go away you'd have negative y squared over 16 is equal to 1. multiply both sides by negative 16. to clear the denominator and the negative you're going to have y squared is equal to negative 16 okay which doesn't have a real solution right if i took the square to this side i'd have y if i go plus or minus the square root of this side okay you're going to have a complex solution there or a non-real complex solution to be specific right one with the imaginary unit i we won't have this on our coordinate plane so we won't have any y-intercepts so that's why if you look at the graph you don't see any points where this guy is going to intersect the y-axis that's what's creating that and that's why we have a shape that is opening to the right and opening to the left right you don't have anything on the y-axis like you had with your ellipse right you have points up here that you connect so that's why there's a drastically different shape between this guy and this guy okay so let's talk a little bit more about our hyperbola in general so i'm just going to go up here to kind of the equations you'd see in your book and run through them real quick so this is the one we're working with where we have the minus in front of the y squared over b squared the other one is where you have the minus in front of the x squared over b squared whatever has the minus in front of it again because of that minus sign in this case you wouldn't have x-intercepts so this guy's going to open up and down okay we'll see an example of that later on in this case we already know we won't have y intercepts so it's going to open to the right and to the left okay so that's what's causing that now your vertices okay your vertices are going to be at a comma 0 and negative a comma 0. and from the center to one of your vertices is going to be a distance of a so that's why i wrote down a equals 3 okay so we know that later on when we come back now b is going to have a purpose here and i'll talk about that in a moment but your foci are at c comma zero and then negative c comma zero and now your formula to get c is a bit different before it was c squared equals a squared minus b squared okay remember that now c squared is going to be a squared plus b squared really easy to remember because it matches the letters and the exact format of the pythagorean formula okay now similarly when we work with the hyperbola that's opening up and down everything is just kind of swapped because we have something that's opening up and down so now the vertices are going to be 0 comma a and 0 comma negative a so they'll be on the y axis in this case and your foci will be on the y axis as well now so 0 comma c and 0 comma negative c notice that a and b have been swapped okay so that's the most difficult thing with these formulas they're swapping around letters and that's where people get confused and so as we move through this lesson i'm going to show you not to use this formula you can just come up with your own okay so let's go back and let's talk about how we can get the rest of this stuff that's graphed here so we know a equals 3 i can erase that here i've already explained why basically you want to think about again our equation is x squared over nine plus y squared over 16 equals one and this is minus not plus okay so with this guy i have these kind of four points that make up a rectangle where does that come from well this is called the fundamental rectangle so let me put fundamental rectangle like that and basically what you do is you think about these as four points as the four corners or the four endpoints of the rectangle and normally depending on your equation you're going to get something like a comma b or you might get b comma a depending again on the equation in your book and then you go through the possibilities of that well i'm going to tell you to not do that okay let's not use those because they get confusing remember this is an x coordinate and this is a y coordinate okay so for the x coordinate go underneath x squared and look at your number here okay take the square root that's what you want okay remember go underneath x squared for x it's all you got to remember go underneath y squared for y take the square root of that number it's 4. then you're going to go through all the positive and negative combinations of that so you've got 3 comma 4 you would have negative 3 comma 4 you would have 3 comma negative 4 and then you would have negative three comma negative four so those four guys there are going to give you your four endpoints so your three comma four your three comma negative four your negative three comma four and your negative three comma negative 4 okay you can think about this guy again this is b squared so the distance of b you're going a distance of a units horizontally and a distance of b units vertically so if i go three units to the right and four units up if i go three is to the right and four units down if i go threes to the left and four units up and then three units left and four units down so that's another way you can think about that we know that c squared is equal to a squared plus b squared okay in this case you don't have to remember what a is and what b is because you can just add these two so 9 plus 16 is 25 so c squared would be 25 and just take the principal square root here that would be 5 okay so c is 5. remember c is the distance from the center to the foci so it's 5 units 5 units to the right and 5 units to the left and your foci are always going to be along the axis where you're opening right so if i'm opening along the x axis okay i'm going to the right and to the left well my foci are going to be on that axis okay so it might be shifted if you get a different type of equation but in this case it's going to be on the x-axis because we're centered at the origin so the foci the foci would be five comma zero and then negative five comma zero okay so pretty easy to remember that as well now the last thing i want to talk about is where we get these lines from basically when you draw the fundamental rectangle you can extend from the diagonals these lines okay these are your asymptotes and basically when you sketch the graph of these branches as we say right each side is a different branch you're going to sketch them to where they approach the asymptote okay they go through the vertex and then approach the asymptote okay so that's how you're going to do that now another way to get these lines or these asymptotes you have this let me just erase this because my screen is going to get too busy you have this formula that says y is equal to plus or minus i'm just going to say rise over run okay and i'm going to tell you why in a minute times x okay so why did i put rise over run well because again they use that b over a or a over b it keeps switching and people get confused so if you write rise over run here you remember that rise is associated with the vertical movement or the y axis okay how much am i going up and down so y i look at y squared i look underneath i get the square root of that number so square root of 16 is 4. so you're going to put 4 there okay for the run i think about x squared under x squared i have 9 square root of that is 3. okay so that's all you need to do so these would be the two equations for these asymptotes so you can get it that way as well basically you would have y equals four thirds x start at the origin go up one two three four to the right one two three there's a point then for the other one y equals negative four thirds x start at the origin go down one two three four to the right one two three there's a point okay so that's basically it when you talk about sketching the graph of a hyperbola they generally want you to find the vertices the foci okay the fundamental rectangle the asymptotes and then sketch the graph okay there's not much more to it it can get shifted like we're going to see in this example here and then it gets a little bit more complex so this one is going to be double complex because i gave it to you in general form so you have to complete the square first so you might get examples like this as well and so let's blow through this real quick we already know how to do this we're going to group our x terms together so negative 16 x squared and then i'm going to go plus my 32x and plus well let me not do that yet so let me just kind of close this down then plus we'll go 9y squared plus 18y and then i'm going to add this 151 to each side so that it cancels over here and goes over here so equals 151 okay and basically i'm doing this because i want to create a perfect square trinomial here okay here and here okay so i'm going to add and subtract the same number inside of the parentheses but before i can do that remember the coefficient for your squared variable needs to be a 1. we're not going to divide anything here we're going to factor okay so i'm going to pull out a negative 16 here i'm going to pull out a 9 here so negative 16 times the quantity x squared in this case it would be minus 2x okay and i'm going to leave a little space there and then plus i've got to pull out a 9 here times y squared this would be plus 2y i'll leave a little space there and then this equals 151. okay so what goes here and here well all you need to do is cut this coefficient here the one on the variable raised to the first power in half meaning you multiply by a half and then square it if you have negative two times a half you get negative 1 negative 1 squared is 1. same thing for 2. if you cut in half you get 1 1 squared is 1. okay so in each case you're going to end up with 1. so you add 1 and subtract 1. add 1 and subtract 1. okay but remember you want only this part here because that's a perfect square trinomial you want only this part here but to do that and kind of get this part outside of the parentheses you've got to distribute the negative 16 and you've got to distribute the not okay so that's very important so what i'm going to do is say that i have negative 16 times the quantity i'm going to close this down and factor it so it's going to be x minus 1 quantity squared and then it's going to be negative 16 times negative 1 which is plus 16 okay and then you're going to have plus your 9 times i'm going to close this down and factor so y and it's going to be plus 1 quantity squared and then 9 times negative 1 would be minus 9. this equals 151 okay so from here it's kind of a piece of cake basically what you want to do is say 16 minus 9 is going to give you 7. so let's just erase this and put a 7 here and if i subtract 7 away from each side of the equation this is going to cancel okay and then over here this would become 144 okay so i can say i have negative 16 there's a quantity x minus 1 squared plus 9 times the quantity y plus one squared equals 144 okay so to finish this up i want this to be a one so divide everything by 144 and you'll be done okay let's scroll down a little bit more so in this case 144 divided by 16 is 9 and you've got the negative there so let's make this a negative out in front and make this a 9 down here and then 144 divided by 9 is 16 so let's make this a 16 down here and we know this is a one now you typically don't want to write it like this whatever you have your minus out in front of you want to put that in the middle so you would want to say that this is the quantity y plus 1 squared over 16 and then minus you have your quantity x minus 1 squared and then over 9 and this equals 1. okay so let's go ahead and copy this all right let's go ahead and paste this in here and we're just going to walk through how we get this the way i like to do these is i like to think about my formula okay my generic formula without the a and the b involved and then i shift them based on what the new center is okay so i pretend like the center is here and then i shift it here okay so i'm going to shift all the points so i just want to think about the fact that this guy is going to open up and down we can see that clearly but if we didn't know that if we didn't have a graph in front of us you have the minus here and it's in front of where you have the x variable involved so again that's how we know that it's going to open up and down okay if the minus was in front of where you had the y involved you would open to the right and to the left now let's talk about how you get the vertices okay we know that with our other formula if we had something like y squared over let's say a squared just for right now then minus x squared over b squared okay this equals 1. well this guy right here again i would take the square root of that whatever that number is okay and i would say that my y-intercepts would be at zero comma a okay and then zero comma negative a okay we talked about that already that's how you find your vertices in this case well here everything is shifted one unit down because of the plus one and one unit to the right because of the minus one so if i started with this guy right here and said okay my vertices would be at zero comma four and zero comma negative four well okay i've got to shift it one unit down so this would be a 3 okay and this would be a negative 5 and then i would go one unit to the right so this would be a 1 and this would be a 1. okay so i'm looking for 1 comma 3. so 1 comma 3. so that's right there let me choose a different color so this is right there and then 1 comma negative 5 so 1 comma negative 5 is going to be right there okay so that's where those two vertices come from of course the center comes from this right here so it's minus whatever this is okay so it's one this is not in that format so you have to change it right if it's a plus one you think about it as minus a negative one okay so it's going to be at one comma negative one so one comma negative one so right there that's your center now let's talk a little bit about your foci okay your foci so we know that c squared is equal to a squared plus b squared but again everything is shifted okay so first you go again with the 16 plus 9 that's 25 okay so you have c squared equals 25 so c is 5. so if this was centered at the origin it would be at 0 comma 5 and 0 comma negative 5. but again everything is shifted one unit to the right so i put this at a one and a one and everything is shifted one unit down so this would be a four and this would be a negative six so you'd have one comma four so one comma four so that's right there and then one comma negative 6. so 1 comma negative 6 so that's right there so that's where those two foci come from okay let's talk about the fundamental rectangle now okay so where do these endpoints come from again i think about x comma y this is how i always write it okay so where does x come from where does y come from okay well x comes from what's underneath the variable x i know in this case it doesn't look like just x squared but just go to where x is go underneath take the square root so the square root of that is three okay then go underneath where y is take square root of that that's four so you got three comma four now things have been shifted because the center's moved okay so you've got to account for that so now what i'm going to say is that let's just kind of go through the possibilities of this and we'll adjust so it would be three comma four it would be negative three comma four it would be three comma negative four and then it would be negative three comma negative four well everything is shifted one unit to the right so this one and this one would be four this one and this one would be negative two okay then also everything has shifted one unit down so this one and this one would be three and this one and this one would be negative 5. okay so that's how i do it personally you may have a different formula that works better but that works for me so you can see the endpoints of this guy are going to be 4 comma 3 so right there and then you've got 4 comma negative 5 so right there and then you've got negative 2 comma 3 so right there and then negative 2 comma negative 5. again all you're really doing is saying from the center okay from the center wherever that is i'm going three minutes to the right and four units up so from the center if i go three units to the right i go one two three and four units up i go there from the center if i go three inches to the left and four units up i go there from the center if i go three minutes to the left and four units down and go there from the center if i go three inches to the right and four units down i go there so it's the same thought process you just have to shift things because the center has moved now you have your fundamental rectangle so you can sketch the graph of the asymptotes and at this point if you're just told that you want to graph this guy you're done okay you don't need to go through and get the equation of these asymptotes but if you do need that equation okay it is a little bit more work because now you're not going to be able to just say that y is equal to plus or minus you know in this case it would be 4 3 times x because i took the 4 from here for the rise the 3 from here for the run okay you can't just do that because the center is not at the origin okay so you have to take a point on this line and you have to use point-slope form okay you know the slope right it's still going to be four-thirds and then negative four-thirds but you don't exactly have the equation so you have to go y minus y sub 1 and i'm going to take the y coordinate from the center which is negative 1. so i'm going to say y minus y sub 1 y minus negative 1 is y plus 1 is equal to i'm going to go with four thirds first times the quantity you're going to have x minus again from the center the x coordinate is one so x minus one and so you would go through and figure this out so you'd have y is equal to subtract one away from each side you'd have four thirds x minus four thirds and then go and say this is minus three thirds so this is four thirds x minus seven thirds so that would be one equation okay for the other one you would need to work with negative four thirds okay so y minus y sub one in this case y sub one is negative one so y plus one equals negative four thirds times the quantity x minus in this case our x value is one so we're going to say x minus one so what you'll have here is y is equal to i'm going to subtract one away from each side so negative four thirds x negative four thirds times negative one is plus four thirds again i subtracted one away from each side so minus three thirds so this would end up being one-third here okay so this is one-third okay so i know this is long and tedious but again this is stuff that you do have to know so let's kind of move this up here and you can check this real quick i know that the y intercept of 1 3 is a little bit difficult to find but let's just go ahead and say that that's right here okay and then the slope is negative 4 3 so if i go down 4 if i go down 1 2 3 4 and i go to the right three so one two three you see that should be about right there again it's not perfect because we're not working with integers but it's very very close okay and then if you look at this one right here four thirds x minus seven thirds again now i'm going with the positive slope so if i look at negative seven thirds let's say that's about right there so if i start here and go up four so one two three four and then go to the right one two three you're gonna be about right there okay it's not perfect but when you work with non-integer values you're always going to be somewhat off okay so that's where everything comes from in this particular case if you're asked for your asymptotes you know how to find them if you're asked for your fundamental rectangle you know how to get it basically you just use the technique that we talked about you just shift things based on where your center has moved okay let's talk about one final topic this is something that you will get and it is a bit frustrating so let's say you're given the vertices and the foci only and they say come up with the equation okay so it's a lot easier than you think the vertices you're given remember from the center of the hyperbola to one of the vertices is the same distance from the center to the other vertex so if i took these two and thought about the midpoint of this line segment okay that would be the center okay of this hyperbola so i have negative 5 and negative 5. that doesn't change so the x value is not changing okay that tells me this guy is opening up and down okay so what i would do is say that the center the center is going to be negative 5 plus negative 5 is negative 10 divided by 2 and back to negative 5 right that doesn't change and then negative 1 plus negative 9 is negative 10 divided by 2 i get negative 5. so my center is at negative 5 comma negative 5. okay so what does that tell me that i'm looking at something like this i'm looking at something like y plus 5 right because it's always minus whatever it is in this case it's negative five quantity squared over let's just say a squared for now then minus you would have x again plus five because it's minus five quantity squared let's say this is over b squared this equals one so we need to find a squared we need to find b squared so what is a a is the distance from the center to one of your vertices okay well if i went from negative 5 comma negative 5 to negative 5 comma negative 1 how far did i travel well this negative 5 and negative 5 that's the same right because i'm traveling vertically i'm going up so all i need to compare is negative 5 to negative 1 how far did i travel well 4 units right so that tells me a is 4 so a squared is 16. okay so i've got that part now i need b squared okay and where this is going to come from is from that formula that a squared plus b squared equals c squared okay and in this particular case you've got your foci that occur at negative 5 comma negative 5 plus square root of 65 and negative 5 comma negative 5 minus the square root of 65 okay you only need one of these but basically c is the distance from the center which again occurs at negative 5 comma negative 5 to one of these guys let's just say the negative 5 comma negative 5 plus square root of 65 okay we know this x coordinate is the same in each case so i'm just comparing the y coordinates so how far did i move in this case up right so i went this is negative 5 and this is negative 5. so just erase those you just have the square root of 65 left so i'm traveling from the center up by the square root of 65 units so that's going to be my c okay so c squared would be 65 so a squared is 16 and c squared is 65 let me make that better so what is b squared well if i subtract 16 away from each side of this equation i would see that b squared is equal to 49 and so that's all i need i don't need to know that b is 7 right i could take the square root but i need b squared so this is 49. so it seems a little bit harder than it is it doesn't take that long but you do have to kind of know what is a that's the distance from the center to one of your vertices what is c that's the distance from your center to one of your foci and then for getting b you need to go through this formula and crank that out so we get the quantity y plus 5 squared over 16 minus the quantity x plus 5 squared over 49 is equal to 1. in this lesson we want to talk about solving nonlinear systems of equations all right so when you get to this section basically what you're looking at are systems where at least one equation is non-linear now solving these it's not a hard thing to do when you get to this section but it is tedious and there's not one general approach so every problem you see has a different approach or different strategy to it so i'm going to do five examples that cover basically all the different types of problems that you're going to see you might get other ones but it's very easy to take one of these examples and apply it to yours all right so let's go ahead and get started the first one's a very easy example we have one that's not linear and one that is so we have 2x squared plus 2y squared minus 21x minus 6y plus 44 equals 0 and x minus 2y equals negative 4. so the strategy here is going to be to solve this linear equation for one of the variables and then plug in to this kind of top equation that's not linear now when you do this you might get extraneous solutions i'm not going to be checking things because this video is already going to be really long because it's tedious but i advise you to check your solutions because when you solve these you can often get extraneous solutions so there are solutions that won't work in the original problem all right so let's go ahead and get started i'm going to solve this bottom equation for x so we'd have x is equal to if i add 2y to both sides of the equation i would have 2y minus 4. so once i have that i know x is equal to this so i can plug this in everywhere there's an x in this top equation okay so i can plug it in there and also there okay so that's what i'm going to do that's going to give me an equation with just y right and then i can solve so i would have 2 times i've got to use parentheses here because this guy plugging in for that and it's squared so 2y minus 4 that quantity is squared and then plus you've got 2 y squared minus 21 times again use parentheses when you're plugging in so we've got 2y minus 4 and then minus 6y plus 44 equals zero okay so the common thing there where you'd make a mistake is to not use parentheses you got to make sure you do that all right so once we have it set up it's very straightforward you just have to do a lot of simplifying which can take some time this guy right here we know how to expand this again this is squared so i would square the first term 2y squared is 4y squared then you have minus 2 times this guy times this guy so 2 times 2 is 4 4 times 4 is 16 then times y is 16y okay and then you would have the last guy the 4 squared so plus 16. now this guy is being multiplied by 2. so let me just wrap this like this we'll just move on we'll clear that in a moment so we have plus this 2y squared i'm going to distribute this to each so i'm going to distribute that to each term there so negative 21 times 2y is going to give me negative 42y and then negative 21 times negative 4 is going to give me positive 84 okay then i have minus 6y plus 44 and this equals 0. all right so i know i need to distribute here so let me just kind of do that 2 times 4y squared would be 8y squared 2 times negative 16y would be minus 32y and then 2 times 16 would be plus 32. so let me just erase this and i'll just kind of drag this up and i'll just slide this down a bit so we have more room to work and let's think about what our like terms are going to be so we see that we have this guy and this guy right nothing else that's squared so 8y squared and 2y squared would give me 10y squared so 10 y squared okay so what else well we have this guy right here this guy right here and this guy right here let me unhighlight these so we can be clear on what we're working with so we have negative 32y minus 42y and negative 6y okay so negative 32 minus 42 is negative 74 and then minus another 6 would be negative 80. so this would give me minus 80 y okay so let me unhighlight those now and then lastly we have the constants so you've got 32 84 and 44. okay so 32 plus 84 is 116 then if i add 44 i get 160. so plus 160 and this equals zero all right let's scroll down and get some room we'll come back up in a minute so what i notice immediately is that this guy right here this guy right here this guy right here they all have a zero at the end of the number so that tells me everything is divisible by 10 so i can make this simpler by just dividing everything in that equation by 10. so this would be y squared this would be minus 8y this would be plus 16 and of course this would still equal 0 because 0 divided by 10 is 0. so from here now it's very easy to solve i can factor this so y and y and we know that this is going to be what it's a perfect square trinomial right because if i factor this it becomes a binomial squared it becomes y minus 4 quantity squared okay so you can write it out the long way like this or again you can put y minus 4 quantity squared so this guy only has one solution it's y equals 4. all right so let me erase everything i'm just going to write y equals 4 here for your notes and i'm going to erase everything so now that we know that y equals 4 i can plug in a 4 for y either in the first equation or the second equation i'm going to pick the second equation because it's simpler so plug in there and you would say x minus 2 times just plug in a 4 for y equals negative 4 of course 2 times 4 is 8 so you can just erase that and put 8. add 8 to both sides of the equation you find that x is equal to 4 as well so the ordered pair solution here would be 4 comma 4 right an x value of 4 and a y value of 4. now i just want to say this one more time when you're solving these you have to make sure that you stop and check your answer in both equations of the system to make sure it works i've already done that so i'm going to be skipping that so i can make this video a little bit shorter all right so the next system is very easy to solve is when you get two parabolas okay so you'll notice that you have negative two x squared plus fourteen x plus y minus fourteen equals zero seven x squared minus forty nine x plus y plus seventy-six equals zero you just have y to the first power and y to the first power no other y's okay but you have your x squared and your x to the first power so if you wanted to you could rewrite these and solve them for y and in each case you're going to have y equals ax squared plus bx plus c okay so let's go ahead and solve each one for y real quick and i'll explain how we can quickly solve this so what i'm going to do is just move everything to the other side since y is already over here so i'll say this is y equals 2 x squared minus 14x and then plus 14. so again kept this on this side moved everything to the other side so all i'm doing is just changing the signs okay that's all i did then for this one i'm going to do the same thing so y is equal to and then i'm just going to move everything to the other side so i'll just be changing the signs so i'll have negative 7x squared plus 49x and then minus 76. so once you've written it in this format remember if i'm solving a system i want to find the points of intersection so where are these guys going to be equal or the same well where they're going to be equal the y values would be the same so if this guy represents y and this guy represents y well i can set those two equal to each other and solve for the x value that makes that happen okay so what i'm going to do is say that 2x squared minus 14x plus 14 again this represents y here is equal to the y from here which would be negative 7x squared plus 49x minus 76. so when you get these they're very very easy to set up just solve each one for y and then equate the two okay so we'll say that we have what if i add 7x squared to both sides i'm going to have 9x squared right here and let's go ahead and just cancel that out i'm going to subtract 49x away from each side of the equation so let's cancel that out there and if i did negative 14x minus 49x i would get negative 63x so let's put minus 63x here then lastly let's go ahead and add 76 to both sides of the equation and i just want to strike that out real quick so 14 plus 76 is going to give me 90 and of course on this side everything's canceled so it's a zero okay so this is equal to zero so now we just have a quadratic equation that we've been solving forever we know that immediately everything here is divisible by 9 right so i can divide everything by 9 and get x squared minus 7x plus 10 equals 0. this is something i can solve with factoring again very very easy let's go ahead and set this up x here and x here and to think about the factors what adds up to negative seven and multiplies to be ten right what two integers well we know we could do negative two and negative five right so negative 2 and negative 5. so if i set each one of these factors equal to 0 and solved i would get x equals 2 and then x equals 5. so x equals 2 and then also 5. okay so let's erase everything so at this point what i want to do is plug in a 2 for x see what y is equal to and then plug in a 5 for x see what y is going to be equal to okay so if i start out with 2 let's just use the top equation so negative two times you'd have two squares let's just go ahead and say this is four and then plus fourteen times two is twenty-eight and then plus y minus fourteen equals zero so we know negative two times four is negative eight okay so negative 8 you might as well just say negative 8 plus 28 is 20 okay and then 20 minus 14 is 6. so let's just say at this point we have y plus 6 equals 0. let's subtract 6 away from each side of the equation and say y is equal to negative 6. so if x is 2 y is negative 6. and again you can check that in each equation make sure it is true okay so i've already checked it so i know that 2 comma negative 6 is a solution so let's erase this and work with 5 now so if i do negative 2 times 5 squared is 25 plus 14 times 5 which is 70 plus y minus 14 equals 0. well i know this is negative 50 right negative 2 times 25 is negative 50. negative 50 plus 70 is 20. and then 20 minus 14 is again 6. so we know that we're going to end up with y is equal to negative 6 again so the other solution would be 5 comma negative 6. so we have an ordered pair 2 comma negative 6 that works and again it works in each you can stop the video and check and then 5 comma negative 6 that works as well again you can pause the video and check that all right the next type of example is one that is best solved using elimination so we have 3x squared plus y squared equals 3 and 4x squared plus 5y squared equals 26. all right so what i'm going to do is eliminate the y variable it's pretty easy to do because i can just multiply equation 1 by negative 5 so i would have what negative 15 x squared minus 5 y squared equals negative 15. again i just multiplied everything there by negative 5. so then this second equation stays the same let me just copy it real quick and we're going to add these two together and what would we get negative 15x squared plus 4x squared would be negative 11x squared this would be gone and negative 15 plus 26 is going to be 11. so let me erase this and let's drag this up here and let me divide both sides by negative 11 and we'll find that x squared is equal to negative 1. okay well if we want to get x by itself we've got to take the square root of each side so i take square to this side and i go plus or minus square root of this side we know the square root of negative one if we want to use the complex number system is i right the imaginary unit so i can just go ahead and say that x is equal to plus or minus i okay so plus or minus the imaginary unit so you can really say that you have x is equal to i and then also negative i okay so we've got to plug each of these in so if i plug in i remember i squared by definition is negative 1. so three times negative one because that's all i squared is plus y squared is equal to three well this is negative three plus y squared equals three add three to both sides and we're going to get that y squared is equal to 6 okay so now if i want to get y by itself i got to take the square root of each side so square to this side and then plus or minus square to this side so y is equal to plus or minus the square root of 6 okay and i can't make that any simpler so if x is i the imaginary unit then y is the square root of 6 and then also if it's i it's the negative square root of 6. so those are two different solutions again if it's i you get square root of 6 or if it's i you also get the negative square root of 6. all right let's think about this other possibility here where x is negative i now we see that if we plug this into this equation here just think about the fact that okay i have negative i being squared but it's going to give me the same result if i think about negative i as negative 1 times i and this is squared well this is what this is negative 1 squared times i squared which is 1 times negative 1 which is negative 1. so it's still going to give me negative 3 here so i'm going to go through and get the same result okay so really i can just say that i would have negative i comma square root of six and then also negative i comma negative square root of six so there's your four solutions for this guy again for this one you want to use elimination and again i've already checked my answers so i know they work all right so the next one we're going to look at we want to do elimination and substitution so these generally take a little bit of time so i have this x squared plus 8x minus y squared plus 10y minus 33 equals 0 and x squared plus 8x minus 5y squared plus 42y minus 97 equals 0. notice how these are the same okay so what i can do here is start by multiplying this bottom equation by negative 1. so what i'd have is let me copy the top one so x squared plus 8x minus y squared plus 10y minus 33 equals 0. for this one if i multiply by negative 1 i'd have negative x squared and then we'd have minus 8x plus 5y squared and then you'd have minus 42y plus 97 equals 0. if i add the left sides together and i set this equal to the sum of the right sides which again would just be 0 in this case well this is going to cancel okay so what i have is negative y squared plus 5y squared which is 4y squared 10y minus 42y would be minus 32 y and then 33 plus 97 is 64. so plus 64 here okay so again we notice that and let me just erase this and let me drag this up so again as i was saying we notice that everything here is divisible by four so just divide through and get rid of it so if i divided this by four i'd have y squared divide this by four i'd have negative eight y and let me make that a little bit better so minus 8y divide this by 4 you'd have 16 and divide 0 by 4 and you still have 0. okay so this is much simpler much easier to work with i can just again just factor this and say that i have y here and y here so i'm going to say this is what this is a perfect square trinomial it factors into y minus 4 quantity squared okay so your solution here for y is just 4. so we know y is 4. so let's figure out what x is all right so i'm just going to plug into the top equation i'm going to plug in a 4 here and here so i would have x squared plus 8x and then i would have minus what's y squared well i'm plugging in a 4 4 squared of 16 and then you have plus 10 times 4 would be 40 and then minus 33 equals 0. so let's go ahead and crank this out real quick negative 16 plus 40 is 24 and 24 minus 33 is negative 9. so i'm just going to erase this and put minus 9 there and then equal 0. okay let me make that a little better again we have something we can solve using factoring so pretty easy overall if you want two integers that sum to 8 and give me a product of negative 9 well i know i want to use positive 9 and negative 1. so the solutions here for x would be negative 9 and positive 1. okay x plus 9 equals 0 the solution is x equals negative nine x minus one equals zero the solution is x equals positive one so i would say that you have one ordered pair that would be negative nine again the solution here and then comma four that y value and another that would be 1 comma 4 that y value ok so two solutions again i've already checked these i know they work once again you always want to check your solutions all right for the last problem we're going to look at it's going to be another one that's extremely tedious and we're going to use a combination of elimination and substitution again so we have 3xy minus 2y squared equals negative 2. we have 9x squared plus 4y squared equals 10. so in this bottom equation you have two variables that are squared in the top equation you have an x to the first power and a y to the first power and then a y squared so the strategy here is to get rid of that y squared and i can do that with elimination okay so that's the first step so i'm going to multiply the first equation by 2. so i'm going to have 6xy and then minus 4y squared this equals negative 4. okay just multiply everything by 2. let me just write this one again so 9x squared plus 4y squared equals 10. if i add these two together then i'd have 9x squared plus 6xy and then this cancels and it would be equal to 6. now before you go any further realize that you could divide everything in this equation by 3. so 9 divided by 3 is 3. so this would be 3x squared 6xy divided by 3 would be 2 xy so plus 2xy and this equals 6 divided by 3 would be 2. okay so i always want to work with something as simple as possible now i can take this guy right here and i can solve it for y okay so the way i would do that is i would subtract 3x squared away from each side [Music] and that would give me 2xy is equal to negative 3x squared and then plus 2. okay so i have that and then i could divide both sides by 2x and let me erase this up here and i'll say what that y is equal to this negative 3x squared plus 2 over 2x okay so we got that figured out now where do we go from here well what i want to do is plug into this equation here i'm going to plug in for y so what that's going to give me is 9 x squared plus 4 times this guy i'm going to put parentheses around it so negative 3x squared and then plus 2 over 2x and this guy squared so be very careful here and then this is equal to 10. so this is going to require some work so let's scroll down a bit so let's think about this for a second forget about the four let's just focus on this part right here so if i had negative 3x squared plus 2 over 2x and again if this was squared then it's really like i had two of them right so negative 3x squared plus 2 over 2x well how do we multiply two fractions two rational expressions all i need to do is multiply the numerators together put that over the product of the denominators so this guy is just negative three x squared plus two quantity squared right so the first one would be squared so the negative three squared is nine x squared squared is x to the fourth power and then after i do that it's 2 times this guy times this guy okay so i know it's minus 2 times 3 times 2 well 2 times 3 is 6 6 times 2 is 12 and then times that x squared okay and then you would have plus this last guy squared which in this case 2 squared is 4 okay then over 2x times 2x is 4x squared okay so we've got that worked out now let me drag this up here and when you work these you want to have a lot of scratch paper because you get a lot of stuff you've got to go through so if i have four times this the fours are going to cancel so let me just erase this let me put equals 10 down here and i'm going to get rid of this and just substitute in what i have to get rid of that and let's just drag this up here now and that's going to be our simplified equation and what i want to do if i want to solve for x i got to get rid of this x squared in the denominator so if i multiply through by x squared what's going to happen is we would have 9x squared times x squared which is 9x to the fourth power okay then it would cancel down here so i'm just left with this numerator here so plus 9x to the fourth power minus 12x squared plus 4. this equals i'd multiply this by x squared so 10 x squared so again all i did was multiply both sides of the equation by x squared to clear that denominator all right so at this point we can solve this for x i'm going to subtract 10x squared away from each side of the equation and i'm going to say that 9x to the 4th power plus 9x to the 4th power is 18x to the 4th power negative 12x squared minus 10x squared would be minus 22x squared and then we have plus 4 and of course this cancels so it's equal to 0. you can immediately see that again we can divide everything by two and so this would be nine x to the fourth power minus eleven x squared plus two equals zero now this can be factored you can use a lot of different techniques to factor this you could easily use substitution where you let u equal x squared and then you can plug in and go through all that because of the length of this video already i'm just going to give you the factored form and say this is going to be 9x squared minus two so that quantity times you have x squared minus one and of course this can further be broken down into x plus one and i forgot the one there times x minus 1 okay so you've got to go through and set each factor equal to 0 and so what will we have well we know for these two it's really easy the solution here would be that x is equal to 1 or negative 1. so that takes care of those two but for this one you need to do a little bit more work so 9x squared minus 2 equals 0 add 2 to both sides of the equation let's just go ahead and say this is 9x squared is equal to 2 divide both sides by 9 and you would have x squared equals 2 9. so if i take the square root of this side and plus or minus the square root of this side i would have what you would have that x is equal to plus or minus the square root of 2 over the square root of 9 is 3. so that's a lot of different things to go through so you have x equals 1 negative 1 square root of 2 over 3 and negative square root of 2 over 3. so we would say square root of 2 over 3 and negative square root of 2 over 3. all right so we come back up here and i'm going to use this that we found earlier that y was equal to this guy right here you can plug into these equations up here this is going to be a lot quicker a lot easier so that's why i'm going to do it this way all right so if i plugged in a 1 we know that we would get what let me kind of do this off to the side y would be equal to 1 squared is 1. so negative 3 plus 2 which is going to be negative 1 over 2 times 1 which is 2. so negative 1 half so if x is 1 y is negative 1 half all right what about for negative 1 well you would have negative 3 times negative 1 squared is 1 so you just have negative 3 plus 2 so that's negative 1. but then down here 2 times negative 1 would be negative 2 so this is positive one-half so if x is negative one then y is positive one-half so let's get rid of these two and let's work with these two now if i had negative three times this guy squared so you have the square root of 2 over 3 this guy squared plus 2 and this is all over 2 times square root of 2 over 3. so if i squared the square root of 2 i would get 2 if i squared 3 i would get 9. so this would be 2 9. so this would be 2 9. if i divide 9 by 3 i get 3. so this would just be negative 1 and this would be 3. so this would turn out to be negative two-thirds so negative two-thirds okay so now from here we have plus two so let me get a common denominator ago i'm just going to multiply this by three over three so this would be six-thirds so plus six thirds and of course this would end up being four thirds so that's my numerator i have four thirds so down here what am i gonna have just two times square root of two over three okay so let's simplify this by just saying that we have four over three times the reciprocal of this which would be three over two times square root of two okay so let's erase this what do we get well this would cancel with this and give me a 2 and this would cancel with this so what i'd have is 2 over the square root of 2 and if you rationalize the denominator there if i multiply this by square root of 2 and this by square root of 2 then what i'm going to end up with is what this would end up being 2 down here and of course i can cancel this so i end up with just the square root of 2. so if i had the square root of 2 over 3 then i would get the square root of 2. okay so now we have one more to check and i know this is tedious but again you're going to get problems like these so you might as well work some so the last one let me just get some room going we would have the negative square root of 2 over 3. if i squared that again i'd have negative 3 times in this case the negative would go away square root of 2 squared is 2 so this would just be 2 9 and we know that this would cancel and become negative two thirds so this is negative two thirds plus two and again we wrote that as six thirds so this became four thirds so that's again the numerator so it's four thirds then over 2 times this guy which is now negative so i would have negative 2 times square root of 2 over 3. okay so let's get rid of this and of course this turns into what 4 over 3 times negative 2 times square root of 2 going in the denominator up here you'd have your 3. okay so let's erase this and let's go ahead and cancel this with this and this with this and say it's a 2 here so what you have here let me just kind of write this over here we'll say this is negative 2 over square root of 2. again you want to rationalize so if i multiply this by square root of 2 over square root of 2 square root of 2 times square root of 2 is 2. and this would be negative 2 times square root of 2. we can cancel this with this and just say we have the negative square root of 2. so this would be negative square root of 2 over 3 and then comma you'd have negative square root of 2. let me erase this and those would be our answers again i've already checked this when you're working these you got to make sure you check them i know it's really really tedious it takes a lot of work to check them but you could report a wrong answer if you don't so here we get 1 comma negative 1 half negative 1 comma 1 half square root of 2 over 3 comma square root of 2 and then negative square root of 2 over 3 comma negative square root of 2. in this lesson we want to talk about systems of inequalities all right so the main idea here if you're solving a system of inequalities you're going to sketch the graph of each inequality and then you're going to look for the overlap of the two graphs that's going to give you your solution for the system because the area of the coordinate plane that satisfies both inequalities of your system so we're going to start with an easy example we have two linear inequalities and two variables we have negative three x plus y is less than or equal to negative five we have y is greater than or equal to x minus four now you can use the test point method or you can solve for y whichever you want to use i'll cover both let's solve for y first because it's a bit easier so what i'm going to do is i'm going to add 3x to both sides of this inequality and this will give me y is less than or equal to i'll have 3x minus 5 okay this one's already solved for y so y is greater than or equal to x minus 4. okay so the reason you solve for y is if you've solved for y you can look at the inequality and tell which direction you need to shade okay in a moment i'm going to talk about a boundary line so once you have that you look at your inequality again if this is solved for y only and you say is this a less than or is this a greater than okay so it could be a less than or less than or equal to that means you shade below the line okay if it's a greater than or greater than or equal to you shade above the line it's very easy to remember a less than or less than or equal to you shade below the line greater than or greater than or equal to you shade above the line now when we talk about this boundary line all it is is we take our inequality and we replace the inequality symbol with an equality symbol so in this case we have y equals 3x minus 5. so we'd graph that in this case we'd have y equals x minus 4. okay so what is the boundary line well basically it separates your coordinate plane into two regions one where all the points all the x comma y ordered pairs are going to satisfy your inequality so if you plug those into the original inequality you would get a true statement the other region which is on the other side of the boundary line none of the points are going to work okay so if you plug it in you're going to get a false statement okay so that's the test point method we'll talk about that in a moment but essentially you graph your boundary line okay so you just sketch the line like if you were normally graphing this guy and then again if you have a less than or less than or equal to you shade below if you have a greater than or greater than or equal to you shade above okay now one last thing when you work with a boundary line if it's a strict inequality meaning you have strictly less than or strictly greater than you want to put a broken or dashed line because in this case your boundary line is not part of the solution set okay it's strictly less than or strictly greater than okay in this case we have non-strict inequalities we have less than or equal to and we have greater than or equal to so because of the or equal to part we want to include the boundary line in our solution set so it's a solid line okay so let's go ahead and graph y equals 3x minus 5 and then we would shade below the line because this is a again less than or equal to so i've already done this to kind of speed it up but again we have y is less than or equal to 3x minus 5. the boundary line is y equals 3x minus 5 and again it's solid because this is non-strict so we know it's in slope-intercept form so the y-intercept is 0 comma negative 5. so go down to 0 comma negative 5. that's your point on the line and then you can say okay well the slope is 3 so go up 1 2 3 to the right 1 up 1 2 3 to the right 1. you know so on and so forth sketch that line now because this is a less than or equal to and because we've solved for y we shade below the line so i'm going to shade everything below the line so that's my shading that's already done for us but again if you're doing this on your own just want to shade below the line okay let's talk a little bit about this test point method if you come back up here let's say you didn't solve this for y you sketch this negative 3x plus y is equal to negative 5 you get the same boundary line okay now if you have this graph let's say you pick a point let's say you pick 0 0 because it's easy to work with if you plug it in it's not going to work how do i know it's not going to work because it's not in the solution region again i talked about the fact that you're either in the solution region or you're in the non-solution region we already know that below the line here is the solution region this zero comma zero at the origin is above the line okay so it's going to be in the non-solution region so if i plugged it in i should get a false statement so if i said what is negative three times zero well that's zero plus zero well that's zero okay and is zero less than or equal to negative five well no zero is greater than negative five and zero is not equal to negative five so that's a false statement okay this is false so if you plug in a point like zero comma zero and it fails right you know you're in the non-solution region so if i use the test point method i plug in that point it doesn't work so i need to shade on the opposite side of that line okay if it does work i shade on that side of the line so that's another way you can get this shading here again below this line okay the other one was y is greater than or equal to x minus 4 again the boundary line is y equals x minus 4. okay and i've drawn these two together and basically what we're going to see here is that let me write this out so y is greater than or equal to x minus 4 the boundary line y equals x minus 4. again to graph this the y intercept is at 0 comma negative 4. so we'll come down here and pick a new color let me pick this color here so that's going to be right there right so 0 comma negative 4. the slope is 1. so i can go up 1 to the right one up one to the right one up one to the right one you know so on and so forth i get this point this point you know how to sketch the graph of that line already now again because this is a greater than or equal to i want to shade above this line so i'm going to shade in this direction okay so i have each of these already done right in terms of the shading for the orange line remember we were shading below okay we were shading below so if we think about the area of the coordinate plane now that satisfies both what do we have what is below the orange line and also above this green or again turquoise line well it's going to be this area right here okay this area right here and to highlight this a little better i have a cleaner graph where this is clearly shown okay so this is the area of the coordinate plane that is below the orange line and above this again i'm just going to call it a green line okay so this is the area of the coordinate plane that is the solution for both okay simultaneously and so it's the solution for your system okay it's got to satisfy both so you can pick any point okay in the shaded area plug it into both inequalities of the system and it's going to give you a true statement in each case let's look at an example where we have a non-linear system so this is where you have at least one non-linear inequality okay so we're going to have to work with a parabola and then we're also going to have a line but you can have many types of shapes you can have a hyperbola and an ellipse you can have a parabola you can have all kinds of different shapes all the things we've worked with so far so we have y is greater than or equal to x squared minus 4x minus 1. we have negative 2x plus y is less than or equal to negative 5. so for this one it's very easy we would just solve it for y so y is less than or equal to you've got your add 2x to both sides so 2x and then minus 5. okay so your boundary line is y equals 2x minus 5 and again it's solid because this is non-strict okay so we would graph this and shade below right because this y is less than or equal to so graph that and shade below so we have y again is equal to 2x minus 5 and y is less than or equal to 2x minus 5. so this is the graph okay this boundary line of y equals 2x minus 5. again my y-intercept is at 0 comma negative 5. my slope is 2. right so i'm going to go up 1 2 to the right 1 up 1 2 to the right 1. you know so on and so forth you get that line and because this is less than or equal to i want to shade below so that's where i got that shading from so we're shading below the line okay let's go back up let's think about this parabola now so this is a little bit more difficult we can just erase this we don't need anymore we have y is greater than or equal to x squared minus 4x minus 1. so when you start getting things that are more complex shapes a lot of times it's easier just to use a test point right because you might not be able to think about where is y greater than in this case well for a parabola it's going to be shading inside of the parabola okay and if it's less than then you want to shade outside the parabola okay but you can also use a test point to show that but for this guy for the boundary you just again use inequality symbol so y equals x squared minus 4x minus 1. to graph this it's easiest to put it in your vertex form okay so we already know how to do this at this point so we would say y is equal to a which in this case is 1 times the quantity x minus h squared plus k so again where does h come from where does k come from h is what the vertex formula tells us this negative b over 2a so what is b b is the coefficient when x to the first power so the negative of negative 4 is 4. let me change that to this and then a is 1 so 2 times 1 is 2. 4 over 2 is 2. okay so i know that h is 2. you plug that in how do we get k well it's f of 2 in this case right i know i have the y notation but basically you're just plugging in a 2 for x and see what you get so 2 squared is 4 then minus 4 times 2 is going to be minus 8. and basically what you have is 4 minus 8 is negative 4 minus 1 is negative 5. right so this is minus 5 here so once we have it in this format we know that the vertex the vertex occurs at two comma negative five okay two comma negative five now from there you can use your step pattern again this is one so you plot the vertex and then you go one three five seven nine remember that step pattern so you go one year to the right then one unit up one use the right three units up one you to the right five units up and again you can reflect across your axis of symmetry to get points on the other side okay so i've already pre-drawn everything let me just copy this real quick and we'll go down to the graph and let me just paste this in so we know what we're working with again this was y equals the quantity x minus two squared and then minus five okay so the vertex again is at two comma negative five so if we go to two comma negative five that's gonna be right here then again you can use your step pattern so you go to the right one up one then to the right one up one two three okay then to the right one up one two three four five then you can reflect across your axis of symmetry which would be that line which is x is equal to 2. okay that's your axis of symmetry so you reflect across you would get this point and this point okay that basically gives you enough to sketch the graph of your parabola it's going to be solid because this is non-strict okay so the same rule holds whether you're graphing a line an ellipse whatever it is if you have a non-strict inequality make it solid okay if it's a strict inequality make it dashed or broken all right so once we have this again if you have a greater than or equal to or greater than you're working with a parabola just shade inside of the parabola so you're going to shade inside the parabola okay but the other way to get this if this is confusing for you and i know a lot of students struggle with that what you can do is you can just take a test point so in this case 0 0 is inside the parabola so you can use that if i plug in a 0 for x and a 0 for y we get 0 is greater than or equal to negative 1 which is true so i know i want to shade inside of the parabola okay if you pick a point outside of the parabola it's not going to work okay so that tells you in the non-solution region okay so you think about this parabola here as separating the coordinate plane into two regions inside the parabola outside the problem okay so if i pick something like let's say negative one comma zero so negative one comma zero it shouldn't work so this would be zero is greater than or equal to negative one squared is one you have negative four times negative one which is four and then minus one well one plus four is five five minus one is four so this says zero is greater than or equal to four which is false okay you can see that's false because it's outside of your solution region okay so this shouldn't work and it doesn't so that's where we get these two graphs okay we put them together so where's the overlap between the two you're shading inside of the parabola okay and you're shading below this line okay you're shading below this line so where's the overlap between the two again if i continue this shading inside you can see it's this small area let me kind of shade this a little bit better it's this small area in here okay and i've made a better graph so that we can see this more cleanly you see this area here that's highlighted is below the orange line and still inside of your parabola so that satisfies both inequalities of the system and so it's your solution in this lesson we want to talk about sequences all right so in this chapter we're going to do a deep dive into sequences and series we're just going to start out with this lesson which is just going to be an introductory lesson to sequences we're just going to learn about the notation and kind of how to plug in for things so i want to start out with just a simple example we have a sub n is equal to 2n so before i go any further let me just tell you that a sequence is nothing more than just a function that computes an ordered list and the domain of this function what's allowed to be plugged in here for n would be the set of natural numbers right or the set of positive integers so remember those if i said n was the set of natural numbers this would be starting with one and going out to positive infinity right you're just increasing by one each time okay so that would be the set of natural numbers or again positive integers so what we could say here is that a sub n is equal to 2n i could relate this to something like f of x equals 2x where i specifically restrict the domain to say that x is a natural number okay i'm just going to put an n there okay so a natural number so we would know that if we wanted to plug in natural numbers for x to kind of generate some ordered pairs or if you wanted to see what the function's value was then i could say okay the first one or the first natural number is 1 so f of 1 would be what we'll just plug in a 1 for x and you would get 2 times 1 which is 2. then if i wanted f of 2 again you'd plug in so 2 times 2 equals 4. so on and so forth f of 3 would be 6 f of 4 would be 8 again so on and so forth now we do the same thing with a sequence but it looks a little bit different because we're using this a sub n notation okay so the way we would say this is we would say a sub 1 okay so a sub 1 is me plugging in a 1 for n it's the same thing as f of 1 here ok so this is a sub 1 that's how we write this so a sub 1 is equal to 2 times 1 okay again here i plugged in a 1 here i plugged in a 1. i'm just plugging in for n whatever is right there it's just like if i said f of 1 is equal to 2 times 1. it's the same thing it's just different notation okay so a common task here is just to go through and write the first five what we call terms of the sequence so this result here is a term okay so if i said a sub 2 that would be 2 times again just plugging in a 2 for n so this is 4. if i said and let me just kind of erase this if i said a sub 3 what is that that's 2 times 3 which is 6. we're just going to be increasing by 2 each time right so we'll say that a sub 4 is going to be 8 and a sub 5 is going to be 10. okay so another common task let me just kind of copy this real quick will be to plot these guys on the coordinate plane let me just kind of clean this up so let's say this is 2 4 6 8 and 10. okay so you can kind of think about these as ordered pairs okay just like if you had x and f of x or x and y but now we have n okay and we have a sub n okay so this is just like if you had x and y or x and f of x so if a sub 1 equals 2 again 2 is called the term okay but really we think about this as again if i had f of 1 equals 2 2 is your y value okay so 1 is what i plugged in okay 1 is what i plugged in that's your x as we think of it 2 is what i got out okay that's my y as i would think about it so now it's going to be the n and the a sub n so i want a 1 going along my n axis okay and i want a 2 going along my a sub n axis so it's going to be right there okay so this is going to be this 1 comma 2 okay then you have 2 4. so again you're just grabbing this and this so 2 and then 4. and we already know how to plot ordered pairs so we're going to spend a lot of time in this but you have 3 and then 6. so we'll have that there and then we have four and eight so four and eight let's see if i got that right okay and then five and ten so five and then ten now what's interesting here is that you're not going to connect these points okay because if i had something like let's say f of x equals 2x well in this case your domain is the set of real numbers so we could connect those points with a line and all the points on the line would satisfy your function okay but in this particular case it doesn't work that way because we've restricted the domain for n to be the positive integers or the natural numbers so you have one comma two right one is n and two for a sub n and then the next guy that you can plug in is 2 and the value for a sub n or the term there is going to be 4 okay so you're going to get these jumps here you're not going to connect this with a lock all right let's just try another example again when you first get to this you're just going to be plugging in and getting used to the notation so we have a sub n equals 4 minus 10 n if i wanted to find the first five terms of the sequence all i'm going to do is i'm going to start out with one okay because that's the smallest natural number that's always what i'm going to start with so i'm going to say what is a sub 1 and all i do is i plug in a 1 for n so 4 minus 10 times 1 okay if i want a sub 2 i'm going to have 4 minus 10 times 2 okay if i want a sub 3 i'm going to have 4 minus 10 times 3. if i want a sub 4 i'm going to have 4 minus 10 times 4. if i want a sub 5 i want 4 minus 10 times 5 okay and you're going to see a pattern here because 4 is always the same really once i get my first number i'm going to decrease by 10 each time okay and that's because this 10 is multiplying this guy here we'll talk about how to go from the results or the terms okay back to the formula later on in the chapter for right now let's just kind of crank these out 4 minus 10 would be negative 6 and then as we get here 10 times 2 is 20 so 4 minus 20 is negative 16 and again you're just decreasing by 10 each time so this would be negative 26 negative 36 and then negative 46. okay let's do one more real quick so we have a sub n equals 1 3 raised to the power of n and then times n minus 1. okay so this one's a little bit more tricky but again all i'm doing is plugging in there and there okay so i have a sub 1 always start with 1 okay then you have 1 let me kind of write this a different way so 1 3 raised to the power of 1 times you're going to have 1 minus 1. we know that 1 minus 1 is 0 and anything times 0 is 0 so we just say this is 0. okay so that's the first guy then a sub 2 would be what well plugging in a 2 there so you'd have 1 3 squared times 2 minus 1 is 1. so 1 3 squared would be 1 9 1 9 times 1 is 1 9. okay so we've got that then a sub 3 let me just kind of write this over here so a sub 3 is what you would have 1 3 cubed so you know that's 1 over 27 and this is times 3 minus 1 is 2. so this will just be 2 over 27 so 2 over 27 and then we'll have a sub 4 okay so i know 1 3 raised to the 4th power all i'm going to have is 1 over 81. so 1 over 81 then times 4 minus 1 is 3. so now i can cancel here i can cancel here because this and this have a common factor of three so this will be one and this will be 27. so this would be one over 27 okay so one over 27 and then for the last guy a sub five what do we get one third raised to the fifth power well three to the fifth power is 243 so one over 243 and then times if you did five minus one that's four so this is going to just be four over 243 okay let's talk a little bit about a recursive definition this is where each term after the first term or could be the first few terms is going to be defined as an expression involving the previous term or terms okay so here what we see is a sub n plus 1 equals a sub n plus 30 specifically if n is greater than 1. so we're still working with the set of natural numbers but specifically this definition here okay that's given to us is only going to work if n is greater than 1 they've given us the fact that a sub 1 is equal to negative 21 so we have the first term of the sequence it's negative 21. if we wanted a sub 2 and a sub 3 and a sub 4 and a sub 5 what we're going to find is that we really just take the previous term okay and add 30. it's all we need to do but how could you show this so let's start out with this so if i wanted a sub 2 i could break this up and show this as a sub 1 plus 1 like this and say okay this is a sub 1 plus 30. okay so now it's clear that a sub 1 is negative 21 so i'm just plugging that in there and this is going to be 9 right and i'm going to rewrite this as 2. i can break this up into a sub 2 plus 1 okay and so what i would say here is that this is equal to my a sub 2 which is 9 just grab that from there plus 30 okay which is going to be 39 okay and again we're going to just keep doing this let me rewrite this as a sub 3. i could write a sub 4 as a sub 3 plus 1 right if i wanted to so again i would just take a sub 3 which is 39 and add 30 okay so this is 69 right so on and so forth here i would just take 69 and add 30 and i'll get 99. right if i wanted the next one i would start with 99 and add 30. you know so on and so forth in this lesson we want to talk about series and also the summation notation all right so in our last lesson we talked about the fact that a sequence was a function that computes an ordered list of numbers so i have a basic example here we have something like a sub n is equal to 2 times n so remember this is the new notation that i introduced to you you could just as easily write this as f of n is equal to 2 times n like this these two are the same okay it's the same this is the typical notation you're going to see with a sequence now when we work with a sequence remember this n here is telling me or reminding me that my domain or what i can plug in for n is the set of natural numbers so this is the set of numbers that starts with 1 so that's the smallest and increases by increments of 1 out to positive infinity so you can say the natural numbers or the positive integers whatever you want to do there but if i ask you for let's say the first four terms you've got to remember that the smallest natural number is 1 so that's what we start with so we say a sub 1 is equal to all i'm going to do is plug in for n i put a 1 there okay right here so i'm going to put a 1 right there so this is 2 times 1. just relate this to what we already know if i had f of 1 what would i do i would plug in a 1 for n so this would be 2 times 1 like this which is 2. same thing different notation okay let me erase this now i said the first four terms so let's go a sub 2 this is 2 times again plugging in a 2 now so this is 4 then i want a sub 3. this is 2 times plug in a 3 now this is 6 and then lastly i'm going to want a sub 4 so this is 2 times 4 and this would be 8. now this could go on forever i just asked for the first four terms right so i got 2 4 6 and 8. but you could do this as long as you wanted to right there's nothing stopping you there's an infinite number of natural numbers so after 4 you go to 5 then 6 then 7 you know so on and so forth out to positive infinity now with a series this occurs when we find the sum of the terms of a sequence okay now in this particular case i've just given you the first four terms okay so i'm going to get a series that's going to be based on these first four terms so what i'm going to do is i'm going to say s for series and then sub 4 because i just have the first four terms so this is going to be equal to and i'm just going to grab these guys so it's going to be 2 okay that's my first term my second term is four my third one is six and my fourth one where i'm stopping is going to be eight and then i just find the sum so two plus four is six six plus six is twelve twelve plus eight is twenty okay so that's an introduction to series now when we work with series we have this special notation that we use and this is known as the summation notation or the sigma notation this is something you have to get used to because when you get to calculus you'll see it a lot you'll also see it in statistics class a lot so something you definitely have to get used to using it's very very easy to use but you do have to get used to it so the first thing you'll notice is this gigantic e looking letter and basically this guy is the greek capital letter sigma okay that's why we said sigma notation and basically on the bottom here you have this i equals one so i'm just going to tell you the easy way to remember this is this is your starting value and this is your ending value okay the i here this is not the imaginary unit i it's officially known as the index of summation i always think about it as just a counter okay a way to keep track of where you are okay because this is where you're going to start and you're going to increment by one okay so that's how i remember it i just think of it as a counter so this is your index of summation officially the one is your starting value or officially in your book you're going to see it's the lower limit of summation okay the lower limit of summation the n is your ending value that's where you're stopping or you can say the upper limit of summation if we go back up here we started with 1 and we ended with 4. so if i come back down here i would have a 1 here and a 4 here okay and then i would just have my rule here which is 2 times i in this particular case we just have a sub i okay so all i would do here is i would plug in a 1 where there's an i i'm just plugging things in just like i did before so i would have a sub 1 then i would increase by 1 okay i'm just using the natural numbers so i'd increase by 1 i'd have a sub 2 then i'd have a sub 3 then i'd have a sub 4 and i would stop at a sub 4 because my a sub n in this case would be a sub 4 okay because n is 4. but if it was 10 i would go out to a sub 10 right so on and so forth so this is your starting value this is your ending value that's what you need to know so if we look at this problem once more what this is telling me to do again just to follow the rule i would start by plugging in a 1 for i okay so i would say what is 2 times 1 then plus i would increment by 1. so i would say what is 2 times 2 then plus again i'm incrementing by 1. so what is 2 times 3 then plus what is 2 times 4 and that's where i stop because this is my ending value this is where i start this is where i am i'm just plugging in for i again all it is is just fancy notation it's the same thing that we did up here okay this is very easy to read but when you get to this notation at first it does seem a bit confusing so we already know this is 20 so let me just kind of reduce this and just say this is 20. all right so let's jump in and look at an example and what we're going to do after this is look at some rules and i'll show you how to get through this more quickly so here we have the summation from i equals 1 to 4 okay of we have this 2 times i minus 9 okay so all i'm going to do here is i'm just going to plug in a 1 to start so i would have 2 times 1 minus 9 okay so what is that 2 times 1 is 2 2 minus 9 is negative 7. so that's my first value and then i increment by 1. i started with 1 now i'm going to 2. so i'm going to have let me just do this down here so 2 times now would be 2 minus 9 and what is this 2 times 2 is 4 4 minus 9 is negative 5. so i'll put a negative 5 here let's just put minus 5. a little bit easier and then let me erase this and now we're going to go to 3. 2 times 3 is 6. 6 minus 9 is going to be negative 3. so let's put minus three and then let's go to four two times four is eight eight minus nine is negative one so let's put minus one there okay so now you just go through and add things okay you can think about this as negative seven plus negative 5 plus negative 3 plus negative 1. so if i did negative 7 minus 5 that's negative 12 then minus 3 is negative 15 the minus 1 is going to be negative 60. so that's your answer okay that's all there is to it again you're just plugging things in okay i start with 1 so that's why i start here and then i increase in increments of 1. just think about the natural numbers they increase in increments of 1 until i get to this value here so i had 2 times 1 minus 9. then i had 2 times 2 minus 9. then i had 2 times 3 minus 9 then i had 2 times 4 minus 9 and you just sum those results and that's how you get your answer let's look at some rules that we're going to need to kind of make this a lot quicker some of these until you practice with them you're not going to make a whole lot of sense of them you might say why do you need them but then you're going to get hit with problems where they're absolutely necessary to get through them in a reasonable amount of time so this first one looks a bit odd so let me just kind of explain this off to the side so let's say and i'm not very good at drawing but let me make this letter here let's say this was five and let's say this is one and let's just say this was the number six what does this mean because i don't have anywhere to plug in things okay so people get confused by that what you see the rule is here is that you would just take this here and multiply by this here so this tells me i would just do 6 times 5 which is 30. but where does that come from well basically even though you don't have an i involved here you've got to go through and do this five times so what you would say is that the first one would be six and then you would increment and say okay now i'm on the second iteration of this so i would do another six then another six then another six then another six so here i is one here i is two here i is three here i is four and here i and i'm messing that up so let me make that better so i is five okay so it's going to give me five iterations of having a value of six i'm summing that together so the quick way to do that is just say okay i just have 5 times 6 which is 30. okay so this is going to come up again i know this is a bit confusing because there's nothing to really plug in there but just know that it's this times this okay that's all you're doing let me erase this for the next one i have c multiplied by a sub i i'm just pulling c outside so i can calculate this sum first and then multiply c by that result okay that's all it's telling me these next two are very easy just think about the fact that you can split this up so instead of having this written like this you can split this up and have the a sub i here and then plus you'll have the b sub i over here same thing with subtraction i have my a sub i here and my b sub i here so this will come in handy and i'll show you this in a moment when we do an example these formulas are really going to come in handy so if you're looking at the summation from i equals 1 to n of just i okay then it's equal to you have n times the quantity n plus 1 over 2 okay so we'll see an example of this in a moment but you have this for i squared and also i cubed so i encourage you especially if you're going to get problems like this if this isn't in your textbook to pause the video and copy down these formulas or to search for these formulas on the internet because they're very very helpful okay a lot of times there's no way you can get through what they're asking you to do in a reasonable amount of time without having these formulas these are the shortcuts to make it happen so let's look at our sample problem and this one you could calculate it it's from i equals 1 okay 210 and we have the summation of 2i plus 5 okay so starting with 1 going to 10. so we know we could go okay 2 times 1 plus 5. then 2 times 2 plus 5. and we can do this all the way until i gets to 10. but i don't want to do that i'm going to show you a quicker way and it's going to take a little bit longer because i'm explaining but in general it's going to be quicker to do this so the first thing i'm going to do is just split this up and again i'm not good at drawing these things but let me try so i'm just going to go like this and that's probably the best i can do so let's say we have a 10 up here and i equals 1 here and i'm just going to say that we have the 2i here and then plus i'm going to make this symbol again or attempt to let me just kind of do that so the 10 here and the i equals 1 here and then a 5. well i already know because i just looked at a problem like this that this is going to be 50 right i have a 5 here and a 10 here if i go back up we know we just saw this right so this is a constant so it's just c times n so we come back here it's 5 times 10 or 50. so at this point you can just erase this and replace it with 50 okay you don't have to worry about it anymore so now i just need to figure out what this is well i can pull this 2 outside okay so i can move this outside and i can calculate this and remember i just showed you the formula for that so we have this n times the quantity n plus 1 over 2 okay so let's go back down so let me do this off this side so you have n which is 10 right so 10 times the quantity 10 plus 1 over 2. so 10 plus 1 is 11 so you would have 11 you would have 11 times 10 11 times 10 is 110 and then divided by 2 would be 55. so you can go ahead and immediately erase this and put 55 okay 55 and you're just multiplying by 2 so 2 times 55 is 110 so this is 110 plus 50 which is going to be 160. so a very quick and easy way to do this it took us longer because i stopped and went back through the formulas but again once you go through and kind of you don't have to memorize them exactly you just have to remember which formulas you can use and then be able to reference them and then you can use them very quickly in your homework or hopefully on a test your teacher will allow you to reference them and not have to memorize them but even if you did have to memorize them they would definitely be very very effective for you because you can find the summation of something like this very very quickly in this lesson we want to talk about arithmetic sequences and series all right so we've already had a basic introduction to the concept of a sequence and now we just want to go a little bit further talk about something more specific known as an arithmetic sequence so i want to start with something very simple and then just get into the definition so suppose we had this list of numbers here we had negative 18 negative 8 2 12 and then we have these three dots so this again tells me that the pattern is going to continue forever so what is the pattern here if i go from negative 18 to negative 8 notice that i'm adding 10. now if i go from negative 8 to 2 notice that i add 10 again then if i go from 2 to 12 notice that i add 10 once again so in each case i have this same number okay that gets added now if i went from 12 to kind of the next term of that sequence i would what i would add 10 i'd get to 22 right if i added 10 again to get to the next term i get to 32 so on and so forth so what we have here is an arithmetic sequence and the reason we have this let me just read the definition to you basically a sequence where each term after your first term is going to be obtained by adding some fixed number to the previous term is known as your arithmetic sequence or some people call it an arithmetic progression now this fixed number in this case it's 10 it's what we keep adding this is known as your common difference okay and we express this with a lowercase d so we'll put d is equal to in this case it would be 10. in most cases when you're working with these you can eyeball your list and figure out what the common difference is just in case you want to be real specific and use a formula i have one that's probably in your textbook so d the common difference is equal to a sub n plus 1 minus a sub n so you've got to take two numbers in that list that are next to each other and the order is going to matter here okay so what you'd want to do if this is a sub n plus 1 minus a sub n so let's just go back up here and just use an example so d equals a sub n plus 1 minus a sub n so let's say i took these two numbers the numbers have to be next to each other and what i'd want is for my a sub n plus 1 remember the way we order this this is a sub 1 this is a sub 2 this is a sub 3 this is a sub 4. so 2 is a sub 3 and then a sub 2 is negative 8. so i want the a sub 3 minus the a sub 2. okay just think about it in terms of this notation again if you're just starting to work with sequences this is a bit confusing but essentially all i'm saying is that if this is 2 okay if this is 2 this is 2 plus 1 which is 3. okay that's all it is so you always if you're pulling two numbers out just take the one on the right in this case that would be two okay and subtract away the one on the left in this case it would be negative eight so you have to do so i would do two minus a negative eight two minus a negative 8 and be careful there because you're subtracting away negative and this becomes 2 plus 8 which is 10 okay again this formula is not super useful because you can eyeball that and clearly see the common difference is 10 but just in case you want to know what it is okay let's jump in and look at an example real quick so here we're given a sub 1 or the first term of the sequence and it equals 1 and we're given d the common difference and this is 4. so we want to find the first five terms of the sequence this is very easy so we already know a sub 1 is 1 and then a sub 2 a sub 3 a sub 4 and a sub 5. so again to get to each one of these i take my first one and i add 4 okay so i take 1 add 4 and get to 5. then as i keep moving i keep doing the same thing okay so as i'm going through each one i'm just taking the previous one and i'm adding four so five plus four is nine nine plus four is thirteen thirteen plus four is seventeen if you wanted a sub six you would add four again and get to twenty you know so on and so forth all right so that's how you would go through and kind of figure out the first few terms of a sequence but in some cases you're going to be asked for the nth term of a sequence okay and you don't have to write out something you know go through all of them until you get to it what if somebody asked you for the 501th term okay for example well here we have a general formula that we can use so we have that a sub n is equal to a sub 1 plus d your common difference times the quantity n minus 1. so let's use this real quick with a little example let me just write this again so we're going to put a sub n is equal to a sub 1 plus you're going to have your d your common difference times the quantity n minus 1. okay so here we're asked to find a sub 37 okay so the 37th term of this sequence and then we want to find a sub n so this is the formula for the general term and i'll show you how to do that so i'm going to start with what i want to find so i'm going to put a sub 37 okay so that's taking the place of a sub n here this equals what is a sub 1 it's 33 so i'm just plugging in then plus d i'm given that it's 2 okay then times the quantity n minus 1. what is n it's 37 okay so it's what i'm trying to find so it's 37 so 37 minus 1 is 36. this is all you need to do this is very very simple don't over complicate things 2 times 36 is 72 and then 72 plus 33 we do 72 plus 33 which is going to be 105. okay so that's my answer for a sub 37. let me just erase this and put this right here like this okay so now let's talk about finding just a sub n so this would be the formula for the general term in the sequence and essentially all i'm going to do is just plug in so i have a sub n is equal to i know what a sub 1 is is 33 then plus i know what d is it's 2 and then times the quantity n minus 1. so you're just going to simplify this and that's it okay so a sub n let me just put a vertical bar there so a sub n is equal to 33 plus 2 times n is 2n and then minus 2 times 1 is 2 and then i can do 33 minus 2 that's 31. so i basically have a sub n is equal to 31 plus 2n okay so we also talked about the fact that a series was when you added all the terms up in a sequence okay so this is going to give you one single number now generally speaking a lot of times they're going to ask you for the sum of the first n terms of a sequence okay so you're going to see this s sub n so if this was s sub 9 i want to find the first nine terms of that sequence and then find the sum okay but we don't have to go through all of that because we have these little quick formulas we can use so you have this one s sub n is equal to n times the quantity a sub 1 plus a sub n over 2. so in this case you have to calculate a sub n okay so if this is s sub 9 you need a sub 9 okay to make this formula work with this one you really don't need a whole lot you just need your common difference and a sub 1 okay so you don't have to actually calculate this so let's look at an example so we're given the following information we have a sub 1 this equals 18 we have d which is 5 and n which equals 35. so basically what we're asking for here is the first 35 terms of the sequence we want the sum of that so this would be our arithmetic series here so basically we're trying to find s sub 35. so this equals what if i go back up and get my formula let's use this one first and then i'll show you this one okay so this is n times the quantity a sub 1 plus a sub n over 2. so we have and let me just kind of slide this down so n times the quantity you have your a sub 1 plus your a sub n over 2. okay so what's n it's 35 so you can go ahead and plug that in what is a sub 1 we know this is 18 okay but what is a sub n we have to calculate that and the way you would do that is you would use that formula that we just worked with so we know if i go back this formula here let me just copy it we'll use it real quick so let me copy this and come back down so in this case because a little bit more work you'd probably want to use the other formula but basically it's not too bad so you would have a sub 35 would be equal to a sub 1 which in this case is 18 and then plus you'd have d which is 5 times the quantity n minus 1 in this case n is 35 so it would be 35 minus 1 which is 34. okay so let's multiply this so 5 times 34 is 170 then plus 18 is 188. so this guy right here this a sub n in this case it would be a sub 35 this would be 188 okay so we're done with that so basically you'd have 18 plus 188 which is going to be 206 so you would have 35 times 206 divided by 2. might as well just divide this by 2 now and so you have 35 times 103 and that would give you 3605 okay so s sub 35 is 3 605. so essentially this is the result from adding the first 35 terms of the sequence that starts with an a sub 1 that's 18 okay so that's your first term and then your common difference is 5. so the next term would be 23 and that would be 28 you know so on and so forth now the other way let me just show you the other formula so we have s sub 35 and this equals you have your n which is 35 times you would have your 2 times a sub 1. a sub 1 is 18 2 times 18 is 36 and then you're going to have plus your common difference times the quantity n minus 1. well n here again is 35 35 minus 1 is 34. so this is 34 times in this case d is going to be 5. and then this is all over 2. okay so 5 times 34 is 170 170 plus 36 is again 206 we already know that 206 divided by 2 is 103 so we multiply this by 103 and again we get 3605 so either way you want to do it both of these formulas are valid this one is a little bit quicker in that scenario because you don't have to go through and calculate what a sub n is so again the answer here for s sub 35 would be 3605 in this lesson we want to talk about geometric sequences and series all right so in our last lesson we talked about the arithmetic sequence and with that guy we found that each term after the first one can be found by taking the preceding term and adding some fixed number to it okay the fixed number was known as the common difference right we represented that with a lowercase d well now we're going to be multiplying okay and so let me give you the official definition a geometric sequence which is also called a geometric progression is just a sequence where each term after the first is obtained by multiplying now the preceding term by a fixed nonzero real number known as the common ratio and we'll use an r to represent the common ratio so you can eyeball this and see what's going on as i go from negative 4 to negative 24 i multiplied by 6. and this is going to be consistent as i go from negative 24 to negative 144 multiply by 6. as we go from negative 144 to negative 864 we multiply by 6. so if you didn't see this by kind of eyeballing it you could take two terms that are next to each other take the term on the right okay and divide by the term on the left so for example if i took this term and this term they're next to each other take the one on the right divide by the one on the left so if i did negative 864 divided by negative 144 i would get positive six okay and that's what this little formula is going to tell you here so r equals you've got your a sub n plus one over your a sub n so if i kind of come back down here if i write r equals in this case this is a sub 4 over in this case this is a sub 3 okay so again you take two that are next to each other the one on the right okay that has a higher value for n is going to go in the numerator the one on the left the one that has a lower value for n is going to go in the denominator okay and the same thing would go if i picked two other ones that were next to each other again i could just take let's say this one and this one if i did in this case this is a sub 3 right a sub 3 over this one's a sub 2. well now i'd get negative 144 over negative 24. let me put r equals this and this would be 6 as well all right now the next thing we're going to talk about you might also be asked to find the nth term of a geometric sequence there's a really easy formula you can use for this so a sub n is equal to a sub 1 times r again the common ratio raised to the power of n minus 1. so again this is going to match this so let's look at an example real quick so we're given this kind of list of terms here we want to find a sub 9 and a sub n so this is the formula for the general term and this is a sub 9. okay so how do we do this well let me just go back up here and grab this formula real quick so a sub n equals a sub 1 times r raised to the power of n minus 1. so let me write that off to the side so we have a sub n okay is equal to a sub 1 and then we're going to multiply this by r raised to the power of n minus 1. okay so let's start with a sub 9. so we have a sub 9 is equal to let me make that a little better and we know that a sub 1 is the first term from the sequence right this is a sub 1 this is a sub 2. you know so on and so forth this is let me go ahead and write this in so this is a sub 3 and this is a sub 4. well we know that a sub 1 is 1 then times what's r well again i can just grab two terms that are next to each other so let me grab a sub 4 and a sub 3. so again if i grab this and this i take the one on the right the one with the higher index value and i divide by the one on the left the one with the lower index value so r is equal to a sub 4 over a sub 3 which is negative 8 over 4 which is negative 2 okay so i'm going to put negative 2 here for r now this is very important when you have a negative you need to wrap it inside of parentheses don't put negative 2 like this because negative 2 the negative and the 2 are raised to the power of n minus 1. so wrap it in parentheses okay make sure you do that now what is n minus 1 well in this case again n this right here matches this right here so 9 is what goes in there so 9 minus 1 is 8. okay so all we need to do now is just crank this out and we're done right so negative 2 to the 8th power is going to be 256 right because again the negative and the two are inside of parentheses so it's all raised to the eighth power so this would be one times 256 which is just 256. so let me write this over here so a sub 9 is equal to 256. okay so what about a sub n and some people get confused by this because you're given this formula here and you're like well a sub n equals this well no you need to plug in things from the problem that you're given right so you need to plug in your a sub 1 and your r okay so all i'm going to do is say that a sub n is equal to my a sub 1 again is 1 and then times your r in this case is negative 2 and if you don't want to put the 1 there it doesn't matter but i'm just putting it for completeness so i'm going to put negative 2 here for my r and then my n minus 1 here okay so whatever term you'd want to find let's say you wanted the 10th term plug in a 10 there plug in a 10 there evaluate and you're good to go okay so you can do that with any term of this sequence okay let's look at something a little bit more challenging i'm going to tell you there's lots of ways to solve these i'm going to show you the way that i do it this is the quickest way that i've found but if you have a better method that's fine too so we have a sub 3 equals 48 and a sub 6 equals negative 3072 we want to find a sub 9 and a sub n so let's start with a sub 9. now we don't have two terms that are next to each other so we can't do the formula for r that we're used to but we can kind of look at our general formula so we know that a sub n is equal to we have a sub 1 times our r raised to the power of n minus 1 okay so what i want to do is just plug some things in real quick and i'm going to show you that you can basically set up a system with two equations and two unknowns and you can solve for what you need okay and again there's other ways to do this but this is how i'm going to do it so i'm going to plug in for a sub 3 just to show you all the steps and put a sub 3 there this would be a 3. 3 minus 1 would be 2 okay and i know what a sub 3 is it has a value of 48 okay so let's put that right there so now we have 48 equals a sub 1 times r squared okay for this one i'm going to do a sub 6 so i'm going to do a sub 6 and i'm just going to skip that step and plug in a negative 3072 and again this would be a 6 6 minus 1 is 5. okay two equations two unknowns right i need to find out a sub 1 i need to find out r so let me solve one of these guys for a sub 1 okay and i'm just going to solve this top one so i'm going to say that a sub 1 is equal to what i would just divide both sides by r squared so 48 over r squared okay so now i can take this which is a sub 1 and plug it in there okay that's all i've got to do so i've got negative 3072 is equal to 48 over r squared times r to the fifth power okay let me just erase all this get this into a cleaner format i can go ahead and get rid of this and say this is to the third power okay so what we'd have is what 48r cubed is equal to this negative 3072. we'll divide both sides by 48 and we're going to get that r cubed is equal to negative 64. if i take the cube root of each side let me make sure i put a 3 there what am i going to get well this on the left would be r and on the right i would have negative 4. so i know r is negative 4. so let me just write here off to the side r equals negative 4. well i still need my a sub 1. so let me erase this real quick we had that solved for a sub 1. so let me just do that again we know that a sub 1 is equal to 48 divided by r squared well we know that r is negative 4 so negative 4 squared again if i'm plugging something in here and i'm squaring i want the negative and the 4 wrapped in parentheses because it's negative 4 that's being squared okay so don't make that mistake so this would be negative 4 squared so this is 48 over negative 4 squared is 16 and 48 over 16 is 3. okay so now i have my a sub 1 and i have my r so i'm ready to go through and find my a sub 9. okay so my a sub 9 again if you use the formula let me just write this out so a sub n is equal to you've got your a sub 1 times r raised to the power of n minus 1. so we know a sub 1 is going to be 3 and we know that this is a 9 here okay and we know this is a 9 here let me make that a little bit better we know this is a 9 here 9 minus 1 is going to be 8. so what is r it's negative 4 let me erase this and put negative 4 raised to the 8th power again if you have a negative there make sure you're wrapped in parentheses all right so this answer is going to be really big if i punch it up on my calculator i get 65 536 so we'll say a sub 9 equals three times again that was 65 536 and then we'll just do this multiplication and that'll give me 196 thousand six hundred eight okay so that's my a sub nine so a sub nine is equal to one hundred ninety six thousand six hundred eight okay so a sub n is really easy to do okay so a sub n let me just write this down here so a sub n equals y again it's a sub 1 which in this case is 3 times your r which in this case is negative 4 raised to the power of n minus 1. all right so now let's move on and talk about some of the problems you'll get with geometric series so this first one is where we're just summing the first n terms of a geometric sequence again this would be a geometric series so you have your s sub n remember the n here is the value that you're kind of stopping at so if it was s sub 10 well then i want to sum the first 10 terms of that geometric sequence so basically to do this you have your a sub 1 here multiplied by 1 minus r raised to this power of n again you've got the same number there and there and then it's over one minus all okay so a real simple formula to use and let's go down here and we've written this in summation notation hopefully you remember how to use this how to read this this is from n equals one to eight it's the summation of three times you have your negative two raised to the power of n minus one so essentially the formula here for the geometric sequence it's a sub n equals three okay that's going to be your a sub 1 and then times your negative 2 that's your r raised to the power of n minus 1. so again this is your a sub 1 and this is your r okay so if i want to use that formula let me go back up to it for a minute you can copy this down real fast and i'll go back down if you want to use the formula what i'll say here is that s sub 8 right because i'm going to that number there is going to be equal to what while i'm plugging in let me just kind of write this formula so it's a sub 1 times it's 1 minus r raised to the power of n okay so let me stop and fill this out so r we know is negative 2 it's negative 2 and we know that n in this case is 8 okay so i'm going to put 8 here let me just close that down and then this is over let me fill in a sub 1 that's going to be 3. so let me put that there it's over your 1 minus r so your 1 minus in this case r again is negative 2. okay so let me copy this real fast and let me just paste this in so we have some room to work okay so this is pretty easy we're just evaluating things so s sub 8 is equal to we have 3 times if i did negative 2 to the 8th power we know that's 256 and then we have 1 minus that okay and this is over 1 minus a negative 2 which is 1 plus 2 or 3. well before i do anything remember this is 3 times this amount here this can be canceled right because this is multiplication so i end up with just 1 minus 256 which is negative 255 okay so the sum of the first eight terms of that geometric sequence would be negative 255. all right so the last thing i want to talk about would be the sum of the terms of an infinite geometric sequence and basically this formula i'm going to give you here is valid as long as the r that you're working with is greater than negative 1 and less than 1. okay if the absolute value of r is greater than 1 then what's going to happen here is you just write that you don't have a sum okay so what we're going to do is just go through this real quick so this is s sub infinity so as you keep summing these terms of your sequence as you keep going forever and ever and ever as you approach infinity you get a value of a sub 1 over 1 minus r okay so pretty easy to work with this formula let's look at an example so we have the summation from k equals one to infinity of negative four thirds times this one-fifth raised to the power of k minus 1. so again this is your a sub 1. this is your r okay so when you look at your r value is it greater than negative 1 and less than 1. well yeah this is one-fifth so this falls in the category where we're going to get an answer from this formula so what we're going to do is just write this out let me just kind of see if i can fit this on the screen my formula again s sub infinity is equal to you've got a sub 1 which in this case is negative four four-thirds okay over you've got one minus r so one minus one-fifth so let me copy this again we're just gonna run out our room here just too much stuff on the screen okay so the fastest way to solve this is just to find the lcd of all the denominators so you've got a 3 and a 5. so your lcd is going to be 15. so i'm going to multiply the top part by 15 and the bottom part by 15. so i'm just multiplying 15 by negative 4 thirds of course the 15 would cancel with a 3 and give you a 5 5 times negative 4 is negative 20. let me write that down here so we have some room then 15 times 1 would be 15 and then minus if i did 15 times one fifth the 15 would cancel with the 5 and give me a 3 3 times 1 is 3. so 15 minus 3 is 12 okay and we know that there's a common factor here of 4. so if i divide this by 4 i get 5. if i divide this by 4 i get 3. so the answer here would be negative five thirds in this lesson we wanna talk about the binomial theorem all right so before we jump in and talk about the binomial theorem we need to understand some different notation that we're gonna come across and we're gonna start with this factorial notation so if you see something like five factorial okay what this is telling you to do is to start with this number five and then you're going to multiply okay by each number going down by one until you hit one okay so what this means is i start with five and again i multiply by each number going down by one so i go down by one to four and then down by one to three and then down by one to two then down by one to one so each number until i get to one just keep going down by one that's all you're doing so you can crank this out really quickly or you can use your calculator you will have this button available to you as long as you're using like a scientific calculator something like a ti 83 or 84 or whatever it is now but basically you do 5 times 4 is 20 times 3 is 60 times 2 is 120 and times 1 is 120 okay so that's pretty easy then when we get to 7 factorial okay we're going to do the same thing we're going to start with 7 and i just keep going down by 1. so i'm going to go down by 1 to 6 then five then four three two one you know so once over so you can crank this out manually or again you can hit the key on your calculator and basically it's going to give you 5040 as an answer so now that we understand the factorial notation it's time to look at a formula we're going to use today for the binomial coefficients okay so we're going to see this more in depth in the next lesson when we talk about combinations but for right now we're just going to take it as a given okay so we have this little n choose r okay so that's how this is written you might see a different type of notation you might see this lowercase n in uppercase c and a lowercase r like this so this and this it's the same thing just different notation okay i prefer to use this one so that's what we're going to go with so this is again and choose r and it's equal to n factorial okay over you have your r factorial times the quantity n minus r factorial okay let's see an example real quick so i have 10 choose 6. so again you could also write this as 10 and then you have the c and then the 6 like this okay so a 10 a big c and a 6. so this is 10 choose 6. that's how that's right in this notation it's read the same way again the same thing different notation so 10 choose 6 is equal to again i take this number here whatever this is it's going to go here and it's going to go here okay so i've set that up already then this number here goes here and also here okay so it's pretty easy to remember once you do it a few times just something you need to work through a few examples with so let's go ahead and crank this out real quick i just want to say in your calculator you have this button available to you as well you would hit 10 and then you would hit the n c r key like this it's a lowercase n an uppercase c a lowercase r and then you would follow that with the six and hit enter and it will give you an answer okay and i'll give you that answer in a minute let's just work it out so what i'm going to do here is go ahead and say 10 factorial is 10 times 9 times 8 times 7 times 6 times 5 times 4 times 3 times 2 times 1. and you see why a calculator is handy here right and then you have 6 factorial so let me write that under here because it's going to cancel 6 times 5 that's 4 times 3 times 2 times 1. 10 minus 6 is 4 so 4 factorial so let's write that over here 4 times 3 times 2 times 1. okay so we know that let me just pick a different color here this will cancel with this so that's gone and you can erase it but i'm just going to leave it there for completeness and then you can start canceling things here so 10 divided by 2 is 5. 9 divided by 3 is 3. 8 divided by 4 is 2 and then really nothing else we can cancel so let's just write here that we have 5 times 3 times 2 times 7 okay everything else is gone and if we punch this up on a calculator 5 times 3 is 15 15 times 2 is 30 and then 30 times 7 is 210 okay and again you can get this more quickly by just using the and choose r key on your calculator okay now let's move on and talk about the binomial theorem so what is this used for it's used to expand something like the quantity x plus y to the fourth power so we already know to kind of work with this we would go x plus y okay times x plus y times x plus y and then times x plus y okay so if we went through and used you know foil and then after we had you know more terms we would just keep going using the distributive property but this would take a really really long time so the binomial theorem is going to allow us to find our answer here really really quickly okay so there's a lot of different techniques and strategies that go with this i'm just going to give you what i've learned over the years i'm not going to go through the book formulas because i feel like they really confuse you so basically the first thing is in your book it's going to call this exponent here n okay in most cases so let's just say this is n so we'll start by saying there's going to be n plus 1 terms okay so if this exponent is a 4 you're going to have 5 terms if it was a seven you'd have eight terms if it was a 12 you'd have 13 terms you know so on and so forth so we know there would be five terms so let me just put five spaces here so a space plus a space plus a space plus a space plus and then a space so one two three four five this is how i always do it now the next thing is you are going to take whatever is in this first position and you're gonna raise it to this power and that's gonna go right here okay so this would be x to the fourth power then you're going to take whatever's in this position okay and you're going to raise it to this power and you're going to put it in the last position okay so this is easy to remember because you're basically saying okay this to this is first right it's in the first position it goes in the first position this to this is in the last position it's in the last position it goes in the last position so that's the first thing you want to do now the next thing is really simple really easy to remember the exponents on your variables as you go to the right okay or as you go to the left they're changing by a value of one okay so what do i mean by that so my exponent on x is going to decrease by one as i move to the right so this is going to be x cubed then it's going to be x squared then it's going to be x to the first power and then here it would be x to the power of 0 which is just 1 so we don't write that now the same thing goes for our y so now we're going this way so we would say this is y cubed we would say y squared we would say y to the first power or just y and here again it would be y to the power of zero or just one okay so you don't have to write that now though either way you look at it as you go to the right okay everything is changing by one so this went down by one down by one down by one okay and this went up by one up by one up by one okay so i started with y to the power of zero now i've got y to the power of one y squared y cubed y to the power fourth so you see the pattern there okay really easy to remember okay so once you have the kind of variable parts worked out and in some cases you're going to have coefficients here so it'll be a little bit more complicated but once you have this part worked out the next part is kind of the tricky one okay so you need to find the coefficients now you can work with the combination formula that we just talked about and i'll show you that in a minute or you can use pascal's triangle so let's do a little side note here we're going to come up here and look at pascal's triangle so you'll see this in your book the main thing to remember if you don't remember anything else about this is that this is row zero okay everyone makes a mistake on that okay you need to know the rows this is zero this is one this is two this is three this is 4 this is 5 okay so it starts with row 0. so what you want to do is find the kind of row that corresponds to that n value or that exponent again that's right here so if this is a 4 then when i look at this i want the fourth row but again i'm starting with zero so really it's the fifth row if you're thinking about it as starting from one but all the books start with zero so it's zero one two three four okay so i want this row right here let me highlight it for you so i'm going to take a 1 a 4 a 6 a 4 and a 1. so let me just copy that real quick i'm going to come back and i'll explain this in a minute so let's just paste this real quick i'm just going to show you how quickly we can be done the one and the one everything just goes in order right so the ones on the ends we don't need to write that because one times anything is just itself this would be a four and this would be a four okay so notice how the second term and the second to last term those are the same and the middle term is just a six okay just that quickly we're done we've expanded this guy we know that it's x to the fourth power plus four x cubed y plus six x squared y squared plus four x y cubed plus y to the fourth power now let me go back and do a little bit of explaining on this pascal's triangle because you might be saying what in the world is this thing basically the way this thing starts is you notice you have a one at the top then you have two ones here okay and basically as you keep going down as you go down the sides notice everything's a one everything's a one okay so on the outsides everything's a one okay but as you get into the middle stuff let me just kind of erase this real quick as you get into the middle stuff think about this number two where does that come from it comes from the sum of this plus this okay so go up to the top left and to the top right if i look at let me just kind of erase this if i look at 3 it's the sum of go to the top left and the top right so those two numbers right so 1 plus 2 is 3. this guy same thing 2 plus 1 is 3. this 4 here 1 plus 3 is 4. this 6 here 3 plus 3 is 6. this 4 here 3 plus 1 is 4. you know so on and so forth so you can draw this thing out yourself okay by remembering that pattern but it's just going to be your preference right i'm going to show you a different way to do this if you have a lot of problems to work through and you're just doing homework or maybe you're on a test and they let you use this thing well i would just use it right because it's a lot faster than what i'm about to show you now let me erase these coefficients and pretend that we don't have them okay so if we don't have them how else can we get them well basically what you would do is you would use that combination formula that i just talked about and you're always going to take this number right here this 4 okay the end value as we say and we're going to put that on the top and let me just kind of slide this down because i'm not going to have enough problem you're going to take the coefficients i'm just going to write it in here so it will be 4 and you basically always take the exponent that's on y or this second value so in this case the exponent on y is zero so it would be zero and then your coefficient here would be four choose one this one would be four choose two this one would be four choose three and this one would be four choose four okay now these on the outsides are always going to be 1 so you can just get rid of them okay you don't need to crank that out again you can use a calculator for this if you do 4 choose 1 you're going to get 4 and if you do 4 choose 3 you're going to get 4. so i get a 4 here and here and we can erase this if you do four choose two you're going to get six okay so you do this there now even if you're using a calculator it is going to be a little bit slower versus having pascal's triangle because again with pascal's triangle i can just go up here i can graph things again this is row zero row one row two row three row four i can just grab the coefficients and i'm done okay i don't have to do anything so it's just a matter of a personal preference in terms of what you want to do okay let's look at a harder example or i would say not harder more tedious so here we have x plus 3y raised to the fourth power let's go ahead and set this up i know i would have five terms right i know i would have five terms so one two three four five okay i would take this guy raise it to the fourth power and then as i go to the right i'm decreasing it by one so x cubed x squared x to the first power and again this is x to the power of 0 but you don't have to write that right that's just 1. then i would take again you've got to be very careful here because whatever's in this position is raised to the 4th power so now you have a number involved okay not just a variable so you've got to be really really careful you would start here with 3y to the fourth power okay then here it would be 3y cubed then here it would be 3y squared then here it would be 3y to the first power and here it would be 3y to the power of 0. again that's 1 so we don't write it so if we simplify this we're just going to write this beneath here so far we have x to the 4th plus this would be 3 times x cubed y and then plus we would have 3 squared is 9 so this would be 9 x squared y squared and then plus 3 cubed is 27 so that would be 27 x y cubed and let me kind of slide this down because we're going to run out of room and then plus we would have 3 to the fourth power which is 81 and then y to the fourth power okay let me just erase this we're just going to keep adding to this we just need our coefficients now and again we just work with something that was raised to the fourth power so you should remember that the coefficients are one four six four one okay so basically again on the outside you can get rid of it here i need to multiply i need to multiply by whatever's there so 4 times 3 is 12 so this would be 12 and 6 times 9 is 54 so this is 54. 4 times 27 is 108. so let's put 108 there and we're done okay so again if you're using this pascal's triangle you just go to the fourth row again it starts with zero i can't say that enough because a lot of students they make that mistake and say this is the fourth row right the one that's labeled with a three okay you want to go to the fourth row grab those coefficients and then just punch those in again if you don't want to do it that way you can use the combination formula so again you start with this so it's four choose zero then it's four choose one then it's four choose two then it's four choose three okay then it's four choose four okay and basically if you crank all these out on your calculator you're going to get the exact same answer okay so our result here is going to be x to the fourth power plus 12 12x cubed y plus 54x squared y squared plus 108xy cubed plus 81y to the fourth power so for this one i have 3x plus 2y this is raised to the fifth power so again now this guy's going to be 5 okay so i'm going to have 6 terms right so it's always this number which again we call this n so n plus 1 so here n is 5 so 5 plus 1 is 6. so you're going to have 6 terms and i'm not going to space it out i'm just going to kind of do it i'm going to take this guy raised to the fifth power so three to the fifth power is 243 and then times x to the fifth power then you would have plus you would have let's just go down we would have three x raised to the power of four so 3x raised to the power 4 3 to the fourth power is 81 so you would have 81 81 x to the fourth power and then plus i'm going down by one so now i would have 3x cubed so 3 cubed is 27 then x cubed then plus let me just kind of drag this down here okay so i'm going to do 3x squared so that would be 9x squared then plus i would have 3x right it'd be raised to the first power and then lastly you would have your 3x raised to the power of 0 which is just 1. okay so you can leave that off now this guy right here again i take 2y i raised to the fifth power that goes last so 2 to the fifth power is 32 so that would be 32y to the fifth power and then as i go this way i would have what 2y to the fourth power so let me write this up here so 2y to the fourth power i know this gets real messy 2 to the 4th power is 16 and then y to the 4th power so you're going to have to multiply here so 16 times 3 is 48 so you would have 48 x y to the fourth power okay and then as we move this way now i'm working with 2 y cubed so 2 cubed is 8. so this would be 8y cubed so 8 times 9 is 72. so you would have 72 x squared y cubed okay and then as i move up here now i'm going to have 2y being squared so it would be 4y squared let me just write y squared there 4 times 27 is again 108. let's write that in and then as we go here now i'm going to have 2y to the first power so i'll just put a y there 2 times 81 is 162. and then here we don't need to write anything you'd have 2y to the power of 0 which is just 1. okay so again you can go through to get the coefficients you would start with you know 5 choose 0 which is one so you can leave that off and then you would go to five choose one and then five choose two and then five choose three and then five choose four and then finally five choose five okay so this would be five choose zero which is one this would be five choose one which is five okay so i'm just going to put a five there this would be five choose two which would be ten then for five choose three you would get ten again and then as we get five choose four it's always going to match this so this is going to be five then as we get to the end we have five choose five again five choose five and five choose zero those are each going to be equal to one so you can get rid of those okay so let's just get rid of this and let's do our multiplication so this stays unchanged five times 162 is 810 so 810 10 times 108 let me just slide this down would be 1080 so 1080. and then 10 times 72 let me just slide this down real quick 10 times 72 would be 720 so this would be 720. 5 times 48 would be 240 so 240 and that's it we're done so we get 243 x to the fifth power plus 810 x to the fourth power y plus 1080 x cubed y squared plus 720 x squared y cubed plus 240xy to the fourth plus 32 y to the fifth power now again if you wanted to use this row here row 5 that's fine you have 1 5 10 10 5 and 1 again what we just saw all right so let's wrap up the lesson and talk about how to find a specific term some books will call this finding the kth term of a binomial expansion so let me just read this problem and i'll give you a simple little formula that you can use so we have the quantity x plus 2y to the seventh power we want to find the third term so you could use the binomial theorem okay and kind of write all the terms out you'd have eight of them right because this is a seven so you know eight terms and then go through and get the third one right but that would be kind of a waste of time because you have this little simple formula that you can use so what i'm going to give you is i'm going to say that this is n again i always refer to that exponent as n this 3 here is going to be a value of k okay that's what we're going to call it so if you're finding the third term k is three if you're finding the tenth term k is ten if you were finding the twenty first term k is twenty one okay so on and so forth so what you're gonna say is that the coefficient is n choose k minus one okay so in our case n is 7 so n let me write this here so n is 7 and then k is 3 so 3 minus 1 is 2. so you can have 7 choose 2. that's the first part then times you're going to have the x which is whatever's in this first position here raised to the power of n minus the quantity okay k minus 1. so this is inside of parentheses here so be very very careful to just plug in and evaluate it with it inside of parentheses okay so if i plug in and say this is x raised to the power of 7 minus the quantity okay you would have 3 minus 1 okay so this is going to be what it's going to be 7 minus 3 minus 1 is 2 right because you do what's inside the parentheses first so you would have 7 minus 2 and 7 minus 2 is 5. okay so that part's done then we would have y raised to the power of k minus 1. now in this case for our y what's our stand in we have 2y right so you have to take whatever's in that second position okay and do this to it so i'm going to take this and say it's times 2 y raised to the power of k is 3 minus 1. so 3 minus 1 is 2 so this would be squared so if i square 2 i get 4 and if i square y i get y squared so this would be 4 y squared okay and then i have 7 choose 2. let's just go ahead and do that you punch that into your calculator you get 21. so you'd have 21 times 4 which is 84. so you'd have 84 x to the fifth power y squared okay let's look at one more of these let's try to do it without the formula i'm going to reference it but basically we have y plus 2x raised to the sixth power okay we want to find the fourth term so again just think about this in your head this is k and this is n okay so just get used to that and just basically say okay what do i need to do i want to do n so this number right here k which in this case is 4 minus 1. so 4 minus 1 is 3. so 6 choose 3. so that's the coefficient part then what's this part here and here well remember i take this guy okay so i take this guy in this case it's y and i raise it to the power of n which in this case is 6 minus and this is important use parentheses the quantity k minus 1. so 4 minus 1 okay so 4 minus 1 is 3 okay so we're going to have 6 minus 3. so 6 minus 3 which is 3. so we'll put y cubed here then the next part is going to be this guy so it's going to be times you have your 2x raised to the power of again k minus 1. so in this case k is 4 4 minus 1 is 3. so 2 cubed is 8 and x cubed is just x cubed so we put times 8 times x cubed okay so let's do 6 choose 3 real quick and that is going to give us 20 okay so this is 20. and so what we would say is 20 times 8 is 160. so i can write this and say this is a 160 okay let me just erase this as i go times y cubed x cubed so that would be our fourth term when we expand this in this lesson we want to talk about combinations and permutations all right so when we get to this section a lot of students are going to have difficulty determining whether to use our formula for permutations or whether to use our formula for combinations basically we're going to start out with one here where we say that there are 20 applicants for three jobs a javascript developer a data scientist and a ui engineer well basically the way that you decide here you're going to be selecting three people out of the 20. so if you have the number of ways of selecting our items out of n items in this case it's 3 out of 20 and repetition is not allowed because the same person can't get the javascript developer job the data scientist job and the ui engineer job it's got to be three separate people well you want to consider the fact that if the order is important you want to use a permutation if the order is not important okay you want to use a combination so here the order is very important to us because it's a different result if i get a job as a javascript developer versus if i get a job as a data scientist okay so because the order is important we want to use our formula for permutations and that's a real easy formula it's going to be p like this and then i'm going to put some parentheses is going to be n comma r okay the n here is going to be the 20 the r here is going to be the 3. we'll fill that in a minute this equals you've got your n factorial over you've got your n minus r factorial okay so in this case we're going to have p and we would say 20 pick three okay that's how you can read that if you want and this is going to be equal to 20 factorial over and i made that a little bit long so let's fix that so 20 factorial over in this case you'd have 20 minus 3 or 17 factorial okay if you want to simplify this you can of course this key is on your calculator if you're using a ti 83 or 84 or whatever the version is you're using but let's just go through this one in the future ones i'm just going to punch it into the calculator and make it a lot quicker let's move this up here this would be what it would be 20 times 19 times 18 times 17 you know so on and so forth but this part here because the 17 factorial is going to keep going down to one you can get rid of that part and just erase this and say it's 20 times 19 times 18 which is going to be what well 20 times 19 is 380 then times 18 is going to be 6840. all right let's look at the next one so we have the senior class at denton high school which contains 60 students wishes to elect a president vice president and secretary so we're choosing three students out of the 60 for these three positions the order is going to matter here right if someone gets selected as a president it's a different result as if they got selected as a secretary so in other words let's say that john got picked to be the president and let's say larry got picked to be the vice president for example and let's say that steve okay steve got picked to be the secretary well if i switch things up i could have the same three students okay but they got different positions that's a completely different result okay so in this case we want to use a permutation and we remember the formula for that so i'm just going to type this into my calculator i'm going to do 60 okay 60 and then you're going to have that key that says n pick r so it looks like this a lowercase n an uppercase p and a lowercase r but i'm going to hit 60 this key right here and then three right because there's going to be three students selected president vice president and secretary and if you run that calculation you're going to get 205 320. so 200 5320 and again if you want to use the proper notation there you've got this p and then you've got the first number that's 60 and then comma 30. okay so this is what this is equal to again if you go on your calculator it's the npr key okay let's look at another example so we have the junior class at bracken high school consists of 250 students they need to elect two representatives for the student council okay well this is a little bit similar to the last question but we're going to end up using a different formula here because now the order doesn't matter anymore okay repetition's still not allowed but the order doesn't matter anymore because let's say john gets the first representative position and steve gets the second one well it doesn't matter if we reverse that and steve got chosen first and then john did because they're both going to just be representatives okay so the order doesn't matter here so we want to use the formula for combinations and we learned this in the last lesson when we talked about the binomial theorem right so we have this formula it's n choose r so we have this c and then n comma r here and this equals your n factorial over your n minus r factorial and then times your r factorial okay so what ends up happening is we are dividing out the duplicates with this extra part here okay this is all you have to remember because if you've got this part down from the permutations formula you're just throwing that extra little bit there in the denominator to divide out the duplicates and that's how you get your combination formula okay so let's erase this real quick let's actually plug in the numbers i'm going to again use my calculator here because you would basically be doing this c here with 250 comma two and i'm not going to set this up you're just going to hit 250 and then go to ncr i'm going to get a result of 31 125 okay let's look at another example so a local high school is having a marathon qualifier there will be a total of 40 runners that will participate the first eight people to finish the race will advance to the state championship okay so do i want a permutation or combination does the order matter here we know that repetition is not allowed right because if i'm in the first position i can't also be in the third position but does the order matter well no not for our result because it just says the first eight people so it doesn't matter if i finished first or seventh okay as long as i'm in that top eight okay i'm going to advance to the state championship so we would use a combination here and again i would just do c and then i'm gonna put 40 comma eight again just punch this into your calculator you're gonna do 40 choose eight and that gives us a pretty big number it's going to be 76 million and then you're going to have 904 000 685 all right let's just look at one more so the state championship marathon consists of 40 runners the top three runners will earn gold silver and bronze medals so can we have repetition here no right because if i finished first i can't also finish second or third okay and then does the order matter well yeah it definitely matters right if i finish in first place that's a different result than if i finish in second place in one case i'm getting gold in the other case i'm getting silver so the order does matter so i want to use the permutations formula so i would just use my calculator or again you've got this p with this 40 comma 3. if you want to say it's equal to you can punch this in on your calculator or again you can use your little formula i just think it's easier to punch it in on a calculator because a lot of these the numbers are just going to be astronomically big so if we go ahead and punch up 40 and then you do your npr key and then 3 you're going to get 59 280 as a result in this lesson we want to look at some probability word problems so at some point in your algebra course or your pre-calculus course you're going to come across a small section on probability and i just want to go through some basic word problems that you're going to see and talk about how we can solve these guys so i'm just going to jump right in i'll just kind of explain things as we go it's a very very easy topic overall so jason flips a coin and then rolls a six-sided die what is the probability that the coin lands heads up and okay i'm going to highlight that and the die shows an even number okay so when you deal with these problems you're going to have problems that say and so some event occurs and some other event occurs in this case one event is flipping a coin and in getting a heads up and the other event is rolling a die and getting an even number so a 2 a 4 or 6. okay when you hear the word and you need to be thinking about multiplication so i would take and use this formula the probability of some event a and some event b happening okay in this case we'd say that coin landing heads up and also the die showing an even number is equal to the probability of this event a happening times the probability of this event be happening okay now this is given the fact that they are independent events meaning the first event does not impact the result of the second event in any way shape or form okay if he flips a coin that does not impact the result of rolling a six-sided die okay so these are independent events we can just use this formula as it's given so what we want to do here is just think about the fact that the probability for flipping a coin and getting a heads up is going to be what it's going to be a half or 50 percent we all know this because we've thought about that throughout life at some point flipping a coin saying hey my odds are 50 50. well how do we show this mathematically where does that 50 50 or the 50 percent come from well essentially what you do is you think about the number of different ways that you can get an outcome from flipping a coin well you can basically get a heads up or tails up right so there's two possibilities so i'll put that in the denominator and then think about what you're looking for what's the result well i want a heads up and that can only happen one way right so i would put that number in the numerator okay and as we do more of these this will make more sense but essentially this is the formula you want to use that's where one half comes from this is basically the number of favorable outcomes in this case a favorable outcome would be landing heads up over the total number of possible outcomes and the possible outcomes here would be landing heads up or landing tails up okay so i know that's one half so let's just say that this satisfies the probability of event a now what about the probability of event b let's say that's the die showing an even number well with a die if it's six sided it can land on a one let me just kind of write this down here in a different color a one a two a three a four a five or six so we know that the two is an even number the 4 and the 6 okay so 2 4 and 6 those are even numbers so how many different favorable outcomes can i have well i have one two three favorable outcomes that are possible out of a total of six so i would put three over six which you can simplify to one half okay and then you just multiply the two together and let me just kind of erase this you multiply the two together and you get one fourth okay now you can leave it as a fraction and say it's 1 4 for the probability or you can convert it to a decimal which is 0.25 and then to a percentage and say it's 25 okay so essentially the probability of this json guy flipping a coin and getting a heads up and also taking a die and rolling it and getting an even number is 25 okay let's look at a slightly more challenging problem and in this case we're going to have dependent events so allison a teacher out of middle school has five boys and six girls in her class allison randomly selects three different students to walk up and give a presentation once the presentation is over the student will leave for the day what is the probability that the first student is a boy the second student is a girl and the third student is also a girl so you see the problem here is that these events are not independent okay once she picks a student and we're looking for her to pick a boy first okay then the next time we want her to pick a girl and then the next time we want her to pick a girl but each time she's picking the class size is changing right so the number of boys is changing the number of girls is changing whatever it is in that scenario but the class size is constantly changing so we have to readjust our probability okay so we're still going to think about multiplying the probabilities together but you use a different formula for this if it was just two things here it's three but let's say it was two things so the probability of a and b okay if these guys are dependent it's just two it would be the probability of a times the probability of b given that a has happened okay so given that a has happened what's the probability of b happening okay and in this case we're going to adjust it because we have one more thing to consider because now we're thinking about the probability that the first student is a boy okay so that happens then the probability that the next student is a girl okay and then the probability that the third student is also a girl so again you have and so you're thinking about multiplication okay so what is the probability on the first grab that that student is a boy well there's going to be five boys okay so there's five possible choices in there that's going to be boys out of a total of 5 plus 6 is 11 11 total students so the first number we're looking at is i'm just going to write this down here 5 over 11. i'm just going to put this as the first okay so the second is going to be what well if i go back up i know that the student leaves he can't be re-chosen now and there's going to now be what well there's one less boy so there's four boys and there's the same number of girls but the total number in the class now is four plus six or ten okay so if i think about picking a girl now it's going to be six girls out of a ten student classroom okay let me just erase this so let's go down here and put that now it's six over 10 okay so this is second you'd say and now what's going to happen for the third choice when we go back up well now there's let me just kind of scratch this out again say this is 4 this is going to go down to 5. there's a total of nine students and there's five girls so there's five over nine okay let me just erase this again and now i'll say this is five over nine and this is going to be the third pick okay so the probability of picking a boy on the first run is 5 over 11. then given the fact that that event has already happened now there's a 6 over 10 probability of picking a girl now given the fact that both of those events happened the probability on the third pick that it's a girl is five over nine okay so again to get the final probability or the probability that all these events happen so this and she picks the first student as a boy and she picks the second student as a girl and she picks the third student as a girl it's going to be the product of these probabilities so before we multiply notice you can cancel this with this and get a 2 here and then 6 divided by 2 would be 3 and then 9 divided by 3 would be 3. okay so i'm going to put a 1 here and basically what you'd have is 5 times 1 or 5 over 11 times 3 which is 33. so you have 5 over 33 which is the probability of picking a boy then a girl then a girl all right so now let's talk about some word problems that involve this or so when event a happens or an event b happens so with this we're going to be adding now so when you hear and you got to be thinking about multiplication when you hear or you've got to be thinking about adding okay but there's a trick to this and let me just kind of go through this with this problem so katie has six nickels and seven dimes in her pocket five of the nickels and one of the dimes are canadian the others are from the us suppose katie randomly selects a coin from her pocket what is the probability that it is a dime or is from the u.s this is the key word here the or okay so what you want to be thinking about is the probability of some event a happening or some event b happening is the probability of the event a happening plus the probability of the event b happening minus okay this is important minus the probability of a and b happening okay i know a lot of people use set notation there so you might see this as well okay so the intersection of a and b however you want to write it doesn't really matter i'm just going to stick to writing and i think that's a little bit easier for us to understand okay in this particular case we don't have something known as mutually exclusive events mutually exclusive events are two events that can never occur simultaneously so in this particular case we don't have that because it says what is the probability that it is a dime or is from the us well we have dimes that are also from the us so we have to use this full formula okay if we didn't have dimes from the u.s we could knock this part out okay this is going to take out the double counting so let's go through real quick and think about what's the probability that it's a dime well we know that she has six nickels and seven dives okay well six plus seven is 13. so 13 coins overall seven of them are dimes so the probability of having a dime is 7 over 13. then plus what's the probability okay that the coin is from the u.s meaning it's not canadian well it tells you that five of the nickels and one of the dimes are canadian so six out of the 13 coins are canadian so 7 out of 13 are american or u.s currency so 7 over 13. 7 plus 7 is 14 14 over 13 is a number larger than one when we deal with probabilities it's either 0 okay meaning it's not going to happen up 2 and including 1. 1 means it's certain to happen anything in between means it may or may not happen okay but you can't get a probability less than 0 or greater than 1 that doesn't make any sense okay so that would be a red flag if you forgot to subtract this out so all we need to do to fix this is subtract out the probability of a and b occurring okay so how many dimes do we have that are also from the u.s well let's think about that we know that there are seven dimes in her pocket okay total of seven and it says that five of the nickels and one of the dimes are canadian so that means that six of the dimes are from the u.s okay so that tells me that i need to subtract away 6 over 13 right because 6 of the dimes are from the u.s so if i crank this out now i get 7 plus 7 which is 14 and then 14 minus 6 which is 8 so i get a result of 8 over 13. so that's going to be the probability that she picks a coin that's going to be a dime or it's from the u.s okay let's look at another example of this and now we're gonna see mutually exclusive events so beth has a fruit basket which contains four apples four peaches and five pairs if she randomly selects a piece of fruit what is the probability that it is an apple or a peach again when you see this keyword or you're thinking about addition okay but in this particular case you don't need that full formula okay because it's not possible for her to select something that's an apple and a peach at the same time so again when you think about the probability of some event a happening or some event b happening it's the probability of a plus the probability of b and then minus the probability of a and b occurring and i'm going to write my a there so a and b well in this particular case you can get rid of this because it's going to be zero again you cannot have a piece of fruit that's an apple and a peach at the same time maybe they're working on that i'm not sure but in this scenario it's not possible so we just think about the probability of a let's say that's where she selects an apple well there's four apples out of a total of four plus four plus 5 4 plus 4 is 8 equals 5 is 13. so the total of 4 over 13 okay that's going to be your probability of picking an apple then plus what's the probability of picking a peach well again it's 4 over 13. so all i have to do is just sum these and i get 8 over 13. all right so let's wrap up the lesson by looking at an example where we use our binomial probability formula so the desks in a classroom are organized into four rows and four columns each day the teacher randomly assigns you to a desk you may be assigned to the same desk more than once over the course of five days what is the probability that you are assigned to a desk in the front row exactly three times okay so if you are dealing with a scenario such as this where you have repeated independent trials okay so this is independent because every day you're randomly assigned to a desk and you can sit where you've sat before and the outcome in each trial is either a success meaning in this case you would get assigned to the front row or a failure okay so if you want to find the probability of r successes in n trials you can use this formula so let me write this down here and we'll come back up in a minute so n choose r okay we remember that and then times p raised to the power of r p is the probability of a success so on a single day what's the probability of you getting seated in the front row i'll get to that in a second then times 1 minus this p raised to the power of n minus r so n is the number of trials so if we go back up we know that this is going to happen for five days so over the course of five days so n is five and the number of successes we're looking for is r right so three times is we're looking for that number to happen so n is five n is five and r is three okay so you're just plugging things in so you have five and three okay then p is the probability of a success on any given day well if we go back we know that there's a total of 16 desks right four rows and four columns four times four 16. so in the front row you're going to have four desks so that means you have a four over 16 or one fourth or 25 percent probability of getting seated in the front row on any given day so we're going to put a 1 4 here you could use a decimal if you want it doesn't matter it's gonna be the same result in the end r again is three and then one minus one fourth if you're using a fraction you would do four fourths minus one fourth which is three fourths and then raised to the power of n which is 5 minus r which is 3 so this is going to be squared okay so once you have this set up you can just erase this and crank this out so on a calculator 5 choose 3 is going to be 10. then if we think about 1 4 cubed that's going to be 1 over 4 cubes of 64. then if we think about 3 4 squared 3 squared is 9 4 squared is 16. can i cancel anything before i multiply i just get rid of this i don't need anymore can i cancel anything well yeah i can cancel a factor of 2 here and here so this would be 32 this would be 5. so 5 times 9 is 45 over what is 32 times 16 that's going to be 512. so 45 or 512 that's going to be your probability of getting seated in the front row exactly three days out of a total of five days