Transcript for:
Function Types Overview

hello and welcome to algebra 2 lesson one in this video we're going to start learning about sets so before we kind of get started with the main topics of algebra 2 i thought it would be very important to cover the topic of sets and the reason for that is a lot of times this gets skipped over in your algebra 1 in your algebra 2 class and there are things that come up that are related to this on standardized tests such as the sat and the act additionally when you get into college level math there are things that you're expected to know that are related to sets so with that being said let's start out by just giving a basic definition for a set so a set is just a collection of things or you could say a collection of just some stuff now it should be clear whether something belongs to the set or not now generally speaking when we're talking about an algebra class when we talk about a set we're talking about a collection of numbers that fit a certain profile or you could say a collection of numbers that fit a certain definition so as an example let's start by just thinking about the whole numbers so this is a set or a collection of numbers that fit a certain profile they start with the number 0 and they increase in increments of 1 indefinitely so generally when we list the whole numbers we'll start with 0 we'll go 0 1 2 3. at some point we're going to stop because we can't list every whole number that's impossible and we'll put three dots now we think about this in regards to being a set we have some more definitions involved so the numbers that are listed here are known as the members or elements of the set and like i said it should be clear whether something is in the set or not in the set so as an example if i asked you is negative 5 a member of the set of whole numbers you would say no because the whole numbers start with 0 and increase in increments of 1 indefinitely so that doesn't fit that profile or that definition so it is clear that negative 5 is not a member of that set now another thing that we do when we have sets we enclose our members or elements inside of set braces so we're going to enclose this inside of set braces so these are set braces okay if you hear that and the last thing i want to say here is that the method that we're using here is known as the roster method okay this is known as the roster method and basically what we're doing is we're listing the members or elements of the set inside of set braces so that's the roster method as another example let's say we talked about the set of integers okay the set of integers so the set of integers if i use the roster method again inside of set braces i would have three dots coming this way comma i'd start with something like let's say negative two then negative one then zero then one then two and comma and then again i'm gonna put three dots and again in case you don't know what the three dots are they're known as an ellipsis and they just tell you that the pattern is going to continue so going this way i know that i'm decreasing by 1. so after negative 2 i'd have negative 3 then negative 4 the negative 5 so on and so forth this way after 2 i'd increased by 1 so i'd have 3 then 4 then 5. so on and so forth so this is the set of integers and again it's clear whether something's in that set or not if i gave you the number 0.675 it's clear that that's not an integer it's not in that set now although generally in algebra we're going to have sets that involve numbers we can have sets that don't involve numbers so let's say as an example i had a set that contained the first five letters of the alphabet well again using the roster method i would list a b c d and finally e right the first five letters of the alphabet inside of my set braces all right so now that we have the basics under our belt let's talk about a couple of other things so the order that we list the elements of a set is not important so let's say for example i have a set whose elements would be 1 6 and 7. this would be equal to the set whose elements would be let's say 6 7 and 1 and it's also equal to the set whose elements would be 1 7 and 6 you know as an example the order is not important what's important are the elements or the members of the set if this set contains 1 6 and 7 and this set contains 1 6 and 7 and this set contains 1 6 and 7 they're equal because the members or elements are the same as another example of this let's think about the states that border louisiana okay my my home state is louisiana and so the states that border it are mississippi m-i-s-s-i-s-s-i-p-p-i okay kind of hard to spell then arkansas and texas so this is the set whose elements are the states that border louisiana and this would be equal to the set with the same elements just reordered so let me reorder this let me write texas first then write mississippi m m-i-s-s i-s-s-i-p-p-i and then let me write arkansas last and again i can change the order as much as i want what's important here are the elements or the members of the set all right so generally speaking we can use capital letters to name a set so we want to let a be the set of whole numbers less than four so we know the whole numbers start with zero and increase in increments of one so i would put a capital a out here and this equals inside of set braces so again the set of whole numbers less than four start with zero then one then two then three it's less than 4 so i'm not going to continue after that so this is my set a again that's got the elements 0 1 2 and 3 involved all right so another big one that you need to know is a set that contains no elements is referred to as an empty set or null set so once we get into algebra 2 we're going to start using something known as solution set notation and a solution set is nothing more than a set whose elements are a solution to your equation so as an example real quick let's say you had something like 2x minus 5 is equal to let's say 27. so add 5 to each side of the equation that would cancel you get 2x is equal to 32 divide both sides of the equation by 2 you get x is equal to 16. so to write this in solution set notation i just literally put 16 inside of set braces that's all it is this is a solution set this is a set whose elements are a solution to the particular equation that you're working on and this isn't particularly that useful when you get into algebra 2 but if you get into higher levels of math where you solve something and you've got a bunch of solutions it's just a neat and clean way to organize your solution okay that's all it's used for but we sometimes get equations that can't be solved okay it'll have no solution involved so a set that contains no elements is referred to as an empty set or null set so what we're going to do is if we have an equation with no solution we're going to use this symbol right here to denote that because that particular solution set has no elements right so if it has no elements it's the null set or the empty set so you can use this symbol here for that or you can also use set braces with nothing inside but one thing i want to warn you of and this is a big thing when you get to college math do not do this this is wrong okay write that down it's a very common mistake that students make do not put this symbol inside of this you don't want to do that that is the set that contains one element and that element is the empty set and i don't want to go into set theory and start confusing you but just know that it's wrong wrong wrong okay you can use this symbol or you can use this symbol but you cannot combine the two so as another example of this let's say i said let b let b be equal to the set of all pigs that are living on saturn now i've never been to saturn chances are you've never been to saturn but i'm going to say that there are no pigs that live on saturn so that set would be an empty set right that set would contain no elements so we can say that this set can be denoted by the symbol for the empty or null set or you could also say b is equal to this but again you cannot say that b is equal to inside of set braces placing this symbol in there that again is wrong that is wrong do not do that no all right so the next thing is that this symbol here which if i draw it let me let me draw it it looks like a sideways u and then you just put a line down the center there so the symbol here is used to denote is a member of the set so as an example we have set z and z is equal to we have the members 3 7 1 and 8. so what i can do is i can take each one and i can say okay 3 is an element of set z it's denoted like that or i could say 7 is an element of set z or i could say 1 is an element of set z or lastly i could say 8. is an element of set z now similarly the symbol here basically we just put a line through the other symbol is used to denote is not let me highlight that is not a member of the set so with the same set z with the members 3 7 1 and 8 i could do something like this i could say something like 6 is not an element of the set z i could really use anything i want other than these four numbers i could say 0 is not an element of set z i could say negative 32 is not an element of set z you know so on and so forth so another important concept would be that of the universal set so this one's a little bit difficult to wrap your head around at first but something after you practice a few of these that you really can't understand so the universal set which is generally denoted with a capital letter u is a set that contains all elements under consideration now if we're talking again about the states that border louisiana let's just name this for the purposes of this example here let's name this as set a so set a has the elements that fit this profile where they're states that border louisiana so we already know that this would be mississippi m-i-s-s i-s-s i-p-p-i we know it would be arkansas and we know that it would be texas okay so three elements in this set but if we think about the universal set here what are all elements under consideration well we're talking about states in the united states of america so all elements under consideration would be all of the states in the united states of america so the universal set here would contain every state now we're not going to list all of them that would be very time consuming but in alphabetical order we could just start with two of them so in alphabetical order it would be alabama there would be alaska and let's put our three dots there and let's finish up with the last two which would be wisconsin and then wyoming now as another example of the universal set let's say we're again talking about the united states so the united states has a president we all know this and we can talk about a set and let's call it set a again whose elements are the presidents that are currently living now this set would change over time we're filming this video in 2015 so in 2015 the living presidents we would have jimmy carter we would have george bush senior so let's just put g period b period and we'll put number one because he was the first george bush to be president of course his son would be george bush ii so then comma after him we have bill clinton and then comma we're gonna have george bush number two or george bush jr so g period b period and then two and then we're gonna have barack obama and to fit this on the screen i'm just gonna put b period o period and then we're going to close up with some set braces at the end so these are all the living presidents as of 2015. now in 2016 we're going to have a presidential election so for sure it's not going to be barack obama because he's going to be out of office so we'll have another president to add to this list but as people pass away as new presidents get elected this set would change over time but what's important is to understand that the universal set the set with the capital letter u would be what it would be all of the us presidents that ever existed okay so we would start with george washington so george washington and then the next one was john adams [Music] and then of course after that you had thomas jefferson and so on and so forth so let's just put three dots here and we'll kind of come in with the final two we had george bush number two or george bush jr and then we had barack obama and again it could be whoever in the next election that wins so this list would be added to if donald trump wins if hillary clinton wins if you know somebody else wins we'll have another president to add to both lists all right so additionally we have finite sets which have a specific number of elements and then we have infinite sets which have an unlimited number of elements so as an example of a finite set let's say i told you that we had a set a which was the whole numbers that were less than six so this would be zero one two three four and then five so that's a finite set i can count one two three four five six elements involved there but if i said give me the set of whole numbers or give me the set of numbers that are larger than let's say five those are each infinite sets because they go on forever and ever and ever and ever you can't possibly count them okay so that's an infinite set all right so let's wrap up the lesson by talking about subsets so this is something that you might get tested on kind of the other stuff we looked at was just kind of you know basic definitions but these are things that you might have to think about a little bit so let's take a look at subsets so if i have set a which contains the elements 1 7 and 3 and set b which contains the elements 1 through 8 then a is known to be a subset of b and more specifically a is a proper subset of b because b contains all elements of a and they're not equal to each other so a is a subset or a proper subset of b and a lot of times this gets confusing because you have a and you have b and this kind of looks like a c so let me make it a little longer i'll also do it in a different color let me do that so it kind of looks like a c but that is a symbol for is a proper subset of again if you look at b it's got one it's got seven and it's got three now if two sets are equal meaning they have exactly the same elements they're known as improper subsets of each other so in this example we have g which has one eight and two and d which has two one and eight these two sets are equal to each other because they have the same elements two and two eight and eight one and one so the exact same elements so they are equal so we can say they are improper subsets of each other so we show that by saying g is an improper subset you see that little bar is an improper subset of d and then also you could say d is an improper subset of g and let me just label this this is g is an improper subset of d all right so we use the symbol here to denote is not a subset of so it's just the proper subset symbol with a line drawn through it so as an example here we have these two sets we have q which has a m and an o we have l which is a d z p o and n so we'd ask the question is q a subset of l well no it isn't and the reason it's not this has a and this has a this has m this doesn't have m so no m this has n and this has n and then this says o and this has o but because it's missing this m here it's not a subset for something to be a subset of another it's got to contain all of the elements you can't be missing something so q q is not a subset of l q is not a subset of l but if i was to change this set and let's say i was to put an m in there then it would be a subset because l would contain all of the elements of q so that's what we mean here when we say q is not a subset of l so lastly let's discuss how many subsets can be made from a given set well the answer to that is 2 raised to the nth power where n is the number of elements that we have so for example if i have p is equal to let's say 5 7 and 9. well the number of subsets that can be made from this is 2 raised to the power of 1 2 3 or 8. you can prove that to yourself by listing the possible subsets so one subset would be just the set five then another would be seven then another would be nine then i would have a subset where i would have five and seven another would be five and nine and another would be seven and nine then another one would contain all of the elements [Music] and what is the last one a lot of people don't get this and it's something you kind of need to be told well the empty set is considered to be a subset of all sets and why is that the case that's one thing that many students struggle with well you have to think of subsets as possible choices from the set so i can choose five i have that i can choose seven i can choose nine i can choose five and seven i can choose 5 and 9. i could choose 7 and 9 i could choose 5 7 and 9 or i can choose nothing i can choose nothing so as an example we can kind of understand let's say that i have a set l so i have a set l and l contains three elements it's the elements that are in my lunch box that i'm bringing to work so let's say i have chips i have soda and i have a sandwich make it real simple so what are the possible subsets that i can make here we know there's three elements so that we know there's going to be eight using our formula but let's go ahead and list them again think about possible choices what can i take out of the lunch box well i could take out just chips i could take out just a soda i could take out just a sandwich now i could take out two items if i wanted to i could take out chips and soda i could take out chips and a sandwich i could take out a soda and a sandwich and then i could take out all three if i wanted to i could take out the chips the soda and the sandwich and then lastly i could choose to just take out nothing right i could choose to open my lunchbox up and say you know what i don't want anything so my choice is to choose nothing so that becomes the empty set choice and so i have my subsets i have 1 2 3 4 5 6 7 8 and that is given to me by my formula which is 2 raised to the number of elements i have i'm going to have 1 2 3 elements 2 to the 3rd power is 8 and i've listed them there for you hello and welcome to algebra 2 lesson 2. in this video we're going to continue to learn about sets so at some point in your math class you've probably heard about something known as a venn diagram if you haven't i can promise you by the time you get out of high school you're going to encounter at some point either in a math class or outside of a math class but a venn diagram is just a way to visually represent the relationship between sets it's actually a very very useful thing even when you're not working with math so as an example let's say we have a universal set which contains the elements one through ten then we have a set a which contains the elements one through three and a set b which contains the elements three through six so the idea here is that we draw a rectangle to represent the universe or the universal set so let me just put a u inside of here to say this whole thing represents set u so all of my elements here should be listed inside the rectangle somewhere now i have two subsets of u i have a which is a subset of u and i have b which is a subset of u and each of these is represented with a circle and notice how the circles are inside of the rectangle right inside of our universe so i can label each circle i can label one of them as a and the other is b doesn't matter which is which and i would list elements inside of the area that it corresponds to and one thing i want to bring your attention to right away i want you to notice how there's a section of overlap this is used to place any element that is part of set a that is also part of set b so a common element and when we start listing things here we see we have one two and three for set a and we have three four five and six for set b three is common to each so that's going to go in this section of overlap now for set a i have one and two that are not common to b so that's just going to go over here somewhere outside of the overlap same thing goes for set b i'm going to put four five and six in here but outside of the overlap now elements that are part of the universal sets that are not part of a and not part of b would just go somewhere inside of this rectangle they just can't go inside of these two circles because again they're not part of a or b so i have 7 i have 8 i have 9 and i have 10 that fit that criteria so this is our first venn diagram let's just go ahead and label it this is our venn diagram and again it's a visual way to represent the relationship between the sets i can look at this right away and i can tell that i have a universal set that's one through ten i have a set a that's one two and three i have a set b that's three four five and six and i can also tell that a and b have a common element of three so some key information that i can pull right away just by looking at a picture versus kind of looking at these sets in the roster method and trying to kind of pull information out that way now a couple of things we want to talk about now the first thing would be the complement of a set so this is represented if i wanted the complement of a i would write a prime or a complement like that with a superscript c on top of the a so a complement contains all of the elements or members of set u that are not elements or members of set a so in this case because set a has one two and 3 i would just exclude those and i would say okay i have 4 5 6 7 8 9 and 10 as the complement of a and again i go back to my venn diagram i can quickly see that if this is set a everything inside this circle i know i didn't color in the lines perfectly but just you know kind of bear with me everything outside of that circle is the complement of a so it's very quickly to get that i can just see that it's 4 5 6. 7 8 9 and 10. very easy to get that information as another example let's take b complement so let's erase this and let's take you can put b prime or b complement like this and in this scenario i could either do it this way i could say okay this contains three four five and six i can mark those out and say it's one two seven eight nine and ten or visually i could look at my venn diagram and if i look at b right here everything outside of this is going to be the complement of b so i'm looking at again 1 2 i have 7 8 9 and 10 7 8 9 and 10. now let's erase this real quick let's now talk about the union of two sets so the union of two sets let's say a and b is denoted with a symbol that kind of looks like a u so we'd say a let me do it in a different color union and then let's return to the other color b is equal to this is going to be the set of all elements of a along with all elements of b but with one key thing here we don't want to double list anything okay so if i put in all the elements from set a that's one two and three then all the elements from set b i've already listed three i don't need to list it again so then just four five and six and again the way you look at the union is if i look at my venn diagram it's everything that's contained in these two circles so everything in these two circles what's not going to go in there would be anything outside of the circle so things like 8 7 9 and 10. things that are outside of the circles you can very quickly look at the venn diagram and get the union now the other thing that we would talk about would be the intersection so let me erase this real quick the intersection just looks like a flipped upside down u so i'm just going to basically reverse that symbol and make it look like this so this is intersection [Music] and the intersection of a and b is going to be the set of all elements that belong to both a and b and again that's where you're going to look at your overlap so if i look at my venn diagram this is the intersection again the set of elements that belong to both a and b so that's going to have one element so a intersect b is equal to we just have one element and that element is three all right let's take a look at another example so here we have a universal set where we have the elements a b c d g h and q and t and everything is in lower case to keep you from confusing it with the set names then we have set a again capital a and this contains the elements g h and n and then set b which contains the elements t q and n so the first question i have here is what is the union between a and b so a union b equals what well let's fill in our venn diagram so let's let this be set a let's let this be set b and in set a i have g i have h and i have n and notice where i put the n it's because it's common a and b so for set b i have t i have q and again i have n n is in the overlap section because it's a member of set a and also set b now in my universal sets i have elements that are not part of a and b i know that a falls outside of those two circles so it is b so does c so does d g doesn't h doesn't n doesn't q doesn't and t doesn't so if i want the union of a and b this is going to be equal to we'll have g h n t and then q and again the main thing here is to make sure you don't double list anything so i have n here and here it only gets listed once when i fill out this information what if i ask for the intersection if i said let me kind of go in here and edit so if i ask for the intersection the symbol looks like this just an upside down u and i know that's not perfect but just pretend it is again i'm looking for the overlap between these two sets so the overlap if i look at the venn diagram it's going to be right here that's real easy to get that's n so that's a intersect b and then what if i ask for some other things let's say i asked for a complement a complement and i could do it this way or again i could do it this way doesn't matter well that's anything that's outside of a so i could just highlight this set right here and i could look for things that are outside of a and what i'd come up with is t q a b c and then d and again it doesn't have to be in alphabetical order the order is irrelevant when you're listing the elements of a set and if i ask for b complement that would be anything that's outside of this circle here so let me highlight that real quick for you so for this case i'm looking at what i'm looking at a b c g h and d all right let's take a look at another one so now we're going to have our universal set which contains 1 3 5 7 9 11 13 and 15. so basically the odd numbers starting at 1 and going through and including 15 then set a has 1 3 and 5 set b has 3 5 and 7 and set c has 9 13 and 15. so let's label this as set u let's label this as set a set b and set c so now what you're going to notice is set a and b have two common elements three and five so those are going to be put in the overlap section and then for set a we have one that's outside of that for set b we have seven that's outside of that and then for set c we have nothing that's common to a and b so notice how there's a gap between these circles right we don't want any overlap because there's nothing that's going to be common so for set c i'm going to write in 9 13 and 15. okay so also in set u we have 11 that's not listed in any of these sets here so that's going to go outside of all those circles so without referencing to this let's answer these questions here so we're just going to look at our venn diagram so we have a union b what is that that's all the elements of a along with all the elements of b again just don't double list anything so looking at the venn diagram that's easy it's a 1 a 3 a 5 and a 7. [Music] then if i want a union c what is that well that's 135 and then 9 13 15. and then b union c is what we have three five seven and then nine thirteen fifteen let's see if we can do some other things here let's erase this and let's ask for the intersection in each scenario now so a intersect b is what again that's your overlap here so that contains the elements 3 and 5. what about a intersect with c there is no overlap those two sets are called disjoint sets they have no elements in common and so this right here would be the null or empty set right because this set contains no elements because there are no elements that are common between set a and set c same thing goes for the intersection of b and c you would put the symbol for the null or empty set now we could also go through and we could do the complement of each so i could do the complement of a the complement of b and the complement of c and again with our venn diagram we can do these very quickly so for a complements i would have 7 9 13 11 and 15. for b complements i would have 1 11 9 13 and 15 and then for c complement i would have 1 3 5 7 and 11. and again just looking at the venn diagram makes things nice and easy to see if i want a complement i could shade a and say i want everything outside of that so that's 7 11 9 13 and 15. if i want b complement shade b and i want everything outside of that so that's 1 11 9 13 and 15 and if i want c complement i could shade c and again i want everything outside of that so that's 1 3 5 7 and 11. all right let's take a look at another one so we have a universal set that contains california oregon washington colorado north dakota new york wyoming texas georgia south carolina maine and wisconsin then i have a set a which is a subset of set u which contains california oregon washington colorado then i have a set b which again is a subset of set u which contains oregon north dakota new york and washington now if i fill this out for set a and set b what they have in common would be washington and oregon so let's say this is set a and this is set b so for set a i would also have california and i would have colorado for set b i would also have north dakota and i would have new york and outside of these sets in set u let me use a different color i would have wyoming i would have texas i would have georgia i would have south carolina i would have maine and i would have wisconsin so once we have our venn diagram built it's easy to go through and answer some questions so let's say i start out by asking you for a union b what's that equal to well that would be california colorado washington oregon north dakota and new york if i asked you for the intersection of a and b that would be the overlap section so that would contain two states washington and oregon if i asked you for a complement again i could highlight this right here and i could say that a complement contained north dakota new york wyoming texas georgia south carolina maine and wisconsin and then lastly if i ask for b complement this would contain the elements i could erase this and i could highlight this okay i can highlight this and i would have california colorado wyoming texas georgia south carolina maine and wisconsin hello and welcome to algebra 2 lesson 3. in this video we're going to learn about solving linear equations in one variable so back in algebra 1 we learned how to solve a linear equation in one variable and hopefully at this point you've completely mastered that concept but in case you haven't we're going to do a full review in this video here and that's going to serve as a stepping stone for the more challenging types of equations that we're going to look at in algebra 2. all right so you will recall that a linear equation in one variable looks like this we have ax plus b equals c so if you're in a textbook following along you'll see something like this and this is a generic form that you get a b and c are just representing real numbers so this right here this right here and this right here so we'll put a b and c are real numbers and then there's one exception to this a which is the coefficient for x cannot be zero so a cannot be zero and obviously that's the case because we want our x to be there we want our variable to be there if i put a zero in for a zero times anything is just zero so this would disappear and we would no longer have a linear equation one variable so we've got to have that restriction there now because this type of equation contains real numbers and a single variable in this case it's x raised to the first power we sometimes refer to this type of equation as a first degree equation so you might hear that in your class so some examples and i know we've seen many of these but 3x plus 5 equals 7. so that's a linear equation in one variable or something like 2 z minus 5 equals i don't know let's say 11. or let's say negative 27 y plus square root of 2 equals negative 15. these are each an example of a linear equation and one variable now something that wouldn't be a linear equation one variable let's say you had two variables involved so something like 2x minus 3y equals 7 or something with an exponent on the variable that's higher than 1. so let's say 3x squared minus 5 equals 2. so these two right here would not be an example of a linear equation in one variable all right so let's take a look at 2x plus 3 equals 9 and we see here that we have x equals 3. so the first thing we're kind of taught when we start learning about equations is that the solution to an equation is a number or it could be more than one number in some cases that when it replaces the variable or you could have variables involved it's going to give you a true statement so what that means is that if x is equal to 3 if i plug a 3 in for x in this equation what's going to happen is the left and the right side will be the same value so to demonstrate this let's say i have 2 and then i substitute a 3 in for x then plus 3 equals 9. so over here 2 times 3 is 6. so i'd have 6 plus 3 equals 9 and then 6 plus 3 is 9 so you get 9 equals 9. this is what you're looking for the same value on the left as on the right so that's how you know you got the right answer right x does equal 3 here and to kind of further expand on this let's say i chose something like x equals 4. i said is that the solution well i'd erase these two and say well 2 times 4 is 8 so i'd have 8 plus 3 which is 11 and no this 11 does not equal 9. this would be a false statement this is this is false so x does not equal 4 that's wrong right we know x equals 3. so how do we go about solving a linear equation in one variable you're not always going to be presented with the answer and just say hey plug this in and see if it works so we use two properties when we first start solving equations the first one is the addition property of equality it tells us we can add the same value to both sides of an equation and we will not affect the solution so if i have something like 2x plus 3 equals 9 we just saw that we know the solution is x equals 3. okay we know that let's say i was to add negative 4 to both sides of the equation so let's say we do 2x plus 3 and i'm going to add negative 4 over here and i'm going to add negative 4 over here and so what's going to happen is 3 plus negative 4 is what that's negative 1. so i have 2x minus 1 over here and then 9 plus negative 4 is 5. this equals 5. so now if i plug a 3 in for x will i still get a true statement some of you might think that you won't but you will i plug a 3 in there i'd have 2 times 3 minus 1 equals 5. we know this would be 6 minus 1 equals 5 and you get 5 equals 5. so i added the same value to each side of the equation and it did not affect my solution my solution was x equals 3 before and it's x equals 3 afterwards now the other property we want to know right away is the multiplication property of equality now you'll recall that this property tells us that we can multiply both sides of an equation by the same non-zero number and it will not affect the solution okay and remember you multiply by zero it becomes zero so that you can't use zero in this it has to be a non-zero value so for example two x plus three equals nine we know this equation has a solution of x equals three if i multiply both sides of the equation by negative two would i still have x equals 3 as the solution let's try it out so 2x plus 3 let's use parentheses here multiply negative 2 by this side let's multiply negative 2 by this side scroll down get a little room going use my distributive property negative 2 times 2x is negative 4x and then negative 2 times 3 is going to be minus 6 and this equals 9 times negative 2 is negative 18. now if i plug in a 3 for x am i going to get a true state let's see negative 4 times 3 minus 6 is equal to negative 18. negative 4 times 3 is negative 12 so you'd have negative 12 minus 6 equals negative 18 and scroll down just a little bit more and what we're going to have is that negative 12 minus 6 is negative 18 so you get negative 18 equals negative eighteen so you can see that before we multiplied both sides of the equation by negative two the solution was x equals three after we multiplied both sides of the equation by negative two the solution was still x equals three so multiplying both sides of the equation by the same non-zero value allowed us to keep the same solution all right so now that we've covered some of the basics let's talk about the procedure to actually solve a linear equation in one variable before i start here i just want you to put this down on a piece of paper your main goal is to isolate the variable so you want x or y or z or whatever the variable is on one side of the equation and you want a number on the other so this is how we go about doing that we want to simplify each side separately so that means you're going to clear parentheses combine like terms on each side okay you just want to completely simplify each side then we want to isolate the variable terms on one side of the equation we're going to do this by using our addition property of equality then we want to isolate the variable and we're going to do this using our multiplication property of equality and then lastly we're going to check the solution in the original equation so we take our solution and we plug it back in for the variable in the original equation and we make sure that the left and the right side are equal all right so let's go ahead and take a look at some practice problems and we're going to start out with we have negative 9 times the quantity p plus 6 this is equal to negative 4 times the quantity 1 plus p so our variable here and i'll just highlight it is p so what we want to do is start out by simplifying each side separately so i'm going to use my distributive property here negative 9 times p is negative 9p then i have negative 9 times 6 and that's negative 54. so then this is equal to i'm going to use my distributive property over here negative 4 times 1 is negative 4 and then negative 4 times p is minus 4 p so now that i've got in this format remember i want to isolate the variable terms on one side of the equation so a variable term just means a term with a variable involved so that would be this term and this term so what i can do is i can either move this over here or i can move this over there doesn't matter what i'm going to do is i'm going to add 4p to both sides of the equation and what is that going to do if i have negative 4p plus positive 4p those two are going to cancel right this becomes 0. now over here i would have negative 9p plus 4p that's negative 5p then minus 54 equals negative 4. now i want to isolate this term on this side of the equation and to do that if i have this minus 54 here i've got to get rid of it and what i do is i want to use my addition property of equality again if i have minus 54 i just add 54 to both sides of the equation negative 54 plus 54 is zero right that gets rid of that and so what i'm left with here is negative 5p by itself it's isolated on the left side is equal to negative 4 plus 54 negative 4 plus 54 is 50. now the third step says we want to isolate the variable how are we going to do that let's take a look at this variable i know a lot of you just know the answer to this because you solved a lot of these already if i want to isolate that variable which is p i've got to get rid of that negative 5. that's the coefficient of p well what's the opposite of multiplication if negative 5 is multiplying p the opposite is division so what i can do is i can divide by negative five and what that does is any non-zero number divided by itself is one so negative five over negative five is 1. 1 times p is the same thing as p now i can do this provided i do it to the other side that's part of the multiplication property of equality it extends to division so i divide this side by negative 5 as well to make it legal and what i have is just p on the left side on the right side 50 divided by negative 5 is negative 10. so this is my solution p equals negative 10 and at this point we're going to start writing things in solution set notation and a solution set is just a set whose elements are solutions to our equation so in this case the set would have one element and that element would be negative 10. so i just put negative 10 inside of set braces and so your teacher is still going to accept this as an answer but more traditionally as we move up we would write it using the set notation so negative 10 is our answer here now what i want to do is i want to show you how to check this let me erase everything okay so what i want to do is i want to plug in a negative 10 everywhere i see a p so i'm going to plug a negative 10 in here and also here and we want to make sure the left and the right side are equal to each other so you'd have negative 9 times the quantity negative 10 plus 6 and this is equal to negative 4 times the quantity 1 plus negative 10. okay so negative 10 plus 6 is negative 4 so you'd have negative 9 times negative 4 and this should be equal to inside of here 1 plus negative 10 is negative 9 so you'd have negative 4 times negative 9. you can already see this is the same value negative 9 times negative 4 is 36 so you'd have 36 on the left is equal to 36 on the right so you have the same value on each side of the equation so that tells you that your solution which is p equals negative 10 is correct all right let's take a look at the next example so we have 4r plus 5r is equal to 5 times the quantity 3 minus 7 r minus 3 times the quantity 5 minus 9r so again i want to simplify each side separately so 4r plus 5r is 9r i'm just combining like terms there this is 9r this is equal to use my distributive property here 5 times 3 is 15 minus 5 times 7r is 35r and then i have i might as well treat this as negative 3. so negative 3 times 5 is minus 15 and then negative 3 times negative 9r is positive 27r and i can combine more like terms here 15 minus 15 is 0. so this would cancel and then negative 35r plus 27r is negative 8r so you get that 9r is equal to negative 8r so now i want to move all my variable terms to one side so in this equation i just have variable terms i have 9r and have negative 8r so what i'm going to do is i'm just going to add 8r to this side of the equation if i have negative 8r plus 8r this becomes 0. make sure you write 0. that's very important so put 0 here and then over here i'll have 9r plus 8r which is 17r now i want to isolate the variable in order to do that again if i want to isolate r and 17 is multiplying r to get r by itself i divide both sides of the equation by 17. this will cancel with this and become one so i'll have one r or just r and this will be equal to zero over 17 which is zero so my solution set contains one element and that element is zero now if you want to check this i'm not going to in the interest of time all you want to do is plug a 0 in for each r in the original equation so they're there there and there and you'll see that the left and the right side will give you the same value and that's how you know your solution is correct all right let's take a look at another one so we have 7 plus 6b plus 1 plus 12b is equal to 4 times the quantity 5b minus 2 minus the quantity 7b minus 6. so on this side i can combine these two and these two seven plus one is eight six b plus 12 b is 18 b over here i use my distributive property 4 times 5b is 20b then minus 4 times 2 is 8. then i'm going to treat this as a phantom negative 1. so instead of having this minus out here i'm going to put plus negative 1 and i'm going to use my distributive property negative 1 times 7b is minus 7b and then negative 1 times negative 6 is plus 6. so over here i can think about 20b minus 7b is 13b okay this is 13 b and then negative 8 plus 6 is negative 2. so now again i want to bring all the variable terms to one side and i want all the numbers on the other so i'm going to subtract 13b away from each side of the equation so that this is gone so 18b minus 13b is 5b and i'm also going to subtract 8 away from each side of the equation so this is gone so i'll put equals negative 2 minus 8 is negative 10. so now i have a variable term on one side and a number on the other all i need to do at this point is isolate the variable so i want to isolate b and again if i want to isolate b and 5 is multiplying b i just divide by 5. okay i divide by 5. this is going to cancel with this and become a 1 1 times b is the same as b so i've isolated the variable so i'll have b is equal to negative 10 over 5 which is negative 2. so my solution set here will contain one element just negative 2. now again if you want to check this in the original equation go everywhere you see a b so here here here and here plug in a negative 2 and you'll see that the left and the right side are equal and so we'll know that our solution here b equals negative 2 is correct hello and welcome to algebra 2 lesson 4. in this video we're going to continue learning about solving linear equations in one variable so now we're going to deal with some special case scenarios so the first thing i want to talk about today is what to do when you have a linear equation in one variable and you have fractions involved so typically to make it easier on ourselves we want to clear the equation of fractions before we begin but as i'm going to show you with this first example you don't have to do that okay you'll get the same answer either way so to clear an equation of fractions we multiply both sides by the lcd that's the least common denominator of all fractions so here is an example i have negative x plus one fifth x is equal to negative 21 fourths plus 14 fifths x plus three fifths minus one half x so to not deal with these fractions i can just figure out what the lcd is so i have a five i have a four another five another five and a two now really the only thing i need to think about is the 5 and the 4. 5 times 4 is 20 so the lcd here would be 20. so what i want to do is i want to multiply both sides of the equation by this number 20. so let me just copy everything so it's negative x plus 1 5 x and then i'll close the brackets there and then i'll put equals put some new brackets we'll have negative 21 fourths and then plus 14 fifths x then plus three fifths and then minus one half x close the brackets and then we're multiplying this by 20. all right let's scroll down get a little room going so i'm going to use my distributive property i'd have 20 times negative x that would be negative 20x and then next i'd have 20 times one-fifth x so let me write that out so plus 20 times one-fifth x so here's where we're going to do some canceling this 20 would cancel with this 5 and give me a 4. then 4 times 1x or 4 times just x is 4x so let me erase this and just put 4x okay then we have equals now next i have 20 that's going to multiply negative 21 4. so let's write that out so negative 21 4 times 20 and of course i can do some canceling here this 20 will cancel with this 4 and give me a 5. 5 times negative 21 is negative 105. let's erase this and put negative 105 then i'll have 20 times 14 fifths x so plus 20 times 14 fifths x and what's going to happen is this 20 will cancel with this 5 and give me a 4. 4 times 14 is 56 then times x is just 56x let's write 56x here and then i have 20 times three-fifths so plus 20 times three-fifths and this will cancel with this and give me a four four times three is twelve so this would be 12. and then lastly i'll have 20 times negative one-half x so i'll put minus one-half x times 20. and again this will cancel with this and give me a 10. 10 times negative 1 is negative 10. so this would be minus 10x so minus 10x so it took a while but we've cleared our equation of fractions and now what we can do is just proceed as we normally do we would simplify the two sides separately on the left side i have negative 20x plus 4x that is negative 16x on the right side i can combine 56x and negative 10x that would give me 46x then over here negative 105 plus 12 is negative 93. so this would be 46x minus 93. and let me scroll down and get a little room going so now what i want to do is put all the variable terms on one side and the numbers on the other really all i need to do is subtract 46x away from each side of the equation so this is going to cancel negative 16x minus 46x is negative 62x and this equals negative 93. one last step for us here we want to divide both sides of the equation by negative 62. so we can isolate our variable x so what's going to happen is this will cancel with this and give me a 1. so i have x by itself and this equals we have negative 93 over negative 62. let's see if we can reduce that any i know that negative 93 i could write this as negative 1 times 31 times 3 and i know that negative 62 i could write this as negative 1 times 31 times 2. so we can see that we can cancel the negative one with the negative one obviously negative over negative is positive we cancel the common factor of 31 and what i'm left with here is three halves so i can erase this and just put x is equal to three halves so again we've been using this solution set notation and so inside of set braces i would just list three halves and we're done there now normally in algebra one i checked everything i would go back and i would substitute three halves in for x and i would show you the left and the right side are equal as these problems get more complex it's more time consuming to do that so my strategy for you is pause the video go ahead and check it now if you're on a timed examination you might say hey what do i do should i check everything and not finish the whole test the answer to that is no if you have a lot to do on your test complete the test don't check anything once you're done where students would normally rush up to hand in their paper and you know run out of the classroom take the extra time to go back and check the ones that you're unsure of you do those first then you go back and you check as many other problems as you can hopefully you can check all of them and you'll catch any silly mistakes that you've made now i want to show you one thing before we kind of move on to the next example let me erase everything so i want to quickly show you that you get the same answer whether you clear the equation of fractions or you don't so over here i have negative x plus one fifth x to get a common denominator going i could just multiply this by five over five and what would that give me it would give me negative five x plus x over five well this would be negative 4x right negative 5x plus x is negative 4x over the common denominator of 5. so that's the left side so negative 4x negative 4x over 5. now this is equal to over here to get a common denominator between here and here i see i have a 5 and a 2. so if i multiply this by 2 over 2 and i multiply this by 5 over 5 what i'm going to have is 28 x over 10 minus over here i'd have 5 x over 10. so i have a common denominator 28 minus 5 is 23 so this would be 23 x over 10. so 23 x over 10 let me erase that and now the other thing we have to deal with is this right here so to get a common denominator here i'll multiply this by 5 over 5 and this by 4 over 4 and what happened is i'd have negative 21 times 5 so that would be minus 105 over 20. and then plus 3 times 4 is 12 so 12 over 20. so if i sum negative 105 plus 12 i get negative 93. so this would be negative or minus 93 over 20. now let me scroll down and get a little room to get all the variable terms on one side i would just simply subtract 23x over 10 away from each side of the equation and of course to do that i need to have a common denominator so i would take negative four fifths times x and multiply this by two over two and that would give me negative eight tenths negative eight tenths x and then minus twenty three x over ten so when we look at this really we think about negative eight minus twenty three and that's negative thirty one so what's going to happen here this is going to become on the left side negative 31 over 10 and this is times x okay this is going to cancel over here and be 0. so this will be equal to negative 93 over 20. now how do i get my solution here let me scroll down a little bit in most cases we see something like this we see you know 5x equals 60 and we divide both sides of the equation by 5 that's the coefficient of x this cancels with this and i get x is equal to 12. but in this case we don't have something that looks like that but we really do i can still divide both sides of the equation by negative 31 over 10. but there's a quicker way to do that you multiply by the reciprocal so when your coefficient for x is a fraction you multiply by the reciprocal remember all i'm trying to do is get x by itself and multiplying something by its reciprocal makes it equal to 1. so if i multiply this by negative 10 over 31 the negatives would cancel and then 10 would cancel with 10. 31 would cancel 31. so this is 1 times x or just x is equal to i've got to multiply by the same thing over here so that's legal so times negative 10 over 31. we know already that the negatives would cancel and we know that if i think about 20 and 10 this would cancel with this and leave me with a 2. between 93 and 31 this would cancel with this and give me a 3. so i end up with x equals 3 halves just like i did okay when we looked at it by clearing the equation of fractions now in this particular case you could make the argument that it wasn't any faster to clear the equation of fractions but i think especially if you have kind of a simpler problem it kind of is so it's a personal preference whether you want to work with the fractions or you don't want to work with the fractions but in either scenario you're going to get the same answer all right for the next one i'm looking at 2 times the quantity x minus five thirds is equal to 11 thirds minus three halves x now if you have a problem okay where there's fractions involved you don't want to clear anything until you have removed all the parentheses and brackets all the grouping symbols that you have to deal with so what i would do here to start the problem multiply 2 times x that's 2x then minus you'd have 2 times 5 thirds that would be minus ten thirds then this equals eleven thirds and this is minus three halves x now it's very easy to clear the fractions now we come through here and we look at the denominators i have a 3 a 3 and a 2. so what is my lcd well it's just 3 times 2 which is 6. now if i multiply both sides of the equation by 6 i'm going to get what so 2x minus 10 thirds equals 11 thirds minus 3 has x multiply this by 6 as well so let's go ahead and crank this out so 6 times 2x is 12x 6 times negative 10 thirds so let's put minus 6 times 10 thirds and we know that the 6 would cancel with the 3 and give me a two two times ten is twenty so this would be minus twenty here and this equals we have six times eleven thirds so six times eleven thirds six would cancel with three and give me a two two times eleven is 22 so this is 22 and then last we have 6 times negative 3 halves x so minus 6 times 3 halves x and this is going to cancel with this and give me a 3 3 times 3 is 9 to be minus 9 x okay so now we're done with that and we have an equation without any fractions involved i want all my variable terms on one side all my numbers on the other so let me add 9x to each side of the equation so this is going to cancel become 0 and then also let me add 20 to each side of the equation so this will become 0 and 12x plus 9x is 21x and this is equal to 22 plus 20 which is 42. so one last step here to solve this i want to isolate x so i divide both sides of the equation by 21 the coefficient of x what's going to happen is this will cancel with this give me an x and then 42 over 21 is 2. so my solution set here contains one element just the number two all right so now let's move on and talk about equations when there's decimals involved so i have here that similarly when an equation contains decimals we can clear the decimals by multiplying both sides by the appropriate power of 10. so what does that mean well back in pre-algebra we learned that if you multiply by 10 or a power of 10 meaning something like 10 to the second power 10 to the third power 10 to the fourth power you know so on and so forth you're going to move your decimal point one place to the right for each zero in that power of 10. so if i have something like three times ten we all know this is thirty but really what i can do is i can start out with three put a decimal point at the end and i can just move it one place to the right because there's one zero there as a more advanced thing let's say i had three times one hundred thousand so i could start out with a three put a visible decimal point and then i've got one two three four five five zeros so this would go five places to the right so one two three four five one two three four five and so i would have here three hundred thousand and a lot of you know that three times a hundred thousand is three hundred thousand just another way to kind of think about doing the operation so when we look at equations with decimals i just need to look for the largest number of decimal places and then our power of 10 is going to need that number of zeros okay so it's just that simple so here's an example we have negative one point eight x minus two point zero six x equals negative two point nine x plus two point four so if you want to clear the equation of decimals and again this is something that you don't have to do it's optional you look for how many decimal places each number is going to have so this one has one this one has two this one has one and this one has one so if i want to clear the decimals from this equation i've got to focus on clearing with regard to the largest number of decimal places so that occurs here and this number right this negative 2.06 so if i take care of this one i take care of the rest so what power of 10 has two zeros well that's a one followed by two zeros or a hundred so if i multiply both sides of the equation by a hundred so i'd have a hundred times inside of parentheses negative 1.8 x minus 2.06 x this is equal to over here inside of parenthesis negative 2.9 x plus 2.4 and again this is times 100 let me scroll down a little bit so i use my distributive property 100 times negative 1.8 x so let me just write this out negative 1.8 x so this is going to move one two places to the right so it's going to be negative 180x then minus i'm going to do 100 times this and again i'm just moving this two places to the right so this and then this so i'd have minus 206x and this equals 100 times this so negative 290x then plus again i have this 100 times this so 240. now i've cleared the equation of decimals so now let me simplify over here and negative 180x minus 206x is negative 386x and this equals over here i have negative 290x plus 240. now let me move all the variable terms to one side all the numbers on the other so let me add 290x to each side of the equation so this is going to cancel let's scroll down a little bit and then negative 386x plus 290x is negative 96x so this is negative 96x is equal to 240 and at this point we know what to do right we have negative 96 that is the coefficient for x so i divide both sides of the equation by negative 96 and we'll get x is equal to if you punch that up on a calculator you get negative 2.5 now here's my rule of thumb if i have an equation that has decimals involved i use decimals as my answer if i have an equation with fractions i use fractions as my answer unless i get an integer whole number in this case it would not be incorrect to simplify the fraction and end up with negative 5 halves so if i wanted to i could factor 240 as 3 times 2 times 2 times 2 times 2 times 5. and then i could factor negative 96 as negative 1 times 2 times 2 times 2 times 2 times 2 times 3. so what's going to happen is this will cancel with this i'll have four factors of 2 that are going to cancel and i'll be left with 5 over negative 1 times 2 or negative 5 halves so either way you want to report your answer is ok right negative 5 halves or negative 2.5 it's the same value in the end so my solution set here i'm going to put negative 2.5 in and again right now i'm going to go back and show you that you get the same answer if you work with the decimals and you don't clear it all right so remember we have that x is equal to negative 2.5 or again it could be negative 5 halves doesn't matter so working with the decimals here if i have negative 1.8 x minus 2.06 x you would end up with negative 3.86 x and this equals negative 2.9 x plus 2.4 i would add 2.9 x to each side of the equation so that cancels so over here on the right i just have 2.4 on the left i'm going to have negative 0.96 x and then i want x by itself so i'm just going to divide both sides of the equation by negative 0.96 and what i'm going to get is that x is equal to negative 2.5 just as i saw over here now you might say man that was really quick why would i ever clear an equation of decimals i would agree with you on that but i'm teaching it because it does come up in your textbook you might have to show your work and you need to fundamentally understand what's going on when you get a section like that in your class but generally speaking if you're using a calculator it is a complete waste of time to clear an equation of decimals i can't think of any scenario where it's going to benefit in a lot of cases with fractions it does make it just easier if you clear the equation of fractions but with decimals in most cases it's not really a productive activity all right let's take a look at another example so i have 1.4 y minus 2.359 and this is equal to negative 0.607 minus 0.06 y all right so i have one decimal place here i have 1 2 three decimal places here one two three decimal places here and i have two here so the highest number of decimal places occurs here and here with three so the power of 10 necessary to clear the decimals from this equation would be a one followed by one two three zeros or the number one thousand so i multiply one thousand by both sides of the equation [Music] okay so i'm going to multiply 1000 times 1.4 and again the decimal point is going to move three places to the right so that would be a one a four and i would have two zeros behind the four so that'd be 1400 and then y then minus if i do 1000 times 2.359 this is going to move three places to the right and i have 2359 and this equals if i do 1000 times this negative zero point six zero seven this is going to move one two three places to the right and i'll have negative six hundred seven and then minus if i do a thousand times zero point zero six y i would have one two and then a third time after that so basically let me write this out because it's a little confusing so point zero six this would go one two put a zero at the end three times you'd end up with the number 60. so minus 60 and then times y okay so now that we have that nothing i can do to simplify on each side i want to move all the variable terms to one side all the numbers to the other so let's add 60y to each side of the equation that's going to cancel and let's let's add 2 359 each side of the equation that's going to cancel so what i'm going to have on the left side is 1460y and this is equal to negative 607 plus 2359 is 1752. so the last thing to do here is to divide both sides of the equation by 1460 and this will cancel with this on the left i just have y and this equals if you punch up 1752 divided by 1460 on a calculator you'll get 1.2 so y here equals 1.2 again in solution set notation inside my set braces i just list one element which is 1.2 all right so the last thing i want to talk about today when you're working with a linear equation in one variable you have three different types of equations that you're going to encounter now most of the time you're going to work with something known as a conditional equation this is an equation that has exactly one solution so let me just write this down conditional and this has exactly one solution so everything we've seen up to this point has been a conditional equation as an example let's say we looked at 5x plus 1 equals 31. this has a solution of x equals 6 because if i plug a 6 in for x i would get 5 times 6 plus 1 equals 31 5 times 6 is 30 so you get 30 plus 1 equals 31 and then 31 equals 31 same value on the left as on the right if i plug anything other than 6 in for x you're not going to get a true statement at the end you'll have some number equals some other number and that would be false so conditional equations have exactly one solution all right so the next situation we want to talk about is when we have an equation with no solution so we have here that an equation with no solution is referred to as a contradiction so i have an example of one of these and they're very easy to spot so if you look at this guy over here if we go to this equation we have negative five times the quantity negative three minus five z then minus four now this is equal to five times the quantity five z minus eight then minus ten so we'll see what's going to happen here is we're gonna end up with some nonsense you'll see what i mean by that so if i use my distributive property to start over here negative 5 times negative 3 is positive 15. then negative 5 multiplied by negative 5z would be positive 25 z now we're done with that we just have this minus 4 here so minus 4 and then this equals on the right side i'm going to use my distributive property so 5 times 5z is 25z then we have 5 times negative 8 that's minus 40. then we have minus 10. so let's continue to simplify on each side so 15 minus 4 is 11. so this side is basically 25z plus 11. then this equals on this side we have 25z and we have negative 40 minus 10 that's negative 50. now what do you see here if you were to pause the video and really think about why you don't have a solution what do you see well what you would see if you really thought about it is the fact that you have 25z here and 25z here if i've got the same thing on both sides of the equation i can just get rid of it because i can always go through and say okay minus 25z minus 25z subtracting the same thing from both sides of an equation this cancels and so does this but where does that leave us it leaves us with this nonsense 11 is equal to negative 50. okay let me write here this is nonsense okay this is a false statement so your variable drops out you're left with nonsense or false statement however you want to think about that you know that you have something known as a contradiction so if you get one of these the way you write your answer you can either put no solution no solution or if your teacher is adamant about you using solution set notation well then you'd say that your solution set is empty right it contains no elements so you use the symbol for the null or empty set right it looks like that or again you can use empty set braces any of these would be acceptable if you wanted to display your solution here all right so the other type of equation the last thing we would deal with is called an identity so an equation with an unlimited okay unlimited or you could say an infinite number of solutions is called an identity okay so infinitely many solutions so here's an example of that so let's say we're looking at six minus eight x so six minus eight x is equal to three times two is six three times negative x is negative three 3x and then you have minus 5x so negative 3x minus 5x is negative 8x and what you basically have here is the same thing on each side of the equation so no matter what value i choose for x doesn't matter what it is if i plug it in for x i'd end up with the same amount right because i'd have 6 minus that same amount no matter what i choose so in this particular case if you don't catch it right away essentially when you go to simplify let's say i subtract 6 away from each side of the equation that's gone and that's gone and then let's say i add 8x to each side of the equation that's gone and that's gone so you end up with 0 equals 0 which is true so you end up with a number equal to another number and the variable just drops out just gone okay so you can just put here that there's an infinite number of solutions or in solution set notation inside of set braces you can simply write all real numbers and there's more ways to kind of notate this but this would be the two easiest ways to write that answer on your test hello and welcome to algebra 2 lesson 5. in this video we're going to learn about converting a repeating decimal into a fraction so back in pre-algebra we learned how to convert back and forth between decimals and fractions but for most of you you're probably never shown how to convert a repeating decimal into a fraction so we'd start out with just asking what is a repeating decimal just in case you don't know so a repeating decimal is one that repeats the same digit okay repeats the same digit or pattern of digits forever so as an example let's say you had something like 0.2677 7 where the 7 is going to repeat forever so we can put the 3 dots after this 7 here to indicate that the 7 will continue on and on and on another way you can notate this you could write this as 0.267 with a bar over the 7. you put a bar over the digit or series of digits that's going to repeat forever now as another example of one of these let's just look at two more let's say i saw something like 0.43 where 4 and 3 repeated forever so then 4 3 4 3 4 three four three you know so on and so forth so again i can use the three dots or i can put 0.43 with a bar on top of the four and the three because that pattern of digits is going to repeat forever a 4 and then a 3 a 4 and then a 3. so on and so forth or as another example let's say i saw something that was kind of longer let's say i saw 0.39176 and let's say the 176 repeats forever so then 176 176 176 you know so on and so forth so i can put the three dots here or again i can write this as zero point three nine one seven six and again i can put a bar on top of the one the seven and the six the three digits that are going to repeat forever okay so let's jump into an example now and for a lot of you this is not going to be very difficult i can tell you that you just write down the steps as we're working through these problems and once you do your own practice problems after five or ten of these you're going to have this down packed it's going to be no issue for you in the future again this is very very simple so let's try to convert 0.97 where the 7 repeats forever again notice the bar on top of the 7 there so the very first thing that you want to do is set a variable like x or y or z or whatever you want to use equal to the value right the repeating decimal so let's just go ahead and use x and say x is equal to 0.97 where the 7 repeats forever and for the sake of this example i'm not going to write it like this i'm going to write it like this and it's just going to make it easier to see what's going on then the next thing i want to do i want to multiply i want to multiply both sides by 10 to the power of n and n is just going to be the number of digits in the repeating string now in this particular case i have one digit that's going to repeat forever so 10 to the power of in this case it would be 1 right 10 to the power of 1 is 10 right we all know that so if i multiply both sides of the equation by 10 what am i going to get on this side i'm going to get 10x on this side i would move my decimal point one place to the right and i would have 9.7 7 7 7. you can write as many sevens as you want just making it clear that the seven repeats forever and go ahead and put your three dots at the end now at this point i have not gotten rid of the repeating decimal it's still right here right that's seven still repeating forever and ever and ever but the step we're going to do now is going to get rid of it so what i'm going to do is i'm basically going to subtract the original equation from this new equation i just got by multiplying both sides by 10. so you might say how is that legal remember these two sides are equal to each other on this side i'm going to subtract away 0.9777 with the three dots there and what's going to happen is this part right here is going to go away it's just going to cancel 7 minus 7 is 0. so all this is gone and i'm left with 7 minus 9 i'd have to borrow to do that so i'd borrow from this this would become eight this would become 17. 17 minus nine is eight put my decimal point eight minus zero is eight so this side is now 8.8 so let's put this equals over here i'm going to subtract away x and remember this is legal because x is equal to this value and i'm basically subtracting away the same thing from each side of the equation so 10x minus x is going to give me 9x now let me do one more thing here let me scroll down and get a little room going i'm just going to multiply each side of the equation by 10 again to clear this decimal point i'm going to multiply this side by 10 and this side by 10 and i'm going to get that 90 x is equal to 88 right this goes one place to the right so now we have something very easy to solve we divide both sides of the equation by 90 the coefficient of x and this is going to cancel with this and i'm going to be left with x is equal to 88 over 90 which can be reduced each is divisible by 2. 88 divided by 2 is 44 and 90 divided by 2 is 45. so there is your solution x is equal to 44 over 45 and remember if we scroll back up here in the beginning we said x was equal to this so if i replace x with this value of 44 over 45 these two have to be equal so that's your value 44 over 45 is equal to zero point nine seven seven seven seven where the seven repeats forever and ever and ever and if you don't believe me punch it up on a calculator if you punch up 44 divided by 45 you will get 0.97 where the 7 repeats forever now your calculator will do one of two things it may truncate so it may stop displaying sevens it's just not display anymore because it just can't or it may round at the end so you might see the final digit as an eight okay but that's not the actual value that seven is going to repeat forever and ever and ever okay so your brain might be a little fuzzy from the first one but hopefully on the second one here you're going to catch up and kind of get in the flow of things so we have 3.895 where the 9 and the 5. that series of digits there is going to repeat forever so we would see something like 3.895 [Music] 9 5 9 5 right so on and so forth and again i want to set this value equal to a variable like x or y or z let's go ahead and just use z kind of spice things up i know we always like to use x and then the next thing i want to do is i want to multiply both sides of the equation by 10 to the nth power where n again is the number of digits in the repeating string so in this case i have one two digits there so n is going to be a two so 10 to the second power is a hundred so one followed by two zeros so this over here if i multiply both sides by 100 this would be 100 z and this would be 3 8 9 point right because this is going to move two places to the right so you'd have 389 point and now i'd have a five and then i'd have my repeating this so i would have nine five nine five and i could do another one nine five and then the three dots okay so now we know that we want to subtract this value from both sides of the equation so i'm going to subtract away 3.8959595 where the 9 5 repeats forever and i'm going to do the same thing over here but remember on this side i want to subtract away z and again that's legal because this is equal to this so i'm subtracting away the same thing from both sides of the equation so over here i'm going to have 99z and over here if i do my subtraction i know from this point on it's going to cancel right this is going to cancel itself out then over here i'm gonna borrow this will become eight 15 minus eight is seven put my decimal point eight minus three is five then i have my eight and i have my three all right let's scroll down get a little room and now let me just clear this decimal from the equation we multiply both sides of the equation by 10 and i'm going to get 990 z is equal to 3857 now i want to just divide both sides of the equation by 990 so z is by itself divide this by 990. all right so let's write this as z is equal to i can factor 3857 into 7 times 19 times 29 and then i can factor 990 as 2 times 3 times 3 times 5 times 11. so between the numerator and denominator i don't have any common factors other than one so there's nothing i can really do my answer is going to be that z is equal to 3857 over 990 and again if we go back up we see that z was equal to this so i can replace this side right here with this i can say that 3857 over 990 is equal to 3.895 and in more convenient form i can put the bar over the 9 and the 5 to show that 9 5 that series of digits is going to repeat forever so again punch this up on a calculator 3857 divided by 990 and you will see this but again your calculator will either truncate and stop at some point with a 5 or it will round up and you'll see a 6 as the very final digit let's take a look at it all so we have 0.104357 where the 4 3 5 7 repeat forever and ever and ever so again let's assign this to a variable let's use y this time so let's say y is equal to again 0.10 i'm going to write it more compact form now that we understand what's going on so 4 3 5 7 with the bar over the 4 the 3 the 5 and the 7. so i'm going to multiply both sides of this equation by 10 raised to the power of n again where n is the number of digits and the repeating string so in this particular case we have one two three four digits so i'd multiply by 10 to the power of 4 so 10 to the power of 4 which is a 1 followed by 1 2 3 4 0's so i'd have 10 000 y is equal to this would move one two three four places to the right so i would have 1043 point i'd have five seven and then what i have is the repeating pattern starting so i would have 4 3 5 7 and i'll put a bar over the 4 the 3 the 5 and the 7 and again now i want to subtract this value away from each side of the equation so if i subtract that away over here i'd have point one zero four three five seven with a bar over the four the three the five and the seven and we know that this part would cancel right that's gone let me scroll down a little bit and over here seven minus zero is seven five minus one is four put my decimal point there bring everything else down three four zero one and this would be equal to over here i'm going to subtract away y remember y equals this so i'm subtracting away the same thing from each side of the equation so that's going to give me 9999 y let me put minus y here now the next thing is i want to just clear this decimal point so let me multiply by a hundred so i can move this two places to the right so that would give me 104 347 then on this side i'm multiplying by 100 as well so it would give me 999 900 y and then for my final step here i'm simply going to divide both sides of the equation by 999 hundred and i'll get my final answer here which is that y is equal to one hundred four thousand three hundred forty seven over 999 900. now 104 347 is a prime number so we're not going to be able to do anything to reduce this fraction so let me just erase some stuff here so we come back up to the top and since y is equal to this i can replace y with this and put 104 347 over 999 hundred again this is equal to what we originally started with we have zero point one zero with the four three five seven repeating forever all right let's take a look at one final problem here so we have 0.8725 or 8725 those three digits repeat forever so again i want to set this equal to a variable so let's pick x again so x equals 0.872 again where there's a bar over the 7 the 2 and the 5. then let's multiply both sides of the equation by 10 raised to the power of n where again n is the number of digits in the repeating string so in this case that's three i have one two three digits so i'd multiply by ten to the third power or one thousand so i'd have one thousand x is equal to this would move one two three places to the right so i would end up with 872.5 and then i would start my 725 repeating forever so let me put the bar over this now what i want to do from here i want to subtract this value from each side of the equation so over here i'm going to subtract away 0.8725 where again there's a bar over the 7 the 2 and the 5 and then over here because x is equal to this i'm going to subtract away x i'm subtracting away the same thing from each side of the equation so on the left i get 999 x on the right this is going to cancel and i'll borrow here 15 minus 8 is 7 and i'll bring down my decimal point and then i'll bring down a 1 a 7 and an 8. so let's scroll down and get a little room going and i'm going to move this to the right one which means i'm going to move this to the right one just multiplying both sides by 10. so i'm going to have 9990x is equal to 8717 now as a final step i'm just going to divide both sides of the equation by 9990 and i'll get that x is equal to 8700 over 9990 now if i factor this number right here i would get 23 times 370 knot and if i factored 9990 i would get 2 times 3 times 3 times 3 times 5 times 37 so they don't share any common factors other than 1 so this would be as simple as we can make it we have 8717 over 9990 and let me just copy this real quick and again because x is equal to this we can say that this value here 8717 over 9990 is equal to 0.8725 where the 7 2 and 5 repeat forever and ever and ever hello and welcome to algebra 2 lesson 6. in this video we're going to learn about solving for a specified variable so in many situations we are given equations with no numbers involved now in this lesson we're going to talk about some basic formulas where we have equations where the variables describe a relationship let's start by looking at the distance formula and you'll probably recall this formula from your algebra 1 course because we use it on our motion word problems so we have that distance which is generally represented with d is equal to the rate of speed which we represent with r times the amount of time traveled which we represent with t now we've all been on a road trip before and let's say you're on a road trip and you're traveling at 80 miles per hour and you do this for three hours well you can plug that into the distance formula and you're going to realize that you went 240 miles so again all i would do is i would plug in i'm going 80 miles per hour that's my rate of speed my time for this trip is three hours so three hours now forget about the units for a second just multiply the numbers if i multiply 80 times 3 i get 240 and now you can think about the units if i'm traveling in miles per hour then i went 240 miles that's going to be my distance so we can see that this distance formula is very very straightforward but one thing that we can do is we can manipulate this formula and we can solve for the rate of speed or we could solve for the time traveled so let's say i came up with a different scenario and i had different inputs let's say that in this scenario that i gave you a distance that was 550 miles so that's how far you traveled and let's say the amount of time it took to go 550 miles was 11 hours so your time is 11 hours now in order to get a solution here for r because the rate of speed right now is unknown what i would want to do is solve this formula for r so what i'm going to think about here is just treating t as if it's the coefficient of r right if i saw something like let's say 12 is equal to 4r if i want to isolate r we know we divide both sides of the equation by 4 and we get 3 is equal to r well it's no different here if i want to isolate r all i'm going to do is divide both sides of the equation by t it doesn't matter that it's a variable in this case so this will cancel with this and i'm going to have that r or the rate of speed is equal to the distance over the amount of time traveled so now that i have it solved for r what i can do let me just kind of rewrite this what i can do is i can plug in my inputs i can plug in 550 miles here and i can plug in 11 hours here and if i think about the numbers there 550 divided by 11 is going to give us 50. so i know my r there is 50. and when we think about the rate in terms of the units we're talking about miles per hour right so miles per hour we have miles here we have hours here so that's going to be miles per hour so my rate is 50 miles per hour now let's say i gave you another scenario let's erase this real quick and let's say that i now gave you a distance let's say it's 300 miles and let's say i gave you a rate let's say the rate is 60 miles per hour and i asked you for a time so now this guy is the unknown so i can solve the formula for time so if i want to solve for time i just divide both sides of the equation by r and i'm going to get that time is going to be equal to distance over rate so once i have that i plug in so i'm going to plug in for distance that's 300 miles i'm going to plug in for the rate of speed that's 60 miles per hour and so 300 divided by 60 is going to give me five so t or time is five and again we have miles and we have miles per hour so we're talking about five hours here so the time is going to be 5 hours so that's just one example of how you have an algebraic formula and you can manipulate it for what you need based on the problem that you're given all right so let's think about another common formula this is the perimeter of a rectangle this is more used in geometry i know by the time you get to algebra 2 you've taken a geometry course maybe you haven't but the perimeter of a rectangle is given as p which stands for perimeter which is equal to two times the length plus two times the width so as an example let's say and this is your width here and here this is your length here and here let's say you have a length that is equal to let's say 8 feet and let's say you have a width that is equal to let's say 4 feet if i told you to calculate the perimeter you just need some basic arithmetic right you're going to plug in so the perimeter is going to be equal to 2 times my length is 8 feet so i'm just going to put an 8 in i'll deal with the units in a minute plus 2 times my width i have 4 feet so i'm going to put 4 in so my perimeter p is equal to 2 times 8 which is 16 plus 2 times 4 which is 8. 16 plus 8 we all know is 24. so the perimeter in this case we think about the units as well would be 24 feet okay 24 feet now that's easy to calculate but what if i gave you a different scenario let me erase this real quick and let's say that i gave you a scenario where i told you what the perimeter was let's say the perimeter is now 40 feet and let's say i gave you one of these let's say i gave you the width and i said that the width is going to be 7 feet well how would i find out what the length is so my l here is unknown well i can solve the formula for length and how do i do that well i want to get this term by itself to start so that means i would subtract 2w away from each side of the equation this would cancel and on the left i would have p minus 2w is equal to 2l now to get l by itself i have a coefficient i have that 2 there and so i'm going to do the same thing i've been doing i'm just going to divide both sides of the equation by 2 so that l can be by itself so this will cancel with this and i'll have l is equal to i'll have p or perimeter minus 2 times w or width over 2. now all i need to do once i've solved for length is just plug in so i know the perimeter is 40 feet so let me scroll down get a little room going so for p i'm going to plug in a 40 and i'll deal with the units in a minute minus 2 times for the width i'm going to plug in a 7 and this is over 2 and this should give me my length so 2 times 7 is 14 so let's just erase this and put 14. 40 minus 14 is 26 so i'd have 26 over 2 which is 13. so my length here is 13. and again you want to use the units so it'll be actually 13 feet right 13 feet now you can check that with the original formula let me kind of erase everything here and if i go to the original formula if i plugged in a 13 here and i plugged in a 7 here when i get a perimeter of 40 and again we're dealing with feet in terms of the units so 2 times 13 is 26 plus 2 times 7 which is 14 yeah that would work out 26 feet plus 14 feet would in fact give me 40 feet so that works out now as another example what if i am given the perimeter and i'm given the length well now i need to solve for the width so let's say this is unknown let's say i give you a length that is 5.5 feet and i give you a perimeter that is 16 feet well now i want to solve the equation for w or the width and it's the same thing it's just a little bit different i'm going to subtract away 2l from each side of the equation so that's going to cancel i'll have p or the perimeter minus 2 times l or the length is equal to 2 times w or the width again divide both sides of the equation by 2 so you can isolate w and what i'm going to have is that w or the width is equal to perimeter minus 2 times the length over 2. so if i just plug in my inputs here if i plug in a 16 here and i plug in a 5.5 here what's that going to give me well i would have 16 minus 2 times 5.5 is 11. so this would be 11. so you'd have 16 minus 11 which is 5 so you'd have 5 halves or 2.5 as a decimal and don't forget the units so this would be 2.5 feet so my width is 2.5 feet let me erase this and we'll put 2.5 feet so again another great example of how we have an algebraic formula we have more than one variable that's involved and we can solve for whatever variable that we want it's just based on the information that's given in the problem and what we need to find all right so here's another great example we have fahrenheit to celsius so if you live in america like i do all of the temperature readings that you get are generally going to be in fahrenheit i know if you take a science class you get a lot of stuff in celsius and you know things like that but in general when we report a temperature here in america we get fahrenheit now if you're in another part of the world you might get it in celsius so you need to be able to convert back and forth so this is the formula if i give you a temperature in celsius you plug it in you multiply by nine fifths you add 32 you get a temperature in fahrenheit so we can try this out let's say i wanted to convert 25 degrees celsius to fahrenheit and i'm not going to write that i'm just going to put to f so what would i do i'd plug in a 25 here plug in a 25 here so my fahrenheit is going to be equal to we'd have nine fifths times 25 plus 32 we know that this would cancel with this and give me 5 5 times 9 is 45 so you're looking at 45 plus 32 and we can do that in our head 5 plus 2 is 7 4 plus 3 is 7. so 25 degrees celsius would be 77 degrees fahrenheit so those two are equal to each other now let's say that i gave you a temperature in fahrenheit and i wanted to go to celsius well i would need to solve the formula for c so how would i do that well i want to isolate c right now f is isolated so let's start out by just isolating this and we'll subtract 32 away from each side to do that so i'll have f minus 32 is equal to nine fifths c and what i want to do here is i have a fraction as the coefficient for c so when i have a fraction i can just multiply by the reciprocal so i can multiply by five ninths and of course what i do to one side i've got to do to the other but this is going to cancel itself out right this is going to be 1. so on the right side i have c on the left side i'd have 5 9 times f minus 5 9 times 32. now of course i can write this more cleanly by saying that i have c is equal to i just switch with sod it's on i like my variable to be on the left over here i would have 5 f minus 5 times 32 is 160 so 160 and this is over the common denominator of 9. now i just showed you that 25 degrees celsius was 77 degrees fahrenheit let's try it out in this formula so let's plug in a 77 here and see if we get 25 so 5 times 77 is 385 so what i'd have is c or celsius is equal to 385 minus 160 over nine if i subtract 385 minus 160 i get 225 so i'd get 225 over 9 and we know this is going to be equal to 25. so again when i go from 77 degrees fahrenheit back to celsius i get 25 degrees celsius so let me rewrite this formula c or celsius is equal to five f minus 160 over nine and let's try just one more we'll go back and forth let's say that i had 37 degrees celsius and i want to convert this to fahrenheit so how many degrees in fahrenheit this is actually a very popular temperature that deals with the human body so if i'm going from celsius to fahrenheit i want to plug into this formula here because it's solved for fahrenheit so what i can do is i can plug in my 37 here and i would have 37 times nine fifths well 37 times nine is 333 so what this would give me is f is equal to 333 over 5 plus you have 32 there now to get a common denominator i know if i multiply 32 times 5 i get 160 so this would be 160 over 5. and so 333 plus 160 is 493 so this would end up being 493 over 5 but in decimal form you'd end up with 98.6 98.6 so 37 degrees celsius is 98.6 degrees fahrenheit now this temperature is significant because for most people a body temperature of 98.6 degrees is normal right if i take a thermometer put it in my mouth under my tongue i'm looking for something around 98.6 degrees so let's just do one more here and all i'm going to do is just convert 98.6 degrees back to show you that this formula does indeed work so let me erase this and i'm going to plug in 98.6 here and 98.6 times 5 is 493. so 493 minus 160 would be 333. so this would be 333 over 9 which we know is 37. so again convert this back and forth you see that again 37 degrees celsius is 98.6 degrees fahrenheit so you can see how solving this for fahrenheit with f and then solving it for celsius allows you to attack any problem they throw at you if they give you a temperature in celsius and ask for fahrenheit you use this one if they give you a temperature in fahrenheit and ask for celsius you use this one so it's an ability to go back and forth and again it's just based on the information given to you in the problem as far as what you want to solve for all right for the last problem let's look at one that's not a generic formula let's just look at something you might see on your sats so you might get something that's like this 2b plus az equals z minus 3g you might say what is this they might say i want you to solve this for z i want you to solve this for b i want you to solve this for g so on and so forth let's just do one and let's just say we're going to solve it for z so let's solve for z so how do we do that well i've got to isolate z on one side of the equation that means i'm going to isolate all the terms with z in it on one side get everything else on the other so i can subtract 2b away from each side of the equation and i could subtract z away from each side of the equation so this is going to cancel this is going to cancel and i will have a z minus z is equal to negative 2b minus 3g now how do i get z by itself this is tricky if you don't have any experience in algebra 1. if you've just gotten into algebra you might go what in the world are we going to do well for most of you that took algebra 1 you know how to factor if you didn't take algebra 1 let me just kind of explain briefly what we're doing here i am able to pull something that's common outside of a set of parentheses so if z is common to each term here i can place it outside of a set of parentheses when i do that what's going to be left in the parentheses is what was multiplying z so here a is multiplying z so it's there then minus here you might say well nothing's multiplying z well yeah you always have a phantom 1 there 1 times z is just z so really if i factor out a z i end up with a quantity a minus 1 and you can check that through multiplication z times a is a z then minus z times 1 is z so you're back to what you got so i just wrote this in a different form and another way you can kind of think about this let me kind of erase this real quick if i had like terms let's say i had 6z minus 3z okay for example here 6z minus 3z would be 3z because i subtracted this minus this 6 minus 3. in this particular case i don't know what a is but if a was 5 for example i'd have 5z minus 1z that would be 4z so if a was 5 5 minus 1 would be 4 i'd have a coefficient out there but i don't have the luxury of that because in this particular case i'm just given a and i don't know what that is so this is the best we can do we factor out the z and we leave it in this form and we say it's equal to negative 2b minus 3g you might say well how do i get z by itself well just like if i had 4z is equal to something i'd divide both sides by 4. well i'm going to do the same thing here i'm just going to divide both sides by this quantity a minus 1. that is what's multiplying z so that's what i'm going to do so i'm going to divide over here by a minus 1 and this is going to cancel with this the quantity a minus 1 over the quantity a minus 1 is 1. so i'm going to have z is equal to i'll have negative 2b minus 3g over a minus 1. okay and that's your answer for z so if you get something like that on a test says solve for z that's exactly what you would do now you might get different scenarios you might have a problem that says solve for z solve for a solve for b solve for g you might have to solve for each one in the end you're just going to isolate whatever they want you to solve for on one side of the equation and everything else is on the other hello and welcome to algebra 2 lesson 7. in this video we're going to have an introduction to applications of linear equations so one of the most difficult things you're going to do in algebra 1 and algebra 2 or even a college algebra course is learn how to set up and solve word problems i know if you've done them in algebra 1 it's it's something where a lot of students just struggle because you have to generate your own equation before you even struggle to solve it but one of the things that can help us if we just start out with some very very easy problems and we learn how to correctly interpret some of the language that we're going to encounter with the word problem i think it's going to put us a step ahead when we start seeing stuff that's very challenging so let's begin by looking at common addition phrases and then we'll move into subtraction and multiplication and division so the first one i have here is the sum of so it's obviously telling us to add we know the sum is a result of adding so if i see something like the sum of let's say a number and 5. how could i represent that well a number is just some unknown number we can use a variable like x or y or whatever you want to use to represent that so if i reread this as the sum of x and 5 that would be x plus 5. then you have something like more than if i saw a phrase that said 3 more than a number so let's say i saw a 3 more than a number and again when we see a number we can use a variable let's use y this time to represent that so if i saw 3 more than y that's just y plus 3 or 3 plus y remember addition is commutative so when you translate stuff with addition the order is not important so i could put 3 plus y or y plus 3 and it's completely irrelevant because i would have the same answer either way what about just plus right the plus that we often hear with addition well i could say something like negative 7 plus a number so i could say negative seven plus a number and let's use z to represent a number so i would have negative seven plus z then we also see something like added two so if i said a number added to negative two so let's do a number added to negative 2. what would that be well i can use again a variable to represent a number let's use i don't know q for example so if i said q added to negative 2 then i would have q plus negative 2 or again i could put negative 2 plus q again addition is commutative so the order is not important what about something like increased by if i saw a number increased by 17. so a number increased by let's say 17. well again a number let's use r in this example so let's say r increased by 17. so we just have r and then if i'm increasing and i'm adding so i'm just adding what it's increased by which is 17. so this is r plus 17. so you can see with addition it's very very easy the translation is straightforward and remember when you're adding things the order is not important so let's talk a little bit about some phrases for subtraction so we have less than so let me scroll down a little bit for less than let's use the example of 6 less than a number so let's put 6 less than a number and i want you to pause the video for a second let's just use x to represent a number and when you pause the video i want you to try to write this phrase now hopefully you tried this and i want to walk you through this real fast 6 less than a number so that means i have some number and i represented that with x and i want 6 less than that so i start out with this and i subtract away 6. okay if i put if i put 6 minus x that would be wrong this is wrong this would be 6 less a number okay with subtraction the wording is important okay because subtraction is not commutative if i have 1 minus 2 that's not equal to 2 minus 1 right i can't just flip the order there it's not like addition whereas if i have 1 plus 2 that's the same as 2 plus 1. okay again subtraction is not commutative now on this one where we have less i can show you let's say i had 6 less a number well in this case this translates to literally 6 minus and then whatever we use for our variable in this case let's use x again so this is 6 minus x so again notice the difference between the two in this example we have 6 less than a number implying that we start out with the number which is x and then we're 6 less than that so we just subtract away 6. in this example it's 6 less just less a number so 6 less a number which would represent it with x so again very important that you understand what you're being asked take your time when you're reading these things especially at first and ensure that you are translating them correctly because if you don't most of the time you will get the wrong answer all right so next we see decreased by so as an example let's say i have a number so a number and it's decreased by let's say 8. so let's say the number here we represent that with y so y decreased by 8 would simply be y and then it's decreased or we're subtracting away eight so very very simple then as another one let's say we see minus we know minus means subtract so let's say i saw four minus a number very straightforward if i use let's say z to represent a number then that's just four minus okay minus z all right for the last one for subtraction let's look at subtracted from so let's say as an example i say a number so a number subtracted from i don't know let's say 11. so a number let's represent that with i don't know let's go with q again so let's say q subtracted from 11. so if it's subtracted from 11 that means i'm starting out with 11 and from that i'm subtracting this number q so that's 11 minus q and again this is one of those ones where it's very easy to make a mistake if you put q minus 11 again that's not going to be right you need to make sure that you translate this as q subtracted from this is the key thing here it's subtracted from 11 so that means i have 11 and then i'm subtracting away q alright so now let's look at multiplication and with the phrases for multiplication again multiplication and addition are commutative so we're not going to have the same issue with subtraction if i have 6 times 3 that's the same as 3 times 6. so the order with multiplication is not important so we start out with something very simple like times so if i have let's say 9 times a number and with a number i can just represent that with let's say x again so 9 times x would just be 9x the next one is also obvious you have multiplied by so if you saw something like a number multiplied by let's say negative 7 with a number i could represent that with anything let's say r so let's say r multiplied by negative 7 that's negative 7 times r which we write negative seven are next we see of and we generally see of when we're talking about fractions right you might see something that says one third of a number so of let's put one of a number and let's represent a number with let's go with x again so one-third of x just means one-third x it'd be like if your paycheck was three dollars and i said give me one third of your three dollars i would get a dollar all right the last one we're gonna look at with multiplication is product of and for this one we could just say the product of a number and let's say negative 15. so the product of a number which i'll represent with let's say y and negative 15 would be negative 15 times y or just negative 15 y all right lastly let's look at some phrases for division again remember division is also not commutative if i have you know 10 over 2 that's not the same as 2 over 10 right these are not equal to each other so when we look at phrases for division we start out with the quotient of so let's say i have the quotient of seven and a number so this would tell me that i have the quotient of seven seven is going to be the numerator and a number a number is going to be in the denominator and i'll represent a number with just x so this would translate into 7 over x the other one we really see with division a lot is divided by which obviously makes sense and you see something like a number divided by let's say negative 2. now a number let's represent this with z so a number so z divided by negative 2 would just be z over negative 2. and again when you encounter phrases with subtraction and division you have to be extra careful to read your wording and interpret it correctly all right so the last thing we want to know before we start looking at some problems is means equals so that's the translation when we're looking at a word problem all right so let's look at some sample problems so here's the first one we have five times a number plus seven times the number is 24. find the number so the first thing i like to do when i encounter these is i represent a number or the number or whatever it is that's unknown with a variable so let's line this out and just put x there let's line this out and just put x there so 5 times x is 5x plus means to add 7 times x is means equals and then 24. very very simple once you do that so i've just translated this to this and now that i have an equation it's very very easy to solve and find the number okay so 5x plus 7x we combine like terms there we get 12x and this equals 24. we divide both sides of the equation by 12 and we get that x is equal to 2 or again using solution set notation we say that this set contains the element 2. now when you solve a word problem you generally want to answer with a nice little neat sentence so we can just make something really simple and say the number is 2 or we can just put the number is 2 like this either way all right let's take a look at another one so we have a number less twice the number is negative five find the number so again everywhere i see something like that i'm going to replace it so let's just line this out and put x or y or z or whatever you want to use so x less twice the number so let's line this up and put x is negative 5. so let's put equals here okay so now i have x less so i know that that means i have x and then i'm subtracting away because i see less twice x so that's 2 times x and then equals for is negative 5. again once you kind of label things like this it's very easy to go from this to this okay so now i have x minus 2x that's negative x and this equals negative five if i multiply both sides of the equation by negative one or divide by negative one whatever you want to do there this cancels with this and i have x is equal to negative five times negative one is five so find the number the number is 5. okay let's look at one that's a little bit more complicated now so when eight is multiplied by the sum of and then we have a number times seven and negative eight the result is we have six times the number plus 36 and then we just want to find the number there's a few things in here that can really trip you up if you've never done any of these problems so the first thing is that we're going to be using our distributive property here i want you to recall if i have something like 5 times let's say i just write 2 plus 3 like this i would do 5 times 2 plus 5 times 3 and that gives me 10 plus 15 which equals 25. i know that if i combine 2 and 3 through addition i'd have 5 5 times 5 is 25. in the case where we have a variable involved we know that we could say we had the sum of let's say 2x plus 3. now i can't combine these because they're not like terms so in this case i do 5 times 2x and i get 10x plus 5 times 3 which is 15. so if i had told you i have 5 times the sum of 2x plus 3 you would have to write it this way because 5 has to be multiplied by the 2x and the 3. i can't just write it like this which is a common mistake right now i'm just multiplying 5 times 2x i'm forgetting about the 3. so it's important to understand that before we start trying to attack this problem so again when 8 is multiplied by the sum of a number and i'll replace this with x and it's being multiplied times seven so this would really be what let's just write this whole thing as 7x makes it nice and simple so let me read when eight is multiplied by the sum of seven x and negative eight so eight is multiplied by the sum of seven x and negative eight seven x plus negative eight or seven x minus but it's important that 8 be multiplied by 7x and also negative 8 which is why i used parentheses here okay very important just like what i just did over there all right so once we get past that we say the result is is so that means equals and then we have 6 times the number we represented the number or a number with x so we have 6 times x which is 6x then plus 36. so again once you start doing this enough you'll recognize what's going on and you'll be able to translate these very quickly the key trap here would not be putting parentheses around the sum of 7x and negative 8 right so 7x minus 8. so 8 times 7x is 56x and then 8 times negative 8 is minus 64. we have this equals 6x plus 36. so now i want all my variable terms on one side all my numbers are on the other so minus 6x minus 6x plus 64 plus 64. so this will cancel and this will cancel so we'll have 50x is equal to 36 plus 64 is gonna give us a hundred we divide both sides of the equation by 50. and we get x is equal to 2. so when we answer this find the number again we can just say the number is 2. all right let's take a look at one more problem so the quotient of negative 11 and 3 is added to 3 times a quantity the result is the number less the quotient of 8 and 3. find the number again these seem kind of difficult they seem kind of really wordy but if you go step by step it's not that bad so the quotient of negative 11 and 3 that's negative 11 divided by 3. very simple so is added 2 so is added 2 just means plus 3 times a quantity so what can we say we can say a quantity is just x so 3 times x is just 3x the result is is means equals the number which we said was x then we have less which means to subtract so we're subtracting away then we have the quotient of 8 and 3 so that's 8 over 3. so again it looks complicated but once you go step by step it's really not so again we just want to find the number so let's scroll down a little bit and i'm going to multiply both sides of the equation by 3 just so i don't have a denominator involved 3 times negative 11 thirds is negative 11 plus 3 times 3x is 9x this equals 3 times x that's 3x minus 3 times 8 thirds would be 8. again i want all my variable terms on one side all my numbers on the other so i'm going to add 11 to both sides of the equation and i'm going to subtract 3x from both sides of the equation and so this is going to cancel and this is going to cancel 9x minus 3x is 6x negative 8 plus 11 is 3. if we divide both sides of the equation by 6 we get that x is equal to 3 over 6 is basically one half or you could say 0.5 so the number is one half hello and welcome to algebra 2 lesson 8. in this video we're going to continue to learn about applications of linear equations all right so in the last lesson we covered how to translate some basic phrases into related mathematical expressions we then looked at some extremely basic word problems so here in this lesson we're going to take the next step and we're going to look at some more challenging problems now fortunately most word problems can be broken down into a type of problem so in this lesson we're going to look at three different types that you're going to encounter in your algebra 1 or your algebra 2 course so let's begin by looking at a general six step method for solving a word problem and this method is generally going to pertain to solving a word problem that involves setting up and solving a linear equation in one variable all right so let's get started with this so we have solving a word problem so the first step and the most obvious step is just to read the problem and determine the question or it could be questions to be answered so it's very important that you take your time and read anything you have extremely thoroughly okay make sure you understand what they're asking you to do the next step is to assign a variable to represent the unknown now in some cases you're going to have more than one unknown in this scenario especially when you're in a chapter where you're solving linear equations in one variable you're going to express any other unknowns in terms of that first unknown that you assigned a variable to and that might be a little confusing with me just saying it but once we get to an example i'll show you what i mean now the next step is to write an equation this can be kind of challenging especially if you don't have a lot of experience with word problems once you go through and you solve enough problems especially since the type of problem doesn't vary that much you're going to get this right away then once you've gotten through this step here you're pretty much home free you know how to solve a linear equation one variable so you're just going to be doing that then you're going to write the answer in terms of the question or questions asked so you make a nice little neat sentence and then the last thing the most important is to check the answer using the words of the problem and you want to make sure that your answer is reasonable if you got a question that asks how many people are on a bus and your answer is negative 3.7 you know you probably did something wrong right you can't have a negative amount of people on a bus so you want to make sure again that the answer is reasonable in terms of the problem that you were given all right let's go ahead and look at the first example so this example deals with finding the unknown individual amounts when the sum of those amounts is given so a very common problem for algebra 1 and algebra 2. so marcy and stephen are siblings marcy is four years older than stephen if their combined age is 34 how old is marcy how old is stephen so we've read the problem i think it's a very simple problem so most of you will understand that we're being asked to find the age of marcy and the age of stephen so that's all we need to find now the next thing is to assign a variable to represent the unknown now in this case again we have two unknowns we need to know how old is marcy and how old is steven so i can pick one of those it doesn't matter which one and i can let a variable like x represent their age so i can just say let x be equal to let's go ahead and say steven's age stevens h and it doesn't matter if i put x equal to marcy's age or if i put x equal to stephen's age and i'll just do it both ways just so you can see that it comes out the same way so if x is equal to stephen's age then how would we model marcy's age well we look at the problem again and it says that marcy is four years older than stephen so if x represents steven's age then x plus four would represent marcy's age now our next step and probably the most difficult one for us is to write an equation so how do i make an equation well think about the information you're given it tells you right here if their combined age is 34. well i've represented steven's age with x so x this is stephen's age and i've represented marcy's age with x plus 4 that quantity this is marcy's age and if i sum these two amounts together it tells me here their combined age is okay or equal to 34. this is their combined h so i've set up an equation and now i can just solve it so let me scroll down get some room going so if i have the equation x plus the quantity x plus 4 and this is equal to 34 i know i can just remove parentheses here this would be 2x plus 4 equals 34. i would subtract 4 away from each side of the equation i would get 2x is equal to 30 divide both sides of the equation by 2 and i'll get x is equal to 15. when we're solving a word problem we don't just put x equals 15 and then turn in the paper and say 100 percent it doesn't work that way we have to answer in terms of the problem now we said that x was stephen's age so that means that if x is 15 stephen is 15 years old so let me erase this and let me write that stephen stephen is 15 years old while marcy is and how old is marcy well marcy's age is x plus 4. if x is 15 i'm just plugging in a 15 here i'd have 15 plus 4 or 19. so while marcy is 19 years old okay so let's erase all this and now let's check to make sure that this answer is reasonable so if i go back up here all you need to do is go back through the problem marcy and steven are siblings so that's not really relevant to us marcy is four years older than stephen if i look at this 19 is 4 more than 15 so that checks out if their combined age is 34 so 15 plus 19 is 34 so everything checks out based on the information in the problem we can say with a hundred percent accuracy that we know that steven is 15 years old while marcy is 19 years old now one thing i want to show you real fast because a lot of you will say well what if i would have put x as marcy's age to begin with okay let's do that let's let x equal marcy's age and then we would have what if marcy is four years older than stephen how do we properly represent stephen's age well marcy's age which is represented with x i'd have to subtract four away now to get stephen's h so then x minus four that quantity would be stephen's age as an example we already know that marcie's age is 19. if i take 4 away from 19 i'd get stephen's age which is 15. and then i do the same thing to set up my equation i'd have x plus the quantity x minus 4 and this equals 34 so i would get 2x minus 4 equals 34 add 4 to each side of the equation i'd have 2x is equal to 38 divide both sides by 2 and i'd get x equals 19. you might say aha x equals 15 before well remember x now is representing marcy's age we change that up and marcy is 19 right while marcy is 19 years old to get stephen's age now i subtract away 4. 19 minus 4 would give me 15. so again either way you set this up you're going to get the correct answer just have to make sure that you model everything correctly again if you're letting x equals marcy's age then steven's age has to be x minus four because marcy is four years older than steven if you're letting x equal steven's age then marcy's age has to be x plus four because again marcy is four years older than stephen all right let's take a look at the next one so during the 16 game 2012 nfl regular season the broncos chargers and raiders combine to win a total of 24 games the chargers won three more games than the raiders but won six fewer games than the broncos find the regular season win loss record for each team okay so the idea here is that at the end of the day we want to find the regular season win loss record for each team so keep in mind that the regular season does not include the preseason and it does not include the playoffs okay so the records would be different if you included those extra games it's also important to note that we're told that the 2012 nfl regular season consists of 16 games so we're gonna use that in a little while but right now let's focus on the other information in the problem what it tells us is that the broncos chargers and raiders combined to win a total of 24 games so that's very important let's highlight the 24 games there then we're also told that the chargers won three more games than the raiders but won six fewer games than the broncos so the idea here is that we could let x be equal to the number of games won by the chargers let me start that off down here let's let x be equal to the number of games we'll say one by let me make this o a little bit better and then we'll go the and then we'll say chargers so that's the first part now the harder part or the part that people struggle with is then modeling how many games the raiders won and how many games the broncos won based on this initial variable that we have that's x again the number of games won by the chargers so coming back up here if i just think about the charters one again that's represented with x three more games than the raiders so that means the raiders won three fewer games than the chargers so if the chargers won x number of games then the raiders won x minus three number of games again three fewer games so we're just gonna come back and say then x minus three let me make this x a little bit better this would be equal to the number of games in this case one by v and we're dealing with the raiders in this scenario okay so the last team would be the broncos and they did better than the chargers it says here the chargers won three more games than the raiders but won six fewer games in the broncos so six fewer games in the broncos that means the broncos won six more games in the charters again if the chargers won x number of games now the broncos won six more than that so just x plus six number of games so let's come back here and let's say then we will say x plus six now would be equal to the number of games in this case one by the we're dealing with the broncos okay so all we have to do to figure out what x is is set up a little equation again if we go back we highlighted this initially it says the broncos chargers and raiders combine to win a total of 24 games so that's the key that means if i take the number of games won by the chargers which is going to be x and i add that to the number of games won by the raiders which is x minus 3 and add that to the number of games won by the broncos which is x plus 6 this sum should be 24. okay let's solve this for x and then we'll go back and figure out how many games the chargers won the broncos one and the raiders one and then we'll figure out the win-loss record for each so x plus x plus x would be three x let's write that first then negative three plus six would be plus three and this equals twenty-four let's go ahead and subtract three away from each side of the equation let's cancel this away and we'll say that 3x is equal to 24 minus 3 is going to be 21. as a final step here let's divide both sides of the equation by 3 and we'll cancel this away and we'll say that x is equal to 21 divided by 3 is 7. so x is going to be equal to 7. so coming back here we know that this would be 7 for the charges so let me just note this that they won 7 games then for the raiders it's 7 minus 3 or four so they won four games then for the broncos it's seven plus six which is 13. so they won 13 games all right so i know how many games they won but how do i figure out how many games they lost well coming back here remember it says specifically the 16 game 2012 nfl regular season so if they played a total of 16 games all i have to do is take that 16 subtract away the number of games they won and that'll give me the number of games they lost so for the chargers again they won seven games so that means 16 minus seven is nine they lost nine games so i'll just say the chargers the chargers one seven games and we'll say they lost nine games and again if you want to make this a little bit shorter you could just say their record was seven and nine most people know what that means we can also say for the raiders they won 4 games so 16 minus 4 would be 12 so their record would be 4 and 12. we'll say maybe a comma here while the raiders we'll say one in this case it would be four games and they lost in this case 12 games and they'll say and the broncos which would be the last team here will say they won again it was 13 games so we'll say 13 games and then 16 minus 13 would be three so we'll say and they lost three games and that's pretty much it i mean you don't need this long of a sentence in most cases something more simple would do but just to be thorough i wrote something like that and just to check things again each team had to play 16 games in the nfl regular season so we have 7 plus 9 that's 16. so that works we have 4 plus 12 that's 16 so that works and then 13 plus 3 that's 16 so that works then additionally you want to check that the sum of the wins would be 24 so this part right here so you have 7 plus 4 is 11 and then 11 plus 13 that is going to give you 24 so that checks out then additionally you want to check this so the chargers won three more games than the raiders the chargers won seven games the raiders won four games so that checks out it also says six fewer games in the broncos the broncos won 13 games the chargers won seven games so that checks out so our solution checks out the chargers won seven games and lost nine while the raiders won four games and lost 12 and then lastly the broncos won 13 games and lost three so the next type of problem we're gonna look at is a percent problem and these problems generally deal with a percent increase or percent decrease and we're going to be expected to find the original amount so mark's retail clothing store currently has an inventory level of 585 000 dollars this was an increase of 150 percent from one year ago how much was the value of the inventory one year ago so this is what we're trying to find here how much was the value of the inventory one year ago so we can set a variable like x be equal to the value of the inventory one year ago so how would i go about setting up an equation once i've determined that x is going to be the value of the inventory one year ago well i've got to look at the keywords here it says this was an increase increase means an addition right so if i started out with x which is the value of the inventory one year ago and i had an increase or i added this amount right here i know i would end up at this amount so x plus the increased amount and how would we get that well it increased by 150 percent so i would take the original value and i would multiply it by 150 percent so in decimal form 150 percent is 1.5 right 150 percent is equal to move this two places to the left you get 1.5 delete the percentage symbol so this is 1.5 and then times x let me label this real quick so it's crystal clear what's going on this is the original and this right here is the increase because it's 150 of the original value so 1.5 times x again x is the original so if i do this x plus 1.5 x i know that in terms of the problem it tells me i should be sitting at 585 000 dollars so these two amounts will sum to 585 000 and then all i need to do is crank out my equation x plus one point five x is two point five x this equals five hundred eighty five thousand we would divide both sides of the equation by two point five and i would get that x is equal to 234 000. and again that doesn't mean i just stop once i've written my answer down and i hand in my test and i say hey got the right answer i've got to translate that remember x is the value of the inventory one year ago so let me erase some stuff so i can fit this on a screen and what we're going to say is that the inventory had a value of 234 000 one year ago all right and we can check that quite easily if we take 234 000 and we add to it the increase of 150 percent again that's 234 000 times 1.5 that would be 351 000. if we sum these two amounts together we do get 585 000. so that is consistent with what we have in the problem so we can say here our answer is correct the inventory had a value of 234 000 one year ago all right lastly we're going to look at a simple interest problem and we looked at this in a previous video where we talked about formulas so gene invested two thousand dollars for three years and earned a total of three hundred dollars in simple interest what was his interest rate now when you get some of these problems there's a formula you're going to use and like with the motion word problems or with simple interest you're going to see these enough to where you're going to have it memorized so for this one you have i is equal to p times r times t where this is the amount of simple interest earned this is your principal this is your rate and this is your time now this is asking us specifically what was his interest rate so in this particular case i want to solve for r because that's what i want to find so what i want to do here is divide both sides of this formula by p t so i can isolate r so i would have r is equal to i over p t now once i have this let me just erase all this so i can kind of tighten everything up here once i have this i'm pretty much good to go on this type of problem because i need to just plug in the amounts given it tells me that gene invested two thousand dollars so that's my principal two thousand for three years so my time is three and it told me that he earned a total of three hundred dollars so i which is the amount of simple interest earned is 300. so now i just have to crank this out and i have an answer so what i'd end up with is 300 over 2000 times 3 we know that's 6 000. and without even using a calculator i can cancel these zeros with these zeros and i get three over sixty now i know sixty is divisible by three it's twenty so this is one over twenty and some of you can do that in your head some of you need a calculator but it ends up being .05 okay .05 or as a percentage i move this two places to the right i'd have five percent so his interest rate was five percent now again to check this let me just erase everything real quick let's go to the original formula so i or the amount of simple interest earned is equal to we know that the principal is two thousand we know that the rate is five percent now so let's put times .05 and we know that the time is three years so let's put times three so let's crank this out and see if it gives us three hundred dollars as an answer so two thousand times three we know that's six thousand six thousand times point zero five yeah we know that's three hundred so this would be equal to three hundred and so we can say our answer here is correct his interest rate was five percent hello and welcome to algebra two lesson nine in this video we're going to continue to learn about applications of linear equations so in this lesson i want to go over three additional types of problems that you're going to encounter we're going to take a look at mixture problems we're going to take a look at some denomination of money problems and we're also going to take a look at some motion word problems so before we get started let's revisit our kind of guidelines for solving a word problem when you have to set up and solve a linear equation in one variable so the first thing you want to do is read the problem and determine the question or questions to be answered then we want to assign a variable to represent the unknown if you have more than one unknown you're going to represent other unknowns in terms of that variable then we want to write an equation then we're going to solve the equation we're going to write the answer in terms of the question or questions being asked and then lastly we're going to check the answer using the words of the problem and again the main thing with checking the answers to make sure that it's reasonable i always use an example where i say if you're asked to find the number of people on a bus or a train or a car and you get like negative 4.3 you know you did something wrong right you're not going to have a negative amount of people on a bus at this level when we talk about linear equations in one variable our mixture of problems will involve basically rates of concentration so let's read through this real quick this is a nice little starter problem for us it's going to teach us how to find the concentration of a substance so an alcohol solution was made by mixing three gallons of an 80 alcohol solution and 4 gallons of a 52 alcohol solution find the concentration of the new mixture so how do we go about doing this well what i want you to do is take a piece of paper and a pencil out if you don't already have one and i want you to write this down let me go to a different page if you want the concentration of a substance so concentration of the substance this would be equal to you need the pure amount of the substance and this is over or divided by the total amount of the mixture so once you understand this right here it's going to be really easy for you to do these mixture word problems i can't tell you how many students i've tutored and they don't even understand this and they're trying to solve the problem you have to know how to get the concentration of the substance before you can do anything so let's go back up and we're going to fill in this information here now the information we're given is that there's 3 gallons of an 80 alcohol solution and 4 gallons of a 52 alcohol solution so let's use this down here so if i've got three gallons three gallons of again an 80 alcohol solution how much pure alcohol is that well i basically just take 3 and i multiply it by 80 or 0.8 as a decimal so 3 times 0.8 gives me what well 3 times 8 is 24 and i have one decimal place between the two factors so this is 2.4 and think about your units there would be gallons so basically 3 gallons of an 80 alcohol solution is 2.4 gallons of pure alcohol so when we think about the pure amount of the substance again from this i'm going to have let me just write it right here i'll have 2.4 going in there so let me erase this real quick and i'm going to put my fraction bar and then in here total amount of the mixture i know this came from a three gallon mixture now i'm going to add to this and i'm going to add to this remember we had another part we were mixing this with 4 gallons of a 52 alcohol solution so 4 times 52 as a decimal is 0.52 4 times 52 is 208 you've got two decimal places between the factors so this is 2.08 so again this is how much in gallons that we have as a pure amount of alcohol so this goes up here 2.08 and the total mixture is 4 gallons so that goes here so what do we have in the numerator again we have the pure amount of alcohol from the first mixture plus the pure amount of alcohol from the second mixture so the numerator is the pure amount of the substance in this case it's alcohol so 2.4 plus 2.08 gives me 4.48 and this is over the total amount of the mixture the first mixture was 3 gallons in total the second mixture was 4 gallons in total so 3 plus 4 would give me 7. now all we need to do is crank this out so 4.48 divided by 7 is 0.64 so we can erase this and put 0.64 and we know as a percentage that's 64 percent so let's go ahead and answer this with a nice little sentence find the concentration of the new mixture the new mixture is 64 percent alcohol alcohol and it's really that simple this is a very easy problem i mean just go through the kind of formula that i gave you but make sure you have that written down because a lot of times you're going to need to know how to find the concentration of a certain substance that comes from you know mixing one or more items all right so now let's take a look at one that's a little bit more challenging so jennifer asks you to make 12 gallons of fruit punch that contains 48 fruit juice by mixing together betty's fruit juice and a calvin's fruit juice now betty's fruit juice is 38 fruit juice while calvin's fruit juice is 53 fruit juice how many gallons of each are needed so a lot of times with mixture problems we get very very confused on what we need to do i want you just to think about the fact that in the end i want a 12 gallon mixture that is 48 fruit juice so that's the goal so let's write that down here our goal is 12 gallons that is 48 percent fruit juice so how are we going to get that well we've got to use betty's fruit juice and we've got to use calvin's fruit juice we don't know how much of each we're going to use but we do know that we're going to take a total of 12 gallons for the mixture so this is my total mixture in gallons so if i let x be equal to the number of gallons of let's say betty's fruit juice [Music] and we'll put used then what can i say about the amount we're going to use from calvin's fur juice so then well if this is in terms of gallons and it's the whole amount we're using remember the total mixture is 12 gallons so whatever i use here can be subtracted away from 12. in other words if i had 12 the total number of gallons from this mixture in the end minus x the number of gallons we're using from betty's fruit juice well that's got to give me the number of gallons of calvin's fruit juice used and of course you could have reversed that and said x is the number of gallons of calvin's fruit juice used and then 12 minus x would have been the number of gallons of baddies for juices doesn't matter you'll get the same answer either way now how do we go about making an equation we've got our we've got our unknowns expressed with some variables how do we make an equation well the one thing is that we know that in the end the amount of pure fruit juice from betty's fruit juice plus the amount of pure fruit juice from calvin's fruit juice has to equal the amount of pure fruit juice in the goal mixture so in other words we know that betty's fruit juice is 38 fruit juice so if i use 0.38 that's 38 as a decimal and i multiply it times x this is the pure fruit juice and this is from betty's now if i add this to and i'm going to do the same thing for calvin's fruit juice so if i add this to his is 53 fruit juice so 0.53 as a decimal times this quantity 12 minus x this is the pure fruit juice in this case it's from kelvins if i sum these two amounts together it's got to be equal to the amount of pure fruit juice in the end mixture so in other words it would be 12 times .48 and this is the pure fruit juice in the final mixture because again the final mixture comes from these two inputs so this has to be true so once we have our equation set up and we understand the logic behind it we just need to solve it we will figure out what x is that will tell us the number of gallons of betty's fruit juice used we can then plug in for here and find out the number of gallons of calvin's fruit juice used and we've solved our problem all right so to solve this we'll have 0.38 times x plus if i use my distributive property 0.53 times 12 is going to be 6.36 then minus 0.53 times x is 0.53 x and this equals 12 times 0.48 is 5.76 all right so if we combine like terms here point three eight x minus point five three x is negative point one five x and then we have plus six point three six and this equals 5.76 of course i will subtract 6.36 away from each side of the equation so that's gone and i'll have negative 0.15 x is equal to 5.76 minus 6.36 is equal to negative 0.6 so now as a final step we divide both sides of the equation by negative 0.15 and what's going to happen is this will cancel with this i'll just have x and that's equal to negative over negative is positive 0.6 divided by 0.15 is 4. so x here equals four now so remember x was the number of gallons of betty's fruit juice used so we're going to use four gallons of betty's fruit juice and then 12 minus four or eight gallons of kelvins for juice so we can say we need four gallons of betty's fruit juice along with eight gallons of calvin's fruit juice and to check this we would think about our equation that we set up let's scroll down a little bit and again the amount of pure fruit juice from betty's fruit juice plus the amount of pure fruit juice from calvin's fruit juice has got to be equal to the amount of pure fruit juice in the final mixture so if i just plug in a 4 everywhere i see an x i can just make sure the left and the right side are equal and i'm going to be good to go [Music] so 0.38 times 4 is 1.52 0.53 times 8 is 4.24 and 12 times 0.48 is 5.76 now if i sum these two right here i do get 5.76 so you get the left side equal to the right side and it's more important here to understand why they're equal again this right here tells me the total amount of pure fruit juice in gallons that i'm going to have from these two inputs i've got 1.52 gallons coming from betty's fruit juice and then i have 4.24 gallons of pure fruit juice coming from calvin's fruit juice so again the sum of that is 5.76 gallons of pure fruit juice and that's what i want because in my final mixture here i've got 12 gallons total that's 48 pure fruit juice so it's 5.76 gallons of pure fruit juice in the final mixture so because these two sides are equal here we know we've found the correct answer again we're going to use four gallons of betty's fruit juice along with eight gallons of calvin's fruit juice all right let's take a look at the next one so for a bill totaling one thousand one hundred five dollars a clerk received hundreds twenties and ones only there were six times as many 20s as hundreds and 25 fewer ones than 20s how much of each type of bill was received so again for this type of problem once you read it and you understand what you're being asked which let's just go ahead and highlight that how much of each type of bill was received you're immediately thinking about what to do in terms of setting up your variable now if you have three things you need to figure out how many hundreds how many 20s and how many ones you really need to think about what's going to be easiest for you to use and again i keep talking about this in these videos if you have something that's used in each comparison then you want to use that because it's easier to model things so if we say that there were 6 times as many 20s as hundreds and 25 fewer ones than 20s well then the amount of 20s that were received or the amount of 20 bills that were received that's used in each comparison so it's easiest to just let the variable like x be equal to that so let's let x be equal to the number of 20s received so then what can we say about the hundreds so it says here there were six times as many 20s as hundreds so that tells me that when i think about the number of hundreds it's one-sixth the amount of twenties so then we can say that one-sixth times the number of twenties received which we represented with x is equal to the number of hundreds received then what can we say about the number of ones well it says that there were 25 fewer ones than 20s well if the number of 20s received is represented with x then we could say then literally x minus 25 would be equal to the number of ones received so now that we've represented each type of bill we want to think about making an equation and what it tells us here is that the bill totaled one thousand one hundred five dollars now a common mistake with these type of problems is just to take what you modeled here and say okay well x plus 1 6 x plus x minus 25 is equal to 1 105. that's not going to get you there because remember this is a value this is the value this right here gives you the quantity of bills in each case x gives you the number or the quantity of 20 bills received that's just a number if i hand you four 20 dollar bills that's not a value you're just saying hey i have four of them you've got to multiply the 4 times the value per bill if it's 20 then 4 times 20 would give you 80. i've got 80 so you've got to match up the information on the left with the information you're given on the right so for each type of bill you got to multiply by the value so for the 20's which is represented with x i'm going to put 20 next to it so this is the value and this is the number of bills again i go back to the example if i hand four 20 dollar bills if you plugged in a 4 there 4 times 20 would give you 80 you know you have 80 dollars there for the number of hundreds i've got 1 6x multiply that by 100 and for the number of ones i've got x minus 25 that quantity make sure you use parentheses let's multiply that by one okay by one and you know one times anything is just itself but again you want to just use parentheses so you get in a good habit of things all right so now let's solve this equation we would have 20x plus 100 times 1 6. well we know they're each divisible by 2 this would be 50 this would be 3. so let's put 50 over 3 times x then plus i could just drop the parentheses here this is just x minus 25 this equals 1 105. all right let's scroll down a little further so 20x plus x is 21x and then we have plus 50 thirds x and then minus 25 equals 1105 let's add 25 to each side of the equation that will cancel and we'll have 21x plus 50 thirds x is equal to 1105 plus 25 would be 1130. i've got a fraction here i can multiply both sides of the equation by 3 or what i can do is i can multiply this by 3 over 3 and i can just work with the fractions it doesn't matter typically we just clear the fractions so let's just do that let's multiply both sides of the equation by three three times 21 would be 63 so this would be 63x plus 3 times 50 over 3 would be 50 so this would be 50x this equals 1130 times 3. well if i think about 3 times a thousand that's 3 000 3 times 100 is 300 3 times 30 is 90. so that would be 3 390. all right so let's continue now 63x plus 50x 3 plus 0 is 3 6 plus 5 is 11. so this would be 113x equals 3390 divide both sides of the equation by 113 and we'll find that x is equal to 30. okay this cancels with this and we got 30. again we're working with a word problem so let's make sense of this again if x is the number of 20 bills received and x equals 30 then i know i got 30 of these and again if you want to just check things 30 times 20 would be 600 so the value from the 20s would be 600 then for the number of hundreds received it's 1 6 times 30. 30 divided by 6 would be 5. so they got 5 of these the value would be 500 and then x minus 25 is the number of ones received 30 minus 25 is five so this is five so this would be five dollars and that's consistent with what we're given in the problem 600 plus five hundred is eleven hundred eleven hundred plus five is one thousand one hundred five again for a bill totaling one thousand one hundred five dollars so we can just state that the clerk received 30 20 dollar bills with five hundred dollar bills and five one dollar bills so again the clerk received thirty twenty dollar bills with five hundred dollar bills and five one dollar bills and again easy to check thirty times twenty is six hundred so that's the value from the twenties received five times 100 is 500 so that's the value from the hundred dollar bills received and five times one is five the value from the one dollar bills received so 600 plus 500 is eleven hundred eleven hundred plus five gives you this 1105 again the total bill that we're told in the problem all right so let's take a look at the next one and now we're going to talk about motion word problems so we saw these in algebra 1 and these use the formula distance equals rate of speed times time traveled so if i'm in a car and i'm going 30 miles an hour for 7 hours i multiply 30 times 7 that gives me 210 and i know i went 210 miles so a passenger plane left rome and flew south two hours later a cargo plane left from the same location and flew 80 miles per hour faster in an effort to catch up after seven hours of flying the cargo plane caught up what was the passenger plane's average rate of speed in miles per hour so for this one again we're going to use our distance formula so distance is equal to rate of speed times time traveled okay this is abbreviated with d equals r times t very common formula very common type of problem so the first thing we want to do again let's go back up we understand that the question here is what was the passenger plane's average rate of speed in miles per hour now to get this first we need to figure out what we need to do to model this and typically you'll say let's let x be equal to what you're looking for well we're looking for again the passenger plane's average rate of speed in miles per hour so let's do that so let's let x be equal to the rate of speed for the passenger plane okay so once we've established that let's talk about the cargo plane's rate of speed so what does it tell us about that well it tells us that the cargo plane left from the same location and flew 80 miles per hour faster it doesn't tell us its exact rate of speed but however fast the passenger plane is going the cargo plane is going 80 miles per hour faster so then i can say let's say then x which is the rate of speed of the passenger plane plus 80 would be equal to the rate of speed for the cargo plane so you might be thinking okay we've got this kind of set up how in the world are we going to get an equation going well what do we know here let's let's make a little table i often don't make tables because in a lot of cases you don't need them but with a motion word problem they really help you organize things so let's do distance equals rate times time like this we'll make a nice little table we have two scenarios so we have our passenger plane and we have our cargo plane now the rate of speed we know already for the passenger plane the r we've modeled it with x that's that's easy enough for the cargo plane we know it's x plus 80. now does it say anything in the problem about the time well yeah it kind of does a passenger plane left roman flew south two hours later so that means that the passenger plane is traveling for two hours more than the cargo plane does so two hours later a cargo plane left from the same location and flew 80 miles per hour faster in an effort to catch up after seven hours of flying so let me just stop there the cargo plane flies for seven hours at the point where they catch up and the problem kind of stops okay the passenger plane again it tells us because of this two hours later thing here was flying for a total of nine hours so nine hours for the passenger plane two hours less for the cargo plane seven hours so let's go back down let's fill in the information here for the passenger plane it's nine hours for the cargo plane it's seven now i want you to realize something they didn't tell us a distance but if i have a rate of speed and i have a time i have a distance because i just multiply the two together if my rate of speed is 30 miles per hour and my time is 7 hours 30 times 7 is 210 i know my distance right it's 210 miles in that case it's the same thing here i have a rate times a time so 9x is the distance i have a rate times the time so 7 times the quantity x plus 80 is the distance you might say okay well that's great how do we get an equation well there's one thing that you might have missed out on here let me go back up and let's read through the problem one more time what could you see that's equal here again a passenger plane left roman flew south two hours later a cargo plane left from the same location so they leave from the same location and flew 80 miles per hour faster in an effort to catch up after seven hours of flying the cargo plane caught up so that means they end at the same location they start at the same location they stop at the same location where the problem kind of stops at the same location you might as well say that but for the sake of the problem they went the same distance so that means the distance that the passenger plane went is the same or is equal to the distance that the cargo plane went now coming back down here i know what the distance is for each one so voila i've got my equation because all i have to do is set this distance which is 9x equal to this distance which is 7 times the quantity x plus 80 and i can solve my equation for x so use my distributive property over here on the right 7 times x is 7x plus 7 times 80. we do 7 times 8 that's 56 put a 0 at the end that's 560. we would subtract 7x from both sides of the equation that cancels you'll have 2x is equal to 560. divide both sides of the equation by 2. and what's going to happen is this will cancel with this i'll have x is equal to 560 divided by 2 is 280 so what does that tell me let me go back up x was the rate of speed for the passenger plane so that tells me that the passenger plane was traveling at 280 miles per hour so the passenger plane was traveling at 280 miles per hour that's their average rate of speed how do we check this well it's actually pretty easy if i take 280 which is the rate of speed and i multiply it by 9 which is how long the passenger plane travels for i get 2520 that's the total number of miles that the passenger plane traveled now with the other guy the cargo plane it goes faster 280 plus 80 is 360. so that's its average rate of speed in miles per hour it does it for seven hours if you multiply 360 by seven you also get 2520 so the mileage syncs up there they travel the same number of miles so the distances are equal when we check them so we know i have the correct answer the passenger plane was traveling at 280 miles per hour hello and welcome to algebra 2 lesson 10. in this video we're going to learn about interval notation so starting on our next lesson we're going to talk about linear inequalities in one variable and most of you know from algebra 1 that we have a method to display our solution for this type of problem and it's known as interval notation interval notation allows us to notate a specific range of numbers on a number line so before we kind of get into interval notation let's just start out with a typical equation i want to show you a few things so we have 2x plus 7 equals 21. so we know how to solve this at this point so 2x plus 7 equals 21. so we would subtract 7 away from each side of the equation i'll have 2x is equal to 14 divide both sides by 2 and i'll get x is equal to 7. now normally this is fine we just put x equals 7. that's a perfectly acceptable way to notate your solution you could also use a solution set so the set contains one element the element is 7. additionally we could show the solution graphically i have a number line here i could just put a filled in circle at seven so these are all ways i could show the solution for this equation then there's another one we're going to talk about this more at the end of this lesson there's something known as set builder notation so i could say the set of all elements x such that x is equal to 7. and if this doesn't make sense now that's okay we're going to cover it at the very end of this lesson but the point is just to show you that there are many different ways to notate the solution for this equation and in every situation it's basically telling us what that x needs to be replaced with 7 for us to get a true statement 2 times 7 would be 14 14 plus 7 is 21. if i replace x with anything else i will not get a true statement right the equation will be false all right so now let's look at an inequality and it's nothing we have to solve we're just looking at x is greater than seven so what can i replace x with here well it's no longer a single value i can replace x with anything that's actually larger than 7. if i put 7 in itself it won't work 7 is not greater than 7 that's false but anything larger than 7 anything to the right of 7 on the number line would make it true eight is greater than seven that's true twelve is greater than seven that's true one thousand is greater than seven that's true one million two hundred sixteen thousand four hundred twenty seven is greater than seven that's true so if i replace x with any value that is larger than seven i will get a true statement so how do we notate this graphically well on the number line we would locate 7 and we want to mark 7 as a boundary point or an area that will separate the solution part from the non-solution part so we do this with two different things we can either use a parenthesis or we can use an open dot so at 7 i can use an open dot like this which a lot of students will see that in their class or i can use a parenthesis that faces towards the solution region now since x is greater than the solution region is to the right of seven so i would face my parenthesis to the right and it would touch seven like that now once we have that what we want to do is we want to shade the number line again to the right of 7 because that's what satisfies this inequality so i would shade all of this and i would shade my arrow to say hey this continues forever and ever and ever and so any number that is larger than 7 would satisfy this inequality now in the situation where you have something known as a non-strict inequality we use a different variation of this we would use a bracket or a filled in dot so in this particular example we have that x is greater than or equal to 5. now because this or equal to remember x can be equal to 5 or x can be greater than 5 for this to be true so now if i replace x with 5 5 is greater than or equal to 5 that's true because this part is true 5 is equal to 5 or any number larger than 5 would also work if i do 216 is greater than 5 that's true so x can be anything that is again 5 or anything larger than five so at five instead of putting a parenthesis or an open dot i'm now going to put either a filled in circle or filled in dot like that or i'm gonna put a bracket there that faces towards the solution region again this is a greater than or equal to so the solution region is to the right so i'll put a bracket here and then just like i did in the last example i'm going to shade the solution region so i'm going to shade everything to the right and i'm going to shade that arrow to say hey this continues forever and ever and ever all right let's try a few examples here so if i saw something like x is less than 3 well i'd find 3 and at 3 i can either put again i can put a an open dot or an open circle got to make sure that there's no color inside of it to indicate that 3 is not part of the solution because this is a strict inequality this is strict x there cannot be three if i put this here and it's x is less than or equal to three well then this would be filled in because now x can be equal to three but that's not what we have here so i'm going to put in an open circle or open dot however you want to think about that or more typically what i use is a parenthesis that faces in the direction of the solution region here it's a less than so the solution region is going to the left so my parenthesis would face to the left and i would just shade everything to the left because all the values to the left of 3 would satisfy this inequality x is less than 3. so if i put 2 in there 2 is less than 3 that's true if i put 0 in 0 is less than 3 that's true if i put negative 7 in negative 7 is less than 3 that's true so anything to the left of 3 would work as a solution okay what about x is less than or equal to negative four so we want to find negative four so that's right here and if it's less than or equal to i've got to include negative four as part of my solution so again i can put a filled in circle like this or i can use a bracket now the bracket faces to the left because it's a less than or equal to so put my bracket in there and then i shade everything to the left what about x is greater than 2 well i'm going to find 2 and again this is a strict inequality so 2 is not included i'm going to put a parenthesis that faces to the right or an open circle so i can do this or i can do this and again it's just a matter of what your textbook or your teacher tells you to do they both mean the same thing so let's let's go with this one for now so x is greater than 2 so anything to the right of 2. so i'm going to shade all this like that and again it's the same thing if i erase this and i do this a parenthesis that's faces in the direction of the solution region we have a greater than so the solution region is to the right all right so now that we understand how to graph an interval on a number line let's talk about interval notation so as i alluded to earlier interval notation is a way to display an interval on a number line now there's some important things that you need to know here the first thing is that we use parenthesis so you have this parenthesis here that's on the left or this parenthesis here on the right to exclude a number from the interval so what exactly does that mean well if i saw something like a parenthesis here and then a value a what it's telling me is that it's anything larger than a a is basically not included so for example if i had x is greater than two my interval would start with a two right here because it's saying anything larger than a two can't be two itself but anything larger than two then you have a comma and let's say i put b there well this is anything smaller than b so in the case of x is greater than 2 well for this part right here x can continue after 2 forever and ever and ever so you use an infinity there now with an infinity because you can't ever touch it you always use a parenthesis but let's say we had something that was a little bit different let's say that we had that x was greater than 2 and x was less than let's say 5. well in this case i want anything smaller than five and i also want anything larger than two so on a number line that would be the range of values between two and five so i have a number line here if i wanted to notate x is greater than 2 and less than 5 i would find 2 on the number line which is here and again we just saw this so i put a parenthesis that's facing to the right because i'm just thinking about x is greater than 2 for right now so x is greater than 2 and then now i want to think about x is less than 5. x is less than 5. so we put a parenthesis at 5 facing to the left and i would shade all the values in between because the values in between are what's going to satisfy this what we call three part inequality if i have a number that's in between 2 and 5 it works if i choose three three is greater than two and also it's less than five if i choose four four is greater than two and also it's less than five if i choose something like six that's outside of this range six is greater than two but it's not less than five it's got to satisfy both conditions so again this is how you would notate something like that so let's look at something like x is greater than negative three so graphically we know that we find negative three it's greater than so our parenthesis would face to the right and then i would shade everything to the right here including this arrow now in interval notation again it's a strict one so i know that negative 3 is not included so i have my parenthesis here i put a negative 3. so i'm saying anything or any value larger than negative 3. this is your cut-off point or your boundary that you're putting in there to say hey it's any value larger than this if i use this exactly it's not going to work but anything larger would then comma this solution continues forever and ever and ever so the endpoint over here we'd put infinity and we always use a parenthesis with infinity so i wouldn't say any value less than infinity because that's that's not ever going to work but typically when you have something on this side we're saying any value that's less than that so as another example let's say i saw let's just erase this real quick let's say i saw that x was greater than negative three and less than let's say seven well in this particular case in interval notation i'd have a negative three next to a parenthesis here comma i'd have a seven next little parenthesis here any value that's larger than this works up to but not including this value because it's any value that's smaller than this that works so it's got to be between the two values so if i look at this graphically at negative 3 i have a parenthesis here at 7 i have a parenthesis here and i shade everything in between what about something like x is less than 7 well again graphically we know we put a parenthesis that faces to the left at 7 and we shade everything to the left pretty simple and then in interval notation i'll have a 7 here and again all this is saying is that it's any value less than 7. 7 is not included because we have the parenthesis next to it now in terms of going this way that solution continues forever and ever and ever it could be anything less than seven so it could be negative whatever however big a number you want to think of let's say negative one million for example or negative 10 trillion whatever you want to think of so we use negative infinity in this particular case anytime i use infinity or negative infinity i always use a parenthesis next to it again because you can't ever touch it infinity is a concept it's not really a number all right so in the case that we want to include a number we use a bracket just like we did when we were graphing so something like this means a or larger so if i had a comma and then infinity that would be let's say i had 3 with a bracket next to a comma infinity this would translate to x is greater than or equal to three x could be three or anything larger or let's say for example i had a bracket here and i had seven comma and let's say we do 9 in another bracket this is telling me that x is greater than or equal to 7 and less than or equal to 9. so it can be 7 it could be 9 it could be anything between also but if you get outside of those ranges if you go less than 7 or greater than 9 you would not have a true statement so when we want to again include a number we use a bracket so on this part right here let me kind of erase this on the right side let's say i have b here for example this is b or smaller so let me put a negative infinity here so this could be for example x is less than or equal to let's say 10. well in this case it'd be negative infinity here and then up to and including 10 use my bracket all right so let's say as an example we had x is greater than or equal to 2 and less than or equal to 9. so we know that x again is greater than or equal to two so let's find two and this number line is a little different because i fit more numbers on it but here's two right here and what i'm going to do is i want to put a bracket that faces to the right i'll put a bracket that faces to the right now x is also satisfying the condition that it has to be less than or equal to nine so we're going to find nine which is right here put a bracket that faces to the left and i would shade everything in between there so x can be 2. i plug a 2 in there it's okay 2 is equal to 2 and it's also less than 9. that works it could be 9 same thing but it can't be anything outside of this range if i plug in a negative 3 for example negative 3 is not greater than 2 so that would fail right it's got to satisfy both conditions now in interval notation i'm just kind of following this right here graphically i know that the smallest possible value that this thing can take on is 2. i put a 2 there to the left and i have a bracket next to it then comma the largest value this thing can take on is a 9. 9 is included so i put a bracket there so this is the smallest value and this is the largest and the only difference between the two is let's say i for example had x is greater than 2 and also it's less than 9 what i do in this scenario is i'm basically doing the same thing i just have parenthesis instead so i would have a parenthesis there 2 9 and then a parenthesis there in this case i'm saying it's anything larger than 2 so to say hey it's not 2 but just anything larger than that then for 9 it's not the largest value because it's not included but it's just anything less than that so it's between the numbers 2 and 9. so it's just paying attention to whether you have a non-strict inequality like here or a strict inequality like here what about something like x is greater than or equal to negative one and less than or equal to ten well again graphically pretty simple i find negative one i put my bracket facing to the right i find ten i put my bracket facing to the left and i shade everything between in interval notation i just basically follow this right i'd have a negative 1 as the smallest value comma a 10 as the largest value and i've got brackets next to each and of course we can have combinations of the two here x is strictly greater than three and less than or equal to five so at three i'd have a parenthesis that's going to face to the right because it's a greater than x is greater than 3. and then at 5 i'm going to have a bracket because it's a less than or equal to 5. so phases to the left and i'm going to shade everything in between here and then in interval notation i just follow this at the left side i'm going to have a parenthesis to say that 3 is not included then comma on the right side i'm going to have a 5 and i'm going to have a bracket to say 5 is included all right for this one we have that x is greater than or equal to 7 and less than 14. so if i find 7 i'd want to put a bracket there facing to the right find 14 i'd want to put a parenthesis facing to the left and i would shade everything in between in an interval notation i'm going to have a bracket and then a 7. again 7 is included and then anything larger up 2 but not including 14 14 is going to have a parenthesis next to it to say that's not included what about x is greater than or equal to negative 2 and less than 12. so here's negative 2 i would have a bracket that faces to the right and then here's 12 i'd have a parenthesis that faces to the left and i would just shade everything in between and again in interval notation i'm going to have a bracket that faces to the right then negative 2 because negative 2 is included comma and then i have a 12 and then 12 is not included x is strictly less than 12 so i'd have a parenthesis there to indicate that 12 is not part of the solution all right so lastly i briefly mentioned something known as set builder notation this is something we will use in math and i didn't really cover it when we talked about sets this is a good point where we can kind of drop it in so set builder notation we basically have a variable that's used as a placeholder this is just saying that it's some real number right and to represent some real number we just use x or y or z so we say the set of all elements x the set of all and this is usually abbreviated by saying the set of all x such that this bar means such that x has a certain property okay so in the example where i have an equation where x is equal to 7 i can notate this by saying the set of all real numbers represented with a variable x such that x is equal to 7. okay very very easy so if i was asked to put something like x is greater than negative 4 in set builder notation i would say the set of all elements x such that and then i would just list the property x is greater than negative 4. as a final one let's say we have x is greater than negative 2 and less than or equal to 5 i could say the set of all elements x such that and then just list the condition x is greater than negative 2 and less than or equal to 5. very very simple hello and welcome to algebra 2 lesson 11. in this video we're going to learn about linear inequalities in one variable so by the time you get to an algebra 2 course hopefully you've had at least some experience with solving a linear inequality in one variable but even if you haven't we're going to cover all the details involved with this process and it's going to serve us well in the next lesson and the lessons after that because we're going to need this as a basis now we start off with a generic example where we have ax plus b is less than c so this is something you'd probably see in your textbook and we can say that a b and c are real numbers so basically anything you can think of at this point could be plugged in for a b and c with one exception a the coefficient of x is not allowed to be equal to zero and that's because if i multiply zero times x i'd get zero and i'd be left with b is less than c now there's one other thing we want to note we have a less than here this is a less than i could replace this with a greater than or a less than or equal to or greater than or equal to all of those would be equally valid in this generic example we just have to choose one of them we just happen to choose a less than so we have ax plus b is less than c where again a b and c are real numbers a is not allowed to be 0 and the less than could be a greater than a less than or equal to or a greater than or equal to so here are just a few examples and we'll get to solving these in a minute so we have 2x minus 7 is greater than 5. we have negative 3x plus 10 is less than or equal to 17. then we have 0.25 x is greater than or equal to negative 2. so the main difference between an equation and an inequality with an equation something like x equals seven x can be one value there it can be seven so on the number line it's basically represented with just this this is x is equal to 7. with x is greater than 7 if i was to think about that that's a range of values i can plug in anything for x that's larger than 7 and i will get a true statement i could plug in an 8 a 9 a 10 a 1 million a 1 billion as long as it's larger than 7 it's good to go now graphically and we covered this in the last lesson at 7 i would put an open circle or i could put a parenthesis facing to the right i typically stick with a parenthesis and i would shade everything to the right what this means is that 7 is not included in the solution that's why i have a parenthesis there because x is strictly greater than 7. if i plug the 7 for x it wouldn't be true 7 is not greater than 7. so that's not included but anything larger than 7 anything to the right of 7 is so that's why i shaded everything to the right and i shaded this hour to say hey this is going to continue forever and ever and ever so anything you can pick that's larger than 7 would work as a solution all right so now let's talk a little bit about how we'd solve a linear inequality in one variable and essentially we use the same techniques as we use when we're solving a linear equation in one variable so you'll recall when we're solving equations we had this property known as the addition property of equality that told us we could add the same value to both sides of an equation and we'd maintain the same solution well we have a similar property when we work with inequalities so it's called the addition property of inequality and it's the same thing it allows us to add the same value to both sides of an inequality now and we don't change the solution so if i see something like x minus 2 is greater than 7 this is a one-step inequality so basically all i need to do to isolate x is just add 2 to both sides of the inequality this would cancel and you would have x is greater than 9. as another example of this if i have x plus 8 is less than or equal to 5 again think about isolating x i just need to subtract 8 away from each side of the inequality and so this would cancel and i would have x is less than or equal to 5 minus 8 is negative 3. all right for the next property we're going to talk about it's a little different than its counterpart that is used when we're solving equations you recall that you have a multiplication property of equality and with this property we can multiply both sides of an equation by the same non-zero value and we will not affect the solution well here with the multiplication property of inequality it's different because if i multiply both sides of an inequality by the same positive value it's the same thing i'll have the same solution i don't have to do anything i have to worry about anything we're basically good to go you can't use zero because of the properties of zero if you multiply by zero you get zero so you can't use zero anything that's negative any value that's less than zero if i multiply both sides by a negative i've got to flip the direction of the inequality so a less than becomes a greater than or greater than becomes a less than so basically your inequality symbol will flip as an example of why this occurs if i have 2 is less than 5 that's true 2 is a smaller value than 5. if i multiply both sides by a positive okay for example let's say 3. so 2 times 3 would be 6 and 5 times 3 would be 15. 6 is still less than 15. this is true but if i took this original 2 is less than 5 and i multiplied both sides by a negative let's just keep it simple and say negative 1. what's going to happen is i'll have negative 2 is less than negative 5. this is now false why is it false because i started out with a bigger number here when i applaud a negative 2 it became a bigger negative right negative 5 is a bigger negative than negative 2. a bigger negative is actually a smaller value you think about the way a number line works a bigger negative is to the left right and as we go to the left numbers actually get smaller and it's a hard concept to kind of wrap your head around because generally speaking until we get into algebra we think of the number five forget about the negative part as being larger than two so that kind of sticks with us for a while until we can shake free of it okay so negative five is a smaller value than negative 2. so for this we need to flip it so we're going to flip and so if i change the direction of the inequality meaning i go from a less than to a greater than this becomes true okay this now becomes true negative 2 is greater than where is a larger value than negative 5. all right so let's look at an example here we have 2x is greater than 10. so we know if we had 2x is equal to 10 we divide both sides of the equation by 2 we get x equals 5. it's the same thing here i want to isolate x and so i want to divide both sides of the inequality by 2 so that i can do that so this cancels with this i get x is greater than 5. because i divided both sides by a positive i didn't have to flip the direction of the inequality symbol right its state is a greater than if i would have had a situation where it was let's say negative 2x is greater than 10 well in this scenario i divide both sides by a negative 2 i've got to flip the direction of the inequality right now it's a greater than it's got to become a less than so it's the symbol flips so this becomes a less than and this will become x is less than negative 5. all right let's say we saw something like negative 1 half x is greater than 30. what would i do to isolate x so negative one-half x is greater than three well i want to get x by itself and what i can do is i can multiply both sides of the inequality by the reciprocal of this so that's negative 2 over 1 or basically negative 2. and so this is going to cancel with this well what i've got to remember again because i multiplied by a negative and i'm working with an inequality i've got to flip the direction of the inequality it's a greater than so it's got to become a less than so the symbol there again it's a greater than it's going to become a less than so this is x is less than 3 times negative 2 is negative 6. all right so now that we have those two properties under our belt it's time to look at some sample problems so we have that 7 is greater than or equal to we have 6x minus 5 minus 4x so just like with an equation i'm going to simplify each side first so i have 7 is greater than or equal to we have 6x minus 4x that's 2x and then minus 5. make this a little cleaner so now i want the variable term on one side of the inequality the number on the other so i can simply add 5 to each side of the inequality 7 plus 5 is 12. so 12 is greater than or equal to we have 2x to get x by itself i would divide both sides by 2. now that's a positive number so i don't need to do anything to the inequality symbol this cancels with this and gives me 6. so that's greater than or equal to and then we have x now typically i like the variable to be on the left and i think it makes it easier when we start graphing and using interval notation so to flip it if your symbol is pointing to x that means x is less than or equal to 6. these are the same 6 is greater than or equal to x or x is less than or equal to 6. it's just easier to read this because typically we read from left to right there's two things i want to show you real quick in our last lesson we showed how to write something in interval notation and we showed how to graph an interval on the number line you might have missed that lesson and that's fine because i'm going to show you really quickly how to notate this solution so the first thing with interval notation is you have basically a value here a comma and a value here so this right here we think of as the smallest and this right here we think of as the largest so we think about x is less than or equal to 6 what is the smallest value that x can take on well there's not going to be one because x is less than or equal to so any value less than 6 that you can think of is going to work and that goes on forever and ever and ever i could do negative 1 trillion that would work so to show this we use infinity and specifically there because it's in the negative direction we use negative infinity now when i use negative infinity or positive infinity what i place next to this is a parenthesis the parenthesis says it's not included okay so the smallest value doesn't exist so we use negative infinity now for the largest possible value we know it's 6. x can go up 2 and include 6. so i put a 6 there and then because 6 is included i use a bracket okay this means is included so six is the largest value and it's included so we use a bracket now graphically it basically looks like this let's go up to a graph let me write my solution here it's x is less than or equal to six so on a number line i would find 6 which is right here and i can put either a filled in circle like this or i can use a bracket that faces towards the solution region the solution region is going to the left because we have a less than let me make that a little better and i want to shade everything to the left so all these values that satisfy that inequality and i can't list all of them so i'm going to shade that arrow to say the solution continues forever and ever and ever now looking at the graph and looking at interval notation you can see some similarities we use a bracket at 6 we use the bracket at 6. again that tells us that 6 is part of the solution and then it continues forever and ever and ever we have this arrow here that's highlighted you can kind of think about this being that that arrow tells us it continues forever and ever and ever in interval notation we express this by saying negative infinity is in place of the smallest value because we don't really have a smallest value so before we kind of move on you know in algebra 1 i checked almost everything in algebra 2 i'm not checking as much stuff just so we can get more in but i want you to know how you can go about checking your answer for this it's a lot more tedious than checking an equation the first thing is is you would replace your inequality symbol with equals so let me erase this real quick i have some room to work and so if i did that i would put 7 is equal to 6x minus 5 minus 4x then just take your solution and do the same thing replace the inequality symbol with equals so if i put a 6 in for x do i get a true statement so this would be 2x minus 5 equals 7. if i plugged in a 6 there 2 times 6 is 12 12 minus 5 is 7. so yes that works out so 6 is your boundary point it separates the solution region from the non-solution region and you can even see that graphically at 6 we have a separation where 6 works and anything to the left of 6 works but anything to the right of 6 doesn't so this is a non-solution region and this right here is a solution region so 6 is your boundary and you'll hear me talk about that a lot moving forward now once you've established that you just check a number on each side so we'll do that with the inequality there and i'm just going to simplify the inequality and make it a little easier for us so 7 is greater than or equal to we know this would be 2x minus 5. if i check 8 it shouldn't work because x has to be less than or equal to 6. so if i plugged in an 8 there i get 16 minus 5 which is 11. so 7 is greater than or equal to 11 that's false that's what we expect h should not work right because we're saying x has to take on a value that's less than or equal to 6. so that's good now the next thing we want to do is we want to check a value that is less than the number six so let's check five so seven is greater than or equal to again this simplifies to two x minus five if i plug in a five for x i get two times five which is ten ten minus five is five so i get 7 is greater than or equal to 5 and that is true 7 is certainly greater than 5. so you can see that it's more tedious to check these you've got to establish the boundary and then you've got to check one on each side of the boundary so one in the non-solution region we checked eight and it didn't work so that's right then one in the solution region we checked five that worked so that was right all right so let's kind of roll through these now so i have one plus 6 times the quantity 6 plus 4 v and this is greater than negative 27 plus 8 v all right so let's simplify each side now so we have 1 plus 6 times 6 is 36 plus 6 times 4 v is 24 v and this is greater than negative 27 plus 8 v so 1 plus 36 is 37 plus 24 v is greater than negative 27 plus 8 v so let's subtract 8 v from each side of the inequality and let's subtract 37 away from each side of the inequality so this is going to cancel and this is going to cancel 24v minus 8v is 16v and this is greater than negative 27 minus 37 is negative 64. so to get v by itself i divide both sides by 16 and i'm dividing by a positive there so i don't have to do anything special so this cancels with this and i get v is greater than we have negative 64 over 16 which is negative 4. so let me erase everything real quick so again we found that v was greater than negative four so typically your book or your teacher is going to ask you to do two things once you get a solution again write it in interval notation and graph it so in interval notation again we have these two blanks where we have the smallest and we have the largest and technically i'm going to put what negative 4 here is the smallest is negative 4 actually the smallest value that v can take on no it's not it's as close as we can get to thinking about it though i can't possibly list the smallest value to show that i'm going to put a parenthesis next to negative 4 to say it's not included but anything larger than negative 4 would work if i had negative three point nine nine nine nine nine nine nine seven eight that would work because looking at a number line that would be larger than negative four but again anything that's negative four or smaller will not satisfy the inequality what about the largest value that this can take on well again because this is going to go forever and ever and ever in the right direction we use infinity and with infinity we always use a parenthesis so again it's not the smallest number technically but we use this concept to notate things okay so anything larger than negative 4 up to infinity now graphically we would show this in a similar way negative 4 is not included so i put a parenthesis that faces to the right and then i would shade everything to the right everything in the solution region including that arrow to say hey this continues forever and ever and ever some people will even put an infinity out here to show that and additionally you might see this in your textbook you might see an open circle like this it's open to show that negative 4 is not part of the solution set all right for the next one we're looking at 6 times the quantity 3x plus 9 is greater than 12 times the quantity 4x minus 8. so 6 times 3x is 18x plus 6 times 9 is 54. this is greater than 12 times 4x is 48x and then minus 12 times 8 is 96. so i want to subtract 48x away from each side of the inequality and i know that this is going to cancel i also want to subtract 54 away from each side of the inequality this is going to cancel 18x minus 48x would be negative 30x and this is greater than negative 96 minus 54 is negative 150 so now to get x by itself i'm going to divide both sides by negative 30. but remember i'm dividing both sides by a negative so this has to be flipped right now it is a greater than i flip the direction it's going to become a less than so x is going to be less than negative 150 over negative 30 is going to be 5. let's erase everything so again we have that x is less than 5. in interval notation again we think about this concept of a smallest value and a largest value and again that's not technically correct but that's how we think about it so x is less than 5. the largest value that this can take on is not 5 but anything smaller than that so nine nine nine 4.99999999 seven eight for example so i put a parenthesis next to five just to say that it's not included what's the smallest value this can take on well it's going to continue forever and ever and ever in the left direction because it's anything less than 5 so negative 1 trillion for example would be a solution so i use negative infinity and again if i use positive or negative infinity i always put a parenthesis next to it now graphically x is less than five find five and put a parenthesis that faces left or put your open circle again to show that five is not included either one and then shade everything to the left so you shade all this in the solution region including the arrow and again some people like to put a negative infinity out here to show that it's just going forever and ever and ever in that direction all right so we also have ones that have fractions involved so just like we saw with equations we can multiply both sides by the lcd now i'm going to simplify before i start so we have b plus 8 3 minus 2b is less than five halves b plus one minus four thirds b plus 17 18. what can i do well i know b minus two b would be negative b negative b plus eight thirds is less than nothing really i can do here without working with fractions so let's just write five halves b plus one minus four thirds b plus 17 18. what would my lcd be here i'm looking at denominators of 3 2 3 and 18 obviously it's going to be 18. so we're going to multiply both sides of the inequality by positive 18. so i don't need to flip anything 18 times negative b is negative 18 b plus 18 times 8 thirds 18 would cancel with 3 and give me 6 6 times 8 is 48 this is less than if i have 18 times 5 halves 18 would cancel with 2 and give me 9 9 times 5 is 45 then times b plus 18 times 1 is 18. then we have 18 times negative 4 thirds b we know that's negative 18 divided by 3 is 6 6 times 4 is 24 so minus 24 b and lastly plus 18 times 17 18 18's would cancel you just have 17. all right so nice and easy now on this side i can do a lot of combining 45 minus 24 is 21. so 45b minus 24b is 21b then 18 plus 17 is 35 and over here i still have my negative 18 b plus 48 so i want all of my variable terms on one side all the numbers on the other so let me go ahead and subtract 48 away from each side of the inequality that's going to cancel and i'll let me also subtract 21b away from each side of the inequality so that's going to cancel if i do negative 18b minus 21b that's going to give me negative 39b and this is less than 35 minus 48 is negative 13. so now we divide both sides of the inequality by negative 39 and again if i'm dividing by a negative or i'm multiplying by a negative this has to be flipped so this becomes greater than from a less than so this cancels and i've got b is greater than negative 13 over negative 39 the negatives cancel 13 over 39 each is divisible by 13 so you get one third so b is greater than one third so again we have that b is greater than one third so we can start out with interval notation again on the left we want the smallest value now one-third is not technically the smallest value but we're putting it in there next to a parenthesis to say that it's not included but anything larger than one-third even if it's a small amount larger than one-third a microscopic amount larger than one-third would work so one-third is not included then comma this continues forever and ever and ever in the right direction so we put infinity here we put a parenthesis there we always use a parenthesis with infinity so graphically i would find one third now i don't have a notch for that so i'm going to create one let's just say one-third is about right there let's say this is one-third and i'm going to put a parenthesis there or again you can use that open circle technique doesn't matter and then i'm going to shade everything to the right everything in the solution region including that arrow and again you can put an infinity out here to say 1 3 is not included but anything larger than 1 3 works as a solution for this inequality all right so the last thing we want to cover is a three-part inequality so we saw these in algebra one and basically what happens is you have this guy in between two inequality symbols so 8 minus 5a is greater than 8 and it's less than 38. so both the conditions have to be met so what's going to happen is you're going to do one of two things you can separate these and solve them and then put it back together in the end which is more work or you can solve it doing the same thing to each part so in other words i want to isolate a in the middle so what i do is i would subtract 8 away from each side to begin so then 8 minus 8 is 0 and this is less than this would be 0 so i'd have negative 5a and then less than 38 minus 8 is 30. now if i want a by itself i would divide each part by negative 5. but remember if i'm dividing by a negative i've got to flip the direction of the inequality right now these are less than so if i flip the direction of that symbol it's going to become a greater than so this would be 0 over negative 5 is 0. i'd have a greater than this would cancel and i'd have a and then a greater than 30 over negative 5 is negative 6. now you don't want to keep it like this you want to put it in the direction of the number line negative six is to the left of zero so you want it lined up that way so we would put negative six is less than a which is less than zero or a is greater than negative six and also it's less than 0. looking at a number line it's a range of values not including negative 6 so i'd put a parenthesis there so anything larger up 2 but not including 0. so i'd put a parenthesis there i would shade everything in between this is just telling me that a is a value that's between negative six and zero in interval notation again we use this concept of the smallest and the largest negative 6 would go in this spot for the smallest it's not included so i put a parenthesis there 0 would go in the spot for the largest it's also not included so i put a parenthesis there so here's interval notation and here's graphically all right let's look at one final problem and again these three part inequalities are no more difficult so we have 7 v minus 5 is greater than or equal to negative 61. and it's less than or equal to negative 40. so again do the same thing to each side so plus 5 here plus 5 here plus 5 here so negative 61 plus 5 is going to give me negative 56 and it's less than or equal to this cancels so i'll just have 7v and it's less than or equal to negative 40 plus 5 is negative 35. so now what i want to do is divide each part by 7 so i can isolate v and what's going to happen is negative 56 over 7 is negative 8 this is less than or equal to we'll have v by itself now because this cancels and this is less than or equal to negative 35 over 7 is negative 5. so v is greater than or equal to negative 8 and it's less than or equal to negative 5. let me drag the solution up so graphically if i located negative 8 it's there so it's included in the solution set so i'm going to put a bracket here that faces to the right same thing with negative 5. it's included in the solution set so i'm going to put a bracket there this one's going to face to the left and i shade everything in between so what this is telling me is that v can be any value starting from negative 8 going up to and including negative 5. now in interval notation again smallest largest this thing about that the smallest value v can take on is negative eight now negative eight is included it's part of the solution so i use a bracket and it goes up to and includes negative five so the largest value be negative five again i'm using a bracket because it's included so your solution here is that v is any value that is greater than or equal to negative eight and less than or equal to negative five so again starting from negative eight including negative eight going up two and including negative five hello and welcome to algebra two lesson twelve in this video we're going to learn about compound inequalities so in our last lesson we had a review of how to solve a linear inequality in one variable now we're going to move on and look at something that's a little bit more challenging and probably new for you in your algebra 2 course we're going to talk about compound inequalities so a compound inequality consists of two inequalities linked by the word and or the word or okay so those are the two different scenarios now the first one we're going to deal with is solving a compound inequality with and now i suggest you get a flash card or if you don't have a flash card a piece of loose leaf write down the procedure for each one because they're a little bit different and then as you're working through your problems you can reference your flash card or your piece of loose leaf whatever you wrote it down on after about 20 or so of these you're going to have this procedure down packed and you won't get confused so the first thing you want to do with the scenario of and you want to solve each inequality separately okay so just treat it as if it's two separate problems the next thing you want to do is find the solution set for the compound inequality now the solution set contains all numbers that satisfy both inequalities both inequalities so if it's not a solution to both it's not a solution for the compound inequality and a good example of this remember when we started talking about sets at the very beginning of algebra 2 we talked about the intersection of two sets so as an example of that i have two basic sets here i have set a which contains the elements one two three and four and i have set b which contains the elements three five seven and nine so if i'm looking for the elements that are common to both i'm looking for something known as the intersection of the two sets so a and b i want the intersection of the two this is the symbol for that and this contains all the elements that are common to both well if you look through here it's just three that's the only element that's common to both so this set contains one element and that element is three and when we're solving this compound inequality with and we're doing the same thing we're looking for the intersection of the two solution sets all right so let's get started with the first problem and we already know how to solve a linear inequality in one variable that part is simple it's just putting the two together all right so if i have 6 minus 3r is less than 27 i can subtract 6 away from each side of the inequality so that's going to cancel i'll have negative 3r is less than 27 minus 6 is 21. divide both sides of the inequality by negative 3. remember if i divide by a negative i've got to flip the direction of the inequality symbol so this is a less than it's going to become a greater than so this would cancel with this and i'd have r and now it's a greater than 21 divided by negative 3 is negative 7. now over here i'm going to solve this one separately let me just put a little line here i'm going to subtract 6 away from each side of the inequality so i'll have 7r is less than or equal to 48 minus 6 is 42. i'm going to divide each side by 7 and i'll get that r is less than or equal to 6. so i have these two solutions i'm going to go down to where there's a number line we're going to do the whole song and dance with the interval notation and graphing and then we're going to put the two solution sets together all right for the first one and let me just write it out here we had 6 minus 3r is less than 27. we found that for this one r was greater than negative 7. so in interval notation what does that look like well it looks like this you'd have a parenthesis to say that negative 7 is not included a comma and then you'd have an infinity symbol with a parenthesis next to that so that's interval notation graphically on the number line at negative 7 it's not included i'd put a parenthesis to show that and then i would shade everything going to the right including that arrow now for the next part of the compound inequality we saw the problem 7 times r plus 6 is less than or equal to 48 and what we found there was that r was less than or equal to 6. so in interval notation since it's less than or equal to 6 we would have a parenthesis next to negative infinity comma and then up to and including 6. so 6 is going to have a bracket next to it all right now graphically at six we put a bracket facing to the left and we shade everything going in this direction again including that arrow now for the compound inequality let me just write this here we have 6 minus 3r is less than 27 and then and we have 7r plus 6 is less than or equal to 48. so what's the solution here well again it's the intersection of the two solution sets and for some people they like to look at it graphically for some people they like to look at it between the two intervals that were created and some people could just look out and say okay i know it's this well i can tell you that if you just look at the two solutions and you put them together r is greater than negative seven and r is less than or equal to six so that's pretty easy to do you could write that as a three part inequality where you say r is greater than negative seven and r is also less than or equal to six in interval notation i would write this with a negative 7 next to a parenthesis to say negative 7 is not included and then comma you have a 6 which is included so that gets a bracket now graphically it's easy to tell what it would have been here's six right here kind of thing about this is the cutoff for the largest value six is included because this has a bracket here and in this situation it's going this way forever and ever and ever so six is included now with negative seven it's not included because in this case it's got a parenthesis there it's not included there okay so that should cut off anything larger than negative seven but kind of anything in between here anything in between would work as a solution for each of them so graphically it's pretty easy to see without even referencing this i know that at negative 7 i would have a parenthesis because it's not included i know at 6 i would have a bracket because it is included and i would shade everything in between everything in between and again the reason we're doing this is it has to be a solution to each right and this is the section of the number line that satisfies both so again our solution for the compound inequality is that r is greater than negative seven which is displayed here it's got to be greater than negative seven negative seven is not included we put a parenthesis there and it's also less than or equal to six so it could be six which is why we have a bracket and then anything less so it's this range here anything greater than negative seven up to and including the number six all right for the next one we're going to work on one with some fractions involved and nothing we're not used to so what we have here is one plus eight thirds n is greater than negative five thirds and we have negative n plus one half is greater than negative seven six again we wanna solve each one separately so i'm gonna start out with one plus we have eight thirds n and this is greater than negative five thirds so i want to clear the denominators so i'm going to multiply both sides here by three and so three times one is three plus three times eight thirds is eight then times n and this is greater than three times negative five thirds is negative five let's scroll down a little bit get a little room going i would subtract three away from each side of the inequality this will cancel i will have 8n is greater than negative 8 divide each side by 8 and i get n is greater than negative 1. okay so again n is greater than negative 1. now over here where i have negative n so i have negative n plus one half is greater than negative 7 6. again i want to clear the denominators the lcd here would be six right i've got a two and a six lcd would be six so i multiply both sides of the inequality by six six times negative n is negative six n plus six times one half is three six cancels with 2 and gives me 3 3 times 1 is 3. this is greater than negative 7 6 times 6 is negative 7. so now let me scroll down a little bit a little room going subtract 3 away from each side this will cancel i will have negative 6n is greater than negative 10. we will divide each side by negative 6. remember if i divide by a negative i've got to flip the direction of the inequality symbol that's a greater than so now it's going to be a less than so this cancels with this and i have n this is less than negative 10 over negative 6 is positive and i know each is divisible by 2. 10 divided by 2 is 5. 6 divided by 2 is 3. so this is positive five thirds so then again this was n is less than five thirds okay so we're just going to use one graph this time and we can put it together and have one solution so we found that for one plus eight thirds n is greater than negative five thirds the solution was n is greater than negative one so in interval notation that looks like this you have a parenthesis a negative one comma infinity and then a parenthesis graphically we know that at negative one we would have a parenthesis and then we would shade everything to the right now i'm going to add something to this i have and so we're connecting this we have negative n plus one half is greater than negative seven six we found that the solution for this was n is less than five thirds so an interval notation that looks like this you have negative infinity next to a parenthesis comma you'll have five thirds and that'll be next to a parenthesis as well because it's not included so graphically if i look at this where's five thirds at well it's basically one and two thirds so it's something like right here let's say there's no notch for but let's let's create one and say this is five thirds so what i'd be doing is i'd be placing a parenthesis there facing to the left okay and then i would shade everything going this way right going this way it's easy to see graphically what the intersection would be it would not include negative 1 it would not include 5 3 but it would include everything in between so let me erase all of this and i'll keep those two parentheses there and i want to shade everything in between this is my solution for the compound inequality and i could also write it with n in between so i could write n in between and we know that it's greater than negative one and it's less than five thirds all right let's take a look at another one so we have 4n plus 7 is less than 9n plus 2 and 13n plus 11 is greater than or equal to 6n plus 11. so let's start out with this one we have 4n plus 7 is less than 9 n plus 2. let's solve that guy so i want to subtract 7 from each side that'll cancel i also want to subtract 9n from each side that'll cancel 4n minus 9n is negative 5n this is less than 2 minus 7 is negative 5. we divide each side by negative 5 and don't forget to flip this so right now that's a less than it's going to become a greater than and this would cancel with this and give me n so n is greater than negative 5 over negative 5 is 1. so for this one we can say that n is greater than 1. for the next one we have 13n plus 11 is greater than or equal to 6n plus 11. i don't have the same thing on each side of an inequality or an equation i could just get rid of it right because i could subtract 11 away from each side it would cancel and now i can just subtract 6n away from each side so i'll have 13n minus 6n that's 7n is greater than or equal to this has all been cancelled so it's 0. now i divide both sides by 7 and i get n is greater than or equal to 0. so this is n is greater than or equal to 0. so again we had 4n plus 7 is less than 9 n plus 2. and 13n 13n plus 11 is greater than or equal to 6n plus 11. so the solution for this guy was that n is greater than one or we could say in interval notation you'd have a parenthesis a one an infinity symbol and a parenthesis graphically i would put a parenthesis at one and i'd shade everything to the right now for this guy 13n plus 11 is greater than or equal to 6n plus 11 we found that n was greater than or equal to 0. so in interval notation you'd have a bracket next to 0 because 0 is included comma and then infinity and you always use a parenthesis there now for this one i'm going to have a bracket that starts at 0 and i'm going to shade everything to the right so it's pretty easy to see what the solution for the compound inequality is anything to the left of 1 and also including 1 doesn't work because it has to work for both and it would not be a solution for this original because here n has to be greater than 1. but anything that's larger than one works for both right because both areas are shaded there and so for the compound inequality let me just erase this part it would essentially be the solution for this guy right so it would be n is greater than one so the solution for the compound inequality is that n is greater than one or in interval notation again the same thing as this one is not included but anything larger is all right let's take a look at one more with and and then we'll start talking about the scenario of or so we have 19v plus 18 is less than 12v minus 10 and 7v plus 11 is less than 8 plus 10v so if i start with this first one here 19v plus 18 is less than 12v minus 10. so i'm going to subtract 12v away from each side that's going to cancel i'm going to subtract 18 away from each side that'll cancel 19v minus 12v is going to give me 7v and this is less than negative 10 minus 18 is negative 28. divide each side by seven and we'll get that v is less than negative four so this is v is less than negative four all right for the second part of this i have that seven v plus 11 is less than 8 plus 10 v again what i can do is subtract 10v away from each side that's going to cancel 7v minus 10v would give me negative 3v i would subtract 11 away from each side that's going to cancel so negative 3v is less than 8 minus 11 is negative 3. i will let me scroll down get a little room going i will divide both sides by negative 3. remember i'm dividing by a negative so i've got to flip the direction here so this becomes a greater than so this cancels with this i have v is greater than 1. so this is v is greater than one and again our problem is 19v plus 18 is less than 12v minus 10 and 7v plus 11 is less than 8 plus 10 v now the solution for this guy right here we know is v is less than negative 4. so in interval notation we would say that we have a parenthesis next to negative infinity because it's anything less than negative four so up two but not including negative four so that's a parenthesis next to that graphically at negative four i put a parenthesis facing to the left and i would shade everything going this way and for this guy over here we found that v was greater than one so in interval notation i put a one it's not included so i have a parenthesis there then comma i have the infinity symbol parenthesis next to that graphically i'd find one i'd put a parenthesis there and i'd shade everything to the right do you see a problem with this scenario there's a solution for each part of the compound inequality but the compound inequality itself has no solution why is that the case because it has to overlap it has to be a solution for each and that doesn't exist there's no solution here no solution because if you look here for this guy right here this 7b plus 11 is less than 8 plus 10v that's satisfied by numbers that are greater than 1. this guy right here this 19v plus 18 is less than 12v minus 10 that's satisfied by numbers that are less than negative 4. so there is no number that is less than negative 4 and also greater than one that doesn't exist and so there is no solution now one thing that your teacher might want you to do remember you have a symbol when you have a set that contains no elements it's a symbol for the null or empty set you might just want to put no solution and then put this symbol if you get this scenario all right now let's talk about solving a compound inequality with the word or so again you want to solve each inequality separately so you always do that part the next part is a little bit easier actually you just say the solution set contains all numbers that satisfy either inequality so it doesn't matter if it's a solution to only one of them or it could be a solution to both either way it's going to be a solution for the compound inequality so as an example with sets i go back to my set a and set b if i want the union of these two sets so the union of a and b is equal to what well everything goes in the only thing you got to make sure of is if you have an element that's common to both you list it only once okay only once so you have one two three only gets listed once then four then 5 then 7 then 9. so the union of these two sets is all the elements combined again nothing gets double listed so each and every element from a along with every element from b is an element of this set all right so let's look at the first problem we have 5k plus 2 is greater than 32 or k plus 5 is less than or equal to 9. so i want to subtract 2 away from each side over here i'll have 5k is greater than 30. i'll divide each side by 5 and i'll get k is greater than 6. so then we have or so we have this one over here subtract 5 away from each side i'll have k is less than or equal to four so again we have five k plus two is greater than 32 or we have k plus five is less than or equal to not so for this guy we found that k was greater than six in an interval notation that looks like this you have six comma infinity with a parenthesis at each end graphically i'd put a parenthesis at 6 that faces to the right and then shade everything to the right then for k plus 5 is less than or equal to 9 we found that k was less than or equal to 4. so an interval notation i'd have a negative infinity comma a 4 with a bracket next to 4 and a parenthesis next to negative infinity so i'd find 4 i'd put a bracket facing to the left because anything less than and i'd shade all of this to the left now we just saw that if we had the connective word and we don't have a solution right because nothing is going to overlap there's no intersection between the two solution sets but with or it's different it can be a solution to this guy or this guy it's a solution for the compound inequality so our solution set is written like this we would say that we have this set so negative infinity up to and including 4 it's the union of this i'm going to put the union symbol with this guy so anything greater than 6 out to infinity so it's the union of the two sets and graphically i don't need to do anything else it's 4 or anything less so that's all part of the solution and then anything greater than six so here's your solution using interval notation and here it is graphically and then additionally you could write it just saying that k is less than or equal to four or just put the word or k is greater than six all right for the next one we're looking at two fifths p minus four thirds is less than two p plus one fourth or three fourths p plus 12 fifths is less than or equal to six fifths p plus three halves all right so let's start out with this guy right here so i have two fifths p minus four thirds is less than 2 p plus 1 4. so i want to clear my denominators and the lcd here would be 4 times 3 times 5 or 60. so let me multiply both sides by 60. 60 times 2 fifths 60 would cancel with 5 and give me 12 12 times 2 is 24 so this would be 24p then minus 60 times four-thirds 60 divided by 3 is 20 20 times 4 is 80. this is less than 60 times 2p is 120p then plus 60 times one-fourth 60 divided by 4 is 15 15 times 1 is 15. so then what i can do is add 80 to each side [Music] and i can subtract 120p from each side so this is going to cancel and this is going to cancel 24 minus 120 is negative 96 so this would be negative 96 p and then it's less than 15 plus 80 is 95. so if i divide both sides by negative 96 remember i've got to flip the direction of this guy this becomes a greater than now i'll have p is greater than 95 over 96 and because this is negative here i'll put a negative out in front so p is greater than negative 95 over 96. so p is greater than negative 95 over 96. now we'll deal with this one so let's scroll down a little bit so i have three fourths p plus 12 fifths is less than or equal to six fifths p plus three halves so the lcd here would be what we have two four five so basically be four times five or twenty so multiply both sides by twenty and so 20 times 3 4 20 would cancel with 4 giving me 5 5 times 3 is 15 so this is 15 p plus 20 times 12 fifths 20 would cancel with 5 and give me 4 4 times 12 is 48 this is less than or equal to 20 times six fifths p 20 would cancel with five and give me four four times six is 24 and times p then plus 20 times three halves 20 would cancel with two and give me 10 10 times 3 is 30. all right so we can now subtract 24p away from each side cancel this you could subtract 48 away from each side cancel this 15 p minus 24 p is negative 9 p and this is less than or equal to 30 minus 48 is negative 18. we divide each side by negative 9. and again we've got to flip the direction of the inequality symbol so this becomes a greater than or equal to so this cancels we have p is greater than or equal to negative 18 over negative 9 is 2. so p is greater than or equal to two so again i had two fifths p minus four thirds is less than two p plus one fourth or the other one was three 4 times p plus 12 fifths is less than or equal to six fifths p plus three halves now for this guy right here we found that p we found that p was greater than negative 95 over 96. so in interval notation that looks like this we would have a parenthesis then negative 95 over 96 and then out to infinity of course you use a parenthesis with that as well for this guy we found that the solution was p is greater than or equal to 2. so in interval notation you have a bracket and then you have 2 and then out to infinity so you put a parenthesis there graphically you can think about this one right here negative 95 over 96 that's going to be a little bit to the right of negative 1 because it's not quite negative 1 it's a little bit larger it's a little bit to the right of it okay if i had negative 96 over 96 that equals negative 1. this is a little bit shy of that so let's just put that right here let's say let's say this is negative 95 over 96 and it's anything larger than that but not including that so i'd put a parenthesis facing to the right and i'd shade all of this now the other guy was anything greater than or equal to two so that would be right here and i would shade everything to the right now for the compound inequality it could be a solution either so we would go with this solution right here because it encompasses both it encompasses this one and this one so i can erase this part right here and just say that and i can shade this in any color i want let's just do red this is the solution for the compound inequality and for this inequality so the solution for the compound inequality is just this guy right here we'll put a parenthesis negative 95 over 96 comma infinity another parenthesis this is graphically and then we can also just say it's p is greater than negative 95 over 96 because this solution contains completely this solution and again if it's a solution to either it's the solution for the compound inequality with or all right let's take a look at one final problem so we have 2x plus 1 is less than negative 19 plus 7x or 3 plus 15x is less than or equal to 10x plus 18. so let me solve this first guy up here so we have 2x plus 1 is less than negative 19 plus 7x what i want to do is i want to subtract 1 away from each side and i want to subtract 7x away from each side so that's going to cancel and we scroll down and get a little room going so 2x minus 7x is negative 5x this is less than negative 19 minus 1 is negative 20. divide each side by negative 5. again flip the direction of this guy so it's now greater than this cancels with this and i'll have x is greater than negative 20 over negative 5 is 4. so here x is greater than 4. all right for this guy right here i could subtract 3 away from each side so that would cancel and i could subtract 10x away from each side so that would cancel 15x minus 10x is 5x and this is less than or equal to 18 minus 3 is 15. we divide each side by five we get that x is less than or equal to three this is x is less than or equal to three all right so we have that 2x plus one is less than negative 19 plus 7x then or we have the other scenario 3 plus 15x is less than or equal to 10x plus 18. so for this guy we found that the solution was x is greater than four so in interval notation we would say that we put a parenthesis at four and then anything out forever and ever and ever so we put infinity here and then graphically at four i'd put a parenthesis facing to the right and shade all of this now for this guy right here we found that x was less than or equal to 3. so in interval notation there's no smallest value so we put negative infinity put a parenthesis next to that and it's going up 2 and including 3 so we put a bracket next to 3. graphically i put a bracket at three facing to the left and then i shade everything to the left and i'm gonna shade this arrow now you can see that there's no overlap between the two solutions so if this wasn't an or if this was an and you would have no solution for the compound inequality but because this is or and it can be a solution to either to be a solution for the compound inequality it's the union of the two solution sets so in other words it is this guy right here so it's negative infinity up to and including three it's the union with this guy we have four that's not included we have the parenthesis there out to infinity graphically it's just this i don't need to change anything i've graphed each one so i don't need to do anything else and then you can also write it as saying x is less than or equal to 3 or x is greater than 4. each of these would show you the solution for that compound inequality hello and welcome to algebra 2 lesson 13. in this video we're going to learn about solving absolute value equations so for most of you in an algebra 2 course this is a new topic most algebra 1 courses do not go into this it's considered a little bit advanced but some of you might have seen it before the main thing here is just to have a fundamental understanding of absolute value everything else we're going to do is very very easy so i want to start with just a brief review talk about things that we learned a long long time ago and you might have forgotten so the absolute value of a number is simply the distance between that number and 0 on the number line i want you to remember that absolute value is either 0 or it's a positive value it's never going to be negative because again it represents a distance it's kind of like if i get in a car and i want to drive somewhere i can drive zero miles or i can drive some positive amount but i cannot go a negative amount right now i don't want to hear the joke where you're going to get in your car and go backwards okay that's not going to work but i have some examples here and just two brief examples here's a number line and let's say we want the absolute value of negative three and again these bars here denote we want to take the absolute value of what's inside so what is the absolute value of negative three well i find negative three on the number line that's here and i just travel to zero how many units would i go well i would go one two three units and i'd end up at zero and it doesn't matter what direction i go and it's how far i need to travel so my absolute value there would be 3. as another example if i wanted the absolute value of 1 well i'm coming from a different direction here but i'm just traveling 1 unit to 0. so the absolute value of 1 is 1. basically i know that all of you remember that the absolute value of a number if it's positive is just the number the absolute value of a number if it's negative is just the opposite of it right just make it positive so if i went through a few and said what is the absolute value of let's say 6 well it's a positive number so it's just the number and i can kind of erase this and look at the number line and say okay well the absolute value of 6 i can find that by just going from 6 to 0 and it's just what it's 1 2 three four five and finally six units traveled to zero okay that's why the absolute value of six is six and the same thing would be true if i went from negative six to zero from here to 0 is also 6 units it's 1 2 3 4 5 and finally 6. so the absolute value of negative 6 i would just make it positive and it's 6. so that leads me to my next point here and that point is that opposites or we call them negatives or additive inverses whatever you want to refer to them as are numbers that have the same absolute value okay the same absolute value very important you understand that for this lesson now i want to show you a couple of examples here and you could come up with as many as you wanted to but something like 3 and negative 3. if i take the absolute value of 3 it's equal to 3. if i take the absolute value of negative 3 it's equal to 3. so 3 and negative 3 are opposites and they have the same absolute value and again you could show that on the number line you can go up here and say okay well this is 3 this is negative 3 and each is three units away from zero this is one two three units away from zero and this is one two three units away from zero as well and again it doesn't matter which direction you're going in you're just counting how far away you are from zero it'd be kind of like if i was north and i traveled 500 miles to a location or if i was south and i traveled 500 miles to a location it wouldn't matter where i came from it's just a matter of it's 500 miles to get there okay so that's the concept of absolute value so these i don't have to show graphically i think we get it at this point we know five and negative five if i take the absolute value of five i get five if i take the absolute value of negative five i would get 5 as well negative 4 and 4. i take the absolute value of negative 4 i get 4 if i take the absolute value of 4 i get 4. so anytime you have opposites the absolute value is the same now let me go to the first example that you're going to see in your textbook or that you're going to come across in class so you'll get something simple like this the absolute value of x equals 6. so the question you have to ask yourself before you read anything before you talk to anybody just thinking on your own what could i replace x with to get a true statement well think about it i want this to be some number that when i take the absolute value of it gives me 6. again there are two numbers that have the same absolute value of six there's six and there's negative six because each is one two three four five six units away from zero on that number line this one the same thing one two three four five six units away from zero on the number line so in this particular case x could be equal to what it could be six or x could be also negative 6. if i plug in a 6 there i take the absolute value i get 6. if i plug in a negative 6 there i take the absolute value i also get 6. as another example let's say i had the absolute value of x is equal to 4. again what could x be here well there's two possibilities it could be 4 or it could be negative 4 because each would have an absolute value of 4 right so it could be x could be 4 or x could be negative 4. again plug in a 4 here absolute value of 4 is 4. plug in a negative 4 here absolute value of negative 4 is 4 as well and again you can just show yourself this on the number line this is 1 two three four units away from zero and this is one two three four units away from zero as well so obviously that's the easiest scenario you're going to come across but you use the same concept when you get into the harder ones to solve an absolute value equation we isolate the absolute value part on one side we then set up a compound equation with or okay remember or is telling you that either solution is going to work so let's look at an example so we have the absolute value of 6n plus 10 is equal to 64. so what we need to do again if this is isolated for us we can say that 6n plus 10 this is inside of absolute value bars can be equal to positive 64. let me put a plus out there for emphasis or it could also be equal to the negative of 64. so 6n plus 10 could be equal to negative 64. and you might say well that i don't know it doesn't make much sense we go back to the last example where we had the absolute value of x was equal to 4. well this is x could be equal to 4 or x could be equal to negative 4. it's the same principle here it's just a little bit more advanced and remember i could plug in a 4 here take the absolute value i get 4. i could plug in a negative 4 here take the absolute value i get 4 both of these would work what i am saying here is i've got to solve for n and i want a value for n that when i plug it in here okay i'm going to multiply it by 6 i'm going to add 10. the result i'm saying could be 64 positive 64 or it could be negative 64. it makes no difference because the absolute value operation will make it positive right if it's positive it stays positive if it's negative it's going to become positive so i don't care if i get 64 or negative 64 because either way i'm going to get a solution to the equation so let's go ahead and solve each one of these and we're going to plug both of them in and i'm going to prove to you that works either way so for this one i'm going to start out by subtracting 10 away from each side and this will cancel i'll have 6n is equal to 64 minus 10 is 54. i will divide both sides by 6 so that n is by itself so this will cancel with this and i'll have n is equal to 54 divided by 6 is not so that's one solution then we have our or so we have another scenario so we're going to subtract 10 away from each side over here as well that will cancel i'll have 6n is equal to negative 64 minus 10 is negative 74. if i divide both sides by 6 here this will cancel with this and i'll have n is equal to so negative 74 over 6 i can reduce this i know it's going to be negative 74 is divisible by 2 74 divided by 2 is 37 and 6 divided by 2 is 3. so there's my other solution so i'm going to put or here and let me erase this part and i'm going to show you that they each work i'm also going to write this in solution set notation again we've got to get more used to doing that so in solution set notation remember i create some set brackets and inside those brackets i'm going to put my solutions as the elements so i have negative 37 thirds comma the other element is 9 that's my other solution so i want to check these two and i want to show you what happens when we plug them in again n is equal to 9 or n is equal to negative 37 thirds so the first one n equals 9. so what happens when i plug in a 9 for n so you have 6 times 9 plus 10 and then we take the absolute value of this and it should equal 64. well 6 times 9 is 54 54 plus 10 is 64. so you get the absolute value of positive 64 which of course is going to equal 64. so the left and the right side here are going to be equal so that's our first scenario that's kind of the easy one to understand the second scenario is the one where it kind of takes a little bit of thought into absolute value to understand so for this one let me kind of back this up a little bit because i'm dealing with a fraction here so it'll be a little messier so i have 6 times negative 37 over 3 that's what i'm plugging in for n then plus 10 and i want the absolute value of this guy and it should equal 64. so i know that this will cancel with this and give me a 2. 2 times negative 37 as we know is negative 74. so this would be the absolute value of negative 74 plus 10 and this should equal 64. negative 74 plus 10 is negative 64. so this would be the absolute value of negative 64 is equal to 64. now again when i take the absolute value of negative 64 it's going to make it positive so the two sides are going to be equal because of the absolute value operation so that's why we end up with two solutions here so this i'm going to scroll down a little bit turns out to be 64 equals 64. and so this is good to go so what we can say here is that this solution is correct as well and again it's an important insight to understand why these two solutions work again when i plugged in a 9 here i got positive 64. took the absolute value i got 64 equals 64. when i plugged in a negative 37 over 3 in for n i got negative 64. took the absolute value of that i got positive 64. so it might be still a little bit confusing but we're going to work a lot of examples here so let's go to the next one so for the next one it's a similar setup we have the absolute value of 10x plus 10 and this is equal to 20. so again what i'm saying is that what's inside the absolute value operation which is 10x plus 10 well that could be equal to 20 but also i'm going to put or this could also be equal to negative 20. so 10x plus 10 could be equal to negative 20. you set up these two scenarios you set up a compound equation with or again because if i find something for x that when i plug it in makes this whole thing equal to 20 i'm good right the absolute value of 20 would be 20. additionally if i find something that i plug in for x and it makes this whole thing equal to negative 20 i'm also good because the absolute value operation will make that negative 20 positive 20. so again you've got to think about those two different scenarios let's go ahead and check this out so we would subtract 10 away from each side of the equation here that would cancel i'd have 10x is equal to 20 minus 10 is 10 divide both sides by 10 and we would get x is equal to 1. over here let's subtract 10 away from each side of the equation that cancels i'll have 10x is equal to negative 30. divide both sides by 10 and we'll get x is equal to negative 3. so what we find here is that x equals 1 or x equals negative 3. in solution set notation i would say negative 3 and 1 are my elements right they're my two solutions let me erase this real quick i'm not going to check all of these but i will check a few just to kind of pound the point home that this value inside can end up being 20 or negative 20 and you're going to end up with the same value on each side right once you take the absolute value so if i start by plugging in a 1 for x the obvious scenario 10 times 1 is 10. so i'd have 10 plus 10 and i take the absolute value of that and that should be 20. and of course 10 plus 10 is 20. so the absolute value of 20 is 20 you get 20 equals 20 so of course we are good to go there the other one again that's the one that confuses people and that's the one that people just don't think about so if i plug in a negative 3 there i would have the absolute value of 10 times negative 3 plus 10. this should equal 20. 10 times negative 3 is negative 30. so i would have the absolute value of negative 30 plus 10 which is negative 20 and this should equal 20. again the absolute value operation here is what allows us to have a negative 20. when i take the absolute value of negative 20 i get positive 20 okay so this becomes 20 and it equals 20. so 20 equals 20 that's true so this works out as a solution as well so here x can be 1 or equally valid x could be negative 3. two solutions for this equation all right so now that we've gotten through the easier ones let's look at a tougher one now what do i do in this scenario i have negative 5 minus 4 and this is multiplication because it's hanging out outside so times the absolute value of negative 8 plus k this is equal to negative 17. well what we want to do is we want to isolate the absolute value operation so we want to get this by itself and we can do that because we happen to be very good at algebra at this point so the first thing i'm going to do is i'm going to add 5 to each side of the equation so this would cancel what i'm left with on the left i'd have negative 4 times i'm going to put a multiplication symbol there so it's obvious the absolute value of negative 8 plus k and this is equal to negative 17 plus 5 which is negative 12. now again this is multiplication i want this thing by itself so if i have something multiplying something else and i want to isolate it i'm going to use division just like if i had 6x equals 30 i want to isolate this i divide both sides by 6. right simple it's the same thing if i want to isolate this i divide both sides by negative 4 because that's what i want to go away so this is left the absolute value of negative 8 plus k and then on the right side negative 12 divided by negative 4 is simply 3. so now we have it in a format that we saw in the last two examples it's all that you need to do just get the absolute value operation isolated on one side you want a number on the other and again you set up your compound equation so this guy inside the absolute value bars can be equal to this or the opposite of that so it is negative eight plus k is equal to three or negative eight plus k again could be equal to negative three if this was three it's good if this was negative three the absolute value operation makes it good as well okay so to solve these very very simple equation add 8 to each side over here you're going to get that k is equal to 11. and over here add 8 to each side and you're going to get that k is equal to 5. so i've got my solution of k equals 11 and then my other solution of k equals 5. so let's erase everything we're going to check that again we have that k equals 11 or k equals 5 and again using the solution set notation 5 comma 11 that's going to be my two elements so let's start by just plugging in a 5. so i plug in a 5 there so i'd have negative 5 minus 4 let me make that negative a little better and then this is times the absolute value of negative 8 plus 5 and this equals negative 17. so negative 8 plus 5 is going to give me a value of negative 3. so this would be negative 3. remember this is multiplication so down here i'd have negative 5 minus 4 times again the absolute value operation makes this positive 3. this should equal negative 17. so i do multiplication before i do anything else you can think about this as plus negative to make it easy negative 4 times 3 is negative 12. so you have negative 5 plus again this is negative 12 and this should equal negative 17 and it does negative 5 plus negative 12 is negative 17. so our solution of k equals 5 is correct all right for the other one we want to try k equals 11. so you'd have negative 5 minus 4 again times the absolute value of negative 8 plus 11 this equals negative 17. so negative 8 plus 11 is 3. so let me replace this with 3. and again the absolute value of 3 is 3 so i'd have negative 5 i'm going to write this as plus negative 4 times 3 equals negative 17. it's the same thing because the absolute value operation whether it's 3 or negative 3 puts out a positive 3. so either way i got the same thing here the rest of it's the same negative 4 times 3 is negative 12 you get negative 5 plus negative 12 equals negative 17. this is negative 17 equals negative 17. so this checks out as well so k here can be 11 or equally valid k here can also be equal to 5. all right so the next three problems i'm going to give you are going to be trap questions that you're going to come across your teacher your textbook whoever you get information from your tutor maybe just this video somebody's going to give you a problem that's going to look something like what you're going to see in the next three problems so a common trap question would be something like this the absolute value of x is equal to negative seven again why is that a trap question go back to a number line and i'll bring one to the screen there tell me what number you have that has a distance from zero of negative seven it doesn't exist you might say oh well seven no it doesn't work that way seven has a distance of seven units away from zero right if i counted it'd be one two three 4 5 6 7. the answer here is there is no such x because as i told you the very first part of this video absolute value is the distance a distance is either 0 or it's positive i get in my car i want to go somewhere i either start the car and it doesn't work and i don't go anywhere and i go damn i'm not going anywhere today or i go some positive amount i do not go negative again not with a joke about backing up that doesn't work so for this particular type of problem there is no solution there is no solution okay you can also put the symbol for the null or empty set right because the solution set contains no elements and you just want to remember that whatever is inside the absolute value bar you just say the absolute value of x for example this is always greater than or equal to zero it can be zero or positive it cannot be negative so this is not going to work all right so let's look at an example so you see something like this it looks nice and normal like okay i'm ready to get my solution so you have negative 7 times the absolute value of 2x plus 2 minus 4 equals 52 okay so i'm going to start by trying to what isolate this i want to isolate this so i'm going to add 4 to each side that's going to give me negative 7 times the absolute value of two x plus two this would cancel it'd be equal to fifty two plus four is fifty-six now if i wanna isolate this again if i'm isolating this and this is being multiplied by that well i want to divide both sides by this so i can get rid of it so this would cancel with this and over here 56 divided by negative 7 is negative 8. so i now have the absolute value of 2x plus 2 is equal to negative 8. is there a problem yes there is i know this is a little bit more complex than me just saying the absolute value of x is equal to negative 7. that's obvious this one right here somebody might start thinking yeah i can make that work i can go 2x plus 2 equals negative 8 or 2x plus 2 equals 8. i see students do this all the time it's not going to work stop and think about it i can find an x here that's going to give me a value of negative 8 when i evaluate so let's go ahead and do that i'm going to show you what happens let's subtract 2 away from each side of the equation and that will cancel you'll have 2x is equal to negative 10. divide both sides by 2 you get x equals negative 5. okay so let's say i plug in a negative 5 here what's going to happen let me just erase this so we have a little room to work we don't even need it anymore so 2 times negative 5 is negative 10. so i have the absolute value of negative 10 plus 2 is equal to negative 8. now finish this up negative 10 plus 2 is negative 8. and you might say oh i nailed it now i got negative 8 on this side negative 8 on this side hold on it's inside of the absolute value operation so it's not going to work once i take the absolute value of this it's positive 8. positive 8 does not equal negative eight so this doesn't work you can't do it that way okay if you isolate your absolute value operation and it ends up being equal to a negative stop there is no solution okay so when we come across a scenario like this and we go through all the work and we isolate it we say okay the absolute value of 2x plus 2 is equal to negative 8. we stop we say nope they're not going to get me on this trap question and there's no solution okay put the symbol for the null or empty set because again no matter what you plug in for x there once you take the absolute value of it it's going to be zero or positive it will never ever ever be negative so you can't possibly get a solution all right so another special case is when you've isolated your absolute value operation and it's equal to 0. so here i have the absolute value of x plus 3 equals 0. well in this case you can get a solution but there's only one possibility because 0 is its own opposite right it's it's absolute value of 0. it's 0 units away from itself by definition if i look at a number line just for reference here's zero how many units is it away from itself zero if i get on google maps and i say how far am i away from where i'm standing i'm zero units away so it's a distance it's zero units away from itself so in this case i would just say well if the absolute value of x plus 3 equals 0 then i just solve one equation x plus 3 equals 0. i don't need to go x plus 3 equals negative 0. i see that i don't test all time 2. please don't do that okay there's no such thing as negative 0. negative 0 is just 0. so then i would just subtract 3 away from each side here and i would get that x is equal to negative 3. and that makes sense because if i plug in a negative 3 there negative 3 plus 3 is 0 the absolute value of 0 is 0. and we could just write this again using solution set notation put a negative 3 inside of some set braces all right so let's look at an example here so we have negative 6 times the absolute value of 3y plus 1 minus 2 is equal to negative 2. again let me isolate my absolute value operation and so to do that i'm going to add 2 to each side so that will cancel and i'll have negative 6 times the absolute value of 3y plus 1 and this is equal to negative 2 plus 2 is 0. now i want this by itself if negative 6 is multiplying that divide both sides by negative 6 to get rid of it okay this is going to cancel with this and i'll have the absolute value of 3y plus 1 is equal to 0 over negative 6 is 0. i have the scenario i just told you about the absolute value operation is isolated and it's equal to zero so there's only one possibility it can only be true that three y plus one equals zero again please don't say or three y plus one equals negative zero the negative or opposite of 0 is just 0. so you have one situation here to solve and i just subtract 1 away from each side i'll get 3y equals negative 1 divide both sides by 3 and i get y equals negative 1 3. okay so let's erase everything okay so y equals negative 1 3. in solution set notation i put a negative 1 3 inside of some set braces and again if you want to check this pause the video plug in a negative 1 3 here and see that you get a true statement you should get negative 2 equals negative 2 and in fact let's just go ahead and do it so i'll have negative 6 times the absolute value of three times negative one third plus one and this is minus two equals negative two so three times negative one third this will cancel with this and give me negative one negative one plus one is zero so this right here would end up being zero so i'd have negative six times remember this is multiplication because it's hanging out outside i'd have negative six times zero then minus two equals negative two zero times anything is zero so this is gone this is zero i'll just have negative two equals negative two so yes this solution is correct y equals negative one third all right so kind of the last special case scenario you're gonna deal with is where you have two absolute value expressions that are equal to each other so before we kind of get into that i want to remind you of one thing that you might have overlooked so two expressions or two numbers can only have the same absolute value if they are equal so that's one scenario 2 has the same absolute value as 2 those two numbers are the same they're equal negative 2 has the same absolute value as negative 2. again those two are the same they're equal negative 19 and negative 19 have the same absolute value or the other scenarios they're opposites like we've been talking about so 14 and negative 14 12 and negative 12 210 and negative 210. so they have the same absolute value because they're opposites or again numbers or expressions that are the same or equal also have the same absolute value so those are your only two scenarios so when we encounter something like this the absolute value of 5x minus 3 is equal to the absolute value of 2x plus 2 so there's two possibilities one is that 5x minus 3 is equal to or is the same as 2x plus 2. so this would kind of be like saying okay the absolute value of 3 equals the absolute value of 3. they're the same same number so of course the absolute value would be the same we're saying that these are the same so that's kind of your first scenario then you have or you have a second scenario where you make one of these into its opposite so for example i have 5x minus 3 is equal to if i want the opposite of this i just put parentheses around it and i put a negative out in front so the negative of or the opposite of the whole expression so 2x plus 2. the common mistake with this a lot of people see in their textbook and they forget about how it was modeled they'll just put a negative out in front this is only making the 2x negative it's not doing anything with this the whole thing we've got to apply this negative there so make sure you use parentheses there so you apply the negative to each term okay very important so this here is the scenario where let's say we have the absolute value of 3 is equal to the absolute value of negative 3. they're opposites so they're still going to have the same absolute value so we've covered each scenario again two expressions can have the same absolute value if they're equal or if they're negatives here they're equal here they're negatives so we start out by solving this guy we add 3 to each side that's going to cancel and we subtract 2x away from each side that's going to cancel 5x minus 2x is 3x and this is equal to 2 plus 3 is 5. divide both sides by 3 and you get x is equal to 5 thirds the other guy over here let's distribute the negative to each term okay just think about it as a phantom negative 1 out there so we have or we will have 5 x minus 3 is equal to negative 2x minus 2 just distributed to the negative let me add 3 to each side and let me add 2x to each side [Music] so this will cancel and this will cancel 5x plus 2x is 7x and this is equal to negative 2 plus 3 is 1. divide both sides by 7 we get that x is equal to 1 7. so my solution set here contains two elements we'll have 1 7 and 5 thirds and let's go ahead and check this one because i want you to see that it works so again we have that x is equal to 1 7 or x is equal to five thirds and if i start by plugging in a 1 7 for x here and here let's see what we get so we have the absolute value of we have 5 times 1 7 minus 3 this should be equal to the absolute value of 2 times 1 7 plus 2. so 5 times 1 7 is 5 7. let me just replace this in here and this is minus 3. so i've got to get a common denominator going multiply 3 by 7 over 7 i'll have 21 over 7 and this ends up being negative 16 over 7 which of course once i take the absolute value of would be 16 over 7. then over here i have 2 times 1 7 plus 2. if i do 2 times 1 7 that's 2 7 that's easy enough and again i've got to get a common denominator going so let's multiply 2 times 7 over 7. 2 times 7 would be 14 so this would be 14 over 7 2 plus 14 would be 16 so this is 16 sevenths and of course if i take the absolute value of that that's just 16 7. so i get 16 7 equals 16 sevenths so we can see that the solution x equals 1 7 does work right we get the same value on the left and the right side and again remember this solution came from when we set them as negatives of each other i took this one and i set it equal to the negative of this one so you have a negative of this one but when we take the absolute value it's the same value so again that's why it works out now for the next part let's check the other guy we're going to put the same value in each side in the absolute value operation because we did this equals this without making one of them the negative of the other so for the other one we have the absolute value of 5 times 5 3 minus 3. then this equals the absolute value of two times five thirds plus two so of course five times five thirds five times five is 25 so this would be 25 thirds so to get a common denominator multiply this by three over three you would end up with nine thirds so 25 minus 9 is 16 so this would be 16 thirds so it'd be 16 thirds over here 2 times 5 is 10 you'd have 10 thirds so ten thirds and over here to get a common denominator multiply this by three over three two times three is six you'd have six thirds and if you add ten plus six you get sixteen over the common denominator of three you'd have sixteen thirds so the absolute value of 16 thirds is equal to the absolute value of 16 thirds of course that's true you get 16 thirds equals 16 thirds so yes this one works out as well this solution came from us just saying 5x minus 3 equals 2x plus 2. so of course when we got down to this stage of checking it it's the same value in here as in here okay it's the same value that we're taking the absolute value of in each case again when we checked this one we had the negatives of each other because we said 5x minus 3 is equal to the negative of this expression 2x plus 2 and so that's why we ended up with this equals the negative of this but again when we take the absolute value it makes everything positive so that's how we got a true statement in that case all right let's take a look at one more so the absolute value of 3x minus 9 is equal to the absolute value of 12x plus 15. so again i want one scenario where they're equal to each other so 3x minus 9 is simply equal to 12x plus 15 or i want another scenario where they're opposites so i can make this guy 3x minus 9 i can make it negative so the negative of that whole thing again make sure you use parentheses is equal to this would stay positive so 12x plus 15. right i won't do anything to that it just stays as it is so this is just straight removing the absolute value bars this one i'm making one into its opposite so that we follow the rule to say if two numbers or two expressions have the same absolute value they're equal to each other or they're negatives that's what we have they're equal they're negatives okay let's solve each one so we add 9 to each side over here and we subtract 12x from each side and so this cancels and this cancels 3x minus 12x is negative 9x and this equals 15 plus 9 is 24. we divide each side by negative 9 and this will cancel we'll get x is equal to 24 over negative 9 is going to be negative each is divisible by 3 so this would be negative 8 over 3. so negative 8 thirds then we have or all right let's deal with this scenario now so the first thing i'm going to do is just apply the negative so negative times 3x is negative 3x and then negative times negative 9 would be positive 9 and this equals 12x plus 15. now what i'm going to do is subtract 9 from each side of the equation and i'm also going to subtract 12x away from each side of the equation so what's going to happen is this will cancel and this will cancel negative 3x minus 12x is negative 15x and this equals 15 minus 9 which is 6. now all i need to do is divide both sides by negative 15. this cancels with this and i've got x is equal to each is divisible by 3. i know it's negative 6 divided by 3 is 2 15 divided by 3 is 5. so this is negative two-fifths so x equals negative eight thirds or x equals negative two-fifths so let's write that up here so again x equals negative two-fifths or x equals negative eight thirds and again in solution set notation we can write this as negative eight thirds comma negative two-fifths inside of some set braces and i'm not going to check this i'm going to leave this up to you to check again you're going to plug in each value for each occurrence of x so i plug in negative 8 3 here and here i would evaluate and make sure i get the same value on the left and the right and i'm going to do the same thing with negative two-fifths what you're going to find is that in one scenario you'll have the same value inside the absolute value operation on each side that makes sense because for one of these we just said remove the absolute value bars 3x minus 9 is equal to 12x plus 15. so that's what we expect there in the other scenario we said the opposite of this was equal to this so in that scenario when you check you're going to end up with opposites inside of the absolute value operation when you take the absolute value you'll end up with the same value on each side hello and welcome to algebra 2 lesson 14 in this video we're going to learn about solving absolute value inequalities so in our last lesson we learned how to solve an absolute value equation so here we're just going to take the next step and learn how to solve an absolute value inequality this isn't any more difficult and the reasoning is the same okay so the main thing i want to make sure that you understand in case you didn't watch the last lesson is the concept of absolute value it's very very basic but if you don't understand it you'll have a hard time understanding the structure of these problems so the absolute value of a number is the distance between that number and 0 on the number line so i have here a sample number line and i just have two quick examples of absolute value so we know that if we see a number inside of these vertical bars that means we want to find the absolute value of it so the absolute value of the number negative 2 is just 2. and so we know that from the number line we can see that negative 2 is here and 0 is here and so what i can do is i can count how far away negative 2 is from 0. it doesn't matter the direction i move it's just how many units i'm going to move so we would travel 1 2 units so you get from negative 2 to 0 so my absolute value is 2. and we all know at this point that if you take the absolute value of a positive number it's just that number if you take the absolute value of a negative number you just make it positive so here i took the absolute value of negative 2 just made it positive so for the absolute value of 5 i just keep the number right it's just 5. and again you can demonstrate this on the number line here's 5 and again here's 0 and i would go 1 2 3 4 5 units to get to 0. and again it doesn't matter which direction i go we're just counting distance it's just the same thing as if you get in your car and you ask your gps how many miles you have to go to a certain location well let's say as this example goes it's 50 miles and let's say i'm in a different location now and it's exactly 50 miles away it would still tell me i'm 50 miles away it wouldn't say it's negative 50 miles an hour wouldn't change based on the direction i'm coming from 50 miles is 50 miles doesn't matter which way i'm coming from so it's the same concept here so absolute value can never be negative it's a distance so it's got to be zero or positive and i think about this with driving again if i tell you we've got to drive somewhere today we can either say no i don't feel like going and we drive zero miles or we can drive some positive amount but we can't not go somewhere or drive backwards or something silly like that and come up with a negative distance okay it just doesn't work that way all right so now that we understand the concept of absolute value let's look at a very simple problem all right so let's take a look at our first sample problem this is a very easy one that you'd see at the very beginning of the section and it's just to understand what's going on when we're asking these questions so the absolute value of x is greater than 2. what are we asking for you might want to pause the video and just think about that for a second what are we asking for well we're asking for values whose absolute value is greater than two or if i started at zero what values are more than two units away from zero on the number line well if i just go 1 2 in this direction i know that anything larger than 2 would work so i'd shade everything to the right here and if i go 1 2 units to the left i know anything less than negative 2 would work so i'd shade everything to the left here so that gives me two solution sets one where it's negative infinity up to but not including negative two and it's the union of this with this other guy anything larger than 2. so thinking about this any number that's involved in either solution set here would satisfy this inequality just pick a number let's say we pick negative 8 plug in a negative 8 there take the absolute value you get 8 8 is larger than 2. let's say you pick 3 over here take the absolute value of 3 you get 3 3 is larger than 2. the only numbers that won't work are from negative 2 to 2 including both if i plugged in a negative 2 there the absolute value of negative 2 is 2 2 is not greater than 2. that's false so we know that this wouldn't work and this wouldn't work and anything in between would work if i chose 1 the absolute value of 1 is 1 1 is not greater than 2. so for this type of problem where we see a greater than we're going to end up with whatever's inside the absolute value bars so in this case it's x and it's going to be greater than 2. so in the first situation you would just remove the absolute value bars that's your scenario x is greater than 2. then you're going to have an or so it's going to be a compound inequality and in the second situation you're going to flip the direction of the inequality so i would have x i'm going to flip this and i'm going to make this part right here whatever's on the right side negative now you might say where does this come from where okay that's a common question you get in class because this one right here this is very straightforward everybody understands that x can be greater than 2. that makes sense but for x to be less than negative 2 yeah you can see it here once you kind of think about it it makes sense but once you get to the next example it's like where did that come from well all you'd have to do is think about absolute value for a second if x let me kind of scroll down a little bit remember i'm plugging something into the absolute value operation i can think about negatives as well so the negative of x the negative of x could also be greater than 2 and if i divide both sides by negative 1 to get x by itself what happens is i have to flip the direction of the inequality symbol so this is going to become a less than right which is what i have there and then this divided by this is just x and then 2 over negative 1 is negative 2. so that's where it comes from and once you see me do that you can remember to do that on your own in the next example and then you won't have to think about what's the other scenario this one's straightforward this one everybody forgets okay so it's x is greater than 2 or x is less than negative 2. if you had a generic example like the absolute value of x is let's say greater than i don't know let's say some variable k well this sets up to x is greater than k or x is less than negative k just following this format here in this case k is just 2 right so x is greater than 2 or x is less than negative 2. so here's another example very similar we have the absolute value of x is greater than 7. so again what we're asking for here is give me numbers whose absolute value or distance from 0 is larger or greater than 7. well i know anything larger than 7 would work and also anything less than negative 7 would work because if i look at 0 here if i count 1 2 3 4 5 six seven yeah anything to the left of negative seven that works and then again one two three four five six seven yeah anything to the right of seven that works as well that's a distance that's greater than seven units from zero but again we don't need to pull out a number line every time if i have a greater than remember use your generic example the absolute value of x is greater than k sets up to x is greater than k or x is less than negative k so x could be greater than 7 or x could be less than negative 7. and again let's do this one more time where does this come from well i can take this part right here and say well the negative of x could also be greater than seven the negative of x could also be greater than seven i divide both sides by negative one this gets flipped and so you'd have this cancels with this x is less than negative 7. so that's where this came from and again you might be asked to write this in interval notation and with or it's the union of the two solution sets so i would have from negative infinity up 2 but not including negative 7 and it's the union with the other solution set which is greater than 7 out to infinity all right so now that we've dealt with the greater than scenario let's talk about the less than scenario so we have the absolute value of x is less than three so here x can be any number that is less than three units away from zero so anything less than one two three units so it can't be 3 but it could be anything less than 3. so 1 2 3 can't be negative 3 but it can be anything greater than that so basically it's the numbers that are between negative 3 and 3 shown here and again the first scenario to do this where x is just simply less than 3 it's very straightforward just remove the absolute value bars and you have x is less than 3. for the other one you have to think about the negative of x is less than 3. well if i divide both sides by negative 1 what's going to happen is this gets flipped this is now greater than this cancels with this and i'll have x is greater than negative three so this sets up a compound inequality with and and really you could write this as a three part inequality where x is in the middle so x is going to be greater than negative three and it's less than three so it's between negative three and three and again you just kind of erase this real quick you could write something like this in interval notation pretty easily you would put negative 3 next to a parenthesis comma 3. so that's your solution set for this now generically if i had the absolute value of x is less than k what i want is remove the absolute value bars you'd have x is less than k then and you would also have remove the absolute value bars so you'd have x flip the direction of this so you'd have greater than and make this negative so negative k in each case you basically do the same thing but in one situation you have or which is when you have a greater than and when you have a less than you have an and so let's take a look at the absolute value of x is less than 10. so following our last example we know that x would be what it would be greater than negative 10 and less than 10. right see how fast i do that so once you've got kind of the rules down you can do these very quickly but again think about this as a concept the absolute value of x is less than 10. so from zero what's less than 10 units away well i would go to negative 10 and i would put a parenthesis facing to the right okay because anything that is negative 10 or further to the left wouldn't work their absolute value or their distance from 0 is 10 units or larger i'm looking for less than 10. so also going this direction i would stop at 10 i'd put a parenthesis there facing to the left again anything that's 10 or larger won't work because the distance from 0 is 10 or more and we want 10 or less so then we want all the values in between here again x can be larger than negative 10 or less than 10. so in interval notation i would just simply write a negative 10 comma 10 with a parenthesis next to each and again just to go through this one more time we know that x could be less than ten the other scenario is the negative of x could be less than ten we divide both sides by negative one this flips and we end up with x is greater than negative 10. so this sets up the compound inequality x is less than 10 and x is greater than negative 10 which we write as a three part x is greater than negative 10 and x is less than 10. okay so that's where all this comes from so now let's look at one that's a little bit more challenging this is something you'll see at the very very first part so we have the absolute value of 2n minus 3 is greater than 13. so the first thing to make sure of is that the absolute value operation is isolated so this is isolated you have an inequality symbol and then some number now the next thing you think about is is it a greater than or is it a less than okay in this case it's a greater than so with a greater than we have or all right we set up a compound inequality with or so the first one is remove the bars so 2n minus 3 is greater than 13 then or the second one is remove the bars so 2n minus 3 flip the sign so it's a less than make this negative so negative 13. again the easy way to remember this go back to your generic example the absolute value of x is greater than k so this turns into x is greater than k or x is less than negative k just use this as an example on the first few and you'll remember the first one is just remove the bars as we have here we just remove the bars we have x is greater than k the second one is remove the bars flip the sign make it negative so we remove the bars we flip the sign from greater than to less than we made 13 negative 13. and again where this comes from if i took the negative of this so let's say i have the negative of this whole expression 2n minus 3 is greater than 13. what i can do is divide each side by negative 1 and i can cancel this divide this by negative 1 but this gets flipped so you get 2n minus 3 which is this flip the sign okay that's why we flip here and then we make this negative so this is negative 13. so again that's where this is coming from all right so let's erase everything let's get our solution so let's scroll down a little bit so i just want to solve this like i normally would so i'd add 3 to each side here that will cancel 2n is greater than 16. divide both sides by 2 and i get n is greater than 8. over here i'll add 3 to each side that'll cancel and i'll have 2n is less than negative 13 plus 3 is negative 10 divide each side by 2 and we'll get that n is less than negative 5. so i get n is greater than 8 or n is less than negative 5. so let's go back up and notate this so your solution here would be again n is less than negative five or n is greater than eight in interval notation i could write from negative infinity up to but not including negative five and it's the union with this other set which is anything larger than 8 out to infinity now graphically at negative 5 i'd put a parenthesis facing to the left and i'd shade everything this way and at 8 i'd put a parenthesis facing to the right and i'd shade everything this way all right let's take a look at another one so we have the absolute value of negative 10x minus 8 is greater than or equal to 12. so again i'm looking at a situation with a greater than it doesn't matter that it's a greater than or equal to it's just going to change the notation a little bit if i have a greater than involved or greater than or equal to it's a compound inequality with or so i remove the absolute value bars so negative 10x minus 8 is greater than or equal to 12 then or remove the absolute value bars negative 10x minus 8 flip the sign so now this is a less than or equal to make this negative so negative 12. and again that comes from the fact that i could have the negative of this be greater than or equal to 12. i divide both sides by negative one this becomes negative and the sign gets flipped it's just that simple so let's go ahead and solve each one here so we'll have plus eight and plus eight that cancels negative 10x is greater than or equal to 12 plus 8 is 20. divide both sides by negative 10 and this will cancel of x is less than or equal to 20 divided by negative 10 is negative 2. so x is less than or equal to negative 2. and again because i divided by a negative i flipped the direction of that inequality symbol okay for this side i'm going to add 8 to each side to begin that'll cancel i'll have negative 10x is less than or equal to negative 12 plus 8 is negative 4. let's go ahead and divide both sides by negative 10. and if i divide by a negative i flip the direction of that inequality symbol this is going to be a greater than or equal to now this cancels x is greater than or equal to negative 4 over negative 10 i know that negatives would cancel so this would be positive each is divisible by 2. 4 divided by 2 is 2 10 divided by 2 is 5. so x is greater than or equal to 2 fifths there so this is an or and let's erase everything and notate this properly so we have x is less than or equal to negative 2 or x is greater than or equal to 2 5. in interval notation again it's the union of the two solution sets so it's anything we have negative infinity out here as the smallest up to and including negative two so i'm gonna use a bracket and then the union with this guy so including two-fifths and anything larger so graphically i can find negative two on the number line that's here put a bracket facing to the left and shade everything going this way now two-fifths doesn't have a notch but two-fifths is 0.4 as a decimal so it's a little bit less than a half so let's just say it's like right there let's say this is two-fifths and again it's anything that includes two-fifths so that value or larger and again if you have this on a test your teacher is going to allow you to make notches where they're on right they understand that so it's not perfectly at two-fifths but you know you've notated where it is that's the best you're gonna be able to do so again anything that is less than or equal to negative two along with anything that is two-fifths or greater would be a solution for this inequality all right let's take a look at another one so we have eight times the absolute value of negative eight x plus two minus two is greater than a hundred ten so common mistake is after you've seen these easy problems you go through and you start saying okay well eight times remove the bars negative eight x plus two minus two is greater than 110 or then you take this and you say well eight times and then inside of parentheses negative eight x plus two minus two is less than negative a hundred ten no okay do not do this wrong you have to isolate the absolute value operation first you want it to look like this the absolute value of something let's just say x doesn't matter what it is is greater than some value k so this x could be anything it could be the absolute value of ax plus b if you want to make it more complicated is greater than k this leads to what x is greater than k or x is less than negative k this the more complicated example just leads to ax plus b is greater than k or ax plus b is less than negative k okay so the main thing here is to isolate the absolute value operation first and then just follow this formula here write it on a flash card and keep using it all right let me erase this real fast and let's go ahead and simplify so i'm going to add 2 to each side i'll have 8 times the absolute value of negative 8x plus 2. this is greater than 110 plus 2 is 112. now i want this by itself i have multiplication here so i'm going to use division to get it by itself divide both sides by eight and so this will cancel with this and scroll down get some room going and i'll be left with the absolute value of negative eight x plus two is greater than what's a hundred twelve divided by eight well most of us don't know that off top of our heads but punch it up on a calculator and you'll get 14. now i have something that is matching again what i saw earlier the absolute value of something let's just say x is greater than k so the absolute value of negative 8 x plus 2 is greater than 14. remember this sets up as or so x is greater than k or x is less than negative k same thing this is or so we have negative 8x plus 2 is greater than 14. all i did was just remove the absolute value bars then or remove the absolute value bars so negative 8x plus 2 flip this direction here so this is a less than make this negative negative 14. so you're basically good to go at this point you just need to find your solution for this so we're going to subtract 2 away from each side and that'll cancel i'll have negative 8x is greater than 12. divide both sides by negative 8. i'm going to flip the direction of this guy this would be a less than this cancels with this i'll have x is less than 12 over negative 8 each is divisible by two in fact each is divisible by four so if i divide this by four i get three if i divide this by four i get negative two so let's put negative three halves and then let's look at the other guy so if i subtract two away from each side over here this is going to cancel i will have negative 8x is less than negative 14 minus 2 is negative 16. divide both sides by negative 8 so this will cancel with this and what i'll have is x and i've got to flip this is greater than negative 16 over negative 8 is 2. so x is less than negative 3 halves or x is greater than two so again we found that x was less than negative three halves or x was greater than two so in interval notation it's the union of the two solution sets so coming from negative infinity up to but not including negative three halves and the union with this other guy after two right anything larger out to infinity graphically i don't have a negative three halves but negative three halves is negative one point five so it's between negative one and negative two so let's just say it's right there this is negative three halves it would be anything less than that so i'll put a parenthesis facing left shade everything to the left and then anything larger than 2 so put a parenthesis at 2 and we'll shade everything to the right so again x can take on any value that is less than negative 3 halves or it can also take on any value that is larger than 2. all right let's take a look at one now where we have a less than so we have the absolute value of negative 6 minus 4n is less than 6. so again if we see something in this format the absolute value of x is less than k it's a compound inequality with and so x is less than k but it's also greater than the negative of k okay so you have x is less than k and x is greater than the negative of k so i'm going to take this part right here and i'm going to say negative 6 minus 4n is less than 6. so the same as it looks right here just drop the absolute value bars then and okay now you have and because you have a less than it's also true that negative six minus four n is greater than flip the sign just like you do before make this negative so negative six so negative six minus four n is greater than negative 6. so we can write this as a 3 part inequality make it easy on ourselves so we have negative 6 minus 4n in the middle it's greater than negative 6 and it's less than positive 6. so i can just erase this and i can drag this up here and again if you're wondering where this came from i've showed it multiple times but it's basically that we take the negative of this so negative of negative 6 minus 4n this is less than 6 because of the absolute value operation that would work as well and then we divide both sides by negative 1. so this cancels that makes me flip the sign and then this becomes negative 6. so we have this guy right here negative six minus four n the sign is flipped and then we have negative six so that's where this part right here comes in let me erase all this now and let's just attack the problem so if i solve this three part inequality i do the same thing to each part i'll add six over here and over here so that'll cancel become zero that'll cancel i'll have negative four n is less than 6 plus 6 is 12. so i divide each part by negative 4 so i can isolate n and what's going to happen is again i've got to still flip the direction when i have a three-part inequality and divide by a negative that doesn't change so this gets flipped and this gets flipped so this is greater than this is greater than this cancels with this and we'll have n so n is less than zero divided by negative four is zero and it's greater than 12 divided by negative 4 is negative 3. so generally we don't leave it in this manner we rearrange it to where it's in the direction of the number line so negative 3 would be here of course that's less than n and we know that n is less than 0. so let's take this solution up and we're going to graph it use it to write an interval so n is greater than negative 3 and less than 0. so here's negative 3. so it's greater than that and it's less than zero so it's between negative three and zero so that's what it looks like graphically and in interval notation i just have the interval greater than negative three but less than zero just like that so n again can take on any value that's greater than 3 but less than 0. all right for the next one we're looking at 7 times the absolute value of 10k plus 3 plus 7 is less than 28. so again i want to isolate the absolute value operation before i do anything don't make the mistake of just trying to apply the rule before you have isolated the absolute value operation so i want to subtract 7 away from each side first i'll have 7 times the absolute value of 10 k plus 3 this is less than 28 minus 7 is 21. so now i can divide both sides by 7 and when i do that this will cancel with this i've isolated my absolute value operation it's the absolute value of 10k plus 3 and this is going to be less than 21 divided by 7 is 3. so now i can go ahead and apply my little formula remove the absolute value bars so i'll have 10k plus 3 and it's just less than 3. and the other scenario would be that 10k plus 3 remove the absolute value bars flip the direction of the inequality symbol so it's greater than and then make 3 negative so negative 3. and again to make this easy on yourself write it as a three part inequality so go ahead and write this as 10 k plus 3 is greater than negative 3 and a less than 3. so let's subtract 3 from each side so this will cancel negative three minus three is negative six this is less than we'll just have 10k now and it's less than three minus three is zero divide each side by ten this will cancel with this and i'll have k in the middle by itself which is what i want and it's going to be greater than negative 6 over 10. i know it's negative 6 is divisible by 2 6 divided by 2 is 3. 10 divided by 2 is 5 so negative 3 fifths and zero divided by 10 is of course zero so then less than zero so k is greater than negative three-fifths and less than zero so in interval notation that's not including negative three-fifths up two but not including zero so it looks like that and then on a number line where is negative three fifths well it's negative point six so if you look between zero and negative one it'd be a little bit past negative point five so let's say about right there we could say this is negative three-fifths so here's zero so what i would do is i would put a parenthesis here facing to the right one here facing to the left it's kind of a small area on our number line we'd shade everything in between so again k can be anything that is greater than negative three-fifths up to and not including zero so anything in that range between again negative three-fifths and zero for the last problem we're going to look at we have five times the absolute value of negative three plus eight x plus two is less than or equal to negative fifty-three so again i want to isolate the absolute value operation first don't just start applying rules to stuff that's a common mistake so let me subtract 2 away from each side of the inequality that'll cancel i'll have 5 times the absolute value of negative 3 plus 8x and this is less than or equal to negative 53 minus 2 is negative 55. so now let me divide both sides of the inequality by 5. so again i can isolate this part right here this will cancel with this and i'll have the absolute value of negative 3 plus 8x is less than or equal to negative 55 over 5 is negative 11. now am i going to continue with this no i am not why think about this the absolute value operation produces a result that is zero or it is positive it cannot ever be less than or equal to negative 11. that's not going to work it can't be less than a negative value it has to be greater than or equal to zero so this has no solution this has no solution and again when you see something with no solution you can say the solution set has no elements so you put the symbol for the null or empty set one more time if i try to plug in something for x multiply by eight and add negative three when i take the absolute value of it the absolute value will never be less than or equal to negative 11. it won't ever happen because the absolute value is either zero or it's positive so that's why again there's no solution hello and welcome to algebra 2 lesson 15. in this video we're going to learn about linear equations in two variables so this is not a new topic for us in algebra two we learned about linear equations and two variables in algebra one but for some of you you never took an algebra 1 course and for others you kind of stumbled through your algebra 1 course so it's important to get a good full review here so the first thing i'm going to talk about is up to this point we've only kind of worked with a linear equation in one variable we also talked about linear inequalities in one variable and some variations on that we looked at some absolute value stuff and some compound stuff you know so on and so forth but the main idea here is that a linear equation in one variable has generally one and only one solution so for this type of problem let's say i have negative 2x minus 5 equals 15 i add 5 to each side of the equation i have negative 2x is equal to 20 i divide both sides by negative 2 and i get a single number i get that x is equal to negative 10. so using a number line i can display that visually by saying this is my solution for this equation negative 2x minus 5 equals 15. and again that means that if i replace x with negative 10 i would get the same value on the left as i have on the right so that at this point we should all know and let me go ahead and write that in solution set notation where i have negative 10 inside of some set braces so now let's kind of get into the main topic here so a linear equation in two variables is of the form and we see ax plus b y equals c so the first thing i want to call your attention to is our variable x and our variable y each is raised to the first power remember if you have a variable or a number and there's no exponent that's visible it's understood to have an exponent of one so i could do this now if you see something like ax squared plus b y squared equals c that's not a linear equation okay so the fact that the exponent is 1 in each case or you don't see an exponent is what makes it a linear equation all right so the next thing that i want to call your attention to we have this a let me kind of highlight this in a different color we have a we have b and we have c so for each of these we kind of say in our book and if you're fumbling through a textbook right now you would see something like a b and c are real numbers with one restriction it's not allowed for a and b to be zero at the same time so in other words a could be zero if b is not b could be 0 if a is not but a and b cannot both be 0 again at the same time so i'll just write here that a and b are not both 0 okay and the reason for that is if they were both 0 think about if i plugged in a 0 here and here well 0 times x would be 0 0 times y would be 0 you'd have 0 equals c well that's only true if c is 0. so your variables would basically drop out right so what would be the point there so a and b are not both zero a b and c are real numbers so let's look at two quick examples so we have two x plus five y equals 7 we have negative 3x minus 7y equals negative 18. so again in each case you can kind of match this up with what i showed you as your generic example so a here is 2 b here is 5. c here is 7. you have x to the first power you have y to the first power if i look at this example below a is negative 3 b is negative 7 and c is negative 18. so very easy to kind of line these up with a generic example and see what we mean when we talk about a linear equation in two variables all right so the main thing that's different is that when we work with a linear equation in two variables it's no longer going to be one solution there's an infinite number of solutions that's going to work so for this reason we generally are going to graph our solution set to the equation to give a visual representation of what's going on now we're not going to graph in this lesson but we're going to do the things that lead up to us being able to graph and you know from algebra 1 if you took an algebra 1 course that graphing is not very difficult right it's actually very very easy so the first thing we need to understand is something called an ordered pair so a solution to a linear equation in two variables is called an ordered pair so generically we work with x and y okay so x is in the first position so it's always first and then y is always second so if i give you an ordered pair like three comma two and i say it's a solution to a certain linear equation and two variables that means that i can plug in a 3 for x and a 2 for y and i would get a true statement right the left and the right side would be equal so the main question you might be asking right now is how do we get an ordered pair or how do we get such a solution for an equation well basically we pick a number and plug in for one of the variables and then we solve for the other so i can pick whatever number i want for x and then i can solve for y or i could pick whatever number i want for y and then solve for x and there's again an infinite number of solutions i can choose any number i want and i'll be able to get a solution for the other one so let's just start out with x equals to zero that's an easy one so let's say x equals zero so i would say 2 times 0 minus 5 y equals 20. and what i'm basically doing is by plugging something in i'm creating a linear equation in one variable that i can solve quite easily so 2 times 0 is 0. so we'll put 0 there negative 5 y equals 20. so we divide both sides by negative 5 that will cancel i'll have y is equal to negative 4. so it's important to note what we did here i plugged in a 0 for x so that means this ordered pair here remember it's x comma y so it would be a 0 for x and it would be a negative 4 for y and just plug in let me just erase everything and start from scratch let's pretend you're given this ordered pair and you say hey does this work as a solution well i'm plugging in a 0 for x okay this is my x value and i'm plugging in a negative 4 for y all i'm doing so 2 times 0 minus 5 times negative 4 does that equal 20. 2 times 0 again is 0 negative 5 times negative 4 is positive 20. so yeah of course this does work 20 is going to equal 20. so yeah that works as a solution all right let's just try something else let's say i picked some number for y doesn't matter what it is let's just say 6 okay that's nice and easy so i plug in a 6 for y well i'll have 2x minus 5 times 6 equals 20. so i'm going to evaluate this negative 5 times 6 is negative 30. so i'd have 2x minus 30 is equal to 20. i would add 30 to both sides of the equation this would cancel i'll have 2x is equal to 20 plus 30 is 50. divide both sides by 2 and finally i find that x is equal to 25. so when y is 6 okay i plugged in a 6 there to start x is 25. so again following this format the x value is first so that's 25 the y value is second so that's 6. and again i can generate as many of these as i'd like pick something for x solve for y pick something for y solve for x all right so let's kind of look at our first kind of sample problem here this is something you're definitely going to see in your textbook as you get into starting to graph these linear equations in two variables the first thing you do or the first thing you're taught is how to create a table of values what this is going to do is it's going to give you some ordered pairs that you're then going to plot on the coordinate plane and then you'll sketch a line through those ordered pairs to create your graph well we're not there yet we're still working on this but i just want you to know where we're going so we have 4x minus 3y equals negative 12. now i'm given a table with some values filled in for me so for example i'm given an x value of 0 here so what they're saying for me to do here is to take this equation so 4x minus 3y equals negative 12 plug in a 0 for x and see what i get for y so 4 times 0 would obviously be 0 minus 3y equals negative 12. so this is gone basically all i need to do to solve this is divide both sides by negative 3 that would cancel and i'd have y is equal to 4. so i can fill in my missing part here and put a 4 there so the ordered pair there would be 0 comma 4. again it's the x value first followed by the y value second and just write that down in your notes just put x comma y so you can refer to it as we're doing this and this is something you're going to remember for the rest of your life x comma y is the ordered pair all right for the next one we're given a y value of 0. so for the y value of 0 i go 4x minus 3 times 0. just plug in a 0 there which corresponds to right here equals negative 12. 3 times 0 is obviously 0 so that's gone so we'll have 4x is equal to negative 12. let's divide both sides by 4 that'll cancel and we'll have x is equal to negative 3. so when y is 0 x equals negative 3 again as an ordered pair x comes first so that's negative 3 comma y go second so that's 0. so this is your x this is your y all right let's take a look at one more so now we're given an x value of three okay of three so let's plug that in so we have four times for x again i'm plugging in a three that's all i'm doing then minus 3y equals negative 12. 4 times 3 is 12 minus 3y equals negative 12. let me subtract 12 away from each side so that cancels scroll down to get a little room so then i'd have negative 3y on the left is equal to negative 12 minus 12 is negative 24. divide both sides by negative 3 so this would cancel and i'll have y is equal to 8. so when x is 3 y is 8. let me erase this so we can get everything back on the screen okay let me scroll up here so as an ordered pair this would be 3 for x comma 8 for y so this is x and this is y so pretty simple exercise overall for sure you'll probably get this on your homework or on a test where they give you a linear equation in two variables and they'll either tell you to make a table with a certain number of ordered pairs or they'll give you an ordered pair where the x value or the y value is missing like we did here let's take a look at one more of these so we have negative x plus seven y equals 28 so i'm given my table of values i have x values i have y values so the first one gives me an x value of 7. so i'm going to plug that in there plug in a 7 there don't forget that negative so put the negative and then plug in a 7 plus 7y equals 28. let's add 7 to each side so that cancels we'll have 7y is equal to 35 we'll divide both sides by 7 and we get y is equal to 5. so when x is 7 y is 5. okay so let me erase this and that's the ordered pair remember the x values first so that's 7 comma the y value of second so that's 5. so this is x and this is y all right for the next one we are given a y value of 3. so we're going to plug in a 3 here and we're going to solve for x so we have negative x plus 7 times 3 is equal to 28. let me [Music] say that 7 times 3 is 21. so negative x plus 21 equals 28 let me subtract 21 away from each side of the equation so this is going to cancel we'll have negative x is equal to 28 minus 21 is 7. we'll divide both sides by negative 1 or multiply by negative 1 whatever you want to do so that'll cancel itself out and i'll have x is equal to negative 7. so when y is positive 3 x is negative 7. so as an ordered pair you could write this as negative 7 for the x value comma 3 for the y value all right for the next one we're given an x value of 0. so i'm going to plug that in right there okay i want a 0 there so negative and then you have 0 plus 7y equals 28 the opposite of 0 is just 0. right so this you just cancel that out and say i'm just dealing with 7y equals 28. we would divide both sides by 7 and we would find that y is equal to 4. so when x is 0 y is 4. so we have 0 comma 4 again 0 is the x value 4 is the y value so pretty simple exercise overall not going to do any more of these for now but you can pretty much tell what you need to do if you're given an x you plug in for x in that equation and you solve for y if you're given a y you plug in for y in that equation and you solve for x and then just report your answer as an ordered pair remember the x value goes first comma your y value so kind of the next thing we need to learn before we start graphing these linear equations and two variables is what to actually do with the ordered pairs what's going to happen is we're going to end up plotting them on something known as the coordinate plane so this is the coordinate plane and for those of you again that took algebra 1 you're very familiar with this we did a lot of work with this in algebra one but if you've never taken that course this is new to you so let me give you a couple of tidbits of information here the first thing is that this thing has many names the official name for it is the cartesian coordinate system and it's named after a guy that was sick in his bed and he watched the fly kind of you know go back and forth so he came up with the idea for this the other name you might hear is the rectangular coordinate system but most often i think that you'll hear it called the coordinate plane but it's interesting to know those three names because those are used most often and if your book uses something different i don't want you to be confused what was he talking about he said coordinate plane they're calling it this cartesian thing so it's important to kind of know all their names for it so you don't get confused the other thing would be how does it work well it's two number lines basically that intersect where they're each at zero so this is the intersection of the two it's the point zero comma zero so the x value is zero the y value is zero and it's the again the intersection of the two this is referred to as the origin and you're going to hear that a lot throughout your study of mathematics whenever you talk about the coordinate plane somebody refers to the origin they're talking about this point right here zero comma zero now another thing that's important is the horizontal number line is referring to x locations so we refer to this as the x axis now similarly the vertical axis refers to y locations so this is referred to as the y axis and to plot an ordered pair is very very simple if i give you something like 4 comma 3 as an example this is your x value this is your y value so that just translates to finding the x location of 4 finding the y location of 3 and just getting the meeting point so an x location of four i look at the x axis again the horizontal axis so i go from the origin one two three four units to the right or i could just start at four either way so that's an x location of four a y location of three i just go up three units to right here and where's the meeting point well i would go one two three four units to the right and i would travel one two three units up so it's right there and it works either way if i want to go one two three units up and then one two three four units to the right i get to the same location so this is the point which i'll label four comma three and four comma three is known as the coordinates of the point it tells us where to go right we go four units to the right and three units up to get to this point now just one more piece of information before we look at some examples of plotting ordered pairs you're probably taught in your textbook and this is very brief about the quadrants so the quadrants go counterclockwise so you kind of think about each section here this is section one or quadrant one so that encompasses this this is quadrant one and then you'd have quadrant two and let me do that in a different color so this is two and this is three and this is four so again counterclockwise so you can highlight this in a different color this is two i'm trying to go over the line i know i'm terrible at drawing but i'm trying and then let's do this one in a different color this is three and then one last color let's do orange for four okay and i know again i went over the line in a couple places but you should get the idea so quadrant one just encompasses all of this right it extends forever and ever and ever in this direction right it goes forever but it's all the values that would be in here and one of the things we can say about quadrant one is what well we know that the x location and the y location are each going to be positive because anything in this quadrant here you can see it highlighted in yellow is going to be to the right of zero and it's going to be on top of zero moving up from zero so the direction going to the right is positive the direction going up is positive so anything in quadrant one is positive positive so quadrant one is positive positive so this is your x and this is your y for quadrant two what can you say well the x locations are going to be negative the y locations are positive because we go to the left of zero and we go north of 0. so that's going to tell me that in quadrant 2. again my x location is negative my y location is positive in quadrant 3 i am negative and negative because my x location is to the left of zero my y location is south of zero so for quadrant three i am negative negative and then finally for quadrant four i am positive in the x values because i'm right of 0 and then i'm negative in y values because i'm south of 0. so this is positive negative and you might be tested on something like this it might say for example what are the values in quadrant two and you might get flustered and say well i didn't memorize that we don't need to just draw yourself a sample coordinate plane they're very easy to draw and once you do that you can just eyeball it and say well i know that anything in quadrant two would be to the left of zero and north of zero so my x values would be negative my y values would be positive all right so let's wrap up this lesson we're just going to plot some ordered pairs so we're going to start out with 0 comma 7. so again it's x location followed by y location so let me just write this real quick it's x comma y and let me just label this too so that we don't get confused this is the y-axis and this let me kind of scroll over this is your x-axis now an x-location of zero is not confusing a lot of you get confused with this if i look at the x-axis just kind of block out the y-axis for a second just just pretend this isn't here if i have a regular number line we've been working with forever and ever and ever where is zero it's right here okay so where is seven as far as a y location now we can erase this and think about seven it's right here so if i go zero units on the x axis i just don't move i start at the origin i don't move and if i go seven units up on the y axis i'm right here so this is going to be 0 comma 7. now for the next one i have the ordered pair 3 comma 5. so i would go 3 units to the right or an x location of 3 and 5 units up up 1 2 3 4 5 so that's a y location of five so this is three comma five what about negative two comma negative seven so i go two units to the left here's negative two i go seven units down here's negative seven so two units to the left seven units down is right here so this is negative two comma negative seven next we have eight comma one so i would find eight on the x axis let's do this one backwards again you can do the y location first if you want so let's find one on the y-axis that's here now let's find 8 on the x-axis that's here so let me go up 1 and to the right 8 so that's here so this is 8 comma 1. all right lastly we have 9 comma negative 6. so i'm going to find 9 on the x-axis that's here negative 6 on the y-axis that's here so i'm going 9 units to the right and 6 units down so that's going to be right here okay so this is 9 comma negative 6. so very very easy in terms of things that you will come across in algebra this is probably one of the easiest things to do to plot an ordered pair you literally just go to the coordinate plane and have to memorize two things the x-axis and the y-axis the x-axis is horizontal the y-axis is vertical you do that just a few times and you've got that down so basically it's just a matter of finding the meeting point of the x location in the y location putting a little dot there and in most cases when you're plotting ordered pairs you just ask you to label it like we did here hello and welcome to algebra 2 lesson 16. in this video we're going to learn about graphing linear equations in two variables so in our last lesson we covered the basics of a linear equation in two variables and if you took an algebra 1 course not only was that a review but this will be a review as well learning how to graph a linear equation two variables is very very simple particularly if you've done it before so i want to kind of just start out by covering some of the things we talked about in the last lesson we're going to go through them very quickly and then we're going to ramp up and start doing some graphing so a linear equation in two variables looks like this it's ax plus b y equals c where a b and c represent real numbers and a and b are not both zero so a can be zero b can be zero but they're not both zero so in this example five x plus two y equals twenty you can see a represents five b represents 2 and c represents 20. now another key point here is that the exponent when each variable is a 1. so i have 5 x to the first power plus 2y to the first power equals 20. i would not have a linear equation if the exponents here were something like 2 or 3 for example so with that behind us now let's talk about a couple of things with the linear equation two variables the first thing is that there's an infinite number of solutions i can choose a value for x let's just say zero for example i plug it in i'd have five times zero plus 2y equals 20. well this is 0. now i have a linear equation in one variable i have 2y equals 20. i know that x is 0 because i plugged a 0 in to start i just need to find out what is y when x equals 0. so i divide both sides by 2 and i get y equals 10. so one solution of the infinite number would be that x is equal to 0 when y is equal to 10. and if you want to you can plug in a 0 for x and a 10 for y and you'll see you get the same value on the left as the right let's do that real quick 5 times 0 plus 2 times 10 equals 20. this is 0 plus 2 times 10 is 20 equals 20. you get 20 equals 20 so this checks out but the main thing here is to realize that i can keep doing that i could choose a different value for x and i would get a different value for y so let's say for example i chose the value of 2 for x so i plug that in now so 5 times 2 plus 2y equals 20. 5 times 2 is 10. so this is 10 plus 2y equals 20. subtract 10 away from each side of the equation this cancels i'll have 2y is equal to 20 minus 10 is 10 divide both sides by 2 and i get y equals 5. so here x equals 2 when y is equal to 5. so this is a big change for us because we're used to solving linear equations in one variable and with those type of equations we generally just have one number that works as a solution okay we don't have an infinite number of solutions that we can just generate but it's good to see this because it leads me to my next point if you have an infinite number of solutions you need a way to kind of visualize the solution set and what we do with this is we graph okay we graph a line on a coordinate plane to kind of say hey here's a visual representation of the solutions for this linear equation and two variables now let's talk about how we would go about graphing something the first thing is that each solution can be written as something known as an ordered pair so we talked about that in the last lesson an ordered pair is x comma y so the order is important i put my x value first followed by y so with x equals 0 y equals 10 the ordered pair would be 0 that's my x value comma and then 10 because that's my y value for this one it would be 2 comma 5 and i could generate more if i wanted to and let's generate one more actually let's see what happens when x equals four so x equals four what is y so plug in a four there we'd have five times four plus two y equals twenty we know that five times four is twenty so you'd have twenty plus two y equals twenty let's subtract twenty away from each side of the equation so that'll cancel we'll have two y is equal to 20 minus 20 is 0. divide both sides by 2 and we'll get that y is equal to 0. so if x is 4 y is 0 or as an ordered pair it's 4 for the x value comma 0 for the y value all right so in the last lesson we talked about how to plot the ordered pairs so let's go down to the coordinate plane and we're going to do that all right so this is our coordinate plane and the coordinate plane consists of two number lines so we have a horizontal number line and a vertical number line so the horizontal number line is known as the x-axis so this is the x-axis the vertical number line the one that goes up and down is known as the y-axis so when we talk about our ordered pairs let me paste those real quick we have 0 comma 10 2 comma 5 and 4 comma 0. all we do is we find the kind of intersection or the meeting point of the two locations one locations for x one locations for y so in other words let me kind of go to this middle one for two comma five remember it's x comma y so an x location of two if i go on the x axis a location of two is just right here for five the y location i go on the y axis a location of five is right here so where's the meeting point between the two well i'm going two units to the right five units up it's right here so this is the point two comma five now when we start talking about zero as a location it gets a little bit complex which you have to kind of think a little bit the lines intersect at zero comma zero this is the meeting point of the x axis and the y-axis so if i look at 0 comma 10 okay that's telling me that i don't want to move at all on the x-axis so i would just sit right here and i would say well the y location is 10 that's here so i'm only going to go up 10 units to arrive at my point this is 0 comma 10. now for 4 comma 0 now i'm going to be on the x axis because 4 is on the x axis it's 4 units to the right the y location is 0 so i'm not going to move up or down at all i'm just going to stay on the x axis this right here is 4 comma 0. now once i have two points i have enough for a line the rule is two points make a line but you always want to use a third one as a check because i can make any two points line up but if i have a third one that's off so i get something that looks like this i've got a line going and then there's a point over here so it's kind of that's not a line so you know you made a mistake somewhere but if you have three points and it just kind of flows into a line you know you got the right answer so what i'm going to do is i'm going to draw a line through our three points here and what we want to do is we want to put an arrow at each end and what does that arrow indicate it indicates that this line continues forever and ever and ever in each direction again there's an infinite number of solutions an infinite number of x y pairs that satisfy this equation every single point on this line every single spot represents a solution to that equation that we started with which was 5x plus 2y equals 20. and a common thing to do would just be to label the line so 5x plus 2y equals 20. this line again represents the solutions to this equation now there's a very important concept that i want to cover with you real quick the concept would be that of an x-intercept and a y-intercept so an x-intercept is where your graph crosses the x-axis so this right here is your x-axis this graph crosses the x-axis right here at this ordered pair 4 comma 0. so that's my x-intercept let me label that real quick for you so this point right here is the x intercept now similarly the point where the graph crosses the y axis this is the y axis going up and down is known as the y-intercept so this let me just label that right there is the y-intercept so a couple of key things here that we want to cover the x-intercept again is where the graph crosses the x-axis so to find this mathematically you're not always going to be graphing stuff you want to plug in a 0 for y and solve for x let's think about this for a second let's go back to the graph why does that make sense where would i cross the x-axis at well on the y-axis i can't go up or down if i'm going to hit the x-axis just think about erasing this axis for a minute if i'm touching the x-axis i've got to be at 0 on the y-axis if i cross here at this point i'm at 0. if i cross here at this point i'm at 0. anywhere you cross the x-axis y will be zero okay so that's very important to understand the second thing would be for the y-intercept again where the graph crosses the y-axis you'd plug in a zero for x and solve for y again looking at this thing we see that at the y intercept the x location is zero you can even see it right there x location of zero at the x intercept the y location is zero so that's how we find these things if i want the x intercept plug in a 0 for y solve for x if i want the y intercept plug in a 0 for x solve for y all right so now that we've graphed our first equation let's talk about one particular method to graph equations now we just used it we didn't we didn't know where we're using it but we did it's something known as the intercept method so it's easy to work with zero as kind of one of your values zero is easy to work with you multiply by 0 it disappears you add 0 you get the same thing you know so on and so forth so the whole thing with the intercept method is i'm going to find the x and y intercepts and then i'm going to find one point as a check now for this one i have negative x plus 4y equals negative 4. so when i look at ordered pairs or if i create a table of values then i would find the y-intercept by plugging in a 0 for x i find the x-intercept by plugging in a 0 for y and then i pick some number arbitrarily to get a check now when you pick a number arbitrarily you want to make sure that you end up with integers the reason for that is you don't want to end up with decimals or fractions and trying to graph this on something where you know the scale isn't integers okay it just you can do it but it's very very tedious so try to work with integers just try your best so i'm going to start by plugging in a 0 for x here i would have negative 0 that's just going to go away plus 4y equals negative 4 divide both sides by 4 you get y equals negative 1. so if x is 0 y is negative 1 again this is the y-intercept when we graph it i'll show you it's going to cross the y-axis there for the other one let's erase this real quick we're going to plug in a 0 for y now find the x intercept so we'd have negative x plus 0 times anything is 0. so i can just mark this out i can have negative x equals negative 4 multiply both sides by negative 1. i get x equals 4. so then this right here would be our x intercept all right let's erase everything now again two points make a line so if you're a hundred percent confident that you got the right answer here stop and do it you can plot zero comma negative one four comma zero draw a line through it and you're done but again if you're like me and you're cautious you want a third point as a check so you want to pick something that's going to produce an integer result so how do you do that well you can kind of eyeball this and tell what would work and what wouldn't if i plugged in a 1 here just think about this for a second you would get 4 times 1 which is 4. so you'd have negative x plus 4 equals negative 4. without doing anything else you know that's going to work because if i subtract 4 away from each side i would have negative x is equal to negative 8. i'm going to end up with an integer result so i'm going to go with y equals 1 and then x if i multiply each side by negative 1 here x is going to give me positive 8. so now i can just label this as a check this is my checkpoint all right so let's erase everything and we have three ordered pairs we have 0 comma negative 1 we have 4 comma 0 and we have 8 comma 1. so let's let's copy these all right so we paste our ordered pairs and again i just want to label this as the x axis and this as the y-axis and we're going to plot each ordered pair so 0 comma negative 1. so 0 is the x-value so i'm not going to move it all left or right negative 1 is the y-value so i'm just going to move down that's right here now remember if x is 0 and y is negative 1 that is the y-intercept and notice how when we graph this it's going to cross right there it's going to touch the y-axis and again that's the definition of the y-intercept it's where it crosses the y-axis now the next one is 4 comma 0. so my x-location is 4 my y location is 0. so i'm not going to move it all vertically so that's going to be right here now notice that this is where the graph is going to cross the x-axis again this is the x-intercept then lastly my point for a check is 8 comma 1. so i find 8 on the x-axis that's right here i find one on the y-axis so think about is going eight units to the right one unit up and everything lines up here so i know that i've got something going okay and again i want to put arrows at each end and the arrows indicate that this line continues forever and ever and ever in each direction so once we've graphed our line again it's common to label it so we'll say this is the graph for negative x plus 4y is equal to negative 4. okay just label it like that and you're not required to label it but it's good practice especially when you begin and again i just want to call your attention to this one more time where y is zero at the location four comma zero this is the x intercept this is where we cross the x axis okay where x is zero which is zero comma negative one this is the y-intercept this is where we cross that y-axis right there okay so again the x-intercept the y-intercept two very important concepts for you to get down something you need to know right away so kind of the next thing that's going to occur is you're going to run across these equations where your intercept occurs at the origin so when this occurs the x and the y-intercept are going to be there and it's known as a line that just passes through the origin you will encounter this as a x plus b y equals c where c is zero so you might see something like four x plus two y equals zero i know this is a line passing through the origin because it's in this format or something like negative 3x minus 7y equals 0. you're looking for something times x plus something times y equals 0 again that's a line that passes through the origin so here's an example we have 3x plus 4y equals 0. so we see it we eyeball it we say i know this passes through the origin so if i try to use my intercept method what's going to happen well the x and the y-intercept occur at the same spot so i gotta pick two additional points not one so if i put zero here i know zero here works and you can eyeball and see if i put a 0 here and a 0 here 3 times 0 is 0 plus 4 times 0 0 0 plus 0 is 0. so once we've identified that we get a point for free we know automatically 0 zero works all right for the next point again just think about using things that will produce integers so a good way in this situation would be to subtract 4y away from each side so i'd have 3x is equal to negative 4y now think about this in terms of what we could do i know if i multiply 3 times 4 i would get something that is going to have a result that's an integer how do i know that well let me show you let's put a 4 here for x and we'll continue plug in a 4 there 3 times 4 is 12 12 equals negative 4 y i know that this divided by this is going to give me an integer because 12 is divisible by 4. so when we think about negative 4 it's divisible by that as well so we divide both by negative 4 12 divided by negative 4 is negative 3. that's easy enough now what i could do is i could switch that around if i'm looking for an easy point i could say what if y was 3 well if y was 3 i would get negative 12 over here and i know negative 12 divided by 3 would give me an integer as well so just look for little tricks like that where you can get yourself an integer so where you're not fighting with a decimal or a fraction so this equals 3x divide both sides by 3. you get x is equal to negative 4. so this is negative 4. and so we have three ordered pairs they're all integers so it's going to be easy for us to graph so we have 0 comma 0 we have 4 comma negative 3 and we have negative 4 comma positive 3. so again come down to my coordinate plane this is x this is y i'll paste my ordered pairs so the first one is zero comma zero so that's at the origin that's right here now remember we said that this was a line that passed through the origin so we expect that point the next point is four comma negative three so four units to the right that's my x location three units down that's my y location four right three down is right here then the other one is negative four comma three so four units to the left three units up so 4 left 3 up is right here and now we just draw a line to connect the points okay make sure you put an arrow at each end and again i'm going to label this as the equation 3x plus 4y equals 0. so again you see something in this format ax plus b y equals 0 you know it passes through the origin and again you can look at that and remember that your x intercept and your y-intercept occur at that point it's passing through both of those okay it's passing through the x-axis and the y-axis when it hits that origin so that's your x and your y intercept all right so pretty easy overall to graph a linear equation two variables i know that the majority of you remember this from algebra 1 and so this is really just a review for you but i want to cover something else before we kind of get into slope and start talking about some things in algebra 2 that we didn't talk about in algebra 1. graphing vertical and horizontal lines this is something that students even though they took algebra 1 they still struggle with it you will see equations such as y equals k remember i told you the definition of a linear equation in two variables was ax plus b y equals c where a b and c were real numbers so zero is a real number and i said that a and b cannot both be zero but one of them could be zero so what i have here is basically the case where this is zero so zero times x plus something times y let's say it's 3y equals let's say 6. so 0 times x is just 0. so this is gone i have 3y equals 6 divide both sides by 3 and i get y equals 2. so this is an equation where y always equals 2 no matter what the value is for x so to graph something like that again this is your x-axis this is your y-axis if y is equal to 2 again i showed you could write this as 0x plus in this case we've already simplified it so let's just say plus y equals 2. anything i plug in for x if i was to make a table of values will become 0. so let's say i just say all right here's x here's y let's say i choose 3 for x well 3 times 0 is 0. so this is going to cancel every time it be 0 i'm just left with y equals 2. so no matter what i plug in for x it becomes 0 and i'm left with y equals 2. so x equals let's say negative 5. y is 2. x equals 7 y is 2. x equals 1 million y is two so y just always equals two so what you do for something like this you just find two on the y axis and you draw a horizontal line that's it that's all you have to do okay and you still want arrows because this goes infinitely in the left and the right direction okay it continues forever and ever and ever so if i gave you something like let's say y equals 5. now let me label this one before we move on this is y equals 2. if i gave you y equals 5 you would just find 5 on the y axis again for any x value you choose y is 5. so if x was negative 3 y would be 5. if x was 6 y would be 5. if x was 9 y would be 5. so you have an infinite number of points you could make like this we would just draw a horizontal line put our arrows and say this is y equals five and let's say i saw something like y equals i don't know negative seven well i just find negative seven on the y axis right there draw a horizontal line okay it's just that simple let me label this as y equals negative 7. all right so additionally kind of the last thing we're going to talk about in this lesson we will see vertical lines so this is where x equals some value so again i go back to my ax plus b y equals c well in this case b is 0 right so a x plus 0 y equals c for simplicity let's just say a is 1 so we have x plus 0 y equals c 0 times anything is just 0. so this simplifies to x equals c or in this case we just put k it just means we're representing k and c it's just some real number okay it can be anything x could be let's say 4 where x could be 6 or x could be 10. whatever it is it's just going to be a vertical line so let me show you that real quick so let's say we had x is equal to 4. again i go back to my table of values if i write this as x plus 0 y equals 4 no matter what i plug in for y x will equal 4. let's say i plug in negative 3. so negative 3. well negative three times zero is zero so this cancels and becomes zero you just have x equals four and that's true for anything it doesn't matter what you plug in so you can pick some points let's say negative three zero and two in every case it's going to be 4 for x so the simple way to do this is not to plot points it's just to find 4 in the x axis which is right here and to draw a vertical line because no matter what the value is for y so all these values for y so at ten x equals four at eight x equals four at five x equals four at three x equals four you know so on and so forth so it just becomes a vertical line okay and you want to make arrows at the top and the bottom just to show that that continues forever and ever and ever in both directions all right so let's try another let's say x was equal to i don't know negative eight for example what would i do there well i'd find negative eight on the x-axis which is right here and i would simply draw a vertical line again make sure to use my arrows let me label everything make it clear so again this was x equals 4 and this is x equals negative 8. and again you can look at things and say well x equals negative 8. so if i have x equals negative eight i could really think about it as x plus zero y equals negative eight well think about some values for y if y is four oh x is negative eight so that's right here if y is 1 oh x is negative 8 so that's right there if y is negative 5 oh x is negative 8 that's right there so it becomes this vertical line okay so if you have x equals some number find that number on the x-axis draw a vertical line hello and welcome to algebra 2 lesson 17. in this video we're going to learn about the slope of a line so for the majority of you you've taken algebra 1 already and you're already very comfortable with the concept of finding the slope of a lot for others you skipped over algebra 1 and just kind of jumped into algebra 2 or you might just be fumbling around trying to find help with your homework on slope i can tell you if you fall into that category you can definitely pick up slope from just watching this video it's a very very easy concept so the slope of a line is nothing more than a measure of its steepness you can think about some real world examples of this you can think about the slope or the pitch as it's called of a roof we all know that different roofs have different steepnesses and so they have different slopes you can also think about the slope of the roads that you drive on this is often called a grade so different roads that you drive on depending on where they are are going to have different steepnesses different slopes or different grades so slope is nothing more than the ratio of vertical change which is known as the rise to horizontal change which is known as the run so we're going to hear this for the rest of our life slope is the rise over the run i want you to write that on a flash card and memorize that you're going to hear it over and over and over again so you've got to kind of know that all right so actually calculating slope is very very easy we're given this formula for when we know two points on the line so to calculate the slope of a line when two points are known we're given this formula again that's known as the slope formula so m stands for slope that's another thing you want to write down because a lot of times you're expected to notate your slope with m so m stands for slope and we have that it's equal to so for the slope formula we have these two inputs for y and these two inputs for x so let me explain real quick we have y and then we have a little two and then y and then a little one so it's a subscript in each case so we say y sub two minus y sub one this is just a way to notate the different y's that we have we have two ordered pairs that we're working with so x comma y and then x comma y so i need to know which x i'm referring to and which y i'm referring to so the way we notate is we say okay this is x sub 1 and this is x sub 2 this is y sub 1 this is y sub 2. and you don't mix and match you don't say this is x sub 1 y sub 2 and this is x sub 2 y sub 1 for example you want to keep those notations consistent right so if this is x sub 1 this is y sub 1 if this is x sub 2 this is y sub 2 okay that's going to help you with your formula now all i do once i've notated things is i just plug into the formula okay it's very very simple so y sub 2 minus y sub 1 is the change in y values or the rise and then x sub 2 minus x sub 1 is the change in horizontal values or the run so again we see this rise over run now one little note before we kind of move on we have that x sub 2 minus x sub 1 cannot be 0 and that's because if this is 0 here we'd be dividing by 0 and that's not allowed all right so let's go ahead and take a look at 3x minus 2y equals negative 6. so normally i would just give you two points and say calculate the slope but let's make it a little bit more interesting so we have 3x minus 2y equals negative 6. just generate any two points that you want and plug them into the formula i just gave you i'm going to go ahead and use the x-intercept and the y-intercept because those are pretty easy to get so for the x-intercept i plug in a 0 for y so for my table of values if i plug in a 0 for y what would i get for x i'd have 3x minus 2 times 0 equals negative 6. this would cancel and be 0. i'd basically have 3x equals negative 6 divide both sides by 3 and i'd get x equals negative 2. so x is negative 2 when y is 0. now for the y-intercept i know that i want to plug in a 0 for x so if i plug in a 0 here i would have 3 times 0 minus 2y equals negative 6 so that would cancel i'd have negative 2y equals negative 6 divide both sides by negative 2 and i'll have y equals 3. so x is going to be equal to zero when y is three so now i have two points and if i have two points i have enough to calculate the slope so i know you probably don't remember the slope format from the other page so let me scroll back and i want you just to copy it down we have m is equal to y sub 2 minus y sub 1 over x sub 2 minus x sub 1. so again that slope formula is m is equal to y sub 2 minus y sub 1 over x sub 2 minus x sub 1. now i have two points here i have negative 2 comma 0 and i have 0 comma 3. so the first question everybody asked and you know i talked about this a lot when i taught algebra 1. what point do i label as x sub 1 y sub 1 what point do i label is x sub 2 y sub 2 this is going to come up in your head and the answer to that is it doesn't matter write that down now because you're going to ask that question again if you haven't asked it already if i start out and say this is x sub 1 and this is y sub 1 and this is x sub 2 and this is y sub 2 i would get the same answer as if i change this and i said this is x sub 2 this is y sub 2 this is x sub 1 this is y sub 1 okay what i want to do is just make sure i'm consistent with my notation so in other words if this is x sub 2 this is y sub 2. okay i have a 2 in each case if this is x sub 1 this is y sub 1. i have a 1 in each case so now that we've got that covered we just plug into the slope formula y sub 2 is 0. so i'm just going to erase this and put a 0. y sub 1 is 3 a raised this input of 3. x sub 2 is negative 2. so erase this and put negative 2. and then x sub 1 is 0. so erase this and put a 0. so zero minus three is negative three negative two minus zero is negative two negative three over negative two is three halves so what does this mean m equals three halves well a slope of three halves means that again i go back to the definition i rise three units for every two units that i run so graphically if i started out at a point on the line and i went up three units and i go to the right two units i'm going to go back to that line so let me graph this line and we're going to show you that so again this is my x-axis this is my y-axis and i had two points i had zero comma three and i also had negative two comma zero and i know that when we were graphing i said to do three points and you can do that if you want but i already know that this is correct so i'm not going to waste your time by getting a third point so zero comma three is right here negative two comma zero is right here okay again this is the graph for 3x minus 2y is equal to negative 6. now we talked about the fact that this had a slope m is equal to three halves so i can start out at any point on this line remember this is the rise over the run okay so now i'm going to show you this if i start out at this point if i rise 3 units so one two three and i run two one two i'm back to the line rise three one two three run two one two back to the line rise three one two three run 2 back to the line it also works a different way if i think about 3 halves as negative 3 over negative 2 negative over negative is positive so it's the same thing really i start at a point i go down 3 so one two three that's my rise and i run negative two so i go to the left two one two back to the line fall three one two three go to the left two right here that's what slope is telling us for every three units i rise i run 2. all right so let me show you one other thing i want to show you how this procedure where we have m equals again y sub 2 minus y sub 1 over x sub 2 minus x sub 1 works on the coordinate plane so again if we use the point 0 comma 3 and we use the point negative 2 comma 0 when i think about y sub 2 minus y sub 1 the y value let's just say from here is three and let's say from here it's zero so three minus zero that gives me the rise or the change in y values from here to here right that's my change in y this distance here so that's three units now my change in x values think about zero minus a negative two which is basically zero plus two or two i'm going from here to here so i'm traveling two units and it's easy to see the slope again it's i go two units to the right and three units up so i rise 3 run to the right 2 or you could say you run 2 and go up 3. either kind of way you want to think about that we typically say rise over run so that's why i would say increase 3 go to the right 2. and again you can reverse that by saying negative over negative some instances you're going to have to do that so you can also say that you fall 3 and go to the left 2. all right so now that we understand kind of the concept behind this let's just crank out a few examples so let's say you get your homework or your test and you're given these two points and they say calculate the slope well again you're going to use the slope formula so m is equal to y sub 2 minus y sub 1 over x sub 2 minus x sub 1. so i just generically label one point is x sub 1 y sub 1 the other is x sub 2 y sub 2. it doesn't matter which is which so this is x sub 1 y sub 1 this is x sub 2 y sub 2. again i could have switched that up it would make no difference in the calculation so for y sub 2 i have negative 18. for y sub 1 i have 2. for x sub 2 i have 19. for x sub 1 i have negative 13. be very careful with plugging in negatives here because you have that minus so you want minus a negative 13. all right so this is equal to negative 18 minus 2 is negative 20. this is over 19 minus a negative 13 is the same thing as 19 plus 13. 19 plus 13 is 32. so you get negative 20 over 32 of course we can reduce this each is going to be divisible by 4. divide this by 4 you get 5 divide this by 4 you get 8. so this equals negative 5 8. so what this means is that my slope or m is negative 5 8 so i would rise by negative 5 for every 8 units that i run and when we think about negatives here there's a couple different things we can think about first off remember a fraction if the numerator is negative and the denominator is positive i can switch that up so when i think about slope i can say that the rise is negative 5 and the run is eight or i could say the rise is positive five and the run is negative eight these two are the same when i think about a negative value for a rise it means i'm going down right vertically i'm decreasing going down when i think about a negative value for a run i'm going to the left just think about your number line and how you increase or decrease if you're on a vertical number line i increase by going up right going north i decrease by going down or going south if i am on a horizontal number line a typical number line as i go to the right i'm increasing as i go to the left i'm decreasing okay so just relate that all right so let's take a look at another example let's say you have 15 comma negative 10 and 10 comma 6. so again let's label these points here and i'm going to switch it up and say this is x sub 2 y sub 2 this is x sub 1 y sub 1 and i just plug into the formula so m the slope is equal to y sub 2 which is negative 10 minus y sub 1 which is 6 over x sub 2 which is 15 minus x sub 1 which is 10. again i'll write the formula here it's y sub 2 minus y sub 1 over x sub 2 minus x sub 1. again rise over run so negative 10 minus 6 is going to be negative 16 over 15 minus 10 which is going to give me 5. so i can't reduce that any further my slope is just negative 16 fifths so again this tells me that i fall 16 units for every 5 that i go to the right or i could switch this up and say that i increase 16 units for every 5 that i go to the left i can have 16 over negative 5 as well all right let's take a look at another example so we have the ordered pair negative 7 comma negative 9 and the ordered pair negative 14 comma 12. so again i'm just going to use my slope formula so m is equal to y sub 2 minus y sub 1 over x sub 2 minus x sub 1. again copy that down and refer to it as we're working the problems so i label each one of these arbitrarily doesn't matter so i'm going to just do x sub 1 y sub 1 and this one is x sub 2 y sub 2. again the only thing that matters is if i put this as a 1 this is a 1. this is a 2 this is a 2. don't mix that part up but you can label either one as x sub 1 y sub 1 and the other is x sub 2 y sub 2. it doesn't matter all right so m is equal to for y sub 2 i have 12 then minus for y sub 1 i have negative 9 so it's minus a negative 9. okay be careful there then over for x sub 2 i have negative 14 and then minus for x sub 1 i have negative 7 so minus a negative 7. so 12 minus a negative 9 is the same as 12 plus 9 that's 21. negative 14 minus a negative 7 that's basically negative 14 plus 7 so that's negative 7. so 21 over negative 7 is negative 3. now typically with slope we'll say negative three over one so we can still refer to rise over run my rise is negative three my run is one so if i'm on a graph and i'm at a point there i can fall three units and run to the right one and i'm back to the line or again another thing i can do i can switch this negative i can say this is three over negative one so i can say the rise is three and the run is negative 1. so now if i'm at a point on the line i can go up 3 and to the left 1 and i'm back on the line all right so pretty easy overall to find the slope of a line if you're given two points if you're not given two points if you're just given the equation then you can just calculate two points then plug it into that slope formula super super easy to do all right so now let's talk about a few special case scenarios so the slope of any horizontal line is always zero so why is that the case well we know a horizontal line just goes straight across so its steepness would be nothing right it's it's not rising at all as it's moving to the right so it would basically be zero over whatever zero over one or zero over twenty five so i go zero units up for every one unit i go to the right or i go zero units up for every 25 units i go to the right and this doesn't look correct zero over one equals zero over 25 yeah zero divided by any non-zero number is zero this is zero equals zero right i have a steepness or slope of 0. so you can use the slope formula to calculate this remember i can generate two points by choosing a value for x arbitrarily let's say it's 3 for example so x is 3 y is 3. x is let's say 10 y is 3. no matter what i choose for x y will always be 3. so then if i plug that into my slope formula m is equal to and let's say this is x sub 1 y sub 1 and this is x sub 2 y sub 2. so m is equal to y sub 2 which is 3 minus y sub 1 which is also 3 that's where you get your 0 from over x sub 2 which is 10 minus x sub 1 which is 3. so 3 minus 3 is 0 over 10 minus 3 which is 7. 0 over 7 is obviously 0. so your vertical change or your rise will always be zero because the value for y for those two points that you choose will always be the same the same number involved in a subtraction problem is always going to give you 0. 5 minus 5 is 0. 20 minus 20 is 0. 2 million minus 2 million is 0. you know so on and so forth so the other special scenario would be the slope of a vertical line so the slope of a vertical line is undefined so why do you think that is well again we go back to our slope formula we have m is equal to y sub 2 minus y sub 1 over x sub 2 minus x sub 1. if x always equals negative 4 for x sub 2 it's going to be negative 4 for x sub 1 it's going to be negative 4. negative 4 minus a negative 4 is negative 4 plus 4 that's 0. so i'm going to have division by 0 which is not allowed it's undefined so that's why we say the slope of a vertical line is undefined and again if you want an example of that let's say you're given a problem where you have something like negative 4 comma 3 and let's say you have negative 4 comma 2. if you see the same value for each x you know you have an undefined slope let's say this is x sub 1 y sub 1 and this is x sub 2 y sub 2 just try plugging it in so for y sub 2 you get 2 for y sub 1 you get 3 for x sub 2 you get negative 4 and for x sub 1 you also get negative 4. and remember minus a negative is plus a positive so really this is negative 4 plus 4. so 2 minus 3 is negative 1 but it's over 0 so again this is undefined all right so a couple of just key points on slope so a positively sloped line rises as we move to the right so if you see a slope that's something like three fifths i know this is positive i know that as x increases so as we move to the right that line's going up so a positively sloped line looks something like this again as we're moving to the right as x is increasing the values for y are increasing so i'm going up then similarly a negatively sloped line falls as we move to the right so as x increases in this case a negatively sloped line would have a y that decreases so you'd see something with a negative slope you have a negative let's say it was negative 3 or let's say negative two-fifths or something like that that type of line again falls as it moves to the right so it looks something like this as we move to the right as x increases the y values are decreasing all right so now let's get into something that is going to probably piss you off slope can be found without using the slope formula it can be found directly from the equation of a line so i know a lot of students they get these equations they go through they figure out these points and they plug in the slope formula they do all this work and then they figure out that it's so easy to just solve for y often they get really really mad so hopefully you're not upset about that it's important to understand all the techniques that we have in math because sometimes you're not given the equation sometimes you have to generate it but i have here that y equals mx plus b you remember that m stands for slope so if i just solve the equation for y the coefficient of x is the slope and i'll prove that to you in a second b is something that's new for us not if we took algebra 1 but if we haven't seen this before b is basically the y coordinate for the y intercept so in other words if i plugged in a 0 for x m times 0 would be 0. so this would disappear i'd have y equals b so the coordinate 0 comma b is your y intercept all right so let's say we're given an equation x minus y equals 28 and i tell you to find the slope so all i'm going to do is just solve it for y so i want to isolate y on one side so the first thing i'm going to do is subtract x away from each side so i'll have negative y is equal to negative x plus 28 then i just want to get rid of the negative so let's divide both sides by negative one and so this will cancel with this i'll have y is equal to you think about this as dividing each by negative 1. so negative x is divided by negative 1 so that's x and then 28 divided by negative 1 that's minus 28. so y equals x minus 28. so the coefficient of x here is understood to be one so that's my slope and my y-intercept would occur where well if i plugged in a zero for x i'd have negative 28 as my result for y so my y-intercept is at 0 comma negative 28 again this is m this is b when you think about b you want to think about the sign that's in front i could really write this as plus negative 28 if i wanted to so that's why we write zero comma negative 28. all right so let me prove to you that this works i'm going to prove to you that the slope is 1 here so let's say you didn't know what i just taught you and you're given this in there your only technique is to use the slope formula so you generate two points let's say that you generate i don't know let's say x is 12 what would y be i'll plug in a 12 there so then i'd have 12 minus y equals 28 i would subtract 12 away from each side of the equation that cancels i have negative y is equal to 28 minus 12 is 16. divide both sides by negative 1 you get y is equal to negative 16. so if x is 12 you get y is equal to negative 16. as another point let's just say we pick how about 14 so if x is 14 what's y so 14 minus y equals 28 we would subtract 14 away from each side of the equation that cancels you get negative y equals 14 multiply both sides by negative 1 or divide by negative 1 whatever you want to do you'll get y equals negative 14. all right so now let's plug in to the slope formula so the slope formula again we have m is equal to y sub 2 minus y sub 1 over x sub 2 minus x sub 1. so let's just say this is x sub 1 y sub 1 and this is x sub 2 y sub 2. so then my y sub 2 is negative 14. my y sub 1 is negative 16. my x sub 2 is 14 my x sub 1 is 12. so negative 14 minus a negative 16 is the same thing as negative 14 plus 16. so negative 14 plus 16 is going to give me 2. 14 minus 12 is also 2 2 over 2 is 1. so you remember when we solved this equation for y we found that the slope was 1. so look how much easier that was this took us forever we had to generate points we had to plug it into the slope formula you know so on and so forth so solving an equation for y is generally going to be the quickest way to find the slope all right let's look at one more of these so suppose you get 6x minus y equals negative 8. again you want to find the slope isolate y on one side the slope is given as the coefficient of x so let me subtract 6x away from each side of the equation that'll cancel you'll have negative y is equal to negative 6x minus 8. we would then just divide both sides by negative 1. so i'd have y is equal to negative 6x divided by negative 1 is 6x negative 8 divided by negative 1 is plus 8. so then i have y equals 6x plus 8. so the coefficient of x which is 6 is my slope this is m all right remember y equals mx plus b my y intercept occurs at 0 comma 8. again plug in a 0 for x 6 times 0 would be 0 i'd have y equals 8. so this is my y-intercept and just one more time let me prove to you that this works so i'm going to write here that m equals 6 and i'll verify this for you let's just say that x was zero so six times zero minus y equals negative eight so this is zero minus y equals negative eight divide both sides by negative one and i'll have y is equal to eight so if x is zero y is eight that's one point for us as another point let's say x was negative two so negative two comma what so six times negative 2 is negative 12 then minus y equals negative 8. let's add 12 to both sides of the equation that'll cancel i'll have negative y is equal to negative 8 plus 12 is going to give me 4 divide both sides by negative 1 and you get y is equal to negative 4. so again let's use our slope formula now and we'll label each point arbitrarily let's say this is x sub 1 y sub 1 this is x sub 2 y sub 2. so i use my formula m equals y sub 2 which is negative 4 minus y sub 1 which is 8 over x sub 2 which is negative 2 minus x sub 1 which is 0. so negative 4 minus 8 is negative 12 over negative 2 minus 0 which is just negative 2 and if i divide negative 12 by negative 2 negative over negative is positive 12 divided by 2 is 6. so i get m equals 6 in the end which is what i found pretty quickly by just solving the equation for y so kind of some wrap-up notes here finding slope is not very difficult at all you can use the slope formula if you know two points on the line so if you're given an equation and you want to use the slope formula generate two points for the line and plug them into the slope formula otherwise if you want to find the slope you can solve the equation for y the slope is given by the coefficient of x hello and welcome to algebra 2 lesson 18. in this video we're going to continue to learn about slope so in our last lesson we talked all about how to find the slope of a line and we learned the slope formula which was m which denotes slope is equal to y sub 2 minus y sub 1 over x sub 2 minus x sub 1. so we learn this formula so if i had an equation for example like this one 2x plus y equals 4 i could find two points i could plug those two points into this formula and i could calculate a slope so let's do that real fast so for this one two points if i choose zero for x my y-intercept i would have zero times 2 so this would go away i just have y equals 4. if i chose 0 for y i would have 2x is equal to 4 i would divide both sides by 2 and i'll get x equals 2. so the x-intercept occurs when y is 0 and x is 2. so i have two points there once i have my two points i label them so they're ordered pairs one is zero comma four the other is two comma 0. so again it doesn't matter which i label as which i'm going to label this one as x sub 1 y sub 1 and this one as x sub 2 y sub 2. you could switch that around and you're going to get the same answer so i plug into this formula i have m is equal to for y sub 2 i have 0 minus for y sub 1 i have 4 and this is over for x sub 2 i have 2 minus for x sub one i have zero so zero minus four is negative four two minus zero is two negative four over two is negative two so my slope m is negative two so we're gonna talk more about what that means in a second but i want to show you something else we talked about this in the lesson also the quicker way to kind of get the slope is to solve the equation for y we put it into a format that is known as slope intercept form so let me just note that m is equal to negative 2 here so slope intercept form looks like this it's y equals m which is your slope times x plus b so this is slope intercept form and if you haven't gotten to this in your class yet you're going to cover this when you talk about different forms of the line okay but it's good to kind of get started with it now because it's just a quick and easy way to get the slope so if i solve this for y meaning i isolate y on one side of the equation by itself i just subtract 2x away from each side this will cancel i'll have y is equal to negative two x plus four now again it's y equals m the slope times x plus b so y equals negative 2 negative 2 is m this is the slope we found that over here by doing a more complex procedure right we had to generate points we had to label the points we had to plug them in you know so on and so forth this is so much faster all i have to do is solve for y and look at the coefficient for x so it's kind of an upgraded procedure so we solve the equation for y we automatically know the slope now the other thing we have is we have the y-intercept remember to get the y-intercept i plug in a 0 for x and i find out what y is well if i plug in a 0 for x i'm always going to have this right here right because 0 times m would give me 0. so this would cancel away and it'd have a y-intercept that occurs at 0 comma b that's why it's called slope-intercept form we know the slope and we know the y-intercept all right so for this particular one i know that the y-intercept would occur at 0 comma 4. now what can i do with this information well again it allows us to know the slope very quickly it allows us to know the y-intercept very quickly but it also allows me to graph very very quickly i don't have to sit there and generate three points and then go plot the points and do all this work i can simply take this point right here zero comma four i can go to my coordinate plane and let me label this this is the x axis this is the y axis so where's zero comma four well that's going to be right here now remember my slope m is equal to negative 2. so what does that mean we talked about that in the last lesson we talked about rise over run so rise over run here is negative 2 over one right because i could take negative two and just put it over one so i rise negative two that means i just fall so i'm gonna go down two units i run one that means i go to the right one so when we talk about running we're talking about horizontal movement so if that's positive i'm going to the right if it's negative i'm going to the left when we talk about rising if it's positive i'm going up if it's negative i'm going down so that's why negative 2 over 1 i went down 2 to the right one and just do that again so down two to the right one and now i have three points here of course i only need two to make a line but it's nice to have three i'm gonna go ahead and just draw my line and this is the graph for the equation 2x plus y is equal to 4. now look how fast i found the slope look how fast i found the y-intercept and look how fast i graph this thing you know in the previous lessons we sat there and we generated points and we you know we did all these complex procedures or not really complex just kind of tedious things now if i just solve this equation for y i can go to my coordinate plane and graph it like two seconds okay so this is the fastest way to graph a linear equation in two variables just solve it for y use the y intercept as your first point and then use your slope to generate one or two or even three more points so that you can sketch the graph all right let's take a look at another example so let's say you saw 5x minus 4y equals negative 12 and i asked you to find the slope find the y-intercept and to graph it okay it's a common example that you might see so let's start out by solving this for a y so this is y here we've got to get it by itself so my first objective is to move this over here and i do that by just subtracting 5x from both sides of the equation so this would cancel itself out i'd have negative 4y is equal to negative 12 minus 5x of course i'm going to switch the order there so i'm just going to write that as minus 5x minus 12. and now to get y by itself all i need to do is divide both sides of the equation by negative 4. so i'm going to divide this side by negative 4 and i'm going to divide this side by negative 4 but i'm going to do it to where i split it up so what i mean by that is i'm going to put negative 5x over negative 4 and then i'm going to space this out and put plus negative 12 over negative 4. so that's going to make it obvious what i need to simplify to here this will cancel with this and i have y and this equals negative 5 over negative 4 i can't really do anything with that but i'm going to write 5 4 because i know the negative over negative will be positive so 5 4 x and then negative 12 over negative 4 i can write that as 3 so plus 3. now what information do i have well i have my slope remember it's y equals m the slope times x plus b so my slope is 5 4. m equals 5 4. now b again if i think about the y intercept it occurs at 0 comma b well b is 3. so the y-intercept will occur at 0 comma 3. so let's use this information to graph the equation so again we found that this was y equals 5 4 times x plus 3. now my y-intercept occurs at 0 comma 3. so 0 comma 3 and again slope m is 5 4. so where is 0 comma 3 that's going to be right here now for a slope of 5 4. again you think about rise over run is equal to five over four and it's very very easy to figure out what's going on here if rise is five that means i just go up five units so one two three four five now if run is 4 then i go to the right 4. 1 2 3 4. that's a point right there and you can check that by plugging it in so that's an x value of 4 a y value of 8. so if i plugged in a 4 here 5 4 times 4 you know the 4s would cancel you'd have 5 plus 3 which is 8. so that works out mathematically all right so now that we see that can i rise 5 and go to the right 4 using what i have here no i'd run off so what i want to do is i want to use a little trick and you'll probably have to do this depending on the scale of what you're using to create your graph but 5 over 4 realize with fractions i can write this as negative 5 over negative 4. negative over negative is positive so the value of that fraction is still positive and i can use that trick to start here and i can say well the rise is negative 5. so i'm going to go 1 2 3 4 5 down the run is negative 4. so horizontally a negative 4 means i'm going to the left so go to left one two three four and that's where i'd be and if you start at this point you can use this to get back to this i'd go up one two three four five go to the right one two three four so you can go back and forth in how you do that okay so if you have a positive value you can go negative over negative to get another point if you need to do that so now that we have three points here let's go ahead and sketch our graph [Applause] okay and let me just label this originally we had it as 5x minus 4y equals negative 12. you could also label it with y equals 5 4 x plus 3. it doesn't really matter you're just saying hey this is the equation for this line that we just graphed all right so now that we understand how to kind of manipulate an equation and we can now graph very quickly we can now find slope very fast let's talk about something that's going to come up usually in the next section so i want to talk about parallel and perpendicular lines so two non-vertical lines that have the same slope are parallel so you'll know that they have the same slope if you solve each equation for y you'll have the same slope but a different y-intercept those are parallel lines and i'll graph an example of that so you can see what that looks like now two perpendicular lines okay perpendicular lines have slopes whose product is negative one so again for that you would just solve each equation for y you'd look at the slopes and you see if you multiply them together if you get negative one those two lines are going to be perpendicular so very very easy topic overall okay let's go ahead and take a look at the first problem so we have 3x plus 5y equals negative 15 along with 6x plus 10y equals 70. so the smart way to attack this problem is to solve each equation for y we want it in the format of y equals m the slope times x plus b the y-intercept again i can't say this enough this is called slope intercept form so when we solve this for y we will know the slope of each equation we can then look at those two slopes and we can see if they're the same if they're the same we have parallel lines if they're different then we're going to check to see if the product of those two slopes would be negative 1 if that's the case we have perpendicular lines and in some cases we're not going to have parallel lines or perpendicular lines you'll see problems where they're just neither all right so let's solve the first one for y here so 3x plus 5y equals negative 15. so 3x plus 5 y equals negative 15. if i want to isolate y then what i'm going to do is i'm going to start by subtracting 3x away from each side of the equation so that's going to give me 5y is equal to negative 3x minus 15. then i'm going to divide each side of the equation by 5 what's that going to give me well we know that this would cancel with this so y is by itself so y is equal to i'll have negative three fifths x so negative three fifths x and then minus we have 15 over 5 which is 3. so let's erase this and i'm just going to drag this over here okay let's take a look at this one now so 6x plus 10y equals 70. so again 6x plus 10 y is equal to 70. to solve this guy for y let me just highlight that i'm going to subtract 6x away from each side of the equation so that's going to give me 10 y is equal to negative 6x plus 70. i want y by itself so let's divide both sides of the equation by 10 and i'll do that over here so there's room so this cancels with this y is by itself and it's equal to negative 6 over 10. well we know that they're both divisible by 2. so it's going to be negative 6 divided by 2 is 3. 10 divided by 2 is 5. so this is negative three fifths then times x then plus we have 70 over 10 which is just seven so let's erase this and let me drag this guy over here now so we see that we have y equals negative three fifths x minus three and y equals negative three fifths x plus seven so what can we conclude here well we can see that the slope here is negative three fifths and the slope here is also negative three-fifths so when two lines have the same slope and they have different y-intercepts this one would have a y-intercept that occurs at zero comma negative three this one would have a y-intercept that occurs at zero comma seven these two lines are parallel right they are parallel lots and let me make these ease a little better what are parallel lines well essentially it looks something like this and i know i didn't draw that perfectly but parallel lines will have the same amount of distance between them no matter how far out you go so you can imagine a coordinate plane just stretch it in your head forever and ever and ever these two lines have the same slope the same steepness so they have the same distance between them always they're never ever ever going to intersect or touch each other in any way so let's go ahead and take these two down to the coordinate plane we're going to graph them and we're going to take a look at what parallel lines would look like all right so for the first equation y equals negative three fifths x minus three the y intercept occurs at 0 comma negative 3. so that's right here and then the slope again rise over run is equal to negative 3 over 5. so i can rise negative 3 and run 5 or i can rise 3 and run negative 5. remember negative over positive is the same as positive over negative so again these are little techniques you can use to get points depending on what the scale of your coordinate plane is so i'm going to go with the negative 3 over 5. so i would go down 3 1 2 3 to the right 5. 1 2 3 4 5. there's a point now let's say i want to make a point somewhere over here so let's use this one let's go up 3 1 2 3 we're going to go to the left 5. 1 2 3 4 5. okay and again this is y equals negative three fifths x minus three or again you could also write 3x plus 5 y equals negative 15 whatever you want to write all right for the next one we have y equals negative 3 5 x plus 7. so same slope different y intercept y intercept there is at 0 comma 7. my slope is again negative three fifths so again i'm going to go down one two three to the right one two three four five and i'll just do that again let's go down one two three to the right one two three four five okay let's go ahead and graph this okay let me label this one as y equals negative three-fifths x plus seven and again you could write that as 6x plus 10y equals 70. it's the same thing in the end right one is solved for y one is not so now that we have this drawn we can observe what parallel lines look like there are two lines that will never ever intersect on the coordinate plane right these are parallel lines [Music] and again you can see that they have the same slope or the same steepness okay but they have different y intercepts so they're in different locations of the coordinate plane and again they're never ever ever going to touch because their slopes are the same let's take a look at the next example so we have x plus 3y equals 12 and we have 4x minus 3y equals 3. so are these lines parallel are they perpendicular or are they just nothing so i would solve each for y again we're looking for y equals mx plus b so for this one i have x plus 3y equals 12. i subtract x away from each side that'll cancel i'll have 3y is equal to let's switch the order here i'll put negative x plus 12 and then i'll divide each side by three so this is going to cancel i'll write y is equal to i'm going to put a 1 there so i put negative 1 3 x plus 12 over 3 is 4. let's go ahead and write this and erase everything here so again y is equal to negative 1 3 x plus 4. all right for the next one let's go ahead and do that over here so 4x minus 3y is equal to 3. let's subtract 4x away from each side that's going to cancel i'll have negative 3y is equal to negative four x plus three and what i want to do now is i want to divide both sides by negative three this will cancel with this i'll have y is equal to negative four over negative three is four thirds and times x and then three over negative three is negative one so minus one let's erase all this and let's write this here so we have y equals four thirds x minus one so looking at the slopes what can we say we have negative one third and we have four thirds so i know they're not the same so they're not parallel lines but would the product be negative one so if i took negative one third which is the slope from here and i multiplied it by four thirds which is the slope from here would that be equal to negative one well if i multiply here negative 1 times 4 is negative 4 3 times 3 is 9. so i end up with negative 4 9. so the answer to that is no they're not parallel they're not perpendicular they're really neither let's take a look at one more so we have 7x plus 4y equals 8 and we have negative 4x plus 7y equals negative 21. so again i want to solve each for y i want y equals mx plus b so if i solve this guy for y i would have again 7x plus 4y equals 8. let's move 7x over to the other side so i subtract away from each side this will cancel i will have 4y is equal to negative 7x plus 8. then let me divide each side by 4 so this will cancel i'll have y is equal to negative 7 4 times x plus 8 over 4 is 2. let's write what this is just drag this up here all right now let's go ahead and do this one now so i have negative 4x plus 7y equals negative 21. and again let me add 4x to each side of the equation that's going to cancel i'll have 7y is equal to 4x minus 21. i'll divide each side by 7. that'll cancel with that i'll have y is equal to 4 7 x negative 21 over 7 is negative 3. so let me erase this and i'll drag this up here so y equals 4 7 x minus 3. so let's look at the slope in each case this one is negative 7 4 this one is 4 7. most of you can see that if i multiply those two together i'm going to get negative 1. right if i take negative 7 4 and i multiply by 4 7 this cancels with this this cancels with this i'm just left with a negative one [Music] so if the product of the slopes is negative one we have something known as perpendicular lines so per pendicular [Music] and what this means is that these two lines will intersect at a 90 degree angle for those of you who have never seen this or taken geometry before i'm going to graph this real quick so that you can see it alright so let's take these two down to the coordinate plane all right so for this one we have a y intercept of zero comma two so that's going to be right here and then my slope my rise over run is negative 7 4. so if i start here and i rise negative 7 i go down 7 units so 1 2 3 4 5 6 7. then i run 4 so i go to the right 1 2 3 4 so that's a point now i'd go off the coordinate plane that i have if i go down 7 and go to the right 4 again so what i'm going to do again i'm going to use a trick if i have negative seven over four that's equal to seven over negative four okay so i could start here and i could rise seven one two three four five six seven and i could run negative four which means i go to the left 1 2 3 4. so let me label this this is y is equal to negative 7 4 x plus two all right so for the other one let me kind of erase this part because it's going to change we have a y intercept that occurs at zero comma negative three so that is going to be right here and then the next part is the slope is now 4 7. so let me change colors for 4 7 i'm going to rise 4 so rise 1 2 3 4 go to the right 7. 1 2 3 4 5 6 7. it's right there and again i can reverse that i can put 4 7 is equal to negative 4 over negative 7. negative over negative is positive so i could start here and i could go down four down one two three four and i could go to the left seven one two three four five six seven again this is going to be y is equal to 4 7 x minus 3. now i want you to look at something real quick again if you've never seen a 90 degree angle or right angle as it's called you want to look at the point of intersection here which is right there it forms a 90 degree angle which is displayed with this symbol here this is a 90 degree angle and if there's 190 degree angle it's such that it's a 90 degree angle at all of them so this is a 90 degree angle this is a 90 degree angle and this is a 90 degree angle and you'll talk more about that in your geometry class i mean we really don't go way into this in algebra but it's something you need to know to kind of understand what perpendicular lines are it's again such that at the point of intersection which is right here you're going to form a 90 degree angle and in fact you're going to form four of them so these are perpendicular lines hello and welcome to algebra 2 lesson 19. in this video we're going to learn about forms of a line so we talked about this already in algebra one we're going to revisit for algebra two when you talk about a linear equation in two variables there are ways you can algebraically manipulate this equation and put it in different forms so we have slope intercept form we have point slope form and then we have standard form that we're going to talk about today so based on what form it's in you can use it for certain things get different information from it and when we start out with slope-intercept form we know the slope and we know the y-intercept and we've already talked about this form in the previous two lessons so we have y equals mx plus b again this is slope intercept form so when i have something like this i know the slope which is denoted as m this is given as the coefficient of x and i know the y intercept is going to occur add to 0 comma b how do i know that well if i want to find the y intercept again i have y equals mx plus b to find the y-intercept i just plug a 0 in for x 0 times anything is 0. so this would go away and i'd have y equals b so that's why i know that the y-intercept occurs at 0 comma b so as an example if i had y equals x plus two i know that the slope is three and the y intercept occurs at zero comma two very very simple if i have something like y equals one fifth x minus seven the slope is one fifth the y intercept occurs at zero comma negative seven super super easy so let's say we wanted to take this information and write an equation in slope intercept form so we have m which is the slope is equal to negative five halves and we have a y-intercept that's zero comma negative four so just match this up we have again y equals mx plus b and do it like this when you first start so you don't forget so you have that in your head and you say okay well y equals i know i know m is slope so m is equal to negative five halves so i'm gonna put negative five halves here then times x then i have plus b the y intercept occurs at zero comma negative 4. so b is really the y coordinate there so just negative 4. so i'm going to put for plus b i'm going to put minus 4. so using this information i write y equals negative 5 halves x minus 4 because negative 5 halves is the slope which is m and the y intercept occurs at 0 comma negative 4. all right let's take a look at another one we have a slope m that is equal to negative 4 thirds we have a y intercept that occurs at 0 comma four so again let's match this up we have y equals mx plus b so this would be y equals for m i have negative four thirds times x and then for b i have four so plus four so y equals negative four thirds x plus four all right what about this one we have m equal to one and we have a y intercept at zero comma three so again if i have y equals mx plus b m is 1 now so y equals 1x is just x and then plus b b is 3 here so plus 3. so y just equals x plus 3. so again slope intercept form is super easy so you know if you take a linear equation two variables and you solve it for y you get slope intercept y equals mx plus b it makes it easy for you to know the slope the y intercept it also makes it easy for you to graph right you have a point and you have the slope so it's super easy to do that as well so now i want to talk about kind of the next form of a line which is point slope form so you're going to use this in two specific scenarios one you know the slope and one point on the line the other is you have two points you're not going to go straight into slope intercept form if you don't have all the right information so with point slope form i plug in for y sub 1 x sub 1 and m and then i can solve for y and i can get slope intercept form so we have y minus y sub 1 equals m times the quantity x minus x sub 1. so let's go down here so let's say we have a slope m that's one half and this line goes through the point two comma four so what we wanna do here again just just copy at the start so point slope form is y minus y sub 1 is equal to m times the quantity x minus x sub 1. so the y sub 1 and the x sub 1 come from the point that you're given so you kind of think about 2 comma 4 this is a point on that line so this is your x sub 1 this is your y sub 1 and this is your m so all that stuff's just going to get plugged in so let me let me highlight m as well so i'm going to have y minus y sub 1 for y sub 1 i'm plugging in a 4. and this equals for m i've got one half times the quantity x minus for x sub 1 i'm plugging in a 2. now if your teacher asks you for point-slope form stop here you've arrived this is point-slope form so if you're asked for that specifically you're done okay but in most cases they want you to solve for y and put it in slope-intercept form this is a scenario where you did not know what the y-intercept was you're just given a point in the slope okay so you've got to go through it this way so now what i'll do is i'll have y minus 4 is equal to one-half times x is one-half x minus one-half times two is just one so i add four to each side of the equation that'll cancel and i'll have y is equal to one-half x negative one plus four is going to give me three so plus three so now it's in slope intercept form and i can also add that the y-intercept occurs at zero comma three so i know the slope which is one-half that matches up there i know a point on the line is two comma four and you could check that plug in a two here two times a half is one one plus three is four so that works out and i also know the y intercept would occur at 0 comma 3. so again y equals 1 half x plus 3. all right let's take a look at another one so suppose i had a slope m that's 1 and this line goes through the point negative five comma negative two so again i'm going to label this x sub 1 y sub 1 and i'm going to plug into the point slope form so that's going to be y minus y sub 1 is equal to m times the quantity x minus x sub 1. so for y sub 1 i have negative 2 for m i have 1 and for x sub 1 i have negative 5. so again point slope form so then y minus a negative 2 be careful there is equal to 1 times the quantity x minus a negative 5. so be careful there as well so again this is point slope form if you're asked for that you can stop in most cases you're asked for slope intercept form so keep going so what i want to do is simplify so this is y plus 2 is equal to i can get rid of the 1 really because 1 times anything is itself so just x minus a negative 5 is plus 5 so i just subtract 2 away from each side of the equation and i get y is equal to x plus 3. so now i know that the y intercept occurs at 0 comma 3. we already knew the slope was 1 and we have a point on the line of negative 5 comma negative 2 right so if i plug in a negative 5 there negative 5 plus 3 is in fact negative 2. so again the equation of this line in slope intercept form is y equals x plus 3. let's take a look at another one so we have m equals two-thirds and it passes through the point negative three comma three so again we have y minus y sub one is equal to m times the quantity x minus x sub one and i'm going to label this as x sub 1 y sub 1 and i'm just going to plug in so i'll have y minus or y sub 1 i have 3 is equal to m which is 2 3 times the quantity x minus for x sub 1 i have a negative 3. so minus a negative 3. [Music] all right so now we have y minus 3 is equal to 2 3 times x is 2 3 x and then we'll have 2 3 times 3 right minus the negative 3 is plus 3. so if i had 2 3 times 3 the 3's would cancel be left with 2 so plus 2. all right so now all i need to do is just add 3 to each side that'll cancel and i'll end up with y is equal to 2 3 x 2 plus 3 is 5. so plus 5. so what can we say here we kind of erase this so i can fit everything on the screen and scroll this up well we can say that the slope is two-thirds the y-intercept occurs at zero comma five and we have a point negative three comma three so if i plugged in a negative three there negative three times two-thirds would be negative two negative two plus five would be three so that does work out all right so here's another scenario you're gonna be given it's kind of a curveball to throw at you let's say you're just given two points so you have through the point negative 2 comma 4 and 4 comma negative 5. so the first thing you got to do is you got to get a slope because you can't make it with this formula you have y minus y sub 1 equals m times the quantity x minus x sub 1. so you're not going to make it you've got to find out what m is so to do that we just use our slope formula so i label one of these points as x sub 1 y sub 1 the other as x sub 2 y sub 2. now here's where it gets confusing the way i label this here has nothing to do with my inputs over here okay i can label this x sub 1 y sub 1 this x sub 2 y sub 2. when i go to plug in for x sub 1 y sub 1 it doesn't matter if i use this point or this point it's irrelevant so knowing that let's do the first step and just find out so m is equal to remember the formula it's y sub 2 minus y sub 1 over x sub 2 minus x sub 1. so m will be equal to y sub 2 is negative 5 minus y sub 1 which is 4 over x sub 2 which is 4 minus x sub 1 which is negative 2 minus the negative 2 is plus 2. negative 5 minus 4 is negative 9 over 4 plus 2 which is 6. so each is divisible by 3 negative 9 divided by 3 is negative 3. 6 divided by 3 is 2. so this is negative 3 halves for the slope let me erase everything and we can put that m equals negative three halves now because we have this information now we can plug in here so y minus something equals negative three halves times the quantity x minus something right for x sub 1. now i can plug in this point as x sub 1 y sub 1 or this point is x sub 1 y sub 1. it does not matter that i labeled them this way to get the slope it's irrelevant now you can kind of choose a point and then forget the other either point would work okay i can't stress that enough because students really get stressed out over this stuff they're like well which one do i plug in now what do i do just pick a point plug it in it'll look different in the point-slope form but if you go into slope-intercept form it's going to be the same and i'm going to prove that to you right now so let's start with this one so let's say we have a 4 here that's a y sub 1 we're plugging in let's say we have minus a negative 2 right here so doing that i have y minus 4 is equal to negative 3 halves times x and then this is plus let me write this as plus 2 and then negative 3 halves times 2 would be minus 3 right because the 2's would cancel all right so if i just add 4 to each side of the equation i'm going to end up with y is equal to negative 3 halves x negative 3 plus 4 is plus 1. now let's say i put this over to the side so y equals negative 3 halves x plus 1. and let's forget about this point let's just forget that it exists let's erase all this and i'm going to show you that it doesn't matter if i choose this point so now i'm going to choose this as x sub 1 y sub 1 and remember this formula is y minus y sub 1 equals m we already know that's negative 3 halves times the quantity x minus x sub 1. so x sub 1 is 4 y sub 1 is negative 5. so minus a negative 5 is plus 5. so we'd have y plus 5 is equal to negative 3 half times x is negative three times x then negative three halves times negative four we know that's positive the four would cancel with the two and give me two two times three is six so as a last step here let's just subtract five away from each side and that's going to give me that y is equal to negative 3 halves x plus 6 minus 5 is 1. same thing here notice that it didn't matter which point i plugged in for in point slope form it looks different because the inputs are different but once i solve for y it's the same okay because it's got the same slope and the same y-intercept because those two points are on the same line all right so let's try another one like that let's say it went through the points negative 1 comma 1 and 0 comma negative 5. again let me calculate slope first so i'm going to label this as x sub 1 y sub 1 and this as x sub 2 y sub 2 and then i'm plugging into my slope formula so m is equal to y sub 2 is negative 5 minus y sub 1 is 1 over x sub 2 is 0 minus x sub 1 is negative 1. so minus the negative 1 is plus 1. negative 5 minus one is negative six over zero plus one which is one so slope is just negative six right m is negative six there so now i plug into my point slope form so y minus y sub one and i can forget about this point now i only need one of them so let me just use this one so let's say this is x sub 1 y sub 1. so y minus 1 is equal to m which is negative 6 times the quantity x minus x sub 1 which is negative 1. so minus a negative 1 is plus 1. so now i'm going to solve this for y so let me say y minus 1 equals negative 6 times x is negative 6x negative 6 times 1 is minus 6. let's add 1 to each side of the equation and we'll get y is equal to negative 6x negative 6 plus 1 is minus 5. so y equals negative 6x minus 5. all right let's take a look at one more and then we're going to jump into standard form so it passes through the point negative 2 comma negative 3 and negative 2 comma 3. so some of you look at this and go wait something's wrong here i've got negative 2 here and i've got negative 2 here so if this is x sub 1 y sub 1 and this is x sub 2 y sub 2 what happens when i plug into the slope formula i get m is equal to y sub 2 is 3 minus y sub 1 so minus a negative is plus a positive so plus positive 3 over x sub 2 which is negative 2 minus a negative 2 right minus x sub 1. so plus 2. well the denominator here is going to be 0. we can't divide by 0 as undefined so this would be 6 over 0 this is undefined what type of line has an undefined slope well a vertical line does and that's x equals some number so you can see that you have the same value for x here and here so x is going to equal negative two okay it's not hard to figure this one out if i saw something where i had a horizontal line i would have the same value for y in each case if i have the same value for x in each case i have a vertical line all right now let's talk about my favorite topic because everyone wants to argue about it the standard form so standard form is going to vary from teacher to teacher textbook to textbook classroom to classroom high school to college everybody says it's something different okay let me give you the best definition i can for it i'm going to tell you that standard form is where we write a linear equation in two variables where a x plus b y equals c like this and a and b are not both zero that's the definition you're going to use if you're in high level math nobody cares if there's integers or fractions or things like that but for the time being while you're in algebra 2 while you're in high school you're probably going to get a much stricter definition you're probably going to be told that a b and c are integers so they don't want to see fractions they don't want to see decimals i don't see anything like that they're also going to tell you that a has to be greater than or equal to zero a and b are obviously not both zero and then lastly a b and c have no common factor except one so let's put a b and c have no common factor except one all right so i would write this down this is more than likely the case of what your teacher wants you to do if you're in high school if you're in a college algebra course or something higher than that and you're just watching this to review i wouldn't worry too much about it you generally just use ax plus b y equals c a b and c can be real numbers the reason we have a stricter definition in high school or at least what i've been told throughout the years is just to practice cleaning up equations you get a lot of practice that you need just manipulating things back and forth so that by the time you get into high level algebra this is not something that you struggle with another thing is it's going to help us have cleaner equations and ones that are more presentable and easier to work with so suppose i gave you this information and said write an equation in standard form so again ax plus b y equals c well i know that the slope is 1 and i know the y intercept occurs at 0 comma 0. so i could start out with y equals mx plus b and i could put it in this format by doing some algebra so y equals 1 which is my slope times x i'm just going to write x plus b here b is 0. so y just equals x well this is easy all i need to do is subtract x away from each side and i'd have negative x plus y equals zero is this in standard form well again it depends on whose definition you're using i told you when i was kind of announcing this that a needed to be greater than or equal to zero in this particular case i basically have a negative one has the coefficient for x so in a stricter version you would say no this is not in standard form and all i would really need to do is multiply both sides of the equation by negative 1. so if i do that negative 1 times negative 1 is 1 so i just have x there negative 1 times y is minus y and this equals 0 times negative 1 is just 0. so i would have x minus y equals 0 and this is considered standard form again usually in a high school setting all right so the next one we have a slope m that's equal to negative three halves and a y intercept that occurs at zero comma negative two so again if i write this in slope intercept form i'd have y equals negative 3 halves x and then minus 2. now i'm going to algebraically manipulate this i want ax plus b y equals c so let me add three halves x to both sides of the equation and that's going to give me three halves x plus y equals negative two now there are some problems with this i want a b and c to be integers right now my a is a fraction so how can i deal with that well i just multiply both sides of the equation by 2 and i clear the denominator so 2 times 3 halves is 3. so this would be 3x plus 2 times y is 2y and this equals 2 times negative 2 is negative 4. so now i've met all the requirements this is greater than or equal to 0 so that's ok a b and c are integers so that's ok and there's no common factor between the three a b and c other than one all right so now we want to look at something a little bit more complex we still want to write it in standard form but the information gets a little trickier it's through this point negative five comma three parallel to y equals negative seven fifths x plus three so if i have a point and i have the slope i can write it in point slope form as we learned a little while ago so m here would equal negative seven fifths now once i extract the information from this equation here i don't need it anymore cross it out okay it's going to confuse you you don't need it okay so get rid of it now i want to plug in a point-slope form so y minus if i label this as x sub 1 y sub 1 then y sub 1 is 3 this equals m which is negative 7 5 times the quantity x minus x sub 1 is negative 5. so minus negative five is plus five so i'd have y minus three is equal to negative seven fifths times x is negative seven fifths x negative seven fifths times five the fives would cancel you'd have minus seven so now if i add 3 to each side this will cancel i'll have y is equal to negative 7 5 x negative 7 plus 3 is negative 4. so now i have it in slope intercept form but i want it in standard form so let me add seven-fifths x to both sides of the equation and what i'm gonna have is seven-fifths x plus y is equal to negative four so as a final step remember i don't want this to be a fraction so i'm going to multiply both sides of the equation by 5 to get rid of that 5 times 7 fifths is 7 so 7 times x or 7x plus 5 times y is 5y is equal to negative 4 times 5 which is negative 20. so in standard form this equation is 7x plus 5y equals negative 20. let's take a look at one more so perpendicular lines have slopes where the product would be negative 1. so 2 times what equals negative 1. well you could use a variable let's say z is such number that you're looking for divide both sides by 2 and you're going to find that z where your unknown number is negative one-half so that's the slope that you would have m here would be negative one-half again because negative one-half times two would give you negative one and perpendicular lines have slopes whose product is going to be negative 1. so again once you've used this get rid of it don't want to think about it anymore it's not relevant to anything else so now if this is x sub 1 y sub 1 and i'll plug into my point slope form so y minus for y sub 1 i've got negative 3 minus the negative 3 is plus 3 equals m which is negative 1 half times the quantity x minus for x sub 1 i have 2. so i have y plus 3 is equal to negative 1 half times x is negative 1 half x then negative 1 half times negative 2 the negatives would cancel that'd be positive one-half times two is one so now i'm going to just subtract three away from each side of the equation i'll have y is equal to negative one-half x one minus 3 is negative 2. all right the last thing i'm going to do is add one half x to each side of the equation so i'll have one half x plus y is equal to negative 2. and then i want to clear this fraction right i don't want that so to do that i'm just going to multiply both sides by 2. multiply this side by 2 multiply this side by 2. 2 times 1 half is one so it's just x plus two times y is two y and this equals negative two times two which is negative four so finally i have this in standard form i have x plus two y equals negative four hello and welcome to algebra 2 lesson 20. in this video we're going to learn about linear inequalities in two variables so once we've mastered the art of graphing a linear equation in two variables you're going to find that it's not much more difficult to master the art of graphing a linear inequality in two variables so we start out with a basic form this is something you'll see in your textbook so a linear inequality in two variables is of the form ax plus b y is less than c where we would say a b and c are real numbers and then a and b are not both 0. now also this less than symbol here could be a greater than it could be a less than or equal to or it could be a greater than or equal to now when we talk about graphing a linear inequality in two variables when we look at our coordinate plane it's going to have three distinct regions so the first region i'm going to talk about is the solution region so this is where we have any ordered pair so we have x comma y that satisfies the inequality the next part i will talk about is the non-solution region so this is any xy or any ordered pair that does not satisfy the inequality so you have two separate regions now these regions are separated by something known as a boundary line so the boundary line is a line that separates the solution region from the non-solution region now the key thing to understand here is that this boundary line can be part of the solution or it can be not part of the solution it's based on what type of inequality you get if you get a strict inequality meaning you get a greater than or a less than right this is strict then the boundary line is not part of the solution okay if you get a non-strict inequality meaning you have greater than or equal to or less than or equal to this is non-strict then in this particular case your boundary line is part of your solution all right so let's talk a little bit about how we actually go about doing this it's quite similar to again graphing a linear equation two variables with just a little bit of a twist to it now there's two kind of ways to do this one way is a little bit slower to involve something known as a test point and then the other way involves just solving for y so i'm going to go through both procedures for you and then you can choose which one is better for you so graphing a linear inequality in two variables the first thing we want to do is replace the inequality symbol with an equality symbol then graph the resulting line this is how we get the boundary line okay when we do this if the inequality symbol again is strict then we want a dashed line because the boundary line in that case is not part of the solution if the inequality is non-strict we want a solid line because in that case the boundary line is part of the solution now the next thing we want to do is we want to choose a test point so this is any ordered pair that is not on a line so you just plug it back in to the original inequality and you see if it works so if it does work you're in the solution region so you shade that region where the test point is if it doesn't work then you're in the non-solution region and you just shade the other side all right so let's start out with the first example we have 2x plus y is less than 3. so the first thing i want to call your attention to is that this is a strict inequality so that tells me that my boundary my boundary line is going to be not part of the solution and so i want a dashed line usually what i do is i'll take and draw either you know gaps in my line or i'll draw a solid line and then kind of erase it depending on what method i'm drawing with if i'm on a computer then i'm going to break it up if i'm drawing it with a pencil then i'll just kind of gap it so we have a strict inequality the boundary line is dashed or broken now the other thing i would call your attention to is that it's not solved for y right now and i'm going to get to that in a minute so all i'm going to do is i'm going to replace this with equals so i'm going to pretend like i have 2x plus y is equal to 3. i'm just going to graph this this is going to be my boundary line this is the boundary so this will separate the solution region from the non-solution region okay so we know how to graph this let's get some points going so if x is 0 y is 3 if let's say i don't want to use the x intercept because that's going to give me a non-integer but let's go ahead and say if x was negative 2 i would have 2 times negative 2 plus y equals 3. so this would be negative 4 plus y equals three add four to each side and i get y is equal to seven so if x is negative two y is seven and let me get one more point going here let's say x is positive two so i'd have 2 times 2 that's 4 plus y equals 3. subtract 4 away from each side of the equation i'd have y is equal to negative 1. so i have three points there three ordered pairs and i'm just going to plot them on the coordinate plane so 0 3 negative 2 comma 7 and 2 comma negative 1. so 0 comma 3. that's going to be right here negative 2 comma 7. so i go 2 units to the left seven units up it's right here and then two comma negative one so that's going to be right here so again this line that i'm going to draw right now should be broken or dashed so because i'm drawing this on a computer i'm just going to draw a straight line then use an eraser to break it up okay and then let me get my eraser and that should be good enough so you just want to make it obvious that you have a broken up or dashed line again a broken line means that the line is not part of the solution if you have a solid line it is part of the solution and i'll give you an example of this if i take 0 comma 3 let's go back up and i plug that in because 0 comma 3 is on the line and this is a strict inequality it should not work so let's see if it does so we have 2 times 0 plus 3 is less than 3. 2 times 0 is 0 so this would cancel you'd be left with 3 is less than 3. well that is not true 3 is not less than 3 3 is equal to 3. but if this was a non-strict inequality let's say for example i had less than or equal to over here well then when it gets to this 3 is less than or equal to 3 well that would work so you can see that your boundary line is part of the solution if you have a non-strict inequality but not part of the solution if you don't okay so in this case we've got a strict inequality so it's not part of the solution now the next thing we want to do is we want to choose a test point so let me go back to the coordinate plane you want to look for something that is not on the line the most common test point you'll use is 0 0 because it's nice and easy to work with so you're going to say is 0 0 a solution for that inequality if it is i'm going to shade everything in that region if it's not i'm going to shade everything in the other region so i go back up here and i say my test point is zero comma zero so i just plug in so i say two times zero plus zero is less than three two times zero is zero so that's gone plus zero is zero so you get zero is less than three that's true so if this is true that means that zero comma zero lies in the solution region so that means that everything over here satisfies the inequality and everything on the other side of the boundary line does not and you could choose a test point over here and see if it doesn't work so something like let's say 4 comma 1. so this is in the non-solution region non-solution region and let's say we look at four comma one so i would have 2 times 4 plus 1 is less than 3. 2 times 4 is 8 8 plus 1 is 9. 9 is not less than 3. so this is false and we expect it to be false because i took a point that was in the non-solution region right it should not work and it didn't so that's good but when we look at this again this is the non-solution region over here is the solution region so typically we will shade this and depending on what you're working with if you're in class right you don't have a computer you would just kind of shade in the direction of the solution region so i'll show you this i'll just take a black marker and just shade here put arrows at the end to say hey this continues forever in this direction anything on this side of the line would satisfy that linear inequality in two variables okay so again you can see three distinct parts non-solution region boundary line which in this case is not part of the solution and then your solution region okay three parts now let me talk a little bit about how to do this in what i would say an easier fashion so let's erase everything here now some textbooks will cover this others will not and again it's a matter of personal preference i don't like using a test point i just think it's very slow what i do personally and what a lot of teachers will have you do is just start out by solving the inequality for y okay so remember if you have an equation solve for y it looks like this y equals m the slope times x plus b in this particular case it's an inequality but we're going to just put it in this format so i'm going to subtract 2x away from each side i'll have y is less than negative 2x plus 3. now i'm going to replace this with an equality symbol so i'm going to have y is equal to negative 2x plus 3. and i'm still going to draw my boundary line this is still the boundary and it's still going to be a dashed line because this is strict but how would i graph y equals negative 2x plus 3. so y equals negative 2x plus 3. well i have a y intercept that occurs at zero comma three so that's right here so i would start out by using that point and my slope is negative two so my rise over my run is negative two over one so i would fall two one two to the right one fall two one two to the right one so i would draw my boundary line based on that so i get the same exact line now here's where the fun part comes in if i've solved this for y i don't need a test point all i do is look at the symbol that i'm given if i have a less than i shade below the line so that's a less than or a less than or equal to let me kind of slide that down less than or less than or equal to shade below the line if i have a greater than or a greater than or equal to i'm going to shade above the line it's very very simple so if i look at this this is a less than i just come back to the graph once i have the line i just shade below it and that's exactly what i see there so look how much faster that would have been versus using kind of the slow method and then on top of that finding a test point going back to plug it in seeing if it works then shading so i think the test point method is quite slow actually and solving the linear inequality and two variables for y is definitely the way you want to go so let's take a look at another one now i'm going to use both methods for one last time and then from here on out i'm just going to solve this guy for y i'm going to kind of blow through these i'm going to do it by solving for y first though so i'm going to subtract 4x away from each side that's going to give me y is greater than or equal to negative 4x minus 1. now to get my boundary line i just replace this with equals so i'll have y is equal to negative 4x minus 1. now i want you to notice something this is a non-strict inequality so in this particular case my boundary line is solid right because points on that line will satisfy the inequality so if i graph y equals negative 4x minus 1 so y equals negative 4x minus 1. my y-intercept occurs at zero comma negative one so that's right here my slope is negative four so again rise over run it's negative four over one so i fall four one two three four i go to the right one i fall four one two three four i go to the right one or i can think about this as four over negative one so i could start here and i could go up four one two three four i can go to the left one so now when i sketch this graph remember this is a solid line now this boundary line is part of the solution [Music] all right now the next thing i'd look at is the fact that this is a greater than or equal to so for a greater than or greater than or equal to i shade above the line so i come back here if i want to shade above the line i would shade this direction just make it completely clear to your teacher or whoever's grading your test that this is the solution region and then this over here is the non-solution region so again you've got three different parts here you've got a non-solution region where ordered pairs do not satisfy the inequality you've got a boundary line which in this case will satisfy the inequality and then you've got a solution region okay so three parts here now you saw how quickly we did that let's erase all of this let's do this the slow way using a test point so i'm just going to start out again by replacing this with equals so i would have 4x plus y is equal to negative 1. and let's get some ordered pairs going so let's say x is 0 we know y equals negative one let's say that x is negative two we'd have four times negative two plus y equals negative one four times negative two is negative eight so we'd have negative eight plus y equals negative one add eight to both sides of the equation that'll cancel you'll get y is equal to seven let's try something simple for x let's try one so if x is one four times one is four plus y equals negative one subtract four away from each side of the equation you get y is equal to negative five all right so i have three ordered pairs there so i've got zero comma negative one i've got negative two comma seven and i've got one comma negative five all right so zero comma negative one is right here negative two comma seven i go two units to the left and seven units up that's right there and then one comma negative five if i go one unit to the right and five units down that's right there so my points are there i would draw my solid boundary line then i would have to choose a test point in this particular case i can still use zero comma zero because it's not on the line so i would go back up here and i would say four times again my test point is zero comma zero so four times zero plus zero is greater than or equal to negative 1. so you'd end up with what 0 is greater than or equal to negative 1 and 0 is greater than negative 1 so that is true so that would tell me that i need to shade above the line and again that's what i already found so using a test point works it's just a little bit slower in my opinion using this method where we solve the inequality for y allows you to graph the boundary line faster right because you can graph based on the slope and the y intercept and it allows you to shade the line faster because you shade based on whether it's a greater than greater than or equal to or less than less than or equal to okay so in all aspects in my opinion it's just simply faster alright so let's look at another one and again i'm not going to use a test point anymore so let's solve this guy for y so we have 5x minus 3y is greater than or equal to 15. if i solve this for y i'm just going to subtract 5x away from each side that's cancelled i'll have negative 3y is greater than or equal to negative 5x plus 15. now very important that you remember this if i divide both sides by negative 3 what do i have to do i've got to flip the direction of the inequality symbol if this is a greater than or equal to i've got to make sure to make this a less than or equal to otherwise i'm going to get the wrong answer i'm not going to shade in the right direction so this cancels with this i've got y is less than or equal to negative 5 over negative 3 we just write 5 3 then times x and this is minus 15 over negative three i've accounted for the negative 15 divided by three is five so minus five so i've got y is less than or equal to five thirds x minus five so let's erase this pull this up here and again i still want to graph my boundary lines so i'm just going to replace this with equals so my boundary line i would have y equals 5 3 x minus 5. now again because this is a less than or equal to it's a non-strict inequality so that tells me that points on the boundary line will satisfy the inequality and so i want a solid line all right so to graph y equals 5 3 x minus 5 i know the y intercept here is at 0 comma negative 5. that's right here my slope is 5 3. so my rise over my run is five over three so up one two three four five to the right one two three then again up one two three four five to the right one two three and let me erase a little part of that it's a little a little long for us so put arrows at each end so this is my boundary line again it's a solid line because we have a non-strict inequality now if i go back to my page i see that when i solve it for y i have a less than or equal to so that means i'm going to shade below the line again it's got to be solved for y you don't want to look at this original guy make sure it's solved for y first y is less than or equal to so that means i shade below the line so if i look if i want to shade below the line i would shade this way so anything shaded that's in the solution region the boundary line is part of the solution here and then anything over here above that line is in the non-solution region so these points will not work and you can always grab a test point to check you can take zero comma zero notice that's in the non-solution region so if we plug it in it should not work so if i go back up here i can take and plug it into this one or the original doesn't matter so 5 times 0 minus 3 times 0 should be greater than or equal to 15. again this shouldn't work so we expect this to be false this will cancel this will cancel you'll get 0 is greater than or equal to 15 which is false 0 is not greater than 15 and 0 is also not equal to 15. but again that's what we expect because 0 0 is in the non-solution region all right so let's talk about some special case scenarios now let's say you came across y is greater than one and you're like what do i do there y is greater than one well i want you to remember a few things here for one if you had y equals one this is a horizontal line this is a horizontal line so to graph this i would go to my coordinate plane and again i'm thinking about y equals one so find one on the y axis it's right here i would just draw a horizontal line so this would be y equals one when we look at this we see that this is a non-strict inequality this is non-strict so we know that points on the line will not satisfy that inequality so what we do is we make this a dashed line so y is greater than one we would have a dashed line so let me take my eraser now and we will break this guy up and again this identifies that points on this line are not part of the solution now if y is greater than one what does that mean in terms of a solution well that means anywhere on the y-axis that's larger than one is a solution so that means i would shade above the line because if y is four four is greater than one if y is eight eight is greater than one anything below the line doesn't work if y is negative one well that's not greater than one so that wouldn't work so this down here would be the non-solution region and anything above the line would be the solution ratio so i would shade this way and again how do you check something like that well formally again if you think about something like 0 x plus y is greater than 1 well if i choose a point where y is greater than 1 it's going to work out so if i choose a y of like let's say 5 for example x can be whatever x could be negative 30. so let's say i have the point negative 30 comma 5. plug in a negative 30 there plug in a 5 here negative 30 times 0 is 0 so that's cancelled you'll get 5 is greater than 1 which is true so it doesn't matter what the value of x is it's irrelevant as long as the y value is larger than 1 we have a true statement which is y on our coordinate plane we just shaded everything above the line again you're just looking at y values so you're only looking at the y axis in this case and anything on the y axis that's larger than 1 going up is going to work as a solution all right so as the other special case scenario let's say you saw something like x is less than or equal to negative 3. so you can almost guess what this is right if i if i graph x is equal to negative 3 what does that look like so x is equal to negative 3. i find negative 3 on the x-axis and i draw a vertical line now in this particular case we have x is less than or equal to negative 3. so the boundary line is a solid line so i don't need to break this up because points on that line will satisfy the inequality if i have x that is actually negative 3 negative 3 is less than or equal to negative 3 that's true because negative 3 is equal to negative 3. then also any x value that is less than negative 3 will work so how do we get less when we're talking about the x-axis well we go to the left so thinking about x values that are less than negative 3 i would shade to the left i would shade to the left just to think about this in terms of x values only because it doesn't matter what the y value is y value can be whatever if i wrote this as x plus 0 y is less than or equal to negative 3 you could choose a y value as high or low as you want let's say you choose 10 for example so if i choose an x value that's let's say negative 6 and a y value of 10 this is going to work because my x value is less than negative 3. plug in a negative 6 there this would cancel whatever it is 0 times anything is 0 so negative 6 would be less than negative 3 so that would work out if you choose any x value that is greater than negative 3 it won't work out in other words if i plug in a 1 here and let's say i use 10 here well 1 comma 10 won't work because again this cancels 1 is not less than negative 3. it's certainly not equal to negative 3. so that won't work so again when you're shading here just think about that any x value that's less than negative 3 or equal to negative 3 would work so you've got a solid line at negative 3 and you've got values to the left of negative 3 all shaded okay so we're just going to wrap up the lesson now by talking about compound inequalities again so we looked at these when we talked about linear inequalities in one variable you will see them when you talk about linear inequalities in two variables but again this stuff is not more complicated it's a little bit more tedious because we've thrown another variable in there but graphing this should not be mentally challenging for you you think about the two different words that you're going to come across you have and and then you also have or when you see a compound inequality with and remember you're looking for the intersection of the two solution sets so in order to say it's a solution for the compound inequality it has to be a solution for both so graphically i'm looking for the section of the coordinate plane that's the overlap right it satisfies both inequalities if i'm thinking about or right the key word or well in that case it could be a solution to either right so it could be a solution to one and not the other or it could be a solution for both it doesn't matter because it could work for either all right so let's just look at this one problem here when we go into our practice set when we go into our test problems we'll see a scenario with or here we're just going to see a scenario with and but again if we looked at or we'd know the difference is it could be a solution for either so we look at x plus y is less than 3 and negative 3x plus y is greater than or equal to negative 1. so again i'm just going to start out by solving each for y i'm going to graph each and then i'm going to look for the overlap so let's start by subtracting x away from each side over here so that's going to give us y is less than negative x plus 3 and then for this guy over here i'm going to add 3x to each side so that's going to give us y is greater than or equal to 3x minus 1. so let's go ahead and take these two down to the coordinate plane we'll just copy them and we're going to graph each one separately and then we're going to look for the overlap of the two graphs so for the first one y is less than negative x plus 3 again to draw the boundary line i would replace this with equals so if i had y equals negative x plus 3 how would i graph that well the y intercept is at 0 comma 3. so that's going to be right here and then the slope is negative one right so that's rise over run of what negative one over one or one over negative one whichever you wanna do so i can fall one go to the right one or i could rise one go to the left one whatever you want to do again if i have enough points to draw a straight line then i'm good to go in this case you have a negative one so i could say negative six over six if i wanted to so i could fall 6 1 2 3 4 5 6 go to the right 1 2 3 4 5 6. that's the point on that line because negative 6 over 6 is negative 1. i can also rise 6 and go to the left 6 if i wanted to but the reason i did that was to get a point down here so i could kind of put my ruler in a row and get a nice clean lot okay you only need two points to make a line but it's just so much easier if you have more points involved so you can make a nice clean line all right so now that i'm going to draw this remember this is a strict inequality so i want a broken or dashed line okay so i'll draw my line to start put some arrows there and what i want to do is just take my eraser now and i'm just going to break it up to let people know if they're looking at this this line is not part of the solution okay anytime you see a dashed or broken line you know it's not part of the solution all right now we have a less than so if we have a less than we're going to shade below the line so i want to shade below the line like that let's look at y is greater than or equal to 3x minus 1. in this case i'm just going to erase this and i'm just going to change this to a stroke eraser i can erase quickly in this case i have y is equal to 3x minus 1 for my boundary so i'd go to 0 comma negative one which is here and my slope is three so rise is three over run which is one so up one two three go to the right one up one two three go to the right one or i could fall one two three go to the left one right because three over one is equal to negative three over negative one again think about those tricks so i'm going to draw this line in green and this line right here is going to be a solid line because this is a non-strict inequality okay so this is a greater than or equal to so i want to shade above the line so i'm going to do that in an orange if i'm looking for the overlap the section on the coordinate plane that satisfies both the thing that i want to do the thing that makes it easier for me i'll take a black marker and i'll start lining out steps that does not work so in the first case y is less than negative x plus 3. we know we have this blue dashed line any point that was below that line worked as a solution so any point above the line doesn't work as a solution for this inequality and therefore doesn't work as a solution for the compound inequality because it has the work for both so i can just take a little black marker or whatever a pencil and just mark this section out and say hey this is not going to be part of the solution for the compound inequality then i'd look at the next part y is a greater than or equal to 3x minus 1. okay well that was this green line here so anything above the line worked anything below the line didn't so this is where it's below the line and that does not work now what's interesting here is that the green line in this case is actually part of the solution with the blue line it's not right it's the difference between a strict inequality and a non-strict inequality but when i look at the overlap i'm just going to highlight this with a normal highlighter now the green line here starting at this intersection point is included so i would highlight all the way down here and at some point i have to stop but this is an infinite area and all the way over here and all the way up here and then all the way here and then this blue line here is not included so i'm just going to highlight that so the green line that's highlighted here that's included the blue line not included and then i'm just going to shade everything here in this area this is the section of overlap or the part of the coordinate plane that satisfies both inequalities and so this is the solution for the compound inequality with and now if you had had a compound inequality with or again you would be highlighting the sections that satisfied either right it wouldn't matter if it satisfied both it would be what satisfied either one but for our problem here we had and so this is a good representation of course you can always make it better if you wanted to if you wanted to get a little bit better we could erase this arrow here and we could kind of extend this line a little more and let me give a highlighter again i'll just kind of highlight a little bit more here just get a better representation a better visual aid here but you're basically saying to your teacher hey anything that's above the green line right starting right here going to that intersection point and including that green line that is also below the blue line not including the blue line is the part of the coordinate plane that satisfies both and so that's your solution for the compound inequality in this case with and hello and welcome to algebra 2 lesson 21 in this video we're going to have an introduction to functions so if you took an algebra 1 course you at least got a basic introduction to the topic of functions in this algebra 2 course we're going to go a little bit further than we did in algebra 1. we're going to start today with a brief introduction and then over the next few lessons we're going to look at some more challenging material that relates to functions so if we're going to talk about functions we've got to start off by talking about relations those of you who know about functions know that all functions are relations but not all relations are functions and we'll get to that later starting out talking about relations it is useful in mathematics to be able to describe one quantity as it relates to another so we talk about this concept of a relation so a relation is just a set of ordered pairs and we talk about a set again that's a collection of some stuff so to get a little bit of an understanding here let's look at an easy example let's say that you're given this scenario so we have that a grocery store cashier is paid 12 dollars per hour worked so 12 dollars per hour worked the amount earned each week is based or dependent on how many hours they work how many hours they work so for anybody that works hourly you know that you take the amount you make per hour and multiply it by the number of hours you work and that gives you the amount of your gross paycheck when i say gross paycheck i mean the amount before any deductions you know taxes anything you have removed from your paycheck so to model this we can use two variables i would let one variable let's say x be equal to the number of hours worked so if i multiplied 12 dollars which is the amount earned in an hour times the number of hours worked i would get a gross paycheck so let's let something like y let's say then y would be equal to the amount of the gross paycheck so i can take 12 again the amount earned per one hour multiply by x which represents the number of hours worked and i should get y which is the amount of the gross paycheck so we have modeled this with an equation so i'm going to say that y is equal to 12 x once again this is the amount earned per hour this is the number of hours worked again you multiply those two together and you get the gross paycheck now to make our example even simpler let's say that the amount that she can work in a week could be zero or any integer value up to and including 40. so she could work zero hours a week one hour a week two three four you know so on and so forth up to the number 40. so for x it could be equal to we'll represent this starting with 0 1 2 3. i'll put my three dots and then i'll end this with 40. so any integer value including 0 and including 40 and anything in between so let's use this information to generate some ordered pairs so i have here y equals 12x i'm going to make a little table for ordered pairs so again x can be an integer beginning with 0 and up 2 and including 40. so the first ordered pair if i put a 0 here for x 0 times 12 is 0 and that makes sense if i don't go to work at all or i work 0 hours i would expect to make nothing now the next ordered pair if i work 1 hour i would get 1 x 12 or 12 for my paycheck and we continue this if i work 2 hours 2 times 12 is 24 and we would go all the way down eventually we get to 40 and that would give us what 40 times 12 is 480. so this relation would look something like this it's a set of ordered pairs so we would have 0 0 we would have 1 comma 12 we would have 2 comma 24 let's put our three dots and we'll put the final one which is 40 comma 480. so this is a relation remember i told you it's just a set of ordered pairs and these ordered pairs relate the number of hours worked to the gross paycheck in this scenario so we can use this example to gain a few insights about relations in general so there are two types of variables we're going to work with the first type is called an independent variable or an input force generally this is going to be x so x here is your input the second type of variable is known as a dependent variable or output so y is generally reserved as the output so this is your dependent variable or your output now why do we say independent variable or inputs why do we say dependent variable or output well the independent variable is something you would choose so let's say i'm a worker and this isn't this isn't 100 accurate but typically i would choose how many hours i would work right i would say i make a choice of how many hours i would work and then i get a paycheck based on that so my independent variable is how many hours i work my dependent variable is my paycheck my paycheck depends on how many hours i chose to work and i know in the real world it's well my employer made me work this or my employer only let me work this i get that but let's just say there's a perfect scenario where you can choose any amount of hours you want to work you can work between 0 and up to 40 hours in one hour increments all right so two other things that we want to talk about we want to talk about domain and we want to talk about range and these are very important concepts for us and they're going to get super complicated as life goes on in math right now they're very very easy to understand so a domain is just the set of allowable values for the independent variable or for x so the domain in this case i told you was the set of integers starting at 0 and going up to and including 40. so this would be your domain then what would the range be the range is the set of allowable values for your dependent variable or y so the range in this case and all we have to do is look at our equation you think about what's plugged in for x well i can plug in a 0 i can plug in a 1 a 2 a 3 up 2 and including 40. starting at 0 and then it's going to increment in multiples of 12 so the next number would be 12 then it would be 24 then it would be 36 right so on and so forth up to and including the number 480. you just have to list enough to where the pattern is clear right i'm increasing in increments of 12. i know after 36 would come 48 i know after 48 would come 60 then 72 then 84 right so on and so forth so this is my domain and this is my range so again you got to remember for domain it's a set of allowable x values or you might hear them say first components okay because in an ordered pair the first component is x the range is the set of y values or second components because in an ordered pair the second component is the y value let's talk about another real world relation that you'll come across suppose you drive at a constant rate of 75 miles per hour the distance you travel is dependent on time so we all know the distance formula right from working with motion work problems we know that distance is equal to rate of speed multiplied by time traveled well in this case the rate of speed is given to us it's 75 miles per hour so distance is equal to 75 times time and to use x and y to make things clearer we can say let x be equal to the amount of time traveled in hours and we could say then y would be equal to our distance if we think about this t here is the independent variable which i'm replacing with x just for simplicity d here is the dependent variable which i'm replacing with y for again simplicity so i'm just going to write this equation as y equals 75 x so x is my input i am choosing an amount of time that i'm going to travel for say i'm on a road trip and i'm like well how long do i want to go today you know i've got a certain number of miles got a certain number of days to make it over do i want to drive for 10 hours 8 hours 6 hours how do i want to stretch out my trip well i choose that value for x i multiply it by 75 and i get a value for y so the distance that is traveled is dependent on the amount of time that i choose to drive okay so again you've got your independent variable x your dependent variable y so let's rewrite this y equals 75x let's go ahead and make some ordered pairs for this as well i'm going to say that x is going to take on a value that is an integer starting with 0 and going up 2 and including 10. so we'll start with 0 and go up 2 and include 10. so that means x can take on these values it could be 0 1 2 you know so on and so forth till we get to 10 y is obtained by taking x and multiplying it by 75. so here it's zero and then we're going to increase by 75 each time so this is 75 this is 150 you know so on and so forth so we get to 10 times 75 which is 750. so if i think about my domain or my set of allowable x values [Music] it's given to me right here right i can choose to drive 0 i can choose to drive 1 2 you know so on and so forth up to and including 10 hours so that's what i can choose as an input i'm limited by my problem and what's told to me and so i can't go over 10 hours in the real world i could drive as much as i want i can drive till i fall asleep and crash into something but for simplicity i'm saying that you can choose not to drive at all which would be zero or you can drive in increments of one hour up to and including 10 right so you can go up to and including 10 hours so that's my domain so what's the range well the range is the set of allowable y values so think about plugging in each option for x well i'd start with plugging in a zero if i drive for zero hours i go zero miles not going anywhere just sitting at the house if i drive for one hour i go 75 miles and again i'm going to increase in increments of 75. so then i'm going to go to 150 and then i would have 225 you know so on and so forth until i get to the maximum amount i can drive which is 10 hours and in that particular case i'm going to go 750 miles all right so now that we've had two basic examples of a relation again this is how one quantity relates to another let's talk about functions because that's really the purpose of this video so a function is a special type of relation all functions are relations let me highlight this a function is a special type of relation each member of the domain must correspond to exactly one member of the range so you might say what in the world does that mean that's super confusing i want you to remember this this is very easy each x value must correspond to or must be associated with or must link up to exactly one y value so if i give you something for x it should be crystal clear what the output is for y in this example when we go back to the grocery store cashier if i choose something for x i know what i'm getting for y if i choose let's say 40 for x i know i'm getting 480 for y so there's a clear association every time i choose something for x i know what the output's going to be now that sounds kind of silly because we really haven't seen stuff yet where you choose something for x and you don't know what you're going to get for y but i'm going to give you some examples here in a minute so i want you just to look at this set of ordered pairs here when you first start looking at functions these are the examples you get they're super super easy to deal with all you're looking for here is duplicate x values okay i'm just going gonna boil it down to that if i look through my ordered pairs i have zero comma three two comma one three comma seven and two comma four do i have a duplicate x value the answer to that is yes i have a two here and a two here this is not a function let me explain why because some of you would probably what do you mean if i think about x values here and i think about y values 0 is my first x value it is linked to or corresponds to 3. so if i have an x value of 0 i know the y value is 3. there's a clear association so that's fine let's skip over this for a second let's go to 3 comma 7. let's go to that ordered pair so 3 is associated with or linked to 7. if i choose an x value of 3 i know i'm getting a y value of 7. here's where there's a problem i have this x value of 2 i get a y value of 1 but i have an x value of 2 that also produces a y value of 4. so it is unclear if i have an x value of 2 whether my y value is 1 or my y value is 4. there is no clear association there and as you get further in math that's going to be a problem for you if you have an input you expect one and only one output if i choose an x there should be exactly one y that is associated with it here i'm choosing an x value of 2 and there's two y's associated with it so this is not a function again it's still a relation but it's not a function all functions are relations but not all relations or functions here you have a relation that is not a function let's look at another example so we have these ordered pairs here we have three comma seven two comma negative six negative five comma three and eight comma four again look for duplicate x values i've got 3 2 negative 5 and 8. if you don't see any you have a function okay so this is a function and again i go back to my example where i make a table so here's x and here's y this is three and it relates to where it's associated with seven this is two it relates to or is associated with negative six this is negative 5 it relates to or is associated with 3 and then this is 8 it relates to or is associated with 4. don't worry about having duplicate values so in other words some of you get confused and say well there's a 3 here and a 3 here that doesn't matter don't don't look at that you're looking for duplicate x values so what would not be a function is let's say i cross this out and put a 3 here so then 3 would be associated with seven but it would also be associated with four you can't have that okay so that wouldn't be a function but that's not our example let me kind of erase all that put it back to what it was three is associated with only seven two is associated with only negative six negative five is associated with three and eight is associated with four so if i choose an x three two negative five or eight then it's crystal clear what i'm gonna get for y all right let's look at another example so let's say i had this relation here you've got the ordered pair 13 comma 7 5 comma 2 6 comma 2 and 7 comma 5. i look for duplicate x values 13 5 6 7. there are no duplicate x values so this is a function this type of example is very very simple to get past one thing that will confuse you especially when you first start it used to confuse me if you write down let's say 13 that's associated with seven five is associated with two six is also associated with two and then seven is associated with five you might say whoa hold on you've got a 2 here and a 2 here that's not a function there it's just not going to work it's okay a y value can be linked to two different x values that's okay it's crystal clear if i tell you x is 5 you know that y is 2. if i tell you x is 6 you know that y is 2. that's what i'm looking for with a function i give you an x you can spit out a y if for example i was to erase this and put this as a five well now we'd have an issue right because this five here would be linked to or associated with two and also five okay it's all about the x value if i give you an x value now of 5 i don't know whether the y value is 2 or 5. there's no clear association so again it's all about the x value let me kind of erase this put this back to our original so we had 7 that was associated with 5. all right let's take a look at one more example like this so we have our relation and it consists of the ordered pairs we have 12 comma 3 6 comma 8 6 comma 2 and negative 10 comma 16. so again what am i looking for with this type of example duplicate x values i have 6 and i have 6 so it fails this is not a function not a function again think about the x values don't worry about the y values if you again look at this with a table 12 is associated with or links to three six is associated with or links to eight six again is associated with or links to a different y value of two that's the problem and we have negative ten that is associated with 16. so it is not crystal clear when i have an x value of 6 what the y value is it could be 8 or it could be 2. there needs to be a clear association for each x value okay for each x value in this case the x value is six there can only be one y value that is associated with it so for six there's not one y value associated with it there's two of them there's eight and there's two so that's why this is not a function just keep saying to yourself for each x there can be one y okay for each x there can be one y that's how you're going to remember this so let me end the lesson by giving you kind of a visual representation of this and again this is a typical example you might see in a textbook a video tutorial or in your class so let's say that you have your domain and this is the set of x values and this just happens to be where you would take off from so take off cities so this is your domain and to keep it simple i'm just going to use three cities let's say there's new york let's say that there's atlanta and let's say that there is new orleans now in my land of y that's my range okay so my range is y this is the destination okay so that's over here this is my land of y so let's say that there's a destination of los angeles let's say there's portland and let's say there's tacoma so think about x values as i'm taking off again y values as destination cities or places that i will arrive so for a function there's a one-to-one correspondence i leave from new york i go to los angeles i leave from atlanta i go to portland i leave from new orleans i go to tacoma so this is an example of a function this is what we want right because we say hey i'm leaving from new york i need you to pick me up in los angeles there's a clear association i know where this flight's going to land i'm leaving from atlanta i know i'm going to end up in portland i'm leaving from new orleans i know i'm going to end up in tacoma so this is a function there's a clear association what's an example where it's not a function well let's say that the flight from new york is associated with los angeles and also portland let's say the one from atlanta is associated with tacoma let's say the one from new orleans is associated with los angeles well where there's a problem now is that this flight from new york is going to land in los angeles and it's also going to land in portland so imagine you get on this plane and you don't know where you're going to land so you're calling people you're like hey uh i might need you to pick me up in los angeles but it also might be landing in portland so can you kind of you know call me right before i land it's not a function because in this particular case there's no clear association i have a value for x which is new york that is associated with two different values for y los angeles and portland so it's not a function because there's no clear association so this would not be a function so not a function hello and welcome to algebra 2 lesson 22. in this video we're going to continue talking about functions so in our last lesson we reviewed the basic definition of a function and when i say reviewed if you took an algebra 1 course the last lesson that you went through here would be nothing new for you we learned about relations in algebra 1 and we also learned about functions but just to give you a quick recap a relation is nothing more than a set of ordered pairs and a function is just a special type of relation and i have here where each x value can be associated with or linked up to one unique y value so the common phrase you'll hear is for each x there is one y now let's look at a quick example and this is something we saw in the last lesson so i have a relation here nothing more than just a set of ordered pairs so we have 3 comma 6 2 8 3 9 and 4 comma 7. so the question i will put to you is this a function or is this not a function and it's really easy to see so let's make a little table with x y values so if i have an x value of 3 i have a y value of 6. if i have an x value of 2 i have a y value of 8. if i have an x value of 3 i have a y value of not well i already had an x value of 3 and that was mapped to a y value of 6. so let's just put 6 comma 9. we'll talk about that in a second then if i have an x value of 4 i have a y value of 7. i'll put the question to you again is this a function the answer is no absolutely not and why is that the case well for each x again there can be one y and you might pause for a minute and be like what does he talk what does that mean well if i ask you what is the value for y if x is 4 here well if x is 4 y is 7. very clear i can shout it out if i say hey what's the value of y if x is 2 well that's easy y is 8. now let me go to the one that's a problem let's say i say what is the value of y if x is 3 well i don't know could be 6 but it also could be 9. so there's no unique value for y given the x of 3 and that's the problem that's why it's not a function for each x value you can have only one y value that corresponds to it so for the x value of three i've got two y values that correspond to it so that's why it's not a function so that's what you need to be thinking about if you get a simple example like this it's really easy i just go through and look for duplicate x values i have an x value of 3 here and here i already know it's not a function that's really simple the other thing that kind of trips students up let's say you had another set of ordered pairs let's just erase this real quick let's just say i was looking at this set right here and it was six comma two and let's say we had three comma two and let's say we had 4 comma 7. something like that is this a function yes it is and you might say well no no no no there's a 2 here and a 2 here but let's look at it let's go through the same reasoning here's the x here's the y so an x value of 6 is corresponding to a y value of 2. an x value of 3 is corresponding to a y value of 2. an x value of 4 is corresponding to a y value of seven so for every x i have i have exactly one y it doesn't matter that different x's correspond to the same y that's okay it's just a matter of if i say hey what is y when x equals six there's a clear answer it's two what is y when x equals three there's a clear answer it's two what is y when x equals four there's a clear answer it's seven so that is the difference at the most basic level of what a function is and what a function is not now let's turn up the heat a little bit so we're done with the algebra 1 kind of easy stuff let's move on into the algebra 2 stuff so you're not going to get a set of ordered pairs kind of driven to you you're not going to get spoon-fed this stuff anymore now you're going to have to look at equations and say this is a function this is not a function so it gets a little harder so what are the tools that we can use for this well one tool i'm going to give you is to graph whatever you're looking at and you're going to use something known as the vertical line test so i'm going to start with y equals 2x plus 7 and i want to look at this on a coordinate plane all right so let's take a look at the graph of y equals y equals two x plus seven and we're going to determine if this is a function or not by doing something known as the vertical line test so what is the vertical line test well the vertical line test tells you if you have a function or not based on if any vertical line will impact the graph in more than one location you might say well why how would that work well think about vertical lines for a second by their nature if i give you a vertical line and let's just say i wanted to graph x equals eight just as an example so for x equals eight i know that x always is eight the y values will change but x always equals eight so for a given x value i have many many many y values an infinite number of them so if a vertical line such as this one were to touch a graph in more than one location let's just say for example i had something that looked like this well if it impacts it here and here well that means i have two different y values for that given x value so that is not a function that's where this vertical line test is going to come into play now let's erase this for a second and let's just think about what we have here we have y equals 2x plus 7. if i drew a bunch of vertical lines as many as i wanted to would i ever hit the graph more than once let's just draw three of them just just for fun so you can look at the lines i drew and in each case they only impact the graph once this one impacts it here this one here and this one here and i could draw as many vertical lines as i want in this case this is the graph of a function so y equals 2x plus 7 is a function in fact anytime you get a linear equation in two variables unless you have a vertical line something like x equals some value it's always going to be a function right and why is x equals some value not a function well as i just showed you when i graphed x equals 8 for that one x value you have an infinite number of y values so that is definitely not going to be a function okay but anything else that you come across even if it's a horizontal line you know if i erase this and i got out let's say a blank sheet and let's say i graphed y equals 6 as an example well i would go to the y axis and i draw a horizontal line through six so i'll put the question to you is this a function the answer is yes because for every x value say for x value of negative 2 i get a y value it doesn't matter that it's the same y value it doesn't matter what x value i choose i always get y equals 6. but for each x i just get one answer i don't say well x is negative 2 y is 6 and then something else it's only 6. so for every x value there is one y value so this meets the definition of a function so every linear equation in two variables that you come across is going to be a function with the exception of when you get x equals some value that is not a function so only a vertical line is going to fail right so i want to talk a little bit more about domain and range while we're looking at this picture so what is the domain here the domain is the set of allowable x values so we can get that from the graph or we can kind of look at the equation figure out if i look at the graph i see that the x values here well because of the arrows there that tells me this continues indefinitely in each direction so for the x values for the y values i know that they're infinite in each case right basically from negative infinity to positive infinity and if i go back and i look at my equation here i could figure that out and how would i go about doing that well what i would say is that what can i plug in for x what would be a possible restriction can i plug in a negative well try it if i plugged in a negative 2 there 2 times negative 2 is negative 4 negative 4 plus 7 is going to be 3. i could plug in as small of a number as i want meaning a really really really big negative number like negative 1 billion i could do that or negative 1 trillion i could go as large of a negative as i want which means a really really really really really small number so that is okay i could go down to negative infinity what about zero could x be zero yeah x could be zero two times zero is zero 0 plus 7 is 7. that's okay the world hasn't blown up everything's fine what about a positive value yeah i could i could make x positive 2. i could have x of 3. 2 times 3 is 6 x plus 7 is 13. so that's okay as well i could go out to positive infinity so x can basically be any real number right so i could write the domain as all real numbers like this i can also write this in interval notation from negative infinity to positive infinity like that what about the range what is y allowed to be remember we think about this in terms of input output x is your input your independent variable y is the output so i plug something in for x i get something for y that's how i have to think about these things well can y be whatever i want it to be the answer to that is yes i can make it as small as i want i can make it zero and i can make it as large as i want just by varying x so the range is going to be all real numbers as well so when interval notation again from negative infinity to positive infinity all right let's look at one that's a little bit more challenging so we have y squared equals x so i'll put the question to you before we look at the graph is this a function so again this is y squared is equal to x does this look like the graph of a function no it does not why because if i use my vertical line test i'm going to find that a vertical line will impact the graph in more than one location let's just take so i draw a line where x is equal to 1. so if x is 1 what could y be well it shows us here that y could be 1 and here that y could also be negative 1. does that make mathematical sense well yeah i've got the squaring operation going on so that means if i take 1 1 squared would give me 1. that works if i take negative 1 and i square that negative 1 squared will also give me 1. so that's why this is not a function for each x there's not just 1 y for an x value of one i've got two y values that are associated one and negative one i take another example of four i've got two y values associated you've got two and negative two i take an x value of nine as another example i've got 3 and negative 3. again it's because of that squaring operation so this is not a function because if i said hey what is the value of y given that x is 9 well you don't know whether answer 3 or whether the answer negative 3 could be either one so not a function now i want to erase all this and i want to talk for a minute about the domain and range what is the domain again it's the set of x values so what can x take on here well if i look at the graph here again this is the x axis do you see any part of this graph that is to the left of 0 no you do not why do you think that's the case it's because again i'm using the squaring operation here so if i square something the result is going to be 0 or it's going to be positive but it's never going to be negative so that means that x can only be zero or positive so let me go back here and we'll write our domain and we'll say that the domain is the set of all x values such that x is greater than or equal to zero and again this is set builder notation so the set of all x values so if i'm thinking about values that i could choose for x it's going to be such that the value of x is greater than or equal to zero it's all we're saying in interval notation you'd put a bracket next to zero then comma infinity okay make sure you have that bracket because zero is included now as far as the range goes when we think about our range what would that be well we go back to our graph think about the y values now this is the y values going up and down forget about the x values it's apparent that i have some that are negative i have one that's zero and i have some that are positive so the arrows here indicate that this would continue going up forever and going down forever so it looks like i can get any value for y that i want and yeah if i'm taking something and squaring it i could square a negative i could square 0 i could square a positive so there's no restriction on the range the range can be whatever so the range would be all real numbers so that's from negative infinity to positive infinity and another way to look at this if you didn't want to waste your time to graph it you said hey i don't want to do that i just want to look at the equation well you could solve this guy for y so i have y squared equals x and we talked about using the square root property in algebra one for those of you that haven't taken algebra one yet you're just kind of jumping in you don't know about the square root property basically if i take the square root of this side i've got to allow for the positive and the negative square root of this side so what would happen is this cancels with this and i have y is equal to plus or minus the square root of x so in other words if i'm making a table of values it would be completely obvious that for each x there's more than one y because whatever i plug in here once i take the square root i've got a positive version and a negative version the only exception to that is going to be 0 right if i have 0 the square root of that is 0 there's no plus or minus 0. it's just 0. so if x is 0 y is 0. that makes sense but for anything else if i plug in a 1 for x i know y would be positive and negative 1. square root of 1 is 1 you'd have plus or minus 1. so you'd have plus 1 and then minus 1. so right there fails the definition of a function if i use something like 4 square root of 4 is 2 so you'd have plus or minus 2. so if x is 4 y is plus 2 or minus 2. as one that's not so clean let me kind of scooch this up a little bit i can even use 5. let's say x is 5. plug in a 5 there i'm going to have the square root of 5. that's not a rational number so what i would just do is list it as a square root of 5. so i could say this is plus square root of 5 and minus the square root of 5. but you get the idea of why this isn't a function that's the main thing here it's just a simple fact that if i have at least one x value that is linked to or corresponds to more than one y value i do not have a function let's take a look at another one so here we have x squared plus y squared equals one those of you that have taken geometry or trigonometry you know this is the unit circle so is a circle going to be a function no think about it a circle looks like this and i know that's not a good drawing but if i draw vertical lines through that it's not going to make it right it's going to impact that graph in more than one location but let's go ahead and look at the graph so this is x squared plus y squared equals 1. again if i draw vertical lines here let's just draw two because it's such a small kind of graph you can clearly see this violates the definition of a function my vertical line hits here and here my vertical line hits here and here so this is not a function now when we start talking about the domain here you're going to really be pressed on this one i'm going to go back and look at the equation and explain why but you look at the domain look at the x values this thing can take on starting at negative 1 and including negative 1 up to positive 1. look at the range starting at negative 1 down here up to and including 1. so we seem pretty limited why is that the case let's look at the equation we're going to figure that out so for the domain and the range we said that it's the interval from negative one to positive one in each case and you can also use set builder notation you could say the set of all x such that x is greater than or equal to negative 1 and less than or equal to 1. you could write it like that or for this one you could say the set of all y such that same thing negative 1 is less than or equal to y which is less than or equal to 1. okay so two kind of different ways to write that a bunch of ways to write it write it however you please with this one i want to solve it for y real quick so we have x squared plus y squared equals one move x squared over here so let's subtract x squared away from each side of the equation and i'm just going to erase it so we'll have y squared is equal to negative x squared plus 1. so y squared is equal to negative x squared plus one okay how do i get y by itself take the square root of both sides again if i take the square root of this side i've got to go plus or minus the square root of this side okay so i'll go plus or minus the square root of this side so that means that y is equal to plus or minus the square root of negative x squared plus 1. i'm just going to flip the order there and i'm going to write 1 minus x squared so looking at what's inside that square root symbol we should see something something should jump out as it explains why the domain and range is so limited the first thing is can i take the square root of a negative number can i have square root of negative 5 no that is not a real number the answer to that is yes but only with the complex number system so if you've worked with imaginary numbers in the past you're probably screaming right now like yeah we can do it you use i blah blah okay we are only working with real numbers so with the real number system the answer to that is no there is no number that results from the square root of negative five it does not exist okay in the real number system so what that means is that the result of this has to be zero or positive it cannot be negative now look at what you're starting with you're starting with one imagine i start with a dollar in my pocket okay what can i spend without going into debt without using credit cards i can spend one dollar okay i can spend less than a dollar but the most i can spend is a dollar so the most that x can be is one because i'm going to take something and square it if it's anything between zero and one i'm okay anything larger than one let's say two for example two squared is four one minus four is negative three the square root of negative three if i had plus or minus the square root of negative 3 i can't do that using real numbers so that's nonsense can't use that okay so the most this can be is going to be one one works one minus one would be zero that's okay anything between zero and one including both works if i have zero there i'd have one minus zero that's one square root of one is one that's okay and then going down into the negative values i know that i can go as far as negative 1 anything past that's going to be a problem because again even though i'm squaring something i'm subtracting it away if i said 1 minus negative 2 squared negative 2 squared is 4 i've got the same problem i've got 1 minus 4 that's negative 3. so the most that x can be is 1 the least it can be is negative 1. so it's got to be from negative 1 up to and including 1 and what would that produce well the same thing in y if on the high side i produce 1 well the square root of 1 is 1 i'm going to have plus or minus this 1 so i have positive 1 and i have negative 1. so that's where you get this domain and range from the inputs the x can be as low as negative 1 as high as 1. the range the output can be as low as negative one or as high as one as well so that's why this is the unit circle this is not a function and your domain and range are both limited from negative one up to and including one all right for the next one we're going to look at y equals 1 over x so obviously before we jump into this thing we all know that we cannot divide by zero so we can't have an x value of zero so we know our domain will be limited to that so i want to look at the graph of this so we have y equals 1 over x as we expect when we have an x value of 0 i want you to notice something we get very very close as we go this way towards zero we get very very close to having an x value that's zero but we will never have an actual x value that zero not this way and not coming from this way as well we have something known as an asymptote here so if i drew a vertical line that was the same as the y-axis let's just say i draw a vertical line here and i know that's not precise but let's just say it is so at x equals zero we have something known as an asymptote so that line that exists at x equals zero my graph will approach it it'll get very very close to it but it will never ever ever touch it now let's explore why we have this phenomenon for a second well we can think about dividing by really really small numbers that are close to zero but not quite zero as i divide one by something close to zero what happens is it becomes larger and larger so if i take one and i divide it by point five this is going to give me 2. if i take 1 and i divide it by something closer to 0 let's say 0.25 this is going to give me 4. notice how this increased if i take 1 and i divided by .05 this would give me 20 right so on and so forth so the closer i get this denominator to zero going this way the bigger the number is going to go so you see how this kind of hockey sticks up so the closer i get an x value to zero the larger y is going to get and i can make it infinitely large as i get infinitely close to an x value of 0. so that's why this guy right here would approach an x value of 0 but it would never ever actually give me one okay so the same phenomenon happens when we think about the negative direction if i'm coming from here going this way coming from the left going to the right this graph will approach an x value of 0 but it will never ever ever give me 1. if i'm dividing 1 by negative 0.25 this is negative 4 right i'm just getting a bigger negative number 1 divided by negative 0.05 is negative 20 right so on and so forth so i can make this value infinitely large or we can think about it as a bigger and bigger and bigger negative by just making this number as close to zero coming from the negative direction so you'd have like negative point zero zero zero zero zero zero zero zero zero zero zero three for example something like that would produce a very very very big negative number all right so as long as we understand what's going on there let's think about our domain and our range graphically we can look at it and it looks like the domain we look at the x-axis here can be anything negative can be anything positive it just can't be zero what about the range well again it looks like the same thing it can be anything positive it can be anything negative but it can't be 0. so you can say you have another asymptote that occurs where y equals 0 right or the x-axis where y equals 0. the graph is going to approach a y value of zero but it's never going to actually give you one okay so it's going to get infinitely larger and larger and larger in terms of an x value here as y approaches zero but it's never going to give you one and it's the same thing here as our y value approaches zero the x value here becomes infinitely negative okay but the y value would never be zero as well so when we go back and write our domain and range we could say that our domain is the set of all x such that x does not equal zero we could say our range is the set of all y such that y does not equal zero now most people understand why x can't be zero you can't divide by zero that's obvious why do you think y can't be zero we'll do a little bit of algebra multiply both sides by x and you'll make it obvious y x y equals one think about that for a second how can two numbers being multiplied together produce one well your multiplicative inverse property means they're reciprocals right so if i had five times one-fifth that's one if i had two times one-half that's one if i had thirty times one over thirty that's 1. so these two values have to be reciprocals now does 0 have a reciprocal no 0 times any number is 0 okay so if x was 0 y no matter what it was when i multiply them together i'd get 0 equals 1. that's false same thing goes for y if i plugged in a 0 for y no matter what x is doesn't matter i'm going to get 0 equals 1 that's false so my domain can be anything other than 0 as well as my range that can also be anything other than 0. so if i think about is this a function or not if i draw vertical lines [Music] so those aren't perfect but you can see in each case that the vertical line only impacts the graph once so here it's obvious here it's obvious here it's obvious this one is where it starts getting tricky because it runs into something that's going very vertical okay so it looks like yeah it's hitting more than once but it's actually not the point of intersection would be something like that okay i don't know what it actually is but it would be something like that it would hit only once for every x here you're just going to get one and only one y so these are kind of tricky to use the vertical line test on but if you plug in something for x you'll get one y now you would say that this function is only defined when x does not equal 0. okay x cannot equal 0 y cannot equal 0. all right so let's take a look at one final one so i have here that y equals the square root of 3 minus x so let's take a look at the graph so we look at this one and it's very easy to see that it is a function now it starts here at 3 in terms of the x value and it goes to the left i don't see anything to the right and i'll explain why in a second but i want to draw some vertical lines here just for the sake of completeness it's pretty obvious that this one is a function won't get any arguments on this now this function is defined for values of x that are three or less so this this this this every vertical line i would draw would only impact the graph in one location now my observation for x is that again x is three or less what about my observation for y do you see any y values down here below zero no you see zero and then this increases and this is an arrow drawn so we know this would increase forever and ever and ever so y could basically be zero or anything larger so let's go back and let's list our domain in our range and let's think about why this is the case again so i have the square root of 3 minus x again this square root thing i don't want to take the square root of a negative number so if i'm starting with three dollars in my wallet what's the most that i can spend without going into debt three dollars so that's why x can be at most three so the domain would be the set of all x such that x is less than or equal to 3. now in this particular case could x be negative yes it can and the reason for that is because i don't have any squaring operation or anything crazy like that if i subtract away a negative i'm adding a positive so if i had the square root of 3 minus let's say negative 6 as an example well that would be the square root of 3 plus 6 which is the square root of 9 okay square root of 9 is 3. so i know that i can have any negative value that i want that would be okay and i can have 0 if i want 3 minus 0 is 3. no problem there i can have any positive value i want up 2 and including the number 3. if i have a 3 here 3 minus 3 is 0 the square root of 0 is 0. now one thing you would notice is that there's no plus or minus out in front this is just the principal square root of 3 minus x so we're okay there that's fine now if i saw something like y equals plus or minus the square root of 3 minus x that's a problem because when i get a result here i'm going to have the positive and the negative so for each x i'm not going to just have 1y so that would be an issue but in this particular case we do have a function now my range is what well we know that the least it can be a 0. we saw that with the graph and why is that the case well the most i can make this is 3. so 3 minus 3 is 0. square root of 0 is 0. so that's where y is 0. and then i can make this grow or have a higher y value by choosing a lower x value because i'm subtracting less away and when i get into negatives i could choose like negative one million and i could make y really really really big so really y can be anything that's zero or larger so the set of all y's such that y is greater than or equal to zero and of course you could write these in interval notation you could say the domain is from negative infinity up to including three and the range is from zero out to infinity all right so let's just kind of wrap this up with some key points here so when you're looking for the domain and the range remember you can't divide by zero you can't take the square root of a negative and get a real number and when we square a number the result is non-negative so if you write down those three things it's really going to help you when you start getting challenged on these questions look for again can't divide by zero can't take the square root of a negative and when you square something it's going to be non-negative hello and welcome to algebra 2 lesson 23. in this video we're going to learn about function notation all right so we're starting by looking at two brief examples here we have y equals x plus 3 and we have y equals x squared minus 5. so at this point we know that these two relations are functions i input something in for x i get a unique output for y so as an example let's say i plug in a 5 for x 5 plus 3 would give me 8 so my output for y is 8. as i change the value of x i'm going to change the value of y so y depends on my input for x so again x is the independent variable y is the dependent variable it's the same with this example here as well if i plug in let's say a 2 for x 2 squared is 4 4 minus 5 is negative 1. if i change the input for x i change the output for y so y depends on x if i say made this 3 3 squared would be 9 9 minus 5 would give me 4. so as i change my independent variable x my dependent variable y is going to change now when we start talking about functions generally we start out with the typical y equals something as we move on we're going to replace that notation of y equals something with f of x equals something and let me tell you this is a very big cause of confusion a lot of students are like where did y go bring y back why can't i have y anymore it's gone and it's not going to come back okay so as you get into higher math particularly college algebra they're not going to spend any time explaining that y is the same thing as f of x they're just going to present it and you're going to have to move on with it so i want to take the time here to make sure you understand this functions are generally named with f g or h so you'll see f g or h and it might be lower case it might be uppercase whatever they want to use and typically what you're going to see is let's say i'm naming the function as f instead of y all you're going to have is f of x so a function of x meaning the function depends on the input that is x so f of x now equals x plus 3. so there's no difference between this and this other than i replaced y with f of x and i could do it over here as well instead of y let me just write let me just write g of x okay g of x so g of x equals x squared minus 5. so i replaced y with g of x so there's really nothing to be confused about here it's something that's very very easy you just have to get used to it i'm just replacing y with f of x or i'm replacing y with g of x or i could have replaced y with let's say h of x okay i'm just saying that i have a function that's dependent on the value that you give me for x so as another example let's say i saw y equals x to the fourth power minus 3x plus 7. i could write this using function notation as f of x is equal to everything is the same just x to the fourth power minus 3x plus 7. super super simple i just replaced y with f of x all right so this notation can immediately be used to evaluate a function for a given x value so in the past we've done this okay it's the same thing we're not we're not learning anything new it's just a different way to notate it so let's say you saw y equals x plus 3 and i said let x be equal to 2. what would y be well you would just say okay y equals i would plug in a 2 for x and i would say y equals 2 plus 3 which is 5. so then we know that if x is 2 y is 5. we've been doing that forever now when i change this up again students go crazy and go what do i do the notation now will be f of x is equal to x plus 3. again i haven't done anything special all i've done is i've taken y and i've replaced it with f of x okay nothing special is going on now the notation to say let x be equal to two we would say f of two okay so you have your f inside the parentheses you have the value that you want x to take on so f of two is equal to replace x with a two then plus three so f of two is five this is just saying what is the function's value when x is two what is the value when x is 2. we can see the function has a value of 5 when x is 2. so let's say i tried another value let me erase all this let's say that i wanted to let x be equal to i don't know 7. so how would i ask for that well i would say i want f of 7. whatever value you put inside the parentheses you're just saying hey if x was seven what is the function's value it's the same thing as saying let x be equal to seven and find out what y is equal to same thing okay so the function's value when x is seven we replace x with a seven okay we just replace x with a 7 you'd have 7 plus 3 that equals 10. so f of 7 is 10. and again you can relate that to your earlier procedure if i just plugged in a 7 for x 7 plus 3 is 10 y equals 10. so y would equal 10 or f of 7 equals 10. the function's value when x is equal to 7 is 10. that's all we're saying all right let's look at a few examples here this is very very easy and you won't get into anything harder when you get to college algebra you're going to talk about the composition of functions we're not going to get into that here i'm going to wait until you get to college algebra to deal with that but i can tell you that it's not much more difficult but you need to know how to do the basics first before you get into the more complicated scenarios so we have f of x is equal to x squared minus 7x plus 2. so if i want f of 0 what am i asking for i'm asking for the function's value when x is zero so i would simply plug in a zero here and here and find out what i get so zero squared is zero minus seven times zero is zero plus two i basically just have 2. so f of 0 or the function's value when x is 0 is 2. then the next one is i want to find f of negative 2. so if i plugged in a negative 2 for x there and there i would have negative 2 and it's inside of parentheses because whatever x is it's squared x is negative 2 so we square the negative and the 2 then minus 7 times negative 2 plus 2. scroll down a little bit we'll have negative two squared is four then we have negative seven times negative two that's positive fourteen so plus fourteen then plus two so four plus fourteen is 18 18 plus 2 is 20. so we can say the function's value when x is equal to negative 2 is 20. all right so next we want f of 5. and i realize you can't see the function anymore so let me just rewrite it f of x is equal to we have x squared minus seven x plus two so f of five would be what plug in a five here and here you'd have five squared minus seven times five plus 5 squared is 25 so you'd have 25 minus 7 times 5 is 35 plus 2. so 25 minus 35 is negative 10 negative 10 plus 2 is negative 8. so f of 5 is negative 8. again the function's value when x equals 5. all right so now we want f of 9 and again i'm just going to write my function f of x is equal to x squared minus seven x plus two so f of nine is what i'm just gonna replace x with a nine here and here so i would have let me kind of write this down here nine squared minus seven times 9 plus 2. 9 squared is 81 minus 7 times 9 is 63 plus 2. so the easy way to do this is to add 81 and 2 first 81 plus 2 is 83 then 83 minus 63 would be 20. so f of 9 or the function's value when x equals 9 is 20. let's take a look at another example so we have f of x is equal to negative 2x cubed minus seven we wanna find f of zero so again all i'm doing is i'm replacing x with that zero whatever's inside of parentheses replace your independent variable with that so i would have negative two times zero cubed minus seven zero cubed is zero negative two times zero is zero so you basically have zero minus seven which is negative seven so f of zero is negative seven so let me rewrite this down here we'll scroll down so f of x is equal to negative 2x cubed minus 7. so let's go ahead and scroll down now and i want f of 2. so f of 2 is what plug in a 2 there so negative 2 times you have two that's cubed minus seven two cubed is eight so i'd have negative two times eight minus seven negative two times eight is negative sixteen so you'd have negative 16 minus seven negative 16 minus 7 is going to give me negative 23. so f of 2 or the function's value when x equals 2 is negative 23. all right so let me paste our function here so f of 10. so what am i going to do again all i want to do is plug a 10 n for x so i'd have negative 2 times we'd have 10 and that would be cubed then minus 7. so i'd end up with negative 2 times 10 cubed is a thousand right it's a one followed by three zeros and then minus seven so i know that i would do negative two times a thousand first that would be negative two thousand and minus seven this is going to end up being negative 2 007 so f of 10 or the function's value when x equals 10 is negative 2007. all right let me throw a little bit of a curve ball at you this is something that really trips students up so i've explained kind of the basics and when you use numbers it's easy to kind of start getting a hold of that can wrap your head around it everything is fine and everything is fine up until you see another variable that's inside the parentheses so let's say you have f of x is equal to 3x minus 1 and you see f of a i can't say how many students go bullets they can go what do i do it's the same thing this is what i want to plug in for this so just like if you had a 3 there if i had f of 3 i would plug a 3 in for x right and then i would just evaluate well if i have f of a plug in a in for x i'd have 3 times a minus one which is just three a minus one all we're saying is that the function's value if x was equal to a would be three a minus one that's easy to see if you have one x there just replace it with an a instead of three x minus one you've got three a minus one okay so it's very very easy to do that so again our function is f of x is equal to three x minus one so what if you see something like f of a plus 2. so you've got a binomial there what do i do again if i had f of 3 i'd plug a 3 in for x if i have f of a plus 2 i plug in an a plus 2 for x so a plus 2. use parentheses because it's going to stop you from making mistakes just use parentheses whenever you substitute things in okay it'll stop you from making silly errors so i would have 3 times the quantity a plus 2 because remember 3 is multiplied by x x now represents this quantity so unless i use my distributive property i'm not going to get the right answer right i don't want to see this because you'll get the wrong answer soon as i'll go okay it's 3a plus two no it's three times the quantity a plus two that's three a plus six okay so that's what you want so then i have minus one there all i did was i substituted in for the x there i plugged in an a plus 2. all i did all right so now use my distributive property 3 times a is 3a then 3 times 2 again is 6 then minus 1. so then 3a plus 6 minus 1 is just so the function's value when x is equal to a plus two is three a plus five again all i did was i replaced x with a plus two that's it no magic no trickery nothing like that all right so one more time we have f of x is equal to 3x minus 1. what if i saw f of z what do i do again take this and plug it in here all i'm doing so i would have 3 plug in a z for x minus 1. all i did was i replaced the x with the z because this is what it told me to do so the value of this function when x is z is 3z minus 1. that's it i just swapped the x with the z let's take a look at another one now so we have f of x is equal to negative 5 x squared minus 3 x minus 7 and i'm given f of z minus 1. so again i'm just going to take this and plug it in here and here and see what i get so i would have negative 5 times i'm going to put this inside of parentheses because i want to multiply by this squared so z minus 1 that quantity has to be squared and then multiplied by negative 5. then minus 3 times this quantity z minus 1. again use parentheses minus 7. so we know how to do this quickly we use our special products formula z minus one if we square that quantity we get z squared minus two z plus one so i'd have negative five times that quantity again z squared minus two z plus 1. do you think about this as negative 3 times z that's minus 3z and then negative 3 times negative 1 is plus 3 and then minus 7. all right so now i want to use my distributive property here negative 5 times z squared is negative 5z squared then negative 5 times negative 2z is plus 10z then negative 5 times 1 is minus 5. then you have minus 3z plus 3 minus 7. so now we're just combining like terms so i start out with negative 5z squared i can't do anything with that and i've got 10z minus 3z that's plus 7z then i've got negative 5 plus 3. that's going to give me negative 2 and then negative 2 minus 7 is minus 9. so the function's value when x is z minus 1 i get negative 5z squared plus 7z minus 9. all right so lastly let's talk about another source of confusion so a function is generally represented in what's known as explicit form so this is when it's solved for the dependent variable so y equals f of x which equals 3x minus 5. i just took y and i replaced it with f of x it's all i've done but y is on one side of the equation the expression 3x minus 5 is on the other solve for y so when we have this it's easy to just evaluate for a given value of x now in other scenarios the function is not solved for the dependent variable so we say that it's an implicit form okay implicit form so this means the function is not solved for the dependent variable remember that you're going to get to calculus you can hear about implicit differentiation and you're going to be like what does that mean what do i do it means the function is not solved for y and you need to find the derivative but you don't want to go through the process of solving it for y first so remember that because when you get the calculus you're going to thank me all right so we have an example here negative 5 equals negative 3x plus 1. so what would i do if i wanted to put this in explicit form well i would just solve it for y just simply add 3x to both sides of the equation and you're done it's an explicit form you would have 3x minus 5 equals y or i could simply say y is equal to 3x minus 5 using function notation i could say f of x is equal to 3x minus 5. again all i'm doing is replacing y with f of x but hopefully you can see the difference in one case in explicit form which we have here it's solved for the dependent variable y when it's in implicit form it's not so y is just kind of jambled up it's not on one side of the equation by itself that's implicit 4. so if you get a problem like this one you've got to do additional work so you might be asked to let's say find f of 3 f of 7 and f of negative 1 4 and we have an implicit function here we have negative x squared plus y equals 4. so what do i got to do i got to solve it for y first then replace y with f of x then do my normal procedure okay it's not much more work so i would add x squared to each side of this equation and i would find that y is equal to x squared plus 4. i can then replace y and say y is a function of x so i use this notation here and it equals x squared plus 4. so if i want to find f of 3 what do i do i just plug a 3 in here so f of 3 is what it's 3 squared which is 9 plus 4 which is going to give me 13. so f of 3 is 13. if i want f of 7 f of 7 plug a 7 in for x so i'd have 7 squared plus 4. 7 squared is 49 49 plus 4 is 53. so f of 7 would be 53 then lastly if i wanted f of negative 1 f of negative 1 well i'd plug a negative 1 in for x remember i'm squaring whatever x is so negative 1 has to be inside of parentheses because i'm squaring the negative and the 1. so negative 1 squared plus 4. negative 1 squared is 1 1 plus 4 is 5. so f of negative 1 is 5. all right let's take a look at one more of these and i made it a little bit harder to solve for y so we want to find f of 0 f of negative 2 and f of 5 4 and we have x to the fourth power minus x squared plus 2y equals 10. again solve for y replace y with f of x very very simple so let me subtract x to the fourth power from both sides of the equation let me add x squared to both sides of the equation we'll just start with that so this is going to cancel i would have 2y is equal to let's write negative x to the fourth power plus x squared plus 10. now to get y by itself what do i got to do i got to divide everything by 2. okay so we divide both sides by 2 this cancels with this and i have y is equal to you'll have a negative and you could write one-half times x to the fourth power plus you can write one half times x squared plus 10 over 2 is 5. so i can go ahead and replace this y with f of x i can say let me just erase all of this we don't need it anymore i can say y is f of x okay they're the same and this is equal to negative and i can write this in a more compact form let me just put negative x to the fourth power over two plus x squared over two plus five let me kind of move these up a little bit okay so now if i want to find f of 0 f of a 0. what do i do plug in a 0 for each x so i know that 0 to the 4th power is 0 the opposite of that would still be 0 divided by 2 i get 0. 0 squared is 0 divided by 2 i get 0. so these two would cancel themselves you're just left with 5. f of 0 is 5. so now i want to find f of negative 2. so f of negative 2 is what i would plug in a negative 2 for each x so i'd have negative and then in parentheses that have negative 2 to the fourth power make sure you use parentheses when you're substituting it'll stop you from making mistakes this is over 2 then plus i'm going to have negative 2 inside of parentheses squared over 2 then plus 5. so negative 2 to the 4th power is 16 but the opposite of 16 would be negative 16. so negative 16 over 2 would be negative 8. so this would be i'm going to make a little line there this would be negative 8 and then negative 2 squared is 4 4 over 2 is 2 so plus 2 plus 5. so negative 8 plus 2 would be negative 6 negative 6 plus 5 is negative 1. so f of negative 2 is negative 1. then lastly we want f of 5. let me kind of scroll down a little bit to get that one going so f of 5. so i would have negative in place of x i plug in a five and that's to the fourth power over two plus i have five that's going to be squared and that's over two then plus five alright let's scroll down get a little room going so five to the 4th power is 625 so i'd have negative 625 over 2 and then 5 squared is 25 so i'd have plus 25 over 2 then plus 5. well i have a common denominator here negative 625 plus 25 would be negative 600 so this would be negative 600 over 2 plus 5. we know negative 600 over 2 would be negative 300 so this would be equal to negative 300 and then plus 5 this would end up giving me negative 295. so f of 5 or once again the function's value when x is 5 is going to be negative 295. hello and welcome to algebra 2 lesson 24. in this video we're going to learn about variation so if you've taken an algebra 1 course you've heard about variation in the past here we're going to talk about the things we talked about in algebra one but we're going to learn about something known as joint variation as well so i want to start by talking about direct variation so this is the easiest type of variation to understand it's the most common type of problem that you're going to get when you talk about variation in an algebra 1 course so y varies directly with x or you might read y varies directly as x if there exists a real number k such that y equals k times x so this k here has a special name it's called the constant the constant of variation and it doesn't change okay the constant variation you might also read this in your textbook or your teacher might say the constant of proportionality it's the same thing okay so y equals k times x so let's get a little example going let's say i had y equals 2x what does this do if we think about this well if i choose a number for x i plug it in i multiply it by 2 and i get a number for y so if i make a little table of values here if i chose let's say 1 for x y would be 2. so x equals 1 y equals 2. if i chose let's say 2 for x y would be 4. 2 for x y is 4. if i chose let's say 3 for x we all know y would be 6. x is 3 y is 6. all right so what can we see from these three ordered pairs here well we see that as x increases by one unit y increases by two units x increases by one y increases by again two so what we can say specifically here if k again that's this right here in this case that's 2 is greater than 0 as x increases y increases as x decreases y decreases okay this is the case with direct variation now to get more specific as x increases by one unit y increases by k units we see that here x increases by one unit it goes from one to two y increases by two units it goes from two to four now we've already seen something like this think about the fact that we talk about slope intercept form all the time if i look at y equals mx plus b if b was 0 and m was 2 i'd have the same thing there y equals 2x so 2 is nothing more than the slope of this line so if i was to graph y equals 2x well i know the y intercept is where i could put plus 0 here and i make it obvious if i plugged in a 0 for x i get a 0 for y so the y intercept is at the origin and my slope is 2. so i would go up 2 to the right 1. up 2 to the right 1. so this would be the graph of y equals 2x and again if i start at a given point if i go to the right one meaning if i increase my x location from one to two what happens well to go back to the line i've got to go up two so i went from one to two my y location goes from two to four so to the right one up two back to the line to the right one increase my x by 1 i got to go up 2 to get back to the line so the right one up 2 to get back to the line so we can clearly see as we increase x by 1 we increase y by 2 or that k if i had y equals 7x as an example if i increase x by 1 i'm increasing y by 7. so as a real world kind of example suppose you go to a local gas station so i have at a local gas station the price per gallon of gas is four dollars and thirty cents so if i wanted to model how much a person would spend on gas there well i could say y is the total purchase and that's just on gas that's not if you stop in and get a candy bar and this would be equal to we'd have four dollars and thirty cents per gallon so let's put four point three zero this is representing k and this would be times x and x would represent the number of gallons okay the number of gallons of gas so if i buy one gallon of gas i just multiply by four dollars and 30 cents and i know i spent four dollars and thirty cents if i buy two gallons of gas i would spend eight dollars and sixty cents because two times four thirty is eight sixty if i did three i would spend twelve ninety you know so on and so forth so as x increases by one unit as x increases by one y increases by 4.30 or 0.30 so again this is what we're talking about when we talk about direct variation there's a constant value here that is multiplying x it will not change all right so there's another type of direct variation we're going to see so y varies directly with the nth power the nth power of x if there exists a real number k such that y equals k times x to the nth power so the only thing different here is we're raising x our independent variable to a power so this is known as direct variation as a power all right so the next one we're going to run across is inverse variation so for inverse variation now y is going to depend on some constant value k divided by x so i have y equals k over x so again this is the constant of variation and so that doesn't change now as an example of this we can think about our distance formula okay now the distance formula is something we use for the motion word problems that we solve distance equals rate of speed times the amount of time traveled so d equals r times t what i have here i have think about using distance as a constant time now varies inversely with the rate so let me just show you this real quick if i have y equals k over x i can make this look like this remember i'm saying that time varies inversely with rate so my time okay is going to be y so what i want to do is get time by itself so divide both sides by r and what i'll have is time is equal to distance over rate now time varies inversely with r just like y varies inversely with x remember we said that distance was going to be constant right so distance is in the numerator there just like k is here and you can think about an easy example of this let me kind of get a little room going here if i think about going on a trip let's say the trip is 500 miles so it's 500 miles so time equals we'll just put 500 up here in the denominator we have our rate of speed does it make sense that this varies inversely time with rate yes it does if i increase my rate my time is going to decrease if i decrease my rate my time is going to increase let's give you an example let's say i decide to drive 50 miles per hour so 500 divided by 50 would give me 10. so that means that my destination took 10 hours to get to if i increase my rate of speed let's say i double it let's say i go 100 miles an hour well now my time is going to decrease it goes from 10 to 5. so we can see with inverse variation as i increased the rate my time decreased okay now if i decrease the rate my time will increase so they're inverse bring one up the other falls with direct variation bring one up the other goes up bring one down the other goes down inverse one goes up one goes down now i should note that this is all reliant on this the constant being positive okay so you gotta note that but if i was to let's say go to a slow speed like let's say 25 miles an hour well now it's going to take me 20 hours to get to my destination so as i decrease the rate of speed the time went up all right so additionally we will also see inverse variation as a power so you might see y equals k over x to the nth power so with direct variation as a power and inverse variation as a power the only difference is your variable x is raised to some power okay the formula is the same other than that all right so the last thing i want to talk about is joint variation and we really didn't cover this in algebra one it's no more difficult it's just a little bit more tedious so this occurs with a product of two or more variables sometimes raised to powers so as an example y varies directly with x and z if there is a constant k such that y equals k times x times z all right so now that we understand the basics of variation i want to talk about how to solve the variation problems that we're going to get they're very very very easy this is one of the easiest sections in all of algebra so just follow this procedure to the letter you're going to be good to go so you want to start by finding the value for k the value for k doesn't change so they'll give you an opening scenario that allows you to find k then what you want to do is you want to rewrite the equation with the known value of k you're going to substitute what is given in the second scenario and then you're going to solve okay so write these steps down pause the video and we're going to use them through our problems and when you get to your textbook problems you're going to fly right through them they're very very easy all right so let's look at the first problem if y varies directly with x and y equals 12 when x equals 4 find y when x equals negative 5. go ahead and write out your equation y varies directly with x so y equals kx now what am i given to start i'm given that y equals 12 so y equals 12 when x equals 4. so if i plug those two things in i can find the value of k now one quick note i can tell you that algebraically we could find k before we even start by dividing both sides by x so this means that k is equal to y over x so i could just plug into that and kind of skip a step right i can say well to find k y would be 12 and x would be 4. so i know that k is 12 over 4 or 3. now traditionally we just plug it in here and say okay y is 12 x is 4 and i go through an extra step and say okay 12 equals 4k then i divide both sides by 4 this cancels and i get 3 equals k or k equals 3. whatever you want to do whatever you're more comfortable with but just know that k is always going to be equal to y divided by x so k is equal to three so now that we have that information we're pretty much good to go we just solve what they want us to solve we're just going to find y when x equals negative five so y is equal to k which we know is 3 times x which we know is negative 5. 3 times negative 5 is negative 15. so we found y when x equals negative 5. when x equals negative 5 y is negative 15. so we're done all right so you saw how easy that was if you didn't take algebra 1 and you were kind of worried about these problems don't be they're very very easy so if z varies directly with w squared and z equals negative 48 when w equals 4 find z when w equals 3. so this is one thing that really trips up students when you start changing letters around like what's z and w doing here should be y and x cross things out if you have to say okay if y varies directly with x squared what would that look like it would be y equals k times x squared or we could say that k is equal to if i divide both sides by x squared y over x squared now here's the kicker i can just erase this now and i can substitute a z for y and i can substitute a w for x so this is z and this is w if you do that a few times you're going to get it z is what varies directly with w squared so z is equal to k times w squared you replace it a few times and you're gonna get it okay it's it's kind of obvious but i i realize when you first see different letters it can really throw you for a loop so it tells us that z equals negative 48 so i could plug in a negative 48 here when w equals 4 so 4 is going to be squared so negative 48 over 4 squared 4 squared is 16 negative 48 over 16 is negative 3. so k will equal negative three k equals negative three so now they want us to find z when w equals three so all i want to do is say z is equal to k is negative three w is three so we'd have three squared so 3 squared is 9 you'd have negative 3 times 9 that's negative 27. so z is going to be negative 27 when w equals 3. all right let's take a look at another one if y varies inversely with x squared and y equals 5 when x equals 2 we want to find y when x equals negative 4. so we think about inverse variation it's y equals k over x now in this particular case y varies inversely with x squared so this would just be x squared here okay that's the only difference so now how would i get k in this case well i could multiply both sides by x squared and this would cancel with this so we can say that k is equal to x squared times y so knowing that let's plug in and get k we know that y equals 5 when x is 2 so y is 5 when x is 2. so k will be equal to 2 squared is 4 times 5 so that's 20. so k is equal to 20. okay let's erase all this let's go back to that original equation y equals k over x squared so find y when x equals negative 4. so all i've got to do is plug in a 20 here and a negative 4 inside of parentheses here because negative 4 is 2b squared so negative 4 times negative 4 is 16 and you think about canceling common factors here so 20 and 16 share a common factor of 4 20 divided by 4 is 5 60 divided by 4 is 4. so basically y would equal 5 4 or 1.25 when x is equal to negative 4. all right let's take a look at a few word problems here we'll just kind of wrap this up so for a given principle the amount of simple interest earned varies jointly with rate and time if nine thousand dollars was earned after six years at three percent annual simple interest how much would be earned after ten years at twelve percent annual simple interest so think about this step by step the first thing is most of you already know the simple interest formula it's i is equal to p times r times t so i is simple interest earned p is the principal or the amount that is invested r is the rate and this is as a decimal and then t is the time and generally this is in years now what are we given here it tells us that for a given principle okay that means the principle is set it's your constant of variation so this guy right here is like your k so if i had y equals k times x times z k would be the constant of variation that's p in this case it's the principal it does not change so then the amount of simple interest earned varies jointly so this is the simple interest earned and it varies jointly with rate and time so here x and z represent rate and time y represents simple interest earned so once we can relate it to our generic formula we pretty much understand what's going on so it tells me that if nine thousand dollars was earned okay so that's the simple interest earned so nine thousand dollars after six years so time is six at three percent annual simple interest so the rate is .03 so i multiply these two together and realize i have that constant amount the principal and my goal right now is to find that okay so let's scroll down get some room going 0.03 times 6 would be what well 3 times 6 is 18 you have two decimal places between the factors so this would be 0.18 times p and this equals 9 000. so let's divide each side by 0.18 and that's going to give me 50 000. so p is equal to 50 000. so that's the principle that is invested that's the constant of variation it's not going to change i'm going to write i is equal to 50 000 because that's my principle times rate times time let's go back up this is what we need to find how much would be earned after 10 years at 12 annual simple interest well this is what i'm looking for to find i and i'm given the inputs so the rate is 12 so as a decimal that's 0.12 and the time is 10 years so this is just 10. so 50 000 times 0.12 is 6000 and 6000 times 10 is 60 000. so i here would equal sixty thousand now we have a word problem so let's answer it with a nice little sentence and we can say sixty thousand dollars would be earned after 10 years all right let's take a look at one more so over a given distance speed varies inversely with time if a car travels a set distance in 10 hours going 110 miles per hour what speed is needed to go the same distance in 4 hours so over a given distance speed varies inversely with time so we know our distance formula is d equals r times t it tells me here that speed or rate varies inversely with time so solve for rate divide both sides by t and you'll see that this cancels with this and yeah you'll have that rate is equal to distance over time so the distance is what's going to be fixed here it's not going to change it's a set distance so that's my constant of variation now if a car travels a set distance in 10 hours so the time is 10 and the rate is a hundred ten miles per hour so if i multiply the two together i get the distance ten times 110 is eleven hundred so the distance here is eleven hundred miles that's a constant it's not going to change now if i take 1100 and i think about what else i need what speed is needed to go the same distance in 4 hours we'll just plug a 4 in here just plug a 4 in here and we'll get our needed rate so 1100 divided by 4 is 275 this is 275 so we need to go 275 miles per hour so let's erase this and we could say that a speed of 275 miles per hour is needed hello and welcome to algebra 2 lesson 25. in this video we're going to learn about solving systems of linear equations by graphing so if you previously took an algebra 1 course you're fairly familiar with how to solve a system of linear equations with graphing substitution and elimination but for those of you who haven't taken an algebra 1 course or you just never ran into this i can tell you it's a fairly easy topic to understand so we just start out with looking at one linear equation and two variables i have y equals 6x plus 3. so it's a linear equation in two variables and i can graph this on a coordinate plane so let's just go ahead and do that real fast so if i have y equals 6x plus 3 graphically what does that look like well i know at this point it's easy to graph something in slope-intercept form my y-intercept occurs at 0 comma 3 so that's going to be right here and my slope is 6. so i would go up six one two three four five six to the right one or i could go down six i could go down one two three or five six and i could go to the left one so let's go ahead and graph this guy okay so this is the graph for y equals 6x plus three so nothing new we just graphed a linear equation in two variables no big deal i want to go back to the previous page and i want to now take a look at this other linear equation of two variables and notice how the variables are the same we have y we have y we have x and we have x so what happens if i graph this equation on the same coordinate plane let's take a look so y equals negative x minus 4. so y equals negative x minus four so the y intercept here occurs at zero comma negative four so that's here and my slope is now negative one right it's negative one so i would go down one to the right one down one to the right one or i could go up one to the left one up one to the left one so let's go ahead and sketch this graph so this is the graph for y equals negative x minus 4. i want to bring your attention to something immediately you notice that this point right here is a point of intersection so that point occurs at negative one for the x value negative three for the y value so this is negative one comma negative three that point is significant because it lies on both lines so let's think about what that means for a second each and every point on this blue line here is a solution to y equals six x plus three each and every point on this green line is a solution to y equals negative x minus 4. so the point of intersection if it lies on both lines that means it's a solution to both of them and therefore it's a solution to our system of linear equations in two variables so we revisit our page and we bring our ordered pair negative one comma negative three so these two equations when we put them together represent a system of linear equations sometimes we'll say a linear system so if i have y equals 6x plus 3 and y equals negative x minus 4 my goal for a solution for a system of linear equations is to have a solution that works in both of the equations so with this ordered pair be a solution to each equation according to our graph it was let's verify so for this one i'm going to plug in a negative 3 for y and a negative 1 for x and i'll see if i get a true statement i'll have negative 3 is equal to 6 times negative 1 plus 3. so i'll have negative 3 is equal to 6 times negative 1 is negative 6 plus 3. so then negative 6 plus 3 is negative 3 so i end up with negative 3 is equal to negative 3. so yeah it works out here all right for the next i'm going to plug in a negative 1 for x and i'm going to plug in a negative 3 for y so negative 3 is equal to you have the opposite of negative 1 so the opposite of negative 1 be careful with that sign and minus 4 so negative 3 is equal to the opposite of negative 1 is 1 so you'd have 1 minus 4 1 minus 4 is negative 3. so you'd have negative 3 equals negative 3. so it works out here as well so we have found a solution for the system because this solution here works in both equations if it works in one but doesn't work in the other it's not a solution for the system so that's why when we're trying to find the solution graphically i go back to this we look for the point of intersection because that point lies on both lines and therefore it satisfies both equations and it's a solution for the system all right let's go ahead and take a look at another example so the first one was pretty easy because both equations were solved for y this one is going to be a little bit more tedious so for myself i like everything to be in slope intercept form just makes it really easy to graph so i'm going to solve all the equations for y so it's going to say y equals m the slope times x plus b because then i have a point on the line and i have the slope so it's very very easy to graph things all right so let's take this guy and let's solve it for y so 8y minus 2x equals 8. let's go ahead and add 2x to both sides of the equation that'll cancel we will have 8y is equal to 2x plus 8 divide both sides of the equation by 8 and what i'm going to have is this will cancel with this y equals 2 over 8 is 1 4. so 1 4 times x plus 8 over 8 is 1. so let's write this here we'll write y equals 1 4 x plus 1. we'll just draw a little arrow to that and let's erase everything all right for this one we'll start out with 3x minus 2y equals 8. let's subtract 3x away from each side of the equation so we know that this is going to cancel we'll have negative 2y is equal to negative 3x plus 8. let's go ahead and divide both sides of the equation by negative 2. so this will cancel i'll have y is equal to negative 3 over negative 2 is just 3 halves times x 8 over negative 2 is minus 4. so now we can write this one as y equals halves x minus four so we'll drag this over here let me erase everything now our two equations are in slope intercept form that makes them super easy to graph so let's go to the coordinate plane we're going to graph each equation and we're going to find the point of intersection so the point of intersection again is going to be our solution for the system all right so for the first equation we have y equals one-fourth x plus one so my y-intercept is at 0 comma 1 and my slope is 1 4. so i would go up 1 to the right 1 2 3 4 up 1 to the right 1 2 3 4. of course i could also say 1 4 is the same as negative one over negative four so i could start from here and i could go down one and to the left four one two three four down one to the left one two three four so we all know that two points make a line but more points will help you to get a more accurate drawing all right so let's go ahead and sketch the graph here all right so this is the graph i'm going to label this of y equals 1 4 x plus 1. now for the other one i have y equals 3 halves x minus 4. so the y intercept occurs at 0 comma negative 4. that's right here the slope for this one is three halves so go up three one two three to the right two one two up three one two three to the right two one two you can see your point of intersection there up three one two 3 to the right 1 2. all right let's go ahead and graph this guy so again this is the graph of y equals you have 3 halves x minus four so the main thing here is to look for the point of intersection that's right here so that's where we have an x value of four a y value of two so the ordered pair four comma two and why is that significant again well because it lies on both lines if a point lies on a line that means it's a solution to that particular equation so if this one lies on both lines at the point of intersection it's a solution to both equations and therefore it's a solution to the system so let's go back and again our ordered pair was 4 comma 2 and we can verify this so i'd plug in a 4 for each x and a 2 for each y and see if the left and the right sides are equal so for y i'm plugging in a 2 for x i'm plugging in a 4. 8 times 2 is 16 minus 2 times 4 is 8. 16 minus 8 is 8. so this first one checks out for the second one now i would have three times for x i'm plugging in a four minus two times for y i'm plugging in a two and this should equal eight three times four is twelve minus two times two is four this does in fact equal eight you get eight equals eight so this one checks out as well so we can say that the ordered pair four comma two is in fact the solution for this system all right so now that we looked at two basic problems i wanna kind of get into some special case scenarios so one scenario you're going to come across is where you have no solution and then another scenario you're going to come across is where you have an infinite number of solutions so when we think about the types of problems we're going to encounter most of the time or i would say the vast majority of the time the two lines will intersect at one and only one point so these are the two examples we just looked at the system is consistent and the equations are independent okay so the system is consistent let me highlight that system is consistent and the equations are independent now here's one of the special case scenarios so you'll get this thrown at you all the time too the graphs do not intersect the reason for that is these are parallel lines remember parallel lines do not intersect by definition so if they don't intersect that means there's not going to be a point that is on both lines and therefore there is no solution for the system so i have here these are parallel lines and there is no solution okay now you'll say that this system is inconsistent okay inconsistent remember the last scenario we had a system that was consistent here there's no solution so the system is inconsistent so the final scenario you're going to deal with is that the graphs are the same line so in this situation there are an infinite number of solutions now you might say what do you mean they're the same line that doesn't make any sense what's going to happen is you're going to start out with two equations that look different but really it's the same equation that they've algebraically manipulated to look different and the way you guard against this is you put everything in slope intercept form that's going to tell you right away if you have one of these special case scenarios so if you put everything in slope intercept form not only you're going to guard against the special case scenarios you're also going to be able to graph anything you're going to deal with very very quickly all right so these equations are said to be dependent okay they're dependent all right so let's look at a special case scenario so we have x plus 2y equals 6 and we have negative 1 half x minus y equals negative 1. so let's solve each one for y so for this one i would say x plus 2 y equals 6. let's subtract x away from each side of the equation this is going to cancel i'll have 2y is equal to negative x plus 6 we want to divide both sides by 2. so what's going to happen is this is going to cancel with this i'll have y is equal to we'll have negative 1 half x plus 6 over 2 is 3. let's erase this real quick and let's drag this up here so now let's solve this one for y as well so we have negative one half x minus y equals negative one so let me add one half x to both sides of the equation [Music] and i'll continue this up here this will cancel i'd have negative y is equal to you'd have one half x minus 1. now all i need to do is multiply both sides of the equation by negative 1 so i can get y by itself or you can divide by negative 1 whatever you want to do let's erase all this so negative 1 times negative y is y and this equals negative 1 times 1 half x is negative one half x and then negative 1 times negative 1 is positive 1. so i want to erase this and drag this up here and i want you to observe something we did a lesson where we talked about parallel lines and hopefully you remember how you can determine if you have parallel lines the two lines have the same slope but a different y-intercept so if i look here again if it's in the format of y equals m the slope times x plus b the y-intercept well the slope is given as the coefficient of x in each case and in each case here i have the same slope negative one-half and negative one-half but the y-intercepts are different here the y-intercept would occur at zero comma three here it would occur at zero comma one so same slope different y-intercept parallel lines so if you're on a test and you solve these for y and you see there's the same slope different y-intercepts you can stop and you can put no solution there's no solution parallel lines will never intersect so there's no ordered pair that's going to work in this one that also works in this one not going to happen you can also put the symbol for the null or empty set now for the sake of completeness let's go ahead and graph these so we can see what they look like for the first one we have y equals negative one half x plus three so the y intercept occurs at zero comma three so that's right here and the slope is negative one half so down one to the right two down 1 to the right 2 or remember negative over positive is the same as positive over negative and the n is just negative so i could put the negative in the denominator and i could say i could rise 1 go to the left 2. all right so let me label this guy this is y equals negative one half x plus three all right the next one was y equals negative one half x plus one so the y intercept for that one will occur at zero comma one and it's got the same slope so down one to the right two down one to the right two down one to the right two or up 1 to the left 2. so let me go ahead and label this one as well this is y equals negative 1 half x plus 1. you notice that you have parallel lines here you can see it these lines will never ever ever intersect the space between them will always be the same so we have parallel lines so for this system there is again no solution all right for the next one we have 3x plus 9y equals 9 and we have x plus 3y equals 3. so again let me solve each one for y so if i solve this for y i would have 3x plus 9y equals 9. let me subtract 3x away from each side of the equation and this is going to cancel we will have 9y is equal to we have negative three x plus nine let's go ahead and divide both sides of the equation by nine this is going to cancel with this we'll have y is equal to negative three over nine is negative one third then times x nine over nine is one so plus one let me erase all of this all right now let me solve this one for y as well so we have x plus three y equals three and so i will subtract x away from each side of the equation that will cancel i'll have 3y is equal to negative x plus 3. divide both sides of the equation by 3 and this will cancel with this i'll have y is equal to negative x over 3 we can basically say this is negative 1 3 x plus 3 over 3 is 1. so let's erase all of this let's drag this up here and what do we notice here we have the exact same equation so whatever ordered pair works here will work here and there's an infinite number of solutions for a linear equation in two variables so because there's an infinite number of solutions for this guy and this guy is the same as this guy there's an infinite number of solutions for the system because the system just contains the same equation twice okay so there's an infinite number of solutions here so we will say there's an infinite [Music] number of solutions hello and welcome to algebra 2 lesson 26 in this video we're going to learn about solving systems of linear equations by substitution so in our last lesson we reviewed how to solve a system of linear equations in two variables using the method known as graphing now if you took an algebra 1 course that was nothing new for you and of course if you took algebra 1 substitution is not new for you and it's also not going to be new when we cover elimination as well but it's important to review these concepts so when we start looking at tougher material it's fresh in your mind now the one thing i'm going to say about graphing you might say why don't we just use graphing we've already learned one method why not just use that forget about the rest graphing is very inefficient if you can imagine having a piece of paper a ruler and a pencil and trying to find a solution that's let's say the ordered pair 200 comma negative 400 okay it's very hard to do that it's not impossible but it's very difficult or let's say you had one half comma negative two-fifths something like that it's very hard to use graphing unless you have a small clean integer answer and those were the examples that i gave and i give them on purpose so that you can just get a conceptual understanding you don't want something where it's just so tedious you you know forget about it i'm not going to do this so return to substitution as an algebraic method this is something you can use no matter how big or small something is fractions decimals whatever you want you can use substitution now i am going to say substitution works best when you have a coefficient for one of the variables that's a one or a negative one okay now with that being said let's kind of dig into this so solving linear systems in two variables with substitution so the first thing that we want to do we want to solve one of the equations for one of the variables so it doesn't matter which equation you use doesn't matter which variable you start out with okay it is going to be helpful as i just mentioned if you have a variable that has a coefficient that is 1 or negative 1 you want to use that because it's easy to solve for that variable but otherwise you can use whatever you want now after you've done that you want to substitute for that variable in the other equation now once you've finished that step you're going to notice that that equation you're working with is a linear equation in one variable something we've been solving forever now so we want to solve the linear equation in one variable and then what that is going to do is give us a solution for one of the variables so once i have that i plug in for the known variable in either original equation doesn't matter which one i use then solve for the other unknown and i've figured out the ordered pair that is a solution for the system and again i say ordered pair because usually that's what you're going to get you know you have special case scenarios where you have no solution and an infinite number of solutions i'm talking about the case where you have one solution all right so the last thing we want to do and it's very important to do this check the result so you want to plug in for x and y in both original equations all right so let's take a look at the first example so we have our linear system here in two variables so we have x plus y equals 5 and we have negative 2x minus 5y equals negative 10. so i would pick one of the equations and solve it for one of the variables now if i look at this first equation and i'm going to label this as equation one so i can refer to it i'm going to label this as equation two so i can refer to it if i look at equation one i notice that the coefficient for x and the coefficient for y is a one so it would make it pretty easy to solve for either one so let's go ahead and just solve it for y so if i have x plus y equals 5 and i solved that for y we've been doing this for a while now i just subtract x away from each side of the equation and so this would become y is equal to i'd have negative x plus 5. now i want you to think for a second what the equals sign means equals is saying is the same as so y is the same as negative x plus five it's like if i have money and i say a dollar is the same as four quarters they are the same in value they don't look the same but they are the same if i give a cashier four quarters if i give a cashier dollar bill i've presented the same value to the cashier and so that is one dollar either way okay so now that we have solved this guy for y i am going to plug into the other equation okay so i'm going to plug in 4y here again y equals this so this is going to get plugged in there that's all i'm doing so i would have what in equation 2 negative 2x minus 5 i've got to use parentheses here because 5 is multiplied by y and y is this why is the quantity negative x plus 5 and this equals negative 10. so notice that i have substituted in for y in the second equation and the result here is a linear equation in one variable i now only have to deal with x and that's pretty easy to do so i have negative 2x i have negative 5 times negative x that's plus 5x and then negative 5 times 5 is minus 25 and this equals negative 10. combine like terms on the left negative 2x plus 5x is 3x then minus 25 equals negative 10. let's go ahead and add 25 to each side of the equation this will cancel we'll have 3x is equal to negative 10 plus 25 is 15. let's scroll down get a little room going we divide both sides by 3 and we get x is equal to 5. now for the ordered pair that is my solution to the system i already know that x is 5. let me erase everything i've done to this point so we know x is 5 so as an ordered pair i'd have 5 comma something now how am i going to get this other something well 5 works as a solution for x in both equations if it's a solution for the system so i can plug a 5n for x here in equation 1 or here in equation 2 and solve for the unknown y so let's go ahead and do it in equation 1 because it's a much simpler equation so we would have 5 plus y equals 5. well we know y equals 0 right but we can go through the steps subtract 5 away from each side of the equation this will cancel you'll get y is equal to 5 minus 5 is 0. so 5 comma 0. now again i want to check this to make sure that my answer is correct so let's erase everything we have x plus y equals 5. so x is 5 plus y is 0 equals 5. we know that's true 5 equals 5. we just did that so the other one you would have negative 2 so negative 2 times plug in a 5 for x minus 5 times plug in a 0 for y equals negative 10. negative 2 times 5 is negative ten of course negative five times zero is zero so i get negative ten equals negative ten so that's true as well so we verify that the ordered pair five comma zero is the solution for this system what if i would have let me just let me just throw this out here what if i would have chosen equation 2 to start out with and let's say instead of solving for y i solved for x would i get the same answer the answer to that is yes let's go ahead and do that so i have negative 2x minus 5y equals negative 10. so i'm going to solve for x so i'm going to add 5y to both sides of the equation so this will cancel i'll have negative 2x is equal to you'll have 5y minus 10 and then let's go ahead and divide both sides of the equation by negative 2. so this is going to cancel i'll have x is equal to five over negative two might as well just put negative five halves times y negative ten over negative two is plus five so this came from equation two so i want to plug into equation one and what do i wanna plug in for i know that x equals x equals this value here negative 5 halves y plus 5. so x is the same as that so i'm going to plug in for x in this equation now and i'm going to plug this in so when i do that what happens i have negative 5 halves y plus 5 then plus y equals 5. so now i have a linear equation in one variable i just have a variable y and again that's very easy to solve for so if i subtract 5 away from each side of the equation this will cancel and this will cancel you'll have negative 5 halves y plus y equals 0. now before i do anything else you know that the answer for y was 0 and you can see the answer for y would be 0 here no matter what when i combine like terms i'm going to have something times y equals 0 that can only be true if y is 0. okay so you don't need to go any further you've proved it to yourself but for the sake of completeness i'm going to go all the way here so let me bring this up and let's go ahead and just multiply both sides of the equation by two that'll be the easiest way to do this so two times negative five halves is negative five and times y two times y is plus two y and this equals 0 times 2 which is 0. so negative 5y plus 2y is negative 3y and this equals 0. and of course to get y by itself i divide both sides by negative 3. this cancels you get y is equal to 0. all right so let's erase everything and of course we know that y equals 0. so what is x again plug in for y in either original equation i can plug into this one or this one doesn't matter let's go ahead and use the second one so negative 2x minus 5 times 0 equals negative 10. so this will go away i'll just have negative 2x is equal to negative 10 divide both sides by negative 2 so this will cancel negative 10 over negative 2 is 5 so you'll have x is equal to 5. so i get 5 comma 0 as the ordered pair as we already found so any way you do it you can solve for any variable and any equation to start and you're going to plug in for that variable and the other equation all right for the next one we have negative 3x plus y equals 9 and we have negative 7x minus 8y equals negative 10. so the first thing i want you to notice here is that you have a coefficient of 1 on your variable y and what i would like to say is the first equation so i'm going to label this as equation 1. this is equation 2. just for reference sake i say hey equation one i'm talking about this one equation two i'm talking about this one so the easiest thing to do here since i have a coefficient of one there is to solve equation one for y so let's go ahead and do that let's say that we have negative 3x plus y equals 9. all i've got to do is add 3x to both sides of the equation and this will cancel i'll have y is equal to 3x plus 9. so that's super simple all right now that i know what y is equal to or what y is the same as i can plug in for y in the other equation so i'm going to plug this again what y is equal to or what y is the same as in for this so i would have negative 7x minus 8 times again it's this whole quantity that y is equal to so 8 is multiplied by that quantity so i'm going to use parentheses 8 times the quantity 3x plus 9 and this equals negative 10. so let's go ahead and solve this equation for x you'll have negative 7x and then you'll have negative 8 times 3x that's negative 24x negative 8 times 9 is minus 72 and this equals negative 10. so negative 7x minus 24x is negative 31x so you'll have negative 31x minus 72 equals negative 10. so let's go ahead and add 72 to each side of the equation that'll cancel you'll have negative 31x is equal to negative 10 plus 72 is 62. so as a last step to get x by itself let's divide both sides by negative 31. this will cancel and we'll see that x is equal to negative two so let's go ahead and erase everything so we already know that x is equal to negative two whereas an ordered pair would be negative two comma and then something so i can plug a negative 2 in for x in either original equation it doesn't matter because in terms of the system i know x is equal to negative 2. so this one's easier to do in equation 1. let's just use that so you would have negative 3 times negative 2 plus y is equal to 9. so negative 3 times negative 2 is 6. so you'd have 6 plus y equals 9. subtract 6 away from each side of the equation and this will cancel we'll have y is equal to 9 minus 6 is 3. so my ordered pair is negative 2 comma 3. let's erase all this and so now we just want to check we want to check an x value of negative 2 and a y value of 3 in each original equation so i'm going to start with equation 1 i'd have negative 3 times x x is negative 2 plus for y i have 3 and this should equal 9. we know this works because we just did it negative three times negative two is six six plus three is nine so you get nine equals nine so this works out there all right for equation two we have negative seven times x again x is negative two then minus 8 times y which is 3 and this should equal negative 10. negative 7 times negative 2 is 14 negative 8 times 3 is negative 24 and again this should equal negative 10 and it does right 14 minus 24 is negative 10 so you get negative 10 equals negative 10 so you can check this one off as well so the ordered pair negative 2 comma 3 or an x value of negative 2 a y value of 3 is the verified solution for this system all right let's take a look at another one so we have 6x plus 3y equals negative 12. we have negative x plus 3y equals 23. so again if i find a variable in my system that has a coefficient of 1 or negative 1 i want to use that as what i solve for so here again with negative 1 being there i'm going to start out by looking at that so again i'm going to label this as equation 1 this is equation two so with equation two i'm going to solve it for x so i have negative x plus three y equals 23 and a couple things i can do here i can swap sides with this one and this one or i can just move 3y over and multiply both sides by negative 1 whatever you want to do doesn't matter so let me just subtract 3y away from each side of the equation that's how i'd like to do it i'll have negative x is equal to negative 3y plus 23 and let me just multiply both sides by negative 1. so negative 1 times negative x is of course x and this equals negative 1 times negative 3y is 3y negative 1 times 23 is minus 23. so i've solved this guy for x so let me erase this and drag this up and if i've solved this for x i can plug in for x in equation 1. so i would have 6 times the quantity 3y minus 23. again x is equal to or x is the same as 3y minus 23. so i can plug in for x there so once i've done that then plus 3y equals negative 12. i have a linear equation in one variable now and it's easy to solve for that variable so 6 times 3y is 18y and then 6 times negative 23 is negative 138 and then plus 3y equals negative 12. so 18y plus 3y is 21y so i'll have 21y minus 138 is equal to negative 12. let's scroll down just a little bit so now if i add 138 to each side of the equation this will cancel and i'll have 21y is equal to negative 12 plus 138 is going to be 126. so as a final step to get y by itself let's divide both sides of the equation by 21 and we'll have that y is equal to 6 right 126 divided by 21 is 6. so y equals 6. all right so let's erase everything so we know that y equals 6 so the ordered pair would be something comma 6. so now i can plug in 4y in either original equation let's go ahead and use equation 2 because it looks a little easier so i would have negative x plus 3 times plug a 6 in for y equals 23. so negative x plus 3 times 6 is 18 equals i can subtract 18 away from each side of the equation this will cancel i'll have negative x is equal to 23 minus 18 is 5. so essentially all i need to do is divide both sides by negative 1 or multiply both sides by negative 1 whatever you want to do you'll get x is equal to 5 times negative 1 is negative 5. so my x value is negative 5 while my y value is 6. so the ordered pair negative 5 comma 6. so now what i want to do i want to go into equation 1 and equation 2 plug in my ordered pair and see if i've got the correct solution for the system so i've got 6 times for x i'm plugging in a negative 5 plus 3 times for y i'm plugging in a 6 and this should give me negative 12. 6 times negative 5 is negative 30 plus 3 times 6 is 18. this should give me negative 12 and it does right you get negative 30 plus 18 is negative 12 and that equals negative 12. so this one checks out now for the next one we have negative and then i'm plugging in a negative 5. so it's basically the opposite of plug in a negative 5 plus 3 times y y is 6 this should equal 23. the opposite of negative 5 is 5 plus 3 times 6 is 18 and this should equal 23 and of course it does 5 plus 18 is 23 so 23 equals 23 so we verified this one as well so our ordered pair here negative 5 comma 6 is the verified solution for this system all right so let's take a look at another one so we have 6x minus 21y equals 2 we have negative 2x plus 7y equals 1. so again i'm going to label my equations this is equation 1 this is equation 2. none of the variables have a coefficient of one none of the variables have a coefficient of negative one so pick whatever you want whatever you think is easiest for you i'm going to go ahead and solve equation one for x because it's the first thing that i see so if i have 6x minus 21y equals 2. if i'm solving for x add 21y to each side of the equation so this is going to cancel i'll have 6x is equal to 21y plus 2. so now i'm going to divide both sides of the equation by 6. this will cancel i'll have x is equal to for 21 and 6 i have a common factor of 3. if i divide 21 by 3 i get 7 if i divide 6 by 3 i get 2. so this is 7 halves times y plus 2 over 6 is 1 3. so x is equal to 7 halves y plus 1 3. okay so let's erase this and so we got this from equation one okay we got this from equation one so i'm going to plug in for x in equation two and i'm gonna plug this in right x equals this so that's going in there so what are we going to get we're going to have negative 2 times the quantity 7 halves y plus 1 3 and then plus 7y equals 1. so negative 2 times 7 halves you can think about negative 2 times 7 halves like that this would cancel with this and you'd have a negative 7. so this would end up being negative 7y and then negative two times one-third would be negative two-thirds and then plus seven y equals one now what happens here negative seven y plus seven y is what that's zero this cancels so what am i left with negative two-thirds equals one well that's not true that's clearly false so what happened here well this happens to be a system that has no solution these are parallel lines so if your variable drops out when you're doing this and you end up with a statement that doesn't make sense like negative two thirds equals one you have no solution you have no solution if the variable were to drop out and you have a true statement then you have an infinite number of solutions now the easiest way to show this is again to solve each equation for y if they're each in slope intercept form you can see that they have the same slope but different y-intercepts and so they're parallel lines so let's put this over here out of the way if i solve this for y 6x minus 21y equals 2. if i subtract 6x away from both sides of the equation this will cancel i'll have negative 21y is equal to negative 6x plus 2. i then want to divide both sides of the equation by negative 21. and so this is going to cancel i'll have y is equal to so for negative 6 over negative 21 negative over negative is positive 6 over 21 they're each divisible by 3 6 divided by 3 is 2 21 divided by 3 is 7 then times x and then 2 over negative 21 we just write minus 2 over 21. so that's this guy right here y equals 2 7 x minus 2 over 21. okay let's erase this and let's now solve this guy for y so negative 2x plus 7y equals 1. we'll add 2x to each side of the equation so this is going to cancel we'll have 7y is equal to 2x plus 1. divide both sides of the equation by 7. so this cancels you'll have and i'll write this up here y is equal to two sevenths x plus one seven so let me erase this and i want you to look at this real quick we know that we have parallel lines when what occurs we have the same slope but different y-intercepts so in this case the slope is 2 7 for each equation and the y-intercepts are different in equation one the y-intercept occurs at zero comma negative two over twenty-one in equation two the y-intercept occurs at zero comma one-seventh so same slope different y-intercepts parallel lines never going to intersect so there's no solution okay these are parallel lines [Music] all right let's take a look at another one so we have negative 2x minus 2y equals 6 and we have 4x plus 6y equals negative 2. so it looks pretty easy if i solve this guy right here for either x or y that looks easiest to me so let's label this as equation 1 and this is equation two and i'll go ahead and solve equation one for let's say x so negative two x minus two y equals six let's go ahead and add two y to both sides of the equation so that'll cancel we'll have negative 2x is equal to 2y plus 6 and as a final step let's divide both sides of the equation by negative 2. so that'll cancel i'll have x is equal to 2 over negative 2 is negative 1 so negative y 6 over negative 2 is negative 3. so minus 3. so x is equal to negative y minus 3. so now what am i going to do i'm going to plug in 4x in equation 2 now i found out that x is equal to or x is the same as negative y minus 3 that quantity so i can plug in for x in the second equation so four times again this quantity here negative y minus three then plus six y is equal to negative two all right so four times negative y is negative four y four times negative 3 is minus 12 then plus 6y equals negative 2. all right so negative 4y plus 6y is 2y then minus 12 equals negative 2. and again if i add 12 to both sides of the equation that will cancel we'll have 2y is equal to negative 2 plus 12 is 10. as a final step i'm going to divide both sides by 2 and we're going to get that and kind of scroll down a little bit run out of room so this will cancel y is equal to 5. let's erase everything so we say that y equals 5. so as an ordered pair something comma 5. all right so i'm going to plug in for y in either original equation equation 1 or 2 doesn't matter let's go ahead and pick equation 2. so i'm going to plug in a 5 there so i would have 4x plus 6 times 5 is 30 so plus 30 equals negative 2. so let's subtract 30 away from each side of the equation this would cancel 4x is equal to negative 2 minus 30 is negative 32. as a final step let's divide both sides by 4 and this is going to cancel we're going to end up with x is equal to negative 32 over 4 is negative 8. so my x value is negative 8 my y value is 5 or again the ordered pair negative 8 comma 5. so let's check this i'm going to go ahead and plug into each original equation so in equation 1 i have negative 2 times negative 8 that's my value for x minus 2 times 5 that's my value for y and this should equal 6. negative 2 times negative 8 is 16. negative 2 times 5 is negative 10 and then we have equals 6. so 16 minus 10 is 6. so you get 6 equals 6 so it works out here and the second equation remember that's what we used to find x we know it already works so let's substitute it in anyway so 4 times we have negative 8 for x plus 6 times we have 5 for y this should equal negative 2. 4 times negative 8 is negative 32 plus 6 times 5 is 30 this should equal negative 2 and of course it does negative 32 plus 30 is negative 2 so negative 2 equals negative 2. so it works out here as well so we verify that the ordered pair negative 8 comma 5 is our solution for this system all right let's take a look at one more problem so we have negative 4x plus 8y equals 16 and we have x minus 2y equals negative 4. so i notice that this equation here and i'm going to label this equation 1 and equation 2. this equation 2 has a coefficient on x that is 1. so when that occurs it's like hitting the lottery right you want to solve for that one i've got to use that because it's easy so i'm going to say okay i have x minus 2y equals negative 4 i'm going to add 2y to both sides of the equation this will cancel we'll have x is equal to 2y minus 4. so now if i know what x equals or what x is the same as i can plug in for x in equation 1. so i'm going to plug this quantity 2y minus 4 in for x there so i would have what i would have negative 4 times the quantity 2y minus 4 plus 8y equals 16. so let me kind of do this off to the side here negative 4 times 2y is negative 8y then negative 4 times negative 4 is plus 16 then plus 8y equals 16. negative 8y plus 8y is 0. this cancels uh-oh my variable dropped out again so i have what i have 16 equals 16 or if i subtract the 16 away from each side of the equation i'd have 0 equals 0. so what i told you when we had that example with no solution if the variable drops out and you're left with a true statement like you are here you have an infinite number of solutions you basically have the same equation for equation one and two it's just algebraically manipulated to look different so let's just write that we have an infinite number of solutions how could we prove that these two equations are the same there's a number of different things we could do the easiest thing you could do is look at it and say well if i multiplied this right here equation 2 by negative 4 on both sides i would get equation 1. so negative 4 times the left side is x minus 2y the right side is negative 4 so multiply that by negative 4 as well so negative 4 times x is negative 4x negative 4 times negative 2y is plus 8y this equals negative 4 times negative 4 is 16. so now i took equation 2 i multiplied both sides by the same non-zero value and i came up with equation 1. negative 4x plus 8y equals 16. negative 4x plus 8y equals 16. so it's the same equation now the other thing we could do is we could solve each equation for y and again you'd see that you have the exact same equation so if i subtract x away from each side of the equation here i'd have negative 2y is equal to negative x minus 4 divide both sides by negative 2. so that would give me y is equal to negative over negative is positive so one half times x negative four over negative two is plus two now for this one up top if i have negative four x plus eight y equals sixteen i would add 4x to both sides of the equation that would cancel i'd have 8y is equal to 4x plus 16. divide both sides by 8 and i would get what 8 over 8 is one so you'd have y is equal to four over eight is one half times x plus 16 over eight is two so when we look at the two equations again they are the same we have y equals one half x plus two y equals one half x plus two same slope same y intercept same equation so of course whatever solution works in equation one also works in equation two because they are the same equation so there's an infinite number of solutions for this system hello and welcome to algebra 2 lesson 27 in this video we're going to learn about solving systems of linear equations by elimination so in the previous two lessons we had a review of how to solve a system of linear equations in two variables first by using the graphing method and then second by using the substitution method so if you've taken an algebra 1 course in the past again none of this is new for you you would have mastered all of these concepts before you got to this algebra 2 course but it's good to review it because in the next lesson we're going to learn how to solve a system of linear equations in three variables so it's good to have these refreshers before you start looking at some tougher material now we already figured out that graphing was essentially just something we learned for conceptual purposes right we're not going to really use that substitution is a good method and it's something we want to use when one of our variables has a coefficient of 1 or negative 1. for elimination we want to use this when we have one pair of variable terms as opposites so in other words if i see something like negative 5x and 5x or negative 10y and 10y if i added those two together they would sum to zero right negative 5x plus 5x is 0. or negative 10y plus 10y is 0. so if i do that that variable will become eliminated okay and that's going to be the whole goal here we're going to eliminate one of the variables from the system so we have a linear equation in one variable that we can solve then we can plug back in to one of the original equations and we can find the other one all right so solving linear systems in two variables with elimination the very first thing you want to do is place both equations in standard form now i know this definition of standard form you know gets argued about and you know oh it's this and oh it's that for the purposes of this video and the purposes of making this very simple let's just agree that standard form would be a x plus b y equals c where a b and c are just real numbers and a and b are not both 0. now once we've done that we want to transform one or possibly both of your equations in a way that one pair of variable terms are opposites okay opposites remember the examples i gave 5x and negative 5x or 10y and negative 10y so you want variable terms that are opposites now one of the easier problems this is given to you you start out that way on the harder problems you've got to transform one of the equations or you've got to transform both of them okay and we'll see some examples of that now once this is done this is where your addition property of equality is going to come into play you're going to add the left sides and set this equal to the sum of the right sides and this will make more sense once we get into the example all right so once you've done your addition step you will have eliminated one of the variables so you're left with a linear equation in one variable so you're going to solve the resulting equation okay once that's done you'll have x equals some value or you might have y equals some value depending on what went on and then you're just going to find the other unknown using substitution so whatever i find i'm going to plug that back in for that particular variable and one of the original equations i'm going to solve for the other unknown now the last thing and the most important is to check the result remember for you to have a solution for a system it's got to work in both okay can't just work in one so plug in for x and y in both let me highlight that in both original equations all right so let's go ahead and get started with the first problem so i have 3x minus y equals 5 and 4y equals 3x plus seven so one thing i like to do i like to number my equations just so i can refer to them so let's say this is equation one and this is equation two so with equation one it already follows this format of ax plus b y equals c so we don't need to do anything there with equation 2 not so much but it's very simple to fix that all i need to do is subtract 3x from both sides of the equation and i would have negative 3x plus 4y is equal to 7. so now it matches this and again we're using the looser definition so let me erase this real quick and i will drag this up here and just say that we have 3x minus y equals 5 and then this guy negative 3x plus 4y is equal to 7. so you can see why we put them both in standard form it's easy to see the values that we have that are lined up on top of each other i've got my 3x on top of negative 3x i've got negative y on top of 4y i've got 5 on top of 7. so everything's lined up in this particular case we have an easy example i want you to notice that we have 3x and negative 3x so we have a pair of variable terms that are opposites so we don't need to do anything further other than proceed to the step where we add the left sides and set that equal to the sum of the right sides now generally we do this vertically but for the first one i want to do it horizontally just to show you something if i start out with 3x minus y and this is equal to 5 so everybody to this point would understand that it's okay to add 2 over here as long as i add 2 over here i added the same thing to both sides of the equation but where people get confused is if i start doing stuff like this let's say i add this to this side so i'm going to add the quantity negative 3x plus 4y based on what we've done so far in our algebra course work we would expect us to add the same exact thing over here but what i'm going to do is i'm going to add 7 over here now you might say is that legal yes i've added the same thing to both sides of the equation it doesn't look the same but i'm told because of this equal sign that they are the same negative 3x plus 4y is equal to 7 so it's the same as 7. so if i add this and this i'm adding the same thing to both sides of an equation and because of the addition property of equality this is mathematically legal now let's go ahead and add now and see what we get so on the left i would have 3x minus y i'm just going to drop the parentheses here so minus 3x plus 4y and this equals 7 plus 5 is 12. over here we have that pair of variable terms you have 3x and negative 3x those are going to cancel and notice how when they cancel the x variable is eliminated okay it's gone so i'm just left with a variable y i have negative y plus 4y that's going to give me 3y and this is of course equal to 12. i can now solve my linear equation in one variable divide both sides by 3. this will cancel you'll get y is equal to 4. so once you've found that out just erase everything so again y equals 4 so i would have as an ordered pair something comma 4. so because 4 is the solution for y in the system that means it works here and here okay in equation 1 and 2. so i can use either equation to find x now so let's go ahead and go with equation 1 because it seems easier we have 3 times x minus plug in a 4 for y and this equals 5. so all i need to do is add 4 to each side of the equation that's going to cancel i'll have 3x is equal to 5 plus 4 is 9. let's go ahead and divide both sides of the equation by 3 and i'll get x is equal to 3. so as my ordered pair goes it would be 3 for the x value comma 4 for the y value now the last thing we want to do is make sure that we didn't make a mistake remember this has to work in both equations to be a solution for the system so i'm going to take 3 and multiply by x x is 3. subtract away 4 because 4 is y and this should equal 5. three times three is nine nine minus four is five you get five equals five so it works out here in the second equation equation number two i would have four times y y is four is equal to 3 times x x is 3 plus 7. we know 4 times 4 is 16 3 times 3 is 9 9 plus 7 is 16 as well so you get 16 equals 16 so it works out here as well so our solution here is the ordered pair 3 comma 4 we're getting an x value of 3 a y value of 4. all right so let's look at one that's a little bit more challenging so we have negative 5x minus 6y equals 13 negative 7 plus x equals negative 6y again i like to number my equations just so i can refer to them this is equation 1 this is going to be equation 2. now i want them each in standard form so let me write this again it's a x plus b y equals c so this one fits the bill that's okay this one not so much what i'd need to do is subtract x away from each side of the equation so i would have negative 7 is equal to negative x minus 6y and of course i can just flip these around and write negative x minus y is equal to negative seven so it matches this and then let me erase all this the other guy equation one is negative five x minus six y is equal to thirteen okay so now that we've got that let's take a look at these two equations now in the last example we were given a nice easy example because we already had a pair of variable terms that were opposites here we're not so fortunate i have negative 5x and i have negative x those aren't opposites i have negative 6y and negative 6y those aren't opposites but it's still kind of easy because if i wanted to i could very easily make these two opposites all i would need to do is multiply one of the equations by negative 1 and that negative 6y would become positive 6y remember i can multiply both sides of an equation by the same non-zero number i can just take this equation here and i can multiply both sides by negative 1 and what i get is negative 1 times negative 5x is 5x negative 1 times negative 6y is 6y negative 1 times 13 is negative 13. so equals negative 13 and this equation stays the same so we have negative x minus 6y equals negative 7. now let me drag this down here what you can see is you have 6y and negative 6y those are now opposites so when we sum the left sides and set that equal to the sum of the right sides that variable is going to drop out so let's go ahead and do that so we want to add the left sides together set this equal to the sum of the right sides now generally we do this vertically in the last example i did it horizontally just to show you so you just put a plus sign out here and a plus sign out here so 5x minus x is 4x 6y minus 6y is 0y right it's just eliminated so you can line this out this is equal to negative 13 minus 7 this is negative 20. so i have 4x is equal to negative 20. of course to solve for x i divide both sides by 4 and i get x is equal to negative 5. so at this point again i can erase everything i just need my two original equations all right so we know x equals negative 5. we don't know the value for y yet so i'm going to plug a negative 5 in for x either in equation 1 or equation 2. it looks way easier to do it with equation 2 so let's do that so we would have negative 7 plus a negative 5 and this equals negative 6y negative 7 plus negative 5 is negative 12 so you'd have negative 12 is equal to negative 6y we would divide both sides of the equation by negative 6. this would cancel you'd have y is equal to 2. so x is negative 5 y is 2. all right let's erase everything let's go ahead and check so we would have negative 5 times plug in a negative 5 for x minus 6 times plug in a 2 for y and this equals 13. so negative 5 times negative 5 is 25 negative 6 times 2 is negative 12 and then this equals 13. we know 25 minus 12 is 13 you would get 13 equals 13 here so it checks out here all right in the next one equation number two we've got negative seven plus for x i've got negative five and this should be equal to negative six times two negative 7 plus negative 5 is negative 12 and this equals negative 6 times 2 is negative 12 as well so it works out here also so this ordered pair negative 5 comma 2 or an x value of negative 5 a y value of 2 is our solution for the system all right for the next one we're going to look at 2x minus 6y equals 0 and then negative 5x minus 3y equals negative 18. so again i like to number my equations this is going to be equation 1 this is going to be equation 2. now we want our equations to be in standard form so ax plus b y equals c this one's in standard form so is this one so we're good to go there now i want one pair of variable terms to be opposites don't have that i have 2x and negative 5x i have negative 6y negative 3y so looking at it what's the easier option do i want to target my x variable or my y variable well it looks like to me it's easier to target y the coefficient of y here is negative six the coefficient here is negative three so what i can do is i can multiply equation two on both sides by negative 2 and what that's going to do is give me a value of positive 6 here so if i have positive 6y here and negative 6y here those are now opposites so let's go ahead and crank that out so for equation 2 i have negative 5x minus 3y equals negative 18 and i'm going to be multiplying both sides by negative 2. so negative 2 times negative 5x is 10x negative 2 times negative 3y is plus 6y and this equals negative 18 times negative 2 which is 36. so now that that's done let me write equation 1 on the top of this so 2x minus 6y equals 0. so now we're at our step where we can add the left sides together set this equal to the sum of the right sides so we're going to add here add here so 2x plus 10x is 12x negative 6y plus 6y is 0y or just 0. this will be eliminated 0 plus 36 is 36. so you get 12x equals 36 of course you can divide both sides by 12 and you'll get x is equal to 3. so let's erase that as an ordered pair 3 will be the x value again we don't know the y value yet so i can plug a 3 in for x in equation 1 or 2 doesn't matter let's just go ahead and plug it in for equation 2. so i'd have negative 5 times 3 minus 3y equals negative 18. i'm going to multiply negative 5 times 3 that's negative 15 and minus 3y equals negative 18. if i add 15 to each side of the equation that will cancel i'll have negative 3y is equal to negative 18 plus 15 is going to give me negative 3. i would divide both sides of the equation by negative 3. this will cancel i'll have y is equal to 1. so x is 3 y is 1. so let's go ahead and check this guy so in the first equation equation one we'd have two times my x value is three minus six times my y value is one and this should equal zero of course this would be two times three or six minus 6 times 1 which is 6. 6 minus 6 is obviously 0. you would get 0 equals 0. so yeah it works out here in the second equation equation 2 you would have negative 5 times x again x is 3 minus three times y y is one this should equal negative eighteen negative five times three is negative fifteen negative three times one is negative three so negative fifteen minus three should equal negative eighteen and again it does you get negative 18 equals negative 18. so it works out here as well so our solution here is the ordered pair three comma one where again an x value of three a y value of one all right let's take a look at another example so we have negative 3 equals 3y plus 3x we have 7 4y equals x plus 15 4. so this is equation 1 and this is going to be my equation 2. if i want it to look like this ax plus b y equals c so for my first equation i could just simply write it as 3x plus 3y is equal to negative 3. so that solves that for the second equation okay for the second equation i can subtract x away from each side so let me write this over here if i have 7 4 y is equal to x plus 15 4 i can subtract x away from each side of the equation that'll cancel so i can reorder this and just put negative x out in front plus 7 4 y is equal to 15 4. now it's not necessary to do this but it's more convenient to work with equations that don't have fractions so go ahead and do yourself a favor multiply both sides of this equation by 4 and get rid of those denominators so 4 times negative x is negative 4x 4 times 7 4 is 7 so this would be plus 7y and this equals 4 times 15 4 is 15. so let's erase all of this and we'll drag this up here okay let me scroll down and get a little room going now what we see is we have 3x and negative 4x 3y and 7y but what would be easier to make into opposites well both of them would be about the same amount of work and what i can do is i'll just go with our target x so to target x we think about 3 and negative 4. the least common multiple between the 2 would be twelve so i want one to be positive twelve and one to be negative twelve well it's really easy to do if i have three x plus three y equals negative three i can multiply both sides of this equation by positive 4. and what that is going to give me 4 times 3x is 12x 4 times 3y is plus 12y this would be equal to 4 times negative 3 which is negative 12. let's erase that then for this one if i have negative 4x plus 7y equals 15 again i want negative 12 here so i would multiply both sides of this equation by 3. so 3 times negative 4x is negative 12x 3 times 7y is 21y and then 3 times 15 is 45. so now i have a pair of variable terms that are opposites i have 12x and negative 12x so we're basically good to go we would add the left sides together we'd set this equal to the sum of the right sides so 12x minus 12x that's 0x are just 0. 12y plus 21y would be 33y and this equals negative 12 plus 45 is 33. if i divide both sides of the equation by 33 i would get that y is equal to 1. all right so our ordered pair would be something for x comma 1 for y now i can plug a 1 in 4y in either equation 1 or 2. looks like it's a whole lot easier to work with equation 1 because again i have to deal with those fractions so let's go ahead and do negative 3 okay negative 3 is equal to 3 times y y is one so three times one plus three x so i'd have negative three is equal to this will just be three plus three x let's subtract three away from each side of the equation that'll cancel i'll have negative 6 is equal to 3x let's now divide both sides by 3 and what i'll have is x is equal to negative 2. so x here is negative two y is one all right so let's go ahead and plug into each original equation we have negative three is equal to three times y y is one plus three times x x is negative two so i'd have negative three is equal to we know three times one is three three times negative two is negative six so if i had three plus negative six that would be negative three so you get negative three equals negative three so it works out here for the next one i have 7 4 times y y is 1 this equals x which is negative 2 plus 15 4. so we know 7 4 times 1 is just 7 4. so just you can erase that for this one i need a common denominator so i would multiply this by 4 over 4. so i'd have negative 2 times 4 that's negative 8 negative 8 over 4. now if i had negative 8 and 15 i get 7 over the common denominator of 4 so yeah this is going to be 7 4 is equal to 7 4. so it works out here as well so our ordered pair negative 2 comma 1 or an x value of negative 2 a y value of one is our solution for the system all right now we've already seen this in the substitution video when the variables are eliminated okay all of them there's nothing left in the equation you're working with when the statement is true there's infinitely many solutions so if you get something like 15 equals 15 29 equals 29 35 equals 35 you know so on and so forth if the statement is false so three equals negative two seven equals negative 21 you know so on and so forth there's no solution okay so in this case right here the second case there are parallel lines okay so they are parallel lines in this case it's the same line it's the same line it's just algebraically manipulated to look different okay and the way you can tell for sure for this type where there's an infinite number of solutions just solve the equations for y you can see you have the same equation for this one where there's no solution solve the equations for y you're going to see you have the same slope but different y-intercepts so you know you have parallel lines parallel lines don't intersect so there's no possible solution all right let's take a look at some special case scenarios so we have 14y plus 21x equals negative 14. you have negative 3x equals 5 plus 2y so let's say this is equation 1 this is equation 2. so for equation 1 again we want to put this in ax plus by equals c or standard form and so all i'm going to do is just switch the order so i would write this as 21x plus 14 y is equal to negative 14. for this one if i have negative 3x negative 3x is equal to 5 plus 2y all i need to do is subtract 2y away from each side of the equation so this would cancel so i'd have negative 3x minus 2y equals 5. so let me let me kind of write that a little better negative 3x minus 2y is equal to 5. all right so now i'm looking to make one pair of variable terms into opposites and again you can pick either one here it's about the same amount of work you have negative 3 and you have 21. so obviously if i multiplied negative 3 by 7 i would get negative 21 there so the coefficient here would be negative 21 the coefficient here would be positive 21. equally as easy if i multiply this guy here by positive 7 negative 2 times positive 7 would be negative 14 negative 14 and positive 14 are opposites so whatever you want to do i'm just going to go with x because that's the first one so if i multiply both sides of this equation by 7 then what i'd have is negative 21x minus 14y is equal to 35. so i can just erase this one and i'll drag this up and now we can add the left sides together set that equal to the sum of the right sides so plus plus now you can see right away 21x minus 21x is 0. 14y minus 14y is 0. so i would have 0 on the left and this equals negative 14 plus 35 is going to give me 21. so is this a nonsensical statement 0 equals 21 yes it is it's obviously not true so this is the case where we have two parallel lines so there's no solution so there's going to be no solution and if you want to prove this to yourself go ahead and pause the video solve each equation for y you're going to find that you have the same slope and different y intercepts in each case right so they're parallel lines so we can also do this and we can say they're parallel lines all right let's take a look at another example so suppose we see 32x minus 8 equals 8y negative 30y equals negative 120x plus 30. so again i want this in standard form so ax plus b y equals c so for this guy again i'm going to label these equations equation 1 and equation 2. for equation 1 if i have 32x minus 8 equals 8y let me add 8 to each side of the equation let me subtract 8y from both sides of the equation so that's going to give me 32x minus 8y is equal to positive 8. then for equation 2 all i need to do is add 120x to each side of the equation i will have 120x minus 30y is equal to 30. all right so this one's a little more tedious and i would just choose the smaller coefficients here so i'm going to work with y so if i have a negative 30 and i have negative eight forget about the signs for a second just just pretend you have eight and you have 30. what's the least common multiple there okay what's the least common multiple well eight is what it's two times two times two thirty is what it's six times five or it's two times three times five so really it would be two times two times two times three times 5 or 8 times 15 which is 120. so if i think about 120 i want one of these to be negative 120 y the other to be positive 120y so let me multiply both sides of this equation by a negative 15 and let me multiply both sides of this equation by a positive four so 4 times 120x is 480x 4 times negative 30y is minus 120y 4 times 30 is 120. for this guy negative 15 times 32 is negative 480 then of course times x negative 15 times negative 8 is positive 120 then times y and this equals 8 times negative 15 which is negative 120. so if i want to add the two left sides here together and i want to add the two right sides here together 480x minus 480x is 0. negative 120y plus 120y is 0. 120 minus 120 is 0. so what do i get i get 0 equals 0 which is true so the variables have dropped out and i'm left with a true statement so there's an infinite number of solutions right these two equations are basically the same they've just been algebraically manipulated to look different so there's an infinite number of solutions and again you can solve each equation for y that'd be the easiest thing you can do and if you solve each equation for y you're going to see that you have the exact same equation in each case and therefore whatever works in equation 1 also works in equation 2 because it's the same equation so there would be an infinite number of solutions for this system hello and welcome to algebra 2 lesson 28 in this video we're going to learn about solving systems of linear equations in three variables so up to this point we've become very familiar with how to solve a system of linear equations in two variables using graphing using substitution and using elimination now we're going to rely on all this knowledge to get us through the next challenge which is solving a system of linear equations in three variables before we learn how to solve that type of system we first need to familiarize ourself with a linear equation in three variables so a linear equation three variables looks like this it's ax plus b y plus c z equals k so just something times x plus something times y plus something times z equals some constant so we would say a b c and k are just real numbers where a b and c are not all zero at the same time because again if you had that your variables would all drop out all right so let's look at an example of this real quick i think if we just use our prior knowledge of working with a linear equation in two variables and extend it to a linear equation in three variables we're pretty much good to go we've just added a variable here so we have 4x plus 2y minus 3z equals negative 14. and i have a sample solution okay we know when we work with a linear equation in two variables we give our solution as an ordered pair now we're going to have a solution as an ordered triple so instead of just an x y it's an x a y and a z and it's in that order so as this example goes i have this ordered triple this one is an x this zero is a y and this 6 is a z so i should be able to plug in a 1 for x a 0 for y and a 6 for z and i should get a true statement so i would have 4 times 1 plug in a 1 for x plus 2 times 0 plug in a 0 for y minus 3 times 6 plug in 6 for z and this should equal negative 14. so 4 times 1 is 4 2 times 0 is 0 negative 3 times 6 is negative 18 this should equal negative 14 and it does you'd basically have 4 minus 18 which is negative 14 so that does equal negative 14. so this works as a solution but just like we saw with a linear equation in two variables a linear equation in three variables is also going to have an infinite number of solutions so i could just generate as many ordered triples as i want as solutions for this equation all right so now that we know what a linear equation in three variables is what if we get a linear system in three variables so this is going to contain three equations with three unknowns so solving a linear system in three variables the first thing we want to do is we want to write each as ax plus by plus cz equals k why do we do this well we're doing it because in the next step we're going to do some elimination with elimination we remember our first step is to write a x plus b y equals c when we have two variables involved well it extends to this with three variables i just have plus c z the reason we do this is because we make it easy to see what we're going to be able to eliminate or what we would need to eliminate a variable all right so now that we've done that we use elimination to eliminate one of the variables from any two equations doesn't matter what variable you pick doesn't matter what two equations you pick but in the next step it does matter you need to eliminate the same variable from any other two equations so if in that last step i eliminated x when i get to this step i've got to eliminate x again what that's going to do is it's going to give me a linear system in two variables if i eliminated x i'm going to have y and z once i have a linear system and two variables we already know how to solve that so solve the linear system in two variables you can use substitution or elimination whatever you want to do use the result to find the third unknown once i know what two of them are i could go back to any original equation and find the third unknown the last step and always the most important check the result in each equation so you're going to plug in for each x y and z in each original equation now the one thing that always comes up and a lot of students ask this they say can you use substitution the whole way through so in other words can i use substitution to generate a linear system in two variables and then solve that with substitution and end up with the answer for my linear system and three variables the answer to that is yes you can do that the reason that we generally don't teach that is because it's super tedious if i solve one of these for one of the variables and then i plug that in to the next two equations i will generate a linear system and only two variables but to do it will be super super tedious and it's very prone to errors because of that so let's start out with what we have we have negative 6x plus 4y plus z equals 20. i'm going to call this equation 1. then i have negative 2x minus 6y minus 5z equals negative 22. i'm going to call this equation 2 and i have 2x minus 2y plus 4z equals 10. i'm going to call this equation 3. now everything is already written for us as we were told to do as ax plus by plus cz equals k so we don't need to do anything on that front we move immediately into picking one variable to eliminate between two equations so if you look at everything it looks like it's pretty easy to eliminate x between equation 2 and 3. i've got negative 2x and i've got 2x so that would work out pretty nicely for us so if i've got negative 2x minus 6y minus 5z equals negative and if i've got 2x minus 2y plus 4z equals 10 i can sum the left sides together and set this equal to the sum of the right sides so negative 2x plus 2x is 0 x has been eliminated negative 6y minus 2y is negative 8y negative 5z plus 4z is negative z and negative 22 plus 10 is negative 12. so i've produced an equation that is x free so let's call this equation four this is negative eight y minus z equals negative 12. all right now the next thing i want to do is i want to get any other two equations so i worked with two and three so it either needs to be one and three or one and two so one of those scenarios i need to eliminate x from those so let's go ahead and just do 1 and 3 because i have a negative 6 and a positive 2. so the coefficients are opposite signs they're not opposites yet but they're opposite signs so that's a little bit easier so we have negative 6x plus 4y plus z equals 20. and then for equation 3 i have 2x minus 2y plus 4z equals 10. now if i want to make these opposites i can multiply 2x by 3 but to make that legal i've got to multiply both sides of the equation by 3. so 3 times 2x is 6x 3 times negative two y is negative six y three times four z is plus twelve z three times ten is thirty all right so let's erase this and we'll just drag this up here and again i'm gonna add the two left sides set this equal to the sum of the right sides negative 6x plus 6x is 0. so again x has been eliminated 4y minus 6y is negative 2y z plus 12z is plus 13z 20 plus 30 is 50. so what this would produce let me kind of move this down to be in line for equation 5 we'll call it this would be negative 2y plus 13 z equals 50. now let me erase this i have produced a linear system in two variables so now we know how to solve this we've been doing this forever now the only other side note i would say is we did that so much quicker than if i let's say solved for z and then i plugged in for z in this equation in this equation again that's the substitution method you could use that would produce a linear system in two variables with just x and y and then i could i could solve that as well but again using that method is just ultra tedious and i want to try to stop you from doing that you could pause the video try it that way and just see how long it takes you to produce a system like that it's going to take you a long time all right so now i have my linear system in two variables it looks like this one is easier to solve with substitution i have a coefficient of negative one there so let's go ahead and solve this for z we have negative 8y minus z equals negative 12. let's go ahead and add 8y to both sides of the equation that would cancel i'll have negative z is equal to 8y minus 12. let's divide both sides by negative 1. and so this is going to produce z is equal to negative 8y plus 12. so i'm going to drag this over here z is equal to negative 8y plus 12. all right so we all know how substitution works we're going to substitute in for z in this other equation so i'm going to plug in here and i'm going to be plugging this in z equals or z is the same as negative 8y plus 12. so i'll have my equation 5 negative 2y plus 13 times in place of z i'm going to plug in negative 8y plus 12 because z is the same as that and this equals 50. so i'll have negative 2y plus 13 times negative 8 is negative 104 so i can go ahead and put minus 104y and then 13 times 12 is 156 and this equals 50. so negative 2y minus 104y is negative 106y then plus 156 and this equals 50. so if i subtract 156 away from each side of the equation this will cancel i'll have negative 106y is equal to 50 minus 156 is negative 106. scroll down and get a little room going as a final step let's divide both sides by negative 106. this will cancel i'll get y is equal to negative 106 over negative 106 is 1. so y equals 1. so we found one of our unknowns so let me erase everything so now i'm going to keep working with this linear system in two variables i'm just going to list that y equals 1 for now so i can plug a 1 back in for y here or here doesn't matter it's easier to work with equation 4 so let's go ahead and do that so in equation 4 i'd have negative 8 times plug in a 1 for y minus z equals negative 12. negative 8 times 1 is going to be negative 8 minus z equals negative 12. let's add 8 to each side of the equation that cancels you'll have negative z is equal to negative 12 plus 8 is negative 4. multiply both sides by negative 1 you'll get z is equal to 4. so y equals one z equals four all right so now we know two of our unknowns and now we can turn to our original system of equations so i go back to equation one two and three so y equals one in every equation z equals four in every equation so i can plug into equation one two or three doesn't matter and find the value of x so let's just go ahead and work with equation three so if i have 2x minus 2 y is 1 so 2 times 1 plus 4 times z z is 4 this should equal 10. so i'd have 2x minus 2 plus 16 equals 10. so i'd have 2x negative 2 plus 16 is 14. this equals 10. let's subtract 14 away from each side of the equation this will cancel i'll have 2x is equal to 10 minus 14 is negative 4. let's divide each side of the equation by 2 and we'll have x is equal to negative 2. all right so i've found my value for x x is negative 2. and let's go ahead and write this as an ordered triple remember it's the x value so that's negative 2. the y value so that's 1 and the z value is 4. so the very last thing we want to do is check i know checking takes a long time and it's tedious but it's very important that you check if you have the time if you're on a timed examination what i always do is or what i always suggest that you do would be to go through everything first make sure you can complete your test then go back and check your answers start with the ones that you think you got wrong and then if you have time check everything all right so let's start with equation one so i would have negative six times for x i'm plugging in a negative 2 plus 4 times for y i'm plugging in a 1 plus z which i'm plugging in a 4 for so this should equal 20. negative 6 times negative 2 is 12. plus 4 times 1 is 4 plus 4 equals 20. 12 plus 4 16 16 plus 4 is 20 so you get 20 equals 20. so it does work out here for equation 2 okay for equation 2 we have negative 2x so x is negative 2 minus 6y y is going to be 1 minus 5z z is again going to be 4 this should equal negative 22. negative 2 times negative 2 is 4 minus 6 times 1 is 6 minus 5 times 4 is 20 this should equal negative 22. 4 minus 6 is negative 2 negative 2 minus 20 is negative 22 you get negative 22 equals negative 22 so it works out here as well and let's check equation three so we'd have two times x x is negative two minus two times y y is one plus four times z again z is four so this equals ten two times negative two is negative four minus two times one is two plus four times four is sixteen this should equal ten negative four minus two is negative 6 negative 6 plus 16 is 10. so you get 10 equals 10 so it works out here as well so we found our solution for our system it took a while but we did find it it's the ordered triple negative 2 comma 1 comma 4 which is an x value of negative 2 a y value of 1 a z value of 4. now i'm not going to check everything moving forward just for the interest of time because it takes forever but i want you guys to pause the video just so you know how to check things plug in for every x plug in for every y plug in for every z and make sure all of the equations are true with our solution all right let's go ahead and take a look at the next example we're going to kind of speed things up just a little bit so i have negative 3x plus 18y equals 19 plus 12z i have negative 9x minus 6z equals negative 19 plus 18y i have negative 6y minus 12z equals 2 plus 18x now this is not in the format that we'd like it in we wanted an ax plus by plus c z equals k do we have to have it in this format no but it makes it convenient for us to see what's going to cancel what we need to do to cancel we don't want to be looking all over the place to find z and x and all these things we want things lined up nice and neat so for the first equation i'm going to write negative 3x plus 18y if i subtract 12z away from each side i could put minus 12z over here and this would be equal to 19. so this would be my equation 1. for equation 2 i have negative 9x i could subtract 18y away from each side then minus 6z and this equals negative 19. okay this is my equation 2. for equation 3 let's subtract 18x away from each side of the equation then minus 6y then minus 12z and this equals 2. so this would be my equation 3. all right so let's scroll down and let's go ahead and eliminate one of the variables from two of the equations now looking at this again everything's nice and neat laid out for me i can see that i have plus 18y and minus 18y so let's go ahead and start with what's easy let's take equation one and equation two and let's eliminate the variable y so we have negative 3x plus 18y minus 12z equals 19. we have negative 9x minus 18y minus 60 equals negative 19. all right so if i add the left sides together and i set this equal to the sum of the right sides negative 3x minus 9x is negative 12x 18y minus 18y is 0. so y is eliminated negative 12z minus 6c is minus 18z 19 plus negative 19 is 0. so i'm going to write equation 4 here as negative x minus 18 z equals zero all right so now what i need to do is pick two other equations and again eliminate the same variable y so let's go ahead and work with equation one and equation three so for equation 1 i have negative 3x plus 18y minus 12z equals 19. for equation 3 i have negative 18x minus 6y minus 12z equals 2. now if i want these to be opposites i need to multiply negative 6 by 3 but i've got to multiply both sides of the equation by 3 to make that legal 3 times negative 18x is negative 54x minus 3 times 6y is 18y minus 3 times 12z is 36z and this equals 3 times 2 which is 6. all right let's bring this up now so let's add the left sides together set it equal to the sum of the right sides negative 3x minus 54x is negative 57x 18y minus 18y is zero y is eliminated negative 12z minus 36z is minus 48z then this equals 19 plus 6 which is 25. all right so let's put this as equation 5 here so we have negative 57 x minus 48 z equals 25. so now i have a linear system in two variables so what i can do is again solve that i can use elimination or substitution really whatever i want so it looks like i'm going to use elimination here so i'm going to choose to eliminate the variable z so how would i do that if i have negative 12x minus 18z equals 0 and i have negative 57x minus 48z equals 25. what's the least common multiple between 18 and 48 so the lcm of 18 and 48. well i know 18 is 6 times 3 so 2 times 3 times 3. what's 48 is eight times six so it's two times two times two times two times three so my least common multiple would be four factors of two and two factors of three so it would be sixteen times nine which is a hundred forty-four a hundred forty-four so i want one of these to be negative 144z the other to be positive 144z so let's multiply this equation both sides by negative eight and then let's multiply both sides of this equation by positive three so i would have negative eight times negative 12x that would be 96x then negative 8 times negative 18z would be positive 144z and this equals 0 times negative 8 which is 0. next we have 3 times negative 57x that's negative 171x then 3 times negative 48z that's going to give me negative 144z and this equals 3 times 25 which is 75. all right so let's erase this i don't need any of this anymore and i'm going to drag this up a little bit so if i add the two left sides together and set it equal to the sum of the right sides what am i going to get 96x plus negative 171x is going to give me negative 75x 144z minus 144z is going to be zero right so z is eliminated zero plus 75 is 75. so i get negative 75x equals 75 i divide both sides by negative 75 and i get that x is equal to negative 1. x is equal to negative 1. okay let's erase all this now if i stay with this system in two variables i'm going to find out what z is so let's plug into equation four we have negative 12 times for x i'm going to plug in a negative one minus 18 times z equals 0. negative 12 times negative 1 is 12. so i would get 12 minus 18z equals 0. i would subtract 12 away from each side of the equation that's gone i'll have negative 18z is equal to negative 12. let's divide both sides of the equation by negative 18. so that'll cancel i'll have z is equal to we know negative over negative is positive but we think about 12 and 18. each is divisible by 6. divide 12 by 6 you get 2. divide 18 by 6 you get 3. so z is going to be two-thirds so now let's look at our equations that have three variables i'm going to take one of them doesn't matter which one plug in a negative one for x plug into two thirds for z i'm going to find out y so let's go ahead and work with equation number one so you'd have negative 3 times x again x is negative 1 plus 18 y minus 12 times z z is 2 3 and this equals 19. negative 3 times negative 1 is 3 plus 18y negative 12 times 2 thirds 12 would cancel with 3 and give me 4. so really i have negative 4 times 2 which is negative 8. this equals nineteen three minus eight is negative five so you'd have negative five plus eighteen y is equal to nineteen add five to each side of the equation that would cancel you'll have 18y is equal to 19 plus 5 is 24. now to get y by itself let's divide both sides of the equation by 18. this will cancel with this i'll get y is equal to 24 and 18 are each divisible by 6 24 divided by 6 is 4 18 divided by 6 is 3. so y is going to be equal to 4 3. so y is equal to 4 3. so as an ordered triple okay as an ordered triple again it's the x value which is negative one the y value which is four thirds and the z value which is two thirds okay so that's going to be your solution to this system in the interest of time i'm not going to go through and check it but i want you to pause the video if you're unsure how to do it plug in a negative 1 for every x of 4 thirds for every y a 2 3 for every z and make sure the left and the right side are equal in each of the three equations all right let's take a look at another one so we have negative x minus 6z equals negative 4y plus 4. we have negative 4x equals 5z plus 3. we have 6y plus 2z equals negative x plus 12. so a lot of students see this and they go well you're you're missing a variable in the second equation that's okay i'm going to show you how that's going to actually be a benefit to you and it's going to take less time to do a problem like this the first thing is i want to write every equation in this format ax plus b y plus c z equals k so for this top equation negative x add 4y to each side so plus 4y minus 6z equals 4. for the next equation negative 4x i don't have a y so i can put plus 0y if i want and then i'm going to subtract 5z away from each side and this equals 3. for the third equation i'll add x to each side of the equation and then plus 6y plus 2z and this equals 12. so just as i always do i'm going to number these this is equation 1 2 and 3. all right now let's talk about how this is an improvement for us the one thing is normally i have to eliminate one of the variables from any two equations then i've got to do that again i've got to eliminate the same variable from another two equations i notice that this equation does not have a y it's got zero y you can basically just take y out so i'm just gonna write this over here as negative four x minus five z equals three if i just take equation one and 3 and i eliminate y i can write this over here as an equation with x and z only and i can basically have a linear system with two variables and i can get my solution very quickly so i can skip a step when this occurs so with equation 1 and 3 i have negative x plus 4y minus 6z equals 4 and i have x plus 6y plus 2z equals 12. now i know some of you will look at that and say oh i've got negative x and x let's just go ahead and eliminate x that's not going to do you any good because over here you have x and z if i eliminate x that have y and z i need a linear system with the same two variables okay so don't think like that all right so the first thing i want to do is think about how i can get 4y and 6y into opposites well you think about the least common multiple between the two and it would be 12. so i want one of them to be 12y the other to be negative 12y so i can multiply both sides of this equation by let's say negative 3. and i can multiply both sides of this equation by let's say positive 2. so negative 3 times negative x would be 3x negative 3 times 4y would be minus 12y negative 3 times negative 6z would be plus 18z negative 3 times 4 would be negative 12. so this equals negative 12. 2 times x would be 2x 2 times 6y would be plus 12y 2 times 2z would be 4z and this equals 2 times 12 which is 24. all right so let me erase this i need this anymore and we're ready to add the two left sides together set that equal to the sum of the two right sides so what's that going to give me 3x plus 2x is 5x negative 12y plus 12y is 0. so y's eliminated 18z plus 4z is 22z this equals negative 12 plus 24 which is 12. so i've got another equation here this is equation 2 so i'll keep that number this will be equation 4. it's 5 x plus 22 z equals 12. all right so now we can just solve for x and z and then plug back into one of the original equations so the easiest thing to do i think in this case would be to use elimination if i have negative 4x minus 5z equals 3 and i have 5x plus 22z equals 12. between negative 4 and 5 the lcm is 20. so i can make one of them positive 20 one of them negative 20 and i'm going to be good to go so i'm going to multiply both sides of this by 5 and both sides of this by 4. so 5 times negative 4x is negative 20x 5 times negative 5z is minus 25z 5 times 3 is 15. so this equals 15. 4 times 5x is 20x 4 times 22z is plus 88z this equals 4 times 12 is 48. let's erase this so let's bring this up all right so let's sum the two left sides set this equal to the sum of the right sides now you have 20x plus 20x is zero negative 25z plus 88z is 63z and this is equal to 15 plus 48 is 63. so divide both sides of the equation by 63. you get that z is equal to 1. so z here equals 1. so i can plug a 1 back in for z in either equation 2 or 4. doesn't matter let's go ahead and use equation 4. so 5x plus 22 times 1 22 times 1 is just 22. this equals 12. let's subtract 22 away from each side of the equation this is going to cancel you'll have 5x is equal to 12 minus 22 is going to be negative 10. let's divide both sides of the equation by 5. we'll get x is equal to negative 2. so x equals negative 2. now we can take these values plug them back in for either equation 1 or equation 3 and get a value for y we can't do that with equation 2 because y isn't there so in equation 1 i could go minus a negative 2 plug in a negative 2 for x plus 4y minus 6z z is 1 so minus 6 times 1. this equals 4. so minus a negative 2 is 2. plus 4y minus 6 times 1 which is 6 and this equals 4. 2 minus 6 is negative 4 so you get 4y minus 4 equals 4. go ahead and add 4 to each side of the equation this will cancel you'll have 4y is equal to 8. divide both sides of the equation by 4 you're going to get that y is equal to 2. okay so y is equal to 2. let's erase all this we'll write this as an ordered triple so it's the x value which is negative two the y value which is two then the z value which is one so negative two comma two comma one or an x value of negative two a y value of two a z value of one that's your solution for this system all right so let's look at one final problem and i know these get super super tedious you know when you start working through stuff like this you can go through an hour's worth of work and it can be only like five or six problems so let's go ahead and start out by just rewriting everything as a x plus b y plus c z equals k for the first equation i have negative 2x negative 2x i'm going to add 4y to each side so plus 4y then plus 4z and this equals 4. so let's go ahead and call that equation 1. for the next equation we'd have 4x minus 4y let's subtract 6z away from each side and this equals negative 18. let's call this equation 2. for the next guy let's subtract 5x away from each side let's have negative 2y next then let's add 4z to each side and this will be equal to negative 2. let's call this equation 3. all right so let's take a look at what we can do here so i want to start with any two equations and i want to eliminate one of the variables so it looks like it'd be really easy to eliminate y if i worked with equation 1 and 2. so if i have negative 2x plus 4y plus 4z equals 4 then i have 4x minus 4y minus 6z equals negative 18. what do we have here if i add the left sides set this equal to the sum of the right sides negative 2x plus 4x is 2x 4y minus 4y is 0. that's eliminated 4z minus 6z is minus 2z this equals 4 minus 18 which is negative 14. so let's say equation 4 here is 2x minus 2z equals negative 14. now i'm going to eliminate the same variable which is y from any two other equations so let's go ahead and use equation one and three so negative two x plus four y plus 4z equals 4. equation 3 is negative 5x minus 2y plus 4z equals negative 2. so i don't quite have what i need if this was negative 4 i'd be good to go so i can accomplish that by multiplying both sides of the equation by 2 2 times negative 5x is negative 10x 2 times negative 2y is minus 4y 2 times 4z is plus 8z and this equals 2 times negative 2 is negative 4. all right so now let's sum the left sides set this equal to the sum of the right sides negative 2x minus 10x is negative 12x 4y minus 4y is 0. y is eliminated 4z plus 8z is plus 12z and then 4 minus 4 is 0. so as equation 5 goes you would have negative 12x plus 12z equals 0. so it looks like it'd be pretty easy to use substitution here let me take equation 4 and solve it for x so i have 2x minus 2z equals negative 14. if i want to solve this for x let me add 2z to both sides of the equation that's going to cancel we'll have 2x is equal to 2z minus 14. let's go ahead and divide both sides of the equation by two that'll cancel we'll have x is equal to two over two is one so it's just z minus fourteen over two is seven so we can say that this relates to this we have x is equal to z minus 7. now i'm just going to plug in 4x in equation 5. so we'd have negative 12 times plug in a z minus 7 for x plus 12 z equals 0. negative 12 times z is negative 12z negative 12 times negative 7 is positive 84 then plus 12 z equals 0. so i want you to notice something you have negative 12z and you have positive 12z those are going to cancel i'm left with 84 equals 0. we've seen this before remember your variable drops out you've got nonsense that means no solution it's the same thing when you're working with a linear system with three variables so there's no solution here no solution all right so we put the symbol for the null or empty set hello and welcome to algebra 2 lesson 29 in this video we're going to learn about applications of linear systems so throughout your study of algebra and i mean algebra 1 algebra 2 college algebra whatever level of algebra you're in you're going to consistently run into word problems and for whatever section you're dealing with they're generally a little bit more challenging than if you just had a problem presented to you the reason for this is you have to dig through the information and then generate the problem and then solve it and go back and answer and figure out if it makes sense so in a lot of cases it will help you substantially if you have just a little procedure a little guideline to refer to when you're working through these problems so i'm going to give you one here and you can make up your own and you can modify this whatever suits your needs so i have here solving applications of linear systems so the first thing and probably the most important thing when you're working any word problem i don't care what it is you want to read the problem and determine the main objective if you don't know what you're being asked to find then how can you find it right so that's the most important thing all right the next thing is to assign a variable to represent each unknown now this changed a little bit versus when we talked about solving a linear equation in one variable right when we looked at word problems for that section in that case i have one equation and one variable if i have more than one unknown those problems are set up in such a way to where i can express the other unknown in terms of that first one that i said hey x is this then this one is x minus this or x plus this or whatever your given scenario is when we deal with a section on linear systems in general the problems are set up to where you just assign a variable to represent each unknown and then you would write a system of equations now for every unknown i have i have to have an equation so if i have two unknowns i need two equations if i have three unknowns i need three equations if i had seven unknowns i would need seven equations so on and so forth now once i've gone through and i've written my system of equations i would simply solve the system and i want to check now with word problems you've got to make sure that when you check your answer it's reasonable okay if i'm on a train and it says how many passengers were on this train with me it can't be negative seven okay that doesn't make sense if i'm calculating a rate of speed it can't be negative 30. okay that doesn't make sense if i'm calculating the amount of distance that was driven and it says negative 2.6 that doesn't make sense so make sure that your answer is reasonable okay in terms of the problem all right so let's take a look at the first example so we have that beth school is selling tickets to a fall musical on the first day of ticket sales the school sold 12 senior tickets and 16 child tickets or a total of 336 dollars the school took in 312 on the second day by selling 15 senior tickets and 11 child tickets find the price of a senior ticket and the price of a child ticket so it's very straightforward here what we need to do we want to what find the price of a senior ticket and the price of a child ticket so we find the two things that we don't know and for each unknown we represent with a variable let's go down to the next page so we will let x be equal to the price of each child ticket and then y will be equal to the price of each senior ticket all right so now we want to go ahead and write a system of equations so let's go back through our little problem we're going to read and gather some information that we can use to set up a system so with these problems they're going to give you information to build your system with this one it tells you about ticket sales so on the first day of ticket sales the school sold 12 senior tickets and 16 child tickets for a total of 336 dollars so let me go to the next page so there was 12 senior and 16 child and it was 336 dollars let's just save that information there let's go back now the other scenario they give us it says the school took in 312 dollars on the second day by selling 15 senior tickets and 11 child tickets so let's go back so now we have 15 senior we have 11 child and the total from this day was 312 dollars so i have two scenarios and this is going to be enough for me to build a system of equations so x is the price of each child ticket now if i sold 16 child tickets and x is the price of each child ticket so let's say x with 10. not saying it is but let's say it was well 16 times 10 will be 160. so i'm going to multiply 16 times x to get the total revenue from selling child tickets that day so in the first scenario i will have 16 times x then i've got to do the same thing for senior tickets so plus 12 times y and this is equal to 336 so this is my first scenario my second scenario over here i'm going to mirror that it's going to give me my second equation so for child tickets it's 11 that we sold times x dollars a piece plus for senior tickets it's 15 sold times y dollars a piece and this equals 312. all right so let's go through this real quick just so everybody understands what's going on this 16 this number here and this 11 here that's how many we sold so that's a quantity the x represents the price per ticket so if i have the amount sold times the price each that gives me the total revenue in this case for child's tickets sold then i have 12 and i have 15. then each one is multiplied by y so in this case it's 12 times y in this case it's 15 times y that's giving me the revenue for senior tickets sold if i take the revenue from child's tickets sold and the revenue from senior tickets sold and i combine them together i should get the total revenue for the day so here that total revenue was 336 here that total revenue was 312. so that's how we went about setting up our system of equations and now it's just a matter of solving the system for x and y and we can report our answer we'll know x is the amount per child's ticket sold and y is the amount per senior tickets sold all right so let's go ahead and use substitution i think that'd be a little bit easier here so i'm going to take equation 1 and i'm going to solve it for y so i have 16 x plus 12 y is equal to 336 so let's subtract 16x away from each side of the equation this would cancel i'll have 12y is equal to negative 16x plus 336. the next thing i want to do is i want to divide both sides of the equation by 12. so this will cancel i'll have y is equal to negative 16 over 12 we know that's negative 16 and 12 are each divisible by 4. so if i divide 16 by 4 i get 4. if i divide 12 by 4 i get 3. so this would be negative 4 3 x and then 336 over 12 is 28 so this would be plus 28. so i have solved equation 1 for y so now i'm going to plug in for y in equation 2. so here's y and i'm going to plug in remember y is equal to or y is the same as negative 4 3 x plus 28 so this will get plugged in here and then i'll have a linear equation in x okay i'll just have one variable so 11x plus now i'm going to substitute in the negative four-thirds x plus 28 like that quantity and again make sure you use parentheses because 15 is multiplied by this whole quantity and this equals 312. so i will have 11x plus i'll do 15 times negative four thirds x plus 15 times 28 equals 312. now let's deal with this so 15 times negative 4 thirds 15 would cancel with 3 and give me 5. 5 times negative 4 is negative 20. so this would be 11x minus 20x plus 15 times 28 is 420 and this equals 312. all right let's go down and get a little bit of room going so 11x minus 20x we know that's negative 9x so let's write negative 9x let's subtract 420 away from each side of the equation that'll cancel so negative 9x is going to be equal to 312 minus 420 is going to give me negative 108. so let's go ahead and divide both sides of the equation by negative 9. so this will cancel with this and i'll have x is equal to 12. all right so now that i know x is equal to 12 let's erase everything let's go back up here i can erase this so x is equal to 12. so i can plug a 12 in for x either an equation 1 or 2 doesn't matter and solve for the unknown y let's go ahead and plug in a 12 for x in equation 2 so i would have 11 times 12 plus 15 y equals 312. 11 times 12 is 132 so i'd have 132 plus 15y is equal to 312. let's subtract 132 away from each side of the equation that's going to cancel we'll have 15y is equal to 312 minus 132 is 180. so let's go ahead and divide both sides of the equation by 15 and we'll find that y is equal to 12. so x and y are the same value they're each 12. so remember x was the price of each child ticket and y was the price of each senior ticket there are each 12 so we can go ahead and answer and say that each senior ticket and each child ticket sold for 12 now can we verify this yeah quite easily if on the first day we sold 12 senior tickets and 16 child tickets well since the price is the same i might as well say okay well i sold a total of 12 plus 16 or 28 tickets now if i took the 28 tickets i sold and i multiplied it by a price of 12 each the total amount that would be received is 336 dollars so that one checks out then in the other scenario the school took in 312 dollars by selling 15 senior tickets and 11 child tickets so 15 plus 11 is 26 so 26 total tickets that day and they're again 12 a piece because a senior taking the child ticket both sell for 12 so 26 times 12 would give me 312 right so they would have taken 312 dollars in that day so the answer is correct here each senior ticket and each child ticket sold for 12 dollars all right so for the next problem a boat traveled 45 miles downstream and back the trip downstream took three hours while the trip back took five hours find the average speed of the boat in still water along with the speed of the current so what do we need to focus on here well there's two things we want to find the average speed of the boat in still water meaning if there was no current how fast would the boat go and then also the speed of the current so we would start by just assigning a variable to each of those so a variable to the speed of the boat in stillwater and to the speed of the current so let's go down let's let x be equal to the speed of the boat and this is in miles per hour in stillwater okay then why would be equal to the speed of the current so the speed of the current and again this is in miles per hour let's go back now in order to answer these two questions we want to think a little bit about what's going on here so the boat traveled 45 miles downstream and back downstream means you're going with the current if you're going with the current you're going to go faster just think about taking like an inflatable raft and just putting it in the water and going with the current even though you don't have a motor or some kind of propeller to move you you're being moved by the current so the current has a general speed that it's going to push you at now if you're going upstream you're fighting the current you know have you ever been in a row boat or anything or even a simple boat and you're trying to fight the current you know it's very hard because you're dealing with a force that's pushing against you now so the trip downstream took three hours while the trip back which would be upstream took five hours so of course if you're going downstream if you're going with the current it's going to take you less time because you have a little helping hand so thinking about this let's go back down this is a motion word problem so for a motion word problem we think about the distance formula right distance is equal to the rate of speed times the amount of time traveled and a lot of times when we have a motion war problem it's usually easiest to just use a little table so you have two scenarios the first one would be downstream the second one would be upstream so you have your distance and that's equal to the rate times the time so let's see if we can fill in this information so downstream when we're doing our trip let's go back up the boat traveled 45 miles so that's the distance so downstream and back the trip downstream took three hours let's just stop there so we're given a distance of 45 miles so we know that 45 the rate of speed we're not given we can use our variables to work with that but the time we are given for the trip downstream we're told the time is three hours so we have that so for the rate if i'm going downstream i have the speed of the boat in stillwater so if there was no current at all that's x and i also have the speed of the current that's helping me right so i would increase the speed by that amount so x plus y x plus y would be my rate of speed here so this times this would give me this now for upstream it tells us what well the distance is the same the only difference is the rate of speed and the time the time now is 5 hours let's go down so the time is 5 the distance is 45 the rate of speed is going to be different here you have the speed of the boat in still water but since you're going upstream you're now fighting the current so you have to subtract away y okay you have to subtract away y and the reason for that is your boat is fighting okay you're in the still water you have a certain speed if you all of a sudden get put against the current you have to subtract away the speed of that current so as an example if i was going six miles an hour and still water and the current was three miles an hour well if i'm fighting the current then now i'm going 6 minus 3 or 3 miles an hour so that's where we get this calculation here now i have a linear system in two variables if i look at my little table again a distance is equal to a rate of speed times a time so for downstream i have this let's label this equation one we have three times the quantity x plus y is equal to for equation 2 i would have 5 times the quantity x minus y is equal to 45. now for this system it's going to be really easy to use substitution inside the parentheses you see you have plus y and minus y so the way we're going to do that let me just kind of move this over here if i have 3 multiplied by the quantity x plus y this is equal to 45 i can divide both sides of this equation by 3. again this is multiplication so this would just cancel so then i'm left with x plus y is equal to 45 over 3 is 15. so this i could write as x plus y equals 15. then for this one i could do the same thing so if i have five times the quantity x minus y equals 45 i can divide both sides of the equation by five this will cancel i'll have x minus y is equal to 45 over 5 is 9. so if i add the left sides together set that equal to the sum of the right sides x plus x is 2x y plus negative y or y minus y is 0. 15 plus 9 is 24. so now i divide both sides of the equation by 2 and i get that x is equal to 12. all right so let's erase all this we know x equals 12 at this point so what is y going to be so x equals 12. so let's plug that back into let's say equation two so we'd have five times the quantity i'd have 12 minus y this equals 45 5 times 12 is 60 so you'd have 60 minus 5y is equal to 45. we would subtract 60 away from each side of the equation that would cancel you'd have negative 5y is equal to 45 minus 60 is negative 15. divide both sides of the equation by negative 5 and you'll get y is equal to 3. so x is 12 y is 3. so i want you to remember that x is the speed of the boat and still water y is the speed of the current so we can say that the speed of the boat in stillwater [Music] was 12 miles per hour the speed of the current was three miles per hour and we could check this pretty easily the trip downstream took three hours okay so remember distance equals rate times time so if i'm going downstream or with the current again i would take my rate of speed in still water which is 12 miles per hour i would add the speed of the current because it's pushing me or helping me by that much so 12 plus 3 is 15. so i'm going 15 miles an hour for 3 hours that means i went 45 miles so this one checks out now the trip back we think about that you're going 12 miles an hour in still water now you're fighting the current so you've got to subtract away what the current speed is so 12 minus 3 is 9 so now we're going nine miles per hour for five hours that's 45 miles so that one checks out as well so again the speed of the boat in still water was 12 miles per hour and the speed of the current was three miles per hour all right let's wrap up our lesson with one that's a little bit more tedious so james won 86 000 in the local lottery he decided to invest the whole amount in three different investment funds one a reit which stands for real estate investment trust pays six percent annual simple interest so remember with simple interest you don't get interest on top of interest so the others a cd certificate of deposit and a bond fund pay two percent and three percent respectively so that means that the cd pays two percent the bond fund pays three percent the total and annual simple interest from the three investments was three thousand nine hundred dollars the annual simple interest earned from the reit was nine hundred dollars less than three times the amount earned from the bond fund and the cd find how much was invested in each so this is kind of a long problem with a lot of information and when that happens you tend to get lost in the mix i just want you to focus on one thing at a time so the first thing is what are we asked to find we're asked to find how much was invested in each there are three things here there's a reit there's a cd and there's a bond fund and we're not given any information as far as how much was invested in each one so we could use a variable to represent each amount so we could say x let x be equal to the amount invested in the reit again that's a real estate investment trust then we could say y would equal the amount invested in the cd or certificate of deposit then we could say z is equal to the amount invested in the bond fund okay so now we've represented these three amounts with variables so we find out what x is find out what y is we find out what z is and we've answered our problem but when you have three unknowns you've got to have three equations to figure them out okay three unknowns you need three equations four unknowns four equations 17 unknowns 17 equations so on and so forth so let's go back through the problem and see if we can figure out three equations so the first one would come from this james won 86 000 in the local lottery he decided to invest the whole amount so all of it that's basically telling me that 86 000 is going to be involved here and three different investment funds so between the reit the cd and the bond fund there's a total of eighty six thousand dollars in them so that tells me that if i sum x y and z that's going to be 86 000. so for equation 1 x plus y plus z is going to be equal to 86 000. and again that's straight from the problem the amount invested in the reit plus the amount invest in the cd plus the amount invested in the bond fund because he only puts money in those three areas is going to sum to 86 000 all right let's see if we can get another equation all right so the next thing we see we have a reit pay six percent annual simple interest the others a cd and a bond fund paid two percent and three percent respectively let me write that down here real fast so let's put that this one pays six percent this one pays two percent and this one pays three percent okay let's go back up now the next part is where we're going to get our next equation the total and annual simple interest from the three investments was 3 900 okay so let's stop now if i had one investment with simple interest what is the simple interest formula let me do this on a little sidebar over here well the simple interest formula in case you don't know it is the simple interest earned which is i is equal to the principal or amount invested times the rate times the time now in this particular case we're talking about annual so that's one year so if time is one but multiplying by one it doesn't change so you just get rid of this the thing about just principal times rate now the principal or the amount invested in each case is listed with a variable so if i take the reit for example that would be x now the rate i'm given is 6 percent you've got to write that as a decimal so that would be 0.06 so let's rearrange this and put .06 so then times x so i'm not given the simple interest earned for just the reit so this .06 x equals i i don't know what i is for that what i know i4 or the 3900 is this guy the reit plus the cd plus the bond fund so my equation would be .06 x plus the cd page two percent so point zero two y plus the bond fund pays three percent so point zero three z and let me erase all this this would be equal to thirty nine hundred and again this comes from the simple interest formula i have a six percent interest rate times an amount invested of x for my reit so if i multiply these two together i'm going to get the amount of interest that's earned from the reits in one year i add that to the amount of interest that's earned from the cd for one year i add that to the amount of interest that's earned from the bond fund for a year and the total as it's told to me in the problem is three thousand nine hundred dollars so this is my equation two let's see if we can get a third equation so we're told here that the annual simple interest earned from the reit so let's stop what was that well we just saw that we put .06 x for that right point zero six is six percent as a decimal it's multiplied by x which is the amount invested in the reits that would give me the amount of simple interest that the reit earns in a year so that's going to be .06 x so this was or in terms of setting up an equation means it's equal to we have nine hundred dollars less than three times the amount earned from the bond fund and the cd so let's start with what's earned from the bond fund and the cd well the cd and the bond fund we have with these two amounts so we can put .02 y plus .03 z now it's three times this amount so three times this quantity and it's 900 less than that so subtract away 900 this would be my third equation let's go back up and make sure this makes sense so again and this is a little complex the annual simple interest earned from the reits which is .06 x was 900 less than three times the amount earned from the bond fund and the cd so we have 900 less than subtracting away 900 there this is the amount earned from the bond fund and the cd and we have three times that amount all right so that's where that equation came from and now we're ready to roll okay so it might take you a little while to get these three equations when you first start you got to read through that problem very carefully but i can assure you if this is in a textbook or on a test the information is enough to get you the equations you need they set it up that way but sometimes it's not so straightforward what you need to do so how do i want to solve this system in most cases with three equations three variables we eliminate a variable and then we solve a system in two variables and then we come back and we figure out the third so let me just start with equation one in equation two so equation one is x plus y plus z equals eighty six thousand equation two we have point zero six x plus point zero two y plus point zero three z equals thirty nine hundred now to make this easy i can multiply both sides of this equation by a hundred every decimal point would go over to the right by 2. so this would be 6x plus 2y plus 3z equals so in this case it'd be two zeros at the end so 390 000. so we would have 6x plus 2y plus 3z equals 390 000. okay so now if i wanted to use elimination here what i would do is just choose a variable doesn't matter what i choose let's go ahead and just choose y so if i want y and 2y to be opposites what would be easiest to make this negative 2y so let's multiply both sides of the equation by negative 2. so negative 2 times x is negative 2x negative 2 times y is minus 2y negative 2 times e is minus 2z this equals 86 000 times negative 2 that's negative 172 000. let's erase this now we don't need it anymore so i can add the left sides together set this equal to the sum of the right sides so 6x minus 2x is 4x 2y minus 2y this is eliminated 3z minus 2z is z so plus z 390 000 minus 172 000 that's going to give me 218 thousand so this would be equation four let's write this on a new sheet so equation four is four x plus z equals two hundred eighteen thousand all right let's go back up all right now let's work with equation two and three and again we've gotta eliminate the same variable this time so we've gotta eliminate y so for equation two i have six x plus two y plus 3z equals 390 000. for equation three it's a little clustered let's kind of clean it up so point zero six x equals three times point zero .02 is .06 then times y plus 3 times .03 is .09 then times z then minus 900. now again i can clear the decimals by multiplying everything by a hundred so this would go two places to the right that would be six x this goes two places to the right it'd be six y this would be nine z and i'd put two zeros at the end here to give me ninety thousand so i can erase this and let me just write it in this format so i'd have 6x let me subtract away 6y from each side let me subtract away 9z from each side and this will equal negative 90 000. okay so let me erase this and so i'll write 6x minus 6y minus 9z is equal to negative 90 000. all right so if i want to eliminate y the easiest thing i can do since i have negative 6 here and 2 here i can multiply 2 times 3 and get positive 6. positive 6 and negative 6 are opposites so i'll multiply this by 3 and the same thing on this side 3 times 6x is 18x 3 times 2y is 6y 3 times 3z is plus 9z this will be equal to 3 times 390 000 is 1 million 170 000. all right so let's erase this and let me scroll down a little bit get some room going so let's sum the left sides set it equal to the sum of the right sides so these are going to cancel 6x plus 18x is 24x and this equals negative 90 000 plus 170 000 is 1 million eighty thousand all right so let's copy this that'll be equation five let me kind of tighten this down a little bit [Music] so it's pretty easy we're just going to solve this equation 5 for x so we would get x is equal to 45 000. now we're going to plug 45 000 for x in equation four so i would have four times 45 000. plus z is equal to two hundred eighteen thousand four times forty five thousand is a hundred eighty thousand plus z equals two hundred eighteen thousand scroll down get a little room going let's go ahead and subtract 180 000 from each side of the equation [Music] that'll cancel we'll have z is equal to 38 000. all right so now i know x is forty five thousand i know z is thirty eight thousand so we know z is thirty eight thousand x is forty five thousand let's just plug that into equation one so we'd have forty five thousand plus y plus for z thirty eight thousand this equals eighty eighty-six thousand so if i had forty-five thousand and thirty-eight thousand i'd get eighty-three thousand should have y plus eighty-three thousand equals eighty-six thousand go ahead and subtract eighty-three 000 away from each side of the equation y is going to equal 3 000. so y equals 3 000. okay so now i want you to remember what x y and z represent x is the amount invested in the reits y is the amount invested in the cd z is the amount invested in the bond fund so let's take this up we're going to say there was 45 thousand dollars invested in the reit with three thousand dollars invested in the cd and thirty eight thousand dollars invested and the bond fund now again if you check this you think about james invested 86 000 total so is that consistent with what you found well 45 000 plus 38 000 is 83 000 plus another 3 000 is 86 000 so it checks out there now the next thing is the total annual simple interest from the three investments is thirty nine hundred dollars does that work out well from the reit you've got forty five thousand dollars invested at six percent that's twenty seven hundred dollars in interest in the cd it's three thousand at two percent so that's sixty dollars and then if you take the bond fund that's thirty eight thousand at three percent that's one thousand one hundred forty dollars so if you add 2700 plus 60 you would get 2760 plus 1140 you do in fact get 3 900 so it checks out here as well now the last one it says the annual simple interest earned from the reit which is twenty seven hundred dollars was nine hundred dollars less than three times the amount earned from the bond fund and the cd so from the bond fund he got eleven hundred forty dollars from the cd he got sixty so combine those two amounts together you get twelve hundred so again it was nine hundred dollars less than three times the amount of twelve hundred dollars three times twelve hundred would be thirty six hundred thirty six hundred minus nine hundred would be twenty seven hundred so that one checks out as well so a very tedious problem overall takes a little bit time to get into it take some time to set up and take some time to check it but worth it in the end because you've got to see a few of these because if you get them on a test you can't just be you know completely rattled okay you've had some experience now in dealing with one of these really tedious problems so again to just state the answer there was forty five thousand dollars invested in the reit with three thousand dollars invested in the cd and thirty eight thousand dollars invested in the bond fund hello and welcome to algebra two lesson thirty in this video we're going to learn about solving linear systems in two variables using matrix methods so in this lesson i want to talk a little bit about another method used to solve linear systems so so far we've learned how to use graphing substitution and elimination but we're going to push ourselves a little bit further and now talk about matrix methods so as we move higher in math things are going to become more complex matrix methods allow for us to solve very large linear systems using a computer very very quickly so we're not going to cover all of the details related to a matrix but what we will do is give an introduction and show one method for finding a solution using a matrix so let's start with the very basics here so a matrix is an ordered array of numbers and it's just a way to store numerical information so you'll see my brackets here and in the brackets you have two rows okay two rows this would be a row going across so you have two of those and you have three columns so the columns are going down so a matrix is named according to the number of rows and columns it contains it's rows first followed by columns so as an example if i see this matrix here you see that you have one two three rows so three going across so three is the first number and you have one two three four columns so four is the second number so this is a three by four matrix similarly with this guy right here you have one two three rows so it's a three rows there and you have one two columns so this is a three by two matrix so another little kind of definition that you'll come across you have a square matrix and let me highlight that a square matrix has the same number of rows as columns so here i have two rows and two columns so this is a two by two matrix as another example of a square matrix we have three rows here and we have three columns so this is a three by three right you can have anything by anything so you have four by four or 27 by 27 as complex as you want to make that all right so i want to start out today with a very basic example so we have the linear system 2x minus 5y equals negative 14. we have 7x minus 7y equals 14. i have here start with an augmented matrix before we get into that i want you to notice that we have each equation already written in standard form so you want to start by putting ax plus b y is equal to c both equations in this form now once you've done that you can write this augmented matrix what this is is i'm going to take the numerical information only so in other words the coefficient for x is a 2. so i'm going to write a 2. the coefficient for y here is a negative 5. so i'm going to write negative 5. the constant is negative 14. so i'm going to write negative 14 there so that's my first row it represents the numerical information from the first equation of the system my second row is the same thing so i have a 7 for x i have a negative 7 for y and i have a positive 14 as my constant now i'm going to put my brackets here and with the augmented matrix we use a vertical bar to separate the coefficients from the constants so my coefficients are here okay these are the coefficients and these are my constants so i'm just going to put a vertical bar right here so from this information here we can get a solution for x and y now in order to do that we're going to use these elementary row operations and there's three of them so let's go ahead and look at that all right so i have here we can manipulate our matrix using row operations these produce matrices that lead to linear systems with the same solution set as the original system all right so here's our row operations the first thing is that we can interchange any two rows that makes sense because if i had an equation and another equation it doesn't matter which one i write on top right i can switch that around as much as i want the second thing is we can multiply any row by a non-zero number well that makes sense i can multiply both sides of an equation by a non-zero number i could divide both sides of an equation by non-zero number and the equation would contain the same solution set so the last row operation we're going to talk about row operation 3 is a little bit challenging to understand so we have here that we can multiply a row by a real number and add this to the corresponding elements of any other row so when people first see this they say well what makes this legal well essentially we're using the same basic principles that we use when we use the elimination method so we're going to quickly look at a different example and i want to just cover why this would work so let me just take an aside from that problem that we're currently working on so just put that to the side take out a fresh sheet of paper and i want you to work through this very basic example with me so if we have 2x plus 3y equals 11. and we have x plus 2y equals 7. so we can easily solve this using elimination at this point so i want to do elimination and i want to do the matrix method i'm not going to go all the way through it i'm just going to go to the part where it's obvious what the solution is so if i wanted to solve this with elimination what would i do let's go ahead and say this is equation 1 and this is equation 2. well i know that i can eliminate x if this was negative 2 here and the way i can do that is i can multiply both sides of this equation by negative 2. so if i did that what i would have is a negative 2x plus negative 4y equals negative 14. so i could write that i have 2x plus 3y equals 11 and i have this other one negative 2x minus 4y is equal to negative 14. okay so now we all know at this point that the x variable would be eliminated so i could add the two left sides of the equation set that equal to the sum of the right sides of the equation so 2x minus 2x would be zero x is eliminated that's gone 3y minus 4y is negative y and this is equal to let me kind of move this down a little bit 11 minus 14 which would be negative 3. so we all know that y is 3. so y is 3. now i can plug a 3 back in here and figure out what x is 2 times 3 is 6 so i would have x plus 6 equals 7. we know x would be 1 right x is equal to 1. now i want to just show you this real quick using a matrix so i can relate the steps so the first thing is we want an augmented matrix so i just told you we take the numerical information again these equations are written in standard form ax plus b y equals c so i have my coefficient for x my coefficient for y and my constant written in that order i'm going to write in the same order here so this is understood to be a 1 so just a 1 a 2 and a 7. i've got a vertical bar that separates the coefficients from the constants now using the numerical information only i can do the same thing that i did here i want to think about how could i get a 0 here so that the coefficient for x is a zero and x is essentially eliminated i would have some number times y equals some value which is what i had here so i'm going to show you that i can use the elementary row operations i can say if i multiplied row 2 here by negative 2 which is exactly what i did here remember i had the second equation i multiplied it by negative 2 and i produced negative 2x minus 4y equals negative 14. so if i multiplied row 2 by negative 2 1 times negative 2 is negative 2. 2 times negative 2 is negative 4. 7 times negative 2 is negative 14. and then i added those elements to row one so negative two plus two would be what that would be zero so i'd have a zero here then negative four plus three would be negative 1 then negative 14 plus 11 would be negative 3. now i don't need to change this row here because i already have what i need there so i can just leave this as 1 2 and 7. okay but it wouldn't matter if i wrote negative 2 negative 4 and negative 14 there but for the purposes of working with a matrix we're not going to do that we're just going to leave this row unchanged now i have the same information here that i have here i know that the coefficient of x is zero so x is eliminated i know that the coefficient for y is negative one so basically i have negative one y is equal to negative three so from this information here i can solve for y okay and then i can go back and figure out what x is now this isn't exactly how we're going to solve using a matrix but it's just an example of how we can manipulate the numerical information to get a solution so i want to return now to the official procedure again this is just a little side note you can you can kind of get rid of this now and let's go back to the original problem what our goal is actually going to be we have this term known as row echelon form and you're also going to hear reduced row echelon form i'm going to cover both of them but i want to focus on row echelon form for now so what you want is ones down the diagonal so starting at the top left you have a one then going this way you'd have another one now beneath the ones you're gonna put a zero okay so this has to be a zero now you're going to have real numbers above and to the side so this is a b and c just representing real numbers if i have something in row echelon form and again these equations are ax plus b y equals c well i know that i have 1x plus a y is equal to b in this case i'm just using generic information and i know here that i have 0x so x is eliminated plus 1y or just y equals c so from this if i know y equals c i can plug in for y here and i can find out what x is again we're just taking the numerical information and finding our solution so again this is row echelon form so i want you to write that down and then once we get this down i'm going to show you reduced row echelon form which isn't any more complicated all right so i want to return now to the system that we were starting with so again it's 2x minus 5y equals negative 14. it's 7x minus 7y equals 14. so let's copy this information and we'll copy this and we'll bring it down to a new sheet of paper so again we have 2x minus 5y equals negative 14. we'll call this equation 1 and then for equation 2 we'll have 7x minus 7y equals 14. all right so again write the augmented matrix just the numerical information that's all we want so we want 2 we want negative 5 and we want negative 14. we want 7 we want negative 7 and we want 14. a vertical bar will go right here to separate the coefficients from the constants and our goal is what again i want ones down the diagonal so i want this to be a one and i want this to be a one so down the diagonal okay that's going to be a one beneath the diagonal i want a zero and then these numbers here would be real numbers so how do we achieve that goal well again we're going to use our row operations so how could i make 2 a 1 using the row operations well i could simply multiply every number in this row by one-half two times one-half equals one so if i do that what would i get two times one-half again is one negative five times one-half is negative five-halves and negative 14 times one-half is going to be negative seven now for this one i have seven negative seven and fourteen i haven't done anything there okay so now that we have this as a 1 we want to make 7 into a 0. it's best to work a column at a time so starting here and working down if i wanted 7 to be a 0 what could i do well i could multiply 7 by 0 but that's not going to work right we can't multiply both sides of an equation by 0 and maintain the same solution that's a violation so what i have to do is i have to multiply this row here by something and add it to this to get this to be 0. so first and foremost what can i add to 7 to make it 0 well obviously negative 7. so that means i need to multiply 1 by some value and it needs to give me negative 7 so that when i add it to 7 i get a value of 0. so 1 times what would give me negative 7 1 times negative 7. so that means i'm going to multiply row 1 by negative 7. and when i do that what am i going to get well 1 times negative 7 is negative 7. negative 5 halves negative 5 halves times negative 7 would be 35 halves and then negative 7 times negative 7 is going to be 49. so i'm going to take that information and i'm going to add it to each element here so this row 1 i'm just going to keep it the same remember i don't have to change it negative 7 plus 7 is 0. so 35 halves plus negative 7 so plus negative seven i multiply this by two over two that would be negative 14 over two and 35 minus 14 would give me 21. so this is 21 halves and then lastly i have 49 plus 14. so 49 plus 14 would give me 63. okay so now let's scroll down the next thing i want to do since i have a 1 here and a 0 here remember i want ones down the diagonal so how can i make this into a 1 well that's easy 21 halves times the reciprocal which would be 2 over 21 would equal 1. so just multiply row 2 by 2 over 21. so multiplying 0 by anything is 0. so let's just start out by saying this is 1 this is negative 5 halves this is negative 7. so that's all the same 0 again times 2 over 21 is still 0. 2 over 21 times 21 over 2 is 1. and then 63 times 2 over 21 well 63 divided by 21 is 3. so this cancels with this and gives me 3. 3 times 2 is 6. now what do i have here well i have row echelon form i have ones down the diagonal i have a 0 below and i have real numbers here here and here so once i'm in that format i have enough information to get a solution remember this is ax plus b y equals c so forget about this top row the bottom row is basically 1y is equal to 6. so i know that y here is equal to 6. now i can take this information and plug it in here for this guy right here i have that 1x or just x plus negative 5 halves times y y is 6 is equal to negative 7. so i know if i do a little bit of math here 6 divided by 2 is 3 negative 5 times 3 is negative 15. so we would have x minus 15 is equal to negative 7. we would add 15 to each side of the equation and we would have that x is equal to 8. so x equals 8 y equals 6. so i'm going to write that x equals 8 y equals 6 or the ordered pair 8 comma 6. now let's check this real quick i'm not going to check any in the future because of the interest of time but i just want to show you that this works so we have 2 times put in an 8 for x minus 5 times plug in a 6 for y this equals negative 14. 2 times 8 is 16 minus 5 times 6 is 30 this should equal negative 14 and it does 16 minus 30 is negative 14 so negative 14 equals negative 14 so it works out here for the next equation of the system we have one we have that 7x so 7 times 8 minus 7y y is 6 equals 14. 7 times 8 is 56 minus 7 times 6 is 42. does that equal 14 yes it does you have 14 equals 14 so it works out here as well so i know you might be a little a little confused on this it's it's okay it's something new it's something that you just have to practice but hopefully you can see the concept and understand what we're doing we're basically just taking the numerical information and we're using the elimination method but we're doing it in a more compact way so let's take a look at the next example so we have negative 9x minus 9y equals negative 27. we have negative 5x minus 4y equals negative 20. so again i'm going to take the numerical information here so a negative 9 another negative 9 and a negative 27. i have a negative 5 a negative 4 and a negative 20. and again you want to make sure that these equations are written in standard form ax plus b y equals c both of them on top of each other verify that the x's are in the same place the y's are in the same place the coefficients are in the same place so that your information is correct so you have your vertical bar okay that separates the coefficients from the constants and so we have our augmented matrix all right so remember we want ones down the diagonal okay so we want this to be a one and we want this to be a one and we want a zero below and then these can be real numbers all right so let's get started how would i make negative 9 into a 1 using the row operations well that's easy i can always multiply negative 9 by the reciprocal which is negative 1 9. if i multiply negative 9 by negative 1 9 i know that this would be 1. if i multiply negative 9 by negative 1 9 again i'm going to get a 1. if i multiply negative 27 by negative 1 9 this cancels with this and i get a 3 here so negative 3 times negative 1 would be 3. and then now i have this row negative 5 negative 4 and negative 20. all right so now i want this next guy here negative 5 remember i work a column at a time so i want this to be a 0. so what can i do well again i want to add something to negative 5 to give me 0. well i know negative 5 plus 5 would give me 0. so in order to do that i would multiply the first row here by 5 right because 1 times 5 would give me 5. so i multiply the first row 1 times 5 would equal 5 1 times 5 would again equal 5 and then 3 times 5 would be 15. so i'm going to take these results here and add them so my first row i'm just going to keep it the same 1 1 and 3. so 5 plus negative 5 is 0. 5 plus negative 4 is 1 and then 15 plus negative 20 is negative 5. now it just so happens that this is an easy example and i've already got what i want which is row echelon form i've got ones down the diagonal i've got a 0 over here and i've got real numbers here so again once we have it in this format we can use substitution to find our solution so i know that from this bottom row here i would have 1y or just y is equal to negative 5. in the top part here i have that x plus y right because i have 1x plus 1y is equal to 3. so just plug in a negative 5 here and you'd find that x minus 5 equals 3 add 5 to each side of the equation you're going to get that x is equal to 8. so the solution for this system would be that x is equal to 8 y is equal to negative 5. right or as an ordered pair you could say 8 comma negative 5. and you can pause the video go back to the system and check it and verify that it's the correct solution i've already done that so i know that 8 comma negative 5 is the solution all right let's take a look at one more example and what i'm going to do in this example i'm going to show you reduced row echelon form as well as row echelon form okay and i'm explaining the difference between the two so we have 5x minus 6y equals 24. we have 2x plus 7y equals negative 28. so again ax plus by equals c so you make sure your equations are in that format again a lot of textbook examples are going to jamble things up so it's not in that format you've got to write it that way so your information is consistent all right so i have a 5 a negative 6 and a 24. the information here i have a 2 i have a 7 and i have a negative 28. so i have my vertical bar that separates the 2 right the coefficients from the constants and now we get to work so again we're going to talk about row echelon form first so that's ones down the diagonal a zero over here and real numbers here so let's start with the first column here how can i make 5 into a 1 well i can multiply row 1 by 1 5th right 5 times its reciprocal of 1 5th gives us 1. so if i multiply five times a fifth i get one if i multiply negative six by a fifth i get negative six fifths if i multiply 24 by a fifth i get 24 fifths and then row 2 stays the same for now so 2 7 and negative 28 okay so let's scroll down now again working at a column at a time how can i make 2 into a 0 well i think about row 1 here to get 2 into a 0 i've got to add 2 to its opposite which is negative 2. so what can i multiply 1 by to get a result of negative two well of course just negative two so if i multiply row one by negative two one times negative two gives me negative two negative six fifths times negative two i know the negatives would cancel you'd basically have 12 fifths and then 24 fifths 24 fifths times negative 2 would be negative 48 fifths so this is negative 48 fifths so for row one i'm keeping that the same so a one a negative six fifths and then a 24 fifths for row two i'm gonna take the result that i got and i'm gonna add so negative two plus two is zero i'm gonna have twelve fifths plus seven so twelve fifths plus seven i could write that as 35 fifths 12 plus 35 is 47 and then the common denominator of 5 is what that's going to be over so that's 47 fifths and then if we think about the next result you'd have negative 48 fifths plus negative 28. so negative 48 fifths plus negative 28 times 5 is negative 140 so i'd have negative 140 over 5. so if i add these two together negative 48 plus negative 140 that would give me negative 188 so i'd have negative 188 over 5. all right so the next thing i want to do is make this a 1. i want ones down the diagonal 0 below i've already got this part i just need this part let's scroll down here and it's very easy i just multiply row 2 by the reciprocal of 47 fifths which is 5 47. so let's keep row one the same so row one stays the same a one a negative six fifths and a twenty-four fifths for row two again if i multiply five over forty-seven by zero i get zero five over 47 times 47 over five is one and then five over 47 times negative 188 over five well i know the fives would cancel and negative 188 divided by 47 is actually negative 4. so at this point i have enough information to get a solution i know that y is equal to negative 4 right so if y equals negative 4 i can plug in for y in this part so i would have x minus 6 fifths y is equal to 24 fifths scroll down a little bit and again y is negative 4. so what we're going to get is that x minus 6 fifths times negative 4 is equal to 24 fifths so we know negative times negative is positive so you get x plus 6 times 4 is 24 so fifths equals 24 fifths so we know x has to be zero right if i subtract 24 fifths away from each side of the equation we would get x is equal to 0. so y equals negative 4 x equals 0. so you can pause the video check and make sure that's the correct solution the ordered pair 0 comma negative 4. now i want to talk about reduced row echelon form so we're going to go back to the row echelon diagram i gave you all right so we know at this point that row echelon form is ones down the diagonal a zero below and you have real numbers a b and c so with this form we have the information to go back and plug into the system and get our solution but with reduced row echelon form we don't have to go back and plug things in we have our solutions let me show you that this is called reduced row echelon form so i would have ones down the diagonal still so this is a one and this is a one but everything in terms of the coefficients other than down the diagonal is going to be a zero so it's a zero here and a zero here and then you have two real numbers here on the right side of your vertical bar so let's just say this is a and this is b so with this guy right here i don't need to use substitution because i know that one x or just x is equal to a so x equals a and then one y or just y is equal to b so i have my solution in this form without going back to plug things in so you might ask the question which one is faster well generally reduced row echelon form is going to be a little bit quicker because we don't have to go back and plug things in right we have x equals something y equals something so i want to just show you on that last example so this is where we kind of stopped and use substitution so let's pretend that we continued so let me just copy this and bring it to a fresh sheet so we already know y equals negative 4x equals zero in order to get that information i want to make this part right here this part right here into a zero i already know that y is negative four i've got that information but i wanna know x is equal to one so how can i make this into a zero well you've got to think about what i could add to negative six fifths to get to zero negative six fifths plus what this is easy six fifths equals zero well since this is a one i just need to multiply this by six fifths so if i multiply row two by six fifths and add it to row one i'm gonna get what i need so let's set this up here so let's set up row two is zero one and negative four so it's already there zero times six-fifths is zero zero plus one is one one times six-fifths is six-fifths six-fifths plus negative six-fifths is zero negative four times six-fifths so negative four times six-fifths negative four times six is going to give me negative 24 so i'd have negative 24 fifths now negative 24 fifths plus 24 fifths is zero so what did i find here now i found that x is equal to 0 which is what it is y is equal to negative 4 which is what it is so with this form i don't need to go back and plug things in okay so in most cases it's just a little bit quicker all right but i want you to understand both forms because you might get tested on so you have row echelon form which is this and you have reduced row echelon form which is this hello and welcome to algebra 2 lesson 31 in this video we're going to learn about solving linear systems in three variables using matrix methods so in the last lesson we learned how to use an augmented matrix to solve a linear system in two variables in this lesson we're going to push ourselves one step further and we're going to work with linear systems in three variables so i want you to recall that our goal is to transform our augmented matrix which is nothing more than the numerical information from the system into a matrix that looks like either this one or this one so the top one is known as row echelon form so we have ones down the diagonal and we have zeros below so you might say well how do i go about getting a solution from this information well i want you to recall that you want to write each equation in standard form before you input the numerical information so i usually would say for a linear system in three variables we want to do ax plus b y plus c z equals d where a is the coefficient for x b is the coefficient for y c is the coefficient for z and d is just a constant so we have our column with coefficients for x a column with coefficients for y a column with coefficients for z and then constants okay so very easy to understand where the information goes now if i know that this is 0 and this is zero that means that x and y are eliminated in this row so then i have one z or just z is equal to some real number f using that information i can go plug into this row here because this corresponds to x being eliminated and only having y and z if i know what z is then it's a piece of cake i can just solve for y once i have y and z i can go into the top row and figure out what is x now if you don't want to go through all that work with the substitution you can use reduced row echelon form with this form you don't need to do anything other than look at your matrix to get the solution for the system you have ones down the diagonal you have zeros above and below so i can simply look at this and say okay well y and z are eliminated here x is some real number c y is some real number e z is some real number f so in each row you'd have a solution for one of the variables so again with this form you don't have to go back and do any messy you know substitution or anything like that you just look at your matrix and you simply have the answer so in this video and all the videos forward we're going to use reduced row echelon form just because it's a little bit cleaner all right so let's go down and we're going to talk about the elementary row operations one more time so the first one is that we can interchange any two rows that should make sense to you if i'm working with a linear system it doesn't matter which equation is on top of the other the second one we can multiply any row by a non-zero number this is the same thing as if i multiplied both sides of the equation by a non-zero number it's the same thing so we can do that with a row as well so the last row operation we're going to talk about row operation 3 is a little bit challenging to understand so we have here that we can multiply a row by a real number and add this to the corresponding elements of any other row so when people first see this they say well what makes this legal well essentially we're using the same basic principles that we use when we use the elimination method so let's take a look at the first example now so you'll notice how i have everything already written for you in standard form so you have ax plus b y plus c z equals d where a b c and d just represent real numbers a is the coefficient for x b is the coefficient for y c is the coefficient for z and d is just some constant everything is lined up and it's very important that you do that so that you put your information in the matrix correctly so we're going to write an augmented matrix so i'm just going to go row by row so in the first row i have negative 1 as my coefficient for x so i'll write a negative 1. then i have 2 as my coefficient for y i have a negative 1 as my coefficient for z and then i have a negative 9 as my constant and i just match that so i have a negative 3 i have a 2 i have a 4 and i have a negative 14. then i have a negative 4 i have a 3 i have a 6 and i have a negative 21. now with an augmented matrix what we want to do is put a vertical bar that separates the coefficients from the constants now i'm going to put my two brackets one on each side and i have my augmented matrix but notice how this works in this first column the column all the way to the left we see that that's all the coefficients for the x variable then moving one to the right all the coefficients for the y variable then all the coefficients for the z variable and all the constants and that's made possible because again everything is in standard form and it's very important that you do that otherwise you're not going to get the right answer all right so once we've done that what we want to do is work one column at a time so i start all the way on the left and i concentrate on this first guy right here remember we want ones going down the diagonal so i want this to be a 1. what row operation can i use to produce that well if i multiplied negative 1 times negative 1 that would give me 1. but i can't just multiply this by negative 1. i've got to do it to the whole row to make it legal so i'm going to do negative 1 times negative 1 that's 1. negative 1 times 2 that's negative 2 negative 1 times negative 1 is 1 negative 1 times negative 9 is 9. now all the other rows would be the same so this is negative 3 this is 2 this is 4 this is negative 14. this is negative four this is three this is six this is negative 21. all right let's scroll down get a little room going so now the next thing if i'm working a column at a time remember one's down the diagonal i want zeros beneath and above depending on what you want to do i'm going to use reduced row echelon form to make it a little easier for us so i want zeros beneath and above the diagonal so i want this to be a zero so what row operation can i use to achieve that well now we're going to think back to our elimination method remember if we were trying to eliminate a variable and we didn't have opposite coefficients already on that variable what we had to do is we had to create that so we would look at transforming one or both of the equations so that one pair of variable terms would cancel in this case all i need to do is look at row one if i look at row one i could multiply row one by three and then i can add the result to row two because three plus negative 3 would give me 0. so if i multiply 1 by 3 i would get 3. so that's where that comes from so multiply row 1 by 3. now i don't have to actually change row 1. i can keep that the same because the information would be the same so i already have my 1 here i don't want to change that i just want to use the information to make this a 0. so multiply row 1 by 3 so i would have 1 times 3 which gives me 3 i would have negative 2 times 3 which gives me negative 6 i would have 1 times 3 which gives me 3 and i'm going to have 9 times 3 which gives me 27. so rule 1 itself again is going to stay the same so 1 negative 2 1 and 9. that doesn't change for row 2 i'm adding the result from the multiplication to what's previously there so negative 3 plus 3 is 0. 2 plus negative 6 is negative 4 4 plus 3 is 7 negative 14 plus 27 is 13. now row 3 just remains the same so we have negative 4 we have 3 we have 6 and we have negative 21. all right now if i'm continuing in this column remember i want this to be a 0 as well so i'm going to do basically the same thing so how can i eliminate this how can i make this a zero well negative four plus four is zero so if i multiply row one by four four plus negative four would give me zero so row one times four we'd have one times four which equals 4 we'd have negative 2 times 4 which equals negative 8 we'd have 1 times 4 which equals 4 and we'd have 9 times 4 which equals 36. so let's go ahead and just replace this row here and i'll leave these other ones intact so negative four plus four would be zero so let's erase this and put zero three plus negative eight would be negative five six plus 4 would give me 10 and then negative 21 plus 36 would give me 15. so this is 15. all right so let's erase this now this column is done so i want to move one column to the right now you might try to start out right here but typically what i like to do if you get ones down the diagonal it's easier to work from there so i started getting a 1 here then i went down if i go this way to this column i want a 1 here it's very easy to make something into a 1 you just multiply by the reciprocal right so negative 4 times negative 1 4 would be 1 and we've got to do it to the whole row so 0 times negative 1 4 0 times negative 1 4 is 0. so i can erase that negative 4 times negative 1 4 is 1. seven times negative one fourth is negative seven fourths so this is negative seven fourths thirteen times negative one fourth is negative thirteen fourths so now that i have a one here i want to get a 0 below and above so that's pretty easy to do if i multiply row 2 by 2 and add the result to row 1 i'm going to make this a 0. right negative 2 plus 2 is zero so zero times two would give me zero one times two would give me two then negative seven fourths times two four cancels with two and gives me two so this equals negative seven halves and then negative thirteen fourths times two again this cancels with this and gives me two this is negative thirteen halves so in the first case i'd have one plus zero which is one so this stays unchanged then i would have two plus negative two which gives me zero so this is zero now then i would have negative seven halves plus one so if i took and erase this and give myself some room so negative 7 halves plus 1 i could write 2 over 2 if i wanted to this would give me negative 5 halves so this will be negative 5 halves and then lastly we will have 9 9 plus negative 13 halves so i'll write 9 is 18 halves minus 13 halves so this will be equal to 5 halves so this will be 5 halves and now let's work on this right here this negative 5 in this column 2. so i can multiply again row 2 by 5 now 5 plus negative 5 would be 0. if i do that when i add the result i'll get 0. so 0 times 5 would be 0. 1 times 5 would be 5 negative 7 4 times 5 would be negative 35 fourths and then negative 13 4 times 5 would be negative 65 4. so we would have 0 plus 0 which would give me 0 then we'd have 5 plus negative 5 which would give me 0. so those two i don't need anymore then next i'd have negative 35 4 plus 10. so negative 35 4 plus for 10 i'm going to write 40 over 4. so negative 35 plus 40 would be 5. so you'd have 5 4 here so 5 4. and then lastly let me erase all this get some room going you'd have negative 65 fourths plus 15. so negative 65 fourths plus for 15 i'm going to write that as 60 over 4 negative 65 plus 60 is negative 5 so this would be negative 5 4. this would be negative 5 4. all right so let's continue so this column's done this column's done so i want to get a one right here again it's easiest if you just do the diagonals with a one first and then make your zeros so for five fourths how do i get that into a one well i multiply by the reciprocal so that's four-fifths zero times four-fifths is zero zero times four-fifths is zero five-fourths times four-fifths is one negative five-fourths times four-fifths is negative one so although we know at this point that z equals negative one we want to continue and get zeros here and here so that we don't need to go back and use substitution so how do i go about getting a zero here well if i multiplied one by seven fourths i could add seven fourths to negative seven fourths and i would have a zero there so if i multiply row three by seven fourths zero times seven fourths equals zero then zero times seven fourths equals zero then one times seven fourths equals seven fourths and then negative one times seven fourths equals negative seven fourths so i'm going to add the results to row two so zero plus zero is zero you can get rid of that zero plus one is one get rid of that seven fourths plus negative seven fourths is zero so that's gone and then i'd have negative seven fourths so negative seven fourths plus negative thirteen fourths negative seven plus negative thirteen would be negative twenty this is over four so this equals negative 5. so this right here is negative 5. so now we know that y okay y because x and z the coefficients are 0 so this would be 1y or just y would be equal to negative 5. so i know z is negative 1 y is negative 5. now we just need to figure out what x is so let's erase all this so my last step here is just to figure out how to get a 0 right here and all i need to do is multiply 1 by 5 halves and add that result to negative 5 halves and i would have a 0 right there and i'm done so i'm going to multiply row 3 by 5 halves 0 times 5 halves is 0. 0 times 5 halves is zero one times five halves is five halves five halves plus negative five halves is zero so this would be zero negative one times five halves is negative 5 halves so this would give me negative 5 halves and i would add that to 5 halves here that would give me 0 as well so what we have now is reduced row echelon form i have ones down the diagonal i have zeros below and zero's above and why this is convenient if i think about this again the information is written as ax plus b y plus c z equals d so i know this is the coefficient for x so 1x plus this is the coefficient for y zero y plus this is the coefficient for z zero z equals then i have a zero so essentially these are gone and i basically have one x or just x equals zero so x equals zero then i know that y which again this coefficient in the middle is the coefficient for y is equal to negative 5 and z is equal to what this is the coefficient for z so 1z equals negative 1. so z equals negative 1. so as an ordered triple if you want to write it that way your solution here is 0 comma negative 5 comma negative 1. it's x y z all right so let me check this first result i'm not going to check any of the ones after this but i do want to check this one all right so for the first equation i'm going to plug in a 0 for x so i'd have negative 0 which is just 0. so don't even worry about that then i'd have 2y so y is negative 5. 2 times negative 5 is negative 10 minus z z is negative one so minus a negative one is plus one this should equal negative nine and it does so it checks out here in the second equation negative three times zero is zero so this would be gone you'd have two y y is negative 5. so that'd be negative 10 plus 4z z is negative 1. so 4 times negative 1 is negative 4. so this equals negative 14 and it does for the final equation we have negative 4x again negative 4 times 0 would be 0. so this would be gone you'd have 3y 3 times negative 5 is negative 15. so this would be negative 15 plus 6z z is negative 1 so 6 times negative 1 is negative 6. so negative 15 plus negative 6 equals negative 21 yes that's true so it checks out in every equation so we verify that this is the solution for the system x equals 0 y equals negative 5 z equals negative 1. all right let's take a look at the next one so we have negative four x minus four y minus four z equals twelve we have negative three x minus three y minus two z equals zero we have negative x minus y minus three z equals twenty six i have everything written in standard form it's ax plus b y plus c z equals d so let's write our information into the augmented matrix so we'd have negative 4 we'd have negative 4 we'd have negative 4 and we'd have 12. we'd have negative 3 negative 3 negative 2 and then 0. we'd have negative 1 negative 1 negative 3 and you'd have 26. all right remember this vertical bar here just separates the coefficients from the constants so we've set up our augmented matrix so what do we do now remember we want ones down the diagonal we want zeros below and above so how do i get negative four changed into a one well i multiply by the reciprocal the reciprocal of negative 4 is negative 1 4. so if i just multiply row 1 by negative 1 4 this would be a 1 this would be a 1 this would be a 1. so i can just erase all of these and just put ones and then this would be what 12 times negative 1 4 would be negative 3. next i want to make this into a 0. so if i want this to be a 0 what can i do well i can multiply row one by three because if this was multiplied by three i'd get three three plus negative three would be zero so if i multiply row one by three i'd have one times three which equals three one times three which equals three one times three which equals three and negative three times three which equals negative nine so i add the results three plus negative three is zero then negative three plus three is zero then 3 plus negative 2 is 1 then negative 9 plus 0 is negative 9. all right so what would i do next well now i want to make this into a 0. so i'm going to multiply row 1 by positive 1 and add it to row 3. so 1 times 1 is 1 1 times 1 is 1 1 times 1 is 1 and negative three times one is negative three so one plus negative one is zero negative one plus one is zero negative three plus one is negative two and negative 3 plus 26 is 23. now before we go any further and waste any more time on this problem i want to point out the fact that this right here is a zero that should signal to you that you have a problem already right now you have one z equals negative nine so z equals negative nine according to row two if i look at row three i have negative 2 z equals 23. if i plugged in a negative 9 here what would i get i'd get negative 2 times negative 9 equals 23 i would get 18 equals 23 which is false okay i end up with a nonsensical statement so this is an inconsistent system when you start to get information that doesn't make sense just stop okay stop do what i just did there you can see obviously this isn't going to work you don't need to go any further now if you wanted to really prove this to yourself you could multiply row 2 by 2 and you could add to row 3. zero times two is zero zero plus zero is zero so this would stay the same zero times two is zero zero plus zero is zero so this would stay the same one times two is two two plus negative two is zero so this would be a zero and then negative 9 times 2 is negative 18 negative 18 plus 23 would give me 5. so what does this tell me here 0x plus 0y plus 0z so basically 0 is equal to the constant 5. that's false okay again you start seeing stuff like this you can stop and say does this make any sense plug in what you know so far and you can obviously see that you get a nonsensical statement and so there is no solution there is no solution [Music] all right let's take a look at one more so we have x minus 3y plus 6z equals 11. we have 6x plus 5y plus 7z equals 13. we have 3x minus 2y minus z equals negative 34. so what you want to do again everything is already in standard form just put the information into the augmented matrix so we'll have a one a negative three a six and an eleven we'll have a six a five a seven and a thirteen we'll have a 3 a negative 2 a negative 1 and a negative 34. put a vertical bar here and then brackets on this side and on this side all right so let's scroll down a little bit now again i want to start out in this first column here and this is already a one so i don't need to do anything remember i want one's down the diagonal that's already a one so i move to the six how can i make that six into a zero well six plus negative six would be zero and i can multiply this one right here by negative six to get negative six so if i multiply row one by negative six and add the result to row two i'll have a zero here so i'll have one times negative 6 that equals negative 6. i'll have negative 3 times negative 6 which would be 18. i'll have 6 times negative 6 which would be negative 36 and i'll have 11 times negative 6 which would be negative 66. so i'm going to add 6 plus negative 6 is 0. so this is 0. 5 plus 18 is 23 negative 36 plus 7 is negative 29 and then negative 66 plus 13 is negative 53. all right so let's erase this so we continue working in this column so how can i get this three to be a zero well i could multiply one times negative three that would give me negative three negative three plus three would be zero so if i multiply row one by negative three it would look like this one times negative three is negative three negative three times negative three is nine six times negative three is negative eighteen and eleven times negative three is negative thirty-three so we add the results to row three three plus negative three is zero nine plus negative two is going to be seven negative eighteen plus negative one is negative nineteen and then negative thirty-four plus negative 33 is going to be negative 67. this is negative 67 here let me erase this all right so the first column is done now i want to go to the second column and again i always want to get a one first so one's down the diagonal that just makes things easier so i can multiply row 2 by 1 over 23 or the reciprocal of 23 to make this a 1. 0 times 1 over 23 is 0. 23 times 1 over 23 is 1. negative 29 times 1 over 23 would just be negative 29 over 23 negative 53 times 1 over 23 would just be negative 53 over 23. okay so now i want to zero above and below so it doesn't matter what you start with let's go ahead and just do the one above first so i would multiply 1 times 3 to get 3 and add 3 to negative 3 to get a 0. so i'm going to multiply row 2 by 3 add the result to row 1. so row 2 multiplied by 3 zero times three equals zero one times three equals three negative 29 over 23 times three is equal to negative 87 over 23 and then negative 53 over 23 times 3 is going to be equal to negative 159 over 23. all right so we're going to add these results to row one so zero plus one is one so i don't need this anymore i can erase this three plus negative three is zero i'm gonna erase this negative 87 over 23 plus 6. so negative 87 over 23 plus what's 6 times 23 well that give me 138 so 138 over 23 and if i did negative 87 plus 138 that would give me 51. so this is 51 over 23. so 51 over 23. let's erase this now lastly we want to do 11 plus negative 159 over 23. so negative 159 over 23 plus 11 times 23 is 253 so 253 over 23. so 253 minus 159 is 94. so this would be 94 94 over 23. and now i want a 0 here so i can multiply row 2 by negative 7 add the result to row 3. so 0 times negative 7 is 0 1 times negative 7 is negative 7 negative 29 over 23 times negative 7 would be 1. we know it's positive and 29 times 7 would be 203 so this would be 203 over 23. and then we look at negative 53 over 23 multiplied by negative 7 again we know it's positive 7 times 53 is 371 so it'd be 371 over 23. all right so now i want to do my addition 0 plus 0 is 0 negative 7 plus 7 is 0 negative 19 plus 203 over 23. so if i get a common denominator here negative 19 times 23 is negative 437 so this would be negative 437 over 23 negative 437 plus 203 would be negative 234. so you would have negative 234 over 23. all right so let's erase these and i've got one more to do so i want to do negative 67 plus 371 over 23. negative 67 times 23 would give me negative 1541 and then this is over 23. so negative 1541 plus 371 would give me negative 1170 over 23. column 1 and column 2 are done the next thing i want to do is work on this column right here again it's always easiest to go down the diagonal and get a 1 here first to get a 1 i would multiply by the reciprocal of negative 234 over 23. multiplying by 0 leaves both of these unchanged don't have to worry about that if i multiply this guy by the reciprocal so negative 23 over 234 we know that this would cancel with this and this with this so everything cancels and just becomes a 1. so that's a 1. now the next one negative 1 170 over 23 multiplied by that negative 23 over 234 it's obvious that the negatives cancel it's obvious that the 23s would cancel but what is 1170 divided by 234 well that's going to give me 5. so let's erase this and put a 5 here so we know what z is at this point z is going to be equal to 5. but let's make this a 0 and this is 0 so we can find out what x and y are so to make this a 0 i can multiply row 3 by 29 over 23 and add it to row 2. so i know that multiplication by 0 gives me 0 and adding 0 doesn't change anything so this one and this one would be unaffected i only really have to do it for these two so i would do one times 29 over 23 this equals 29 over 23 and if i had 29 over 23 with negative 29 over 23 i would get 0. that's easy the next thing i would want to do is multiply 5 times 29 over 23 and that would give me 145 over 23. so this right here i would add so 145 over 23 145 over 23 plus negative 53 over 23 what's that going to give me well it's going to give me 92 over 23 which is equal to 4. so this is a 4. so now i know that z is 5 y is 4. so let's continue to figure out what x is let's erase this so how can i make this into a 0 well i can multiply row 3 by negative 51 over 23 add the result to row 1. all right so i already know if i multiplied row 3 here by negative 51 over 23 that the result from this would be 0 and this would be 0. if i add 0 to a number it remains unchanged so this one and this one will be unaffected i would take 1 and multiply it by negative 51 over 23 and get negative 51 over 23 and i would multiply 5 by negative 51 over 23 and i would get negative 255 over 23. now negative 51 over 23 plus 51 over 23 that's zero that's pretty easy so this is gone i just need to figure this one out so i'd have 94 over 23 plus negative 255 over 23. so this equals negative 161 over 23. so if i divide negative 161 by 23 i will get negative 7. so we're finally done and i know it takes a long time especially when you're new to this but once you kind of get good you can kind of skip a few things here and there some of the calculations like what i just did in the last one you can do that over and over again and it shortens your time this really is faster than if you're working with the traditional method so now that this is in reduced row echelon form i know that one x or x is equal to negative seven so x is equal to negative seven i know one y or y is equal to 4 and i know 1z or z is equal to 5. so as an ordered triple this is negative 7 comma 4 comma 5. so that's your solution for the system again if you want to stop the video at this point check it in the original system you'll verify that again x is negative seven y is four and z is five hello and welcome to algebra 2 lesson 32 in this video we're going to learn about solving systems of linear inequalities so now that we've finished talking about solving systems of linear equations we talked about two variables three variables we learned about graphing substitution elimination and matrix methods i want to revisit something that we talked about in algebra one it's basically the same thing but i want to make sure that you understand it before we get into college algebra so when we talk about a system of linear inequalities it looks like this we have x plus y is greater than negative two and six x plus y is less than three the solution for this would be where the two graphs overlap or the area of the coordinate plane that satisfies both so all i need to do is simply graph each inequality and look for the overlap of the two graphs so we know the quickest way to do this is to solve the inequality for y so for this first one i would have y is greater than subtract x away from each side so minus x minus 2. for this one i would subtract 6x away from each side so y is less than negative 6x plus 3. so once i have this information i can just take it to the coordinate plane let's just copy it and so what we're going to do we know that the boundary line in each case would be 1. it would be a broken line because we have a strict inequality here y is strictly greater than y is strictly less than so in that case the boundary line is not part of your solution so i think about y is equal to negative x minus two i graph this without broken line so the y intercept occurs at zero comma negative two so that's there and my slope is negative one so i'd fall one go to the right one fall one go to the right one or i could rise one go to the left one so let's draw a line remember to break this guy up so let's just kind of break it up okay so that's my line and where would we shade i can erase this boundary line i don't need that anymore y is greater than so greater than means i shade above the line so my solution for this guy right here i'm shading this way so all of this anything above that line would work as a solution now for the second inequality we have y is less than negative 6x plus 3. so again another broken or dashed boundary line so we think about y is equal to negative 6x plus 3. the y-intercept occurs at 0 comma 3. let me erase this we know where the solution is for the first one so it's right here and then negative six is the slope so fall six one two three four five six go to the right one fall six one two three or five six go to the right one all right so let's graph that guy okay and again this has to be broken up so let me take my eraser out and let me just change the size of it we'll just take some chunks out of it okay that's good just make sure everybody knows that it's a broken line and we want to shade below this line so if i'm shading this way here below the line and i'm shading above the line here let me show you this one in black so this was shaded above the line okay that's good enough so anywhere that is below this green line but also above the blue line would work as a solution for the system let me kind of get a highlighter going i'll highlight this as best i can okay went over the line a little bit but i think you get the idea anything above the blue line and below the green line okay anything that fits that criteria so it's this area of the coordinate plane here all right let's take a look at the next example so we have 3x minus 2y is greater than 10 negative 6x plus 4y is greater than or equal to 16. so for this one again we want to solve each one for y let's go ahead and subtract 3x away from each side of the inequality so you'd have a negative 3x over here plus 10. then i'm going to divide each side of the inequality by negative 2. now remember we haven't worked with inequalities in a while if you divide both sides of an inequality or you multiply both sides of an inequality by a negative what do you have to do remember remember remember you have to flip the direction of that inequality symbol that's going to become a less than so you'd have y is less than negative 3 over negative 2 i can write as 3 halves and then times x 10 over negative 2 is negative 5. so minus 5. so that one is y is less than 3 halves x minus 5. for this one what am i going to do i can add 6x to each side so i'd have 4y is greater than or equal to 6x plus 16. we can then divide both sides by 4 and what are we going to have this will cancel with this i'll have y is greater than or equal to 6 over 4 each is divisible by 2 6 divided by 2 is 3 4 divided by 2 is 2. so 3 halves x plus 16 over 4 is 4. so let's write this as y is greater than or equal to 3 halves x plus 4. now what do we see here before we do anything same slope different y-intercepts now think about this a little bit more this is a less than y is less than so i'm going to be shading below this line this is a greater than or equal to i'm going to be shading above this line will this system have a solution no it will not and i'm going to show you that for the first one you have a strict inequality so that means your boundary line is going to be a dashed or broken line so we graph y equals three halves x minus five the y intercept occurs at zero comma negative five the slope is three halves so up one two three to the right one two up one two three to the right one two up one two three to the right one two okay and this is supposed to be a broken up line because this is a strict inequality here so let's get our eraser out take some pieces out of this thing again just enough to make it clear that hey this is a dashed line the line is not part of the solution it's all you need to do okay so i think that's obvious all right so the next thing is it's a less than so that means i would be shading below the line so i would shade in this direction okay now for the next one i have a non-strict inequality so that means the boundary line is going to be solid because the boundary line in this case is part of the solution y could be greater than or y could also be equal to three halves x plus four so for that or equals 2 you have to have a solid boundary line so my y-intercept here occurs at 0 comma 4 my slope is again 3 halves so up 3 1 2 3 to the right 1 2. up 3 one two three to the right one two or i could go down three one two three to the left one two down three one two three to the left one two okay now this is a greater than or equal to so i would shade above the line so i'm going to shade in this direction now is there any area of the coordinate plane that satisfies both no below this line satisfies the first one the line in above the line satisfies the second one there's no area of overlap so there's no solution there's no solution here and again we found that out right away and we're going to have the same slope but different y-intercepts and then we investigated further and we looked at the symbols we knew this one was a less than and this was a greater than or equal to so we knew this was going to be the scenario where we're going to come across right where there'd be no overlap and so we wouldn't have a solution all right for the next system we're going to look at we have x plus 2y is greater than or equal to negative 2. we have 3x plus 2y is less than 2. we have 2y is greater than or equal to subtract x away so negative x minus 2 and then we would divide both sides by 2 so that would give me y is greater than or equal to you'd put negative 1 half x minus 1. now for this guy right here let's subtract 3x away from each side of the inequality you will have 2y is less than negative 3x plus 2. let's divide both sides by 2. you'll have y is less than negative 3 halves x plus 1. so y is less than negative 3 halves x plus 1. if i look at this first one here the boundary line will be solid i'm basically going to graph y equals negative 1 half x minus 1. so the y intercept would occur at 0 comma negative 1 and the slope is negative 1 half so down one to the right two down one to the right two we're up one to the left two up one to the left two all right so this is a greater than or equal to so i'm going to shade above this line so i would shade everything going this way now for the next guy i have a dashed boundary line so i have y is less than negative three halves x plus one so i would graph y equals negative three halves x plus one let me just erase this temporarily the y intercept occurs at zero comma one so that's here the slope is negative three halves so down three one two three to the right two one two down three one two three to the right one two or i could go up three one two three to the left one two all right so let's break this up a little eraser so let me reshade the first one we had a greater than or equal to this was the first one here the line in black we were supposed to shade above the line we did but we had to erase it so we could draw this one so let me reshade this so we're shading above the line now for the second one we want to shade below the line that's a less than so let's shade this in a green so you can see where it overlaps so anything below the blue line and anything at the black line or above the black line anything that fits into that region let me kind of highlight it so the black line itself is included here up to that point right there and then let me kind of highlight this the blue line here would not be included because it's not part of the solution okay i know i went over the line but just pretend i didn't okay so i'll shade all this so that's where the two graphs overlapping this is an infinite area we continue out here forever and ever and ever but we just look at a small portion just to get a general idea of where the solution for the system would be so anything below this blue line here not including that blue line going down that's also including the black line but going up so anything that i've highlighted in yellow that region of the graph would satisfy the system hello and welcome to algebra 2 lesson 33 in this video we're going to have a review of the rules of exponents so i know for the majority of you you thoroughly covered the rules of exponents when you were in your algebra 1 course now we're about to start working with polynomials again so it's very very important that we understand the rules of exponents so that's where we're going to review them in this lesson and then in the next lesson we'll start working with polynomials again so the first rule i want to cover is the product rule for exponents super super easy to remember what it says is if you have like bases and you're multiplying what you do is you keep the base the same and you add the exponents so as an example we have x to the power of a times x to the power of b so the base is the same in each case we have x and we have x so what we do is we say this is equal to x base stays the same and we add the exponents exponent here is a so i have a plus the exponent here is b so i have b so x to the power of a plus b and as an example of this we have 2 cubed times 2 squared so the base in each case is the same it's a 2 here and it's a 2 here so i would keep two the same and i would add the exponents i would add three plus two which would give me five so this is two to the fifth power now when we work with exponents we can leave this as two to the fifth power because it's nice and convenient or you could write 32 you can evaluate it if you want really if you're in this section on your textbook a lot of answers will be kept in that form all right so let's scroll down a little bit for the next example we see x to the seventh power times x squared so the base x is the same so i'll keep it the same and i'll add the exponents here i have a seven here i have a two so this would be x to the power of seven plus 2 which is x to the power of 9. for the last example i have 6x squared times 3x to the fourth power so what i'm going to do is i'm going to write this as 6 times 3 times x squared times x to the fourth power we know that 6 times 3 is 18 and then x squared times x to the fourth power we have the same base that's x so we keep that the same and we add the exponents 2 plus 4 is 6. so this would be 18x to the sixth power all right so the next rule i want to deal with relates to the power of zero so what happens when we raise a non-zero value to the power of zero well it's 1. right so if i have some x raised to the power of 0 as long as x is not equal to 0 the value is always going to be 1. so it doesn't matter what i plug in here for x as long as it's not zero and let me state that we'll say x does not equal zero then the value here is going to be a one so for example three to the power of zero is one this whole amount four x or four times x it's in parentheses and it's raised to the power of zero so this is one then down here let me kind of scroll down a little bit we have negative 316 741 raised to the power of zero everything's in parentheses there so this equals 1. so again it doesn't matter what it is it could be 1 billion it could be 1 trillion it could be whatever you want as long as it's not 0 when the value is raised to the power of 0 it's equal to 1. so the next thing we want to talk about would be negative exponents so what do we do when we run across these so if i have x to the power of negative m this is equal to 1 over x to the power of m so basically what you're doing here is you're taking the reciprocal of the base and you're making the exponent positive so the reciprocal of x would be 1 over x so that's how we have 1 over x here and the exponent goes from negative m to positive m another way you can kind of think about this if i have x to the power of negative m and i just write it over 1. if i drag something across the fraction bar i just have to change the sign of the exponent so if i drag this across the fraction bar here what's going to happen is i'm going to have a 1 up here and then down here since i drag this down i'd have x raised to since this was negative it will now be positive so here's a perfect example of that i have x to the power of negative m over y to the power of negative n as i drag this across the fraction bar you'll see here it's in the numerator y is the same it's just that the exponent has changed its sign it goes from negative to positive same thing here x gets dragged across the fraction bar it's now in the denominator and notice how the sign on the exponent has changed here it's negative here it's positive so if you drag something across the fraction bar right from numerator to denominator or from denominator to numerator you just need to change the sign of the exponent so as an example with some numbers let's say you had something like 6 to the power of negative 2 and this is over let's say 2 to the power of negative 3. what we can do is we can take this guy right here we can bring it into the numerator we just have to make the exponent positive so this would be 2 cubed and then we could take this guy right here bring it into the denominator we just have to make the exponent positive so that would be 6 squared you could verify that with a calculator i can simplify this 2 cubed is 8 6 squared is 36 and we know they have a common factor this would end up being 2 over 9 in its simplest form and you could punch this up on a calculator and verify you can do 6 to the power of negative 2 divided by 2 to the power of negative 3. now you're going to end up with a decimal that's 0.2 with a 2 repeating forever and ever and ever but it just verifies that you have the same answer here as the fraction 2 9. so i also have here 1 over x to the power of negative m is equal to x to the power of m so if you have 1 over something where that something has a negative exponent again i can just drag this up into the numerator i can think about this as being over 1 and i can change the sign of the exponent another way to think about this kind of a longer way to think about this if i have 1 over x to the power of negative m remember i could take the reciprocal of the base make the exponent positive so i could write this as 1 over 1 over x to the power of m so if 1 is divided by this fraction really i could write 1 over 1 divided by 1 over x to the power of m which turns into 1 over 1 times the reciprocal of this which would be x to the power of m over 1 which equals x to the power of m over 1 which is just x to the power of m so that's how i got that kind of in a longer format all right so two quick examples let's say you see 6 to the power of negative 2. all this is is it's 1 over 6 squared take the reciprocal of the base make the exponent positive the reciprocal of six is one over six if i make the exponent positive instead of being negative two it's positive two you could leave it like this so you could write one over thirty six and again another way to think about this or an easier way to get the answer is just to set up a little fraction bar here and say okay i'm going to drag this down here i know there'll be a 1 up here now and so if i drag this down here 6 will be the same but the sign of the exponent is going to be positive now so this would be 1 over 6 squared all right then for the next example we have 7 times x to the power of negative 4 over y squared so again with this part right here i could separate this and say this is 7 over y squared times x to the power of negative 4 you could put it over 1 if you wanted to it doesn't matter if i take the reciprocal of this what would that give me so it would be 1 over x to the fourth power so this would be 7 over y squared times 1 over x to the fourth power which would be 7 over you'd have y squared times x to the fourth power now another way to kind of think about this and what i always do if i have 7 x to the power of negative 4 over y squared you're just going to bring this down here and again when i cross the fraction bar the exponent is going to be positive instead of negative so i can erase this up here and either way i do it this is 7 over y squared times x to the fourth power all right the next rule let's talk about the quotient rule for exponents so if we have the same base and division is involved we keep the base the same and we subtract exponents so a to the power of m over a to the power of n is equal to a to the power of m minus n let's look at a couple of examples here all right for the first example we have six cubed over six to the fourth power so we have the same base here of six so what i'm going to do is keep that base the same and i'm going to subtract exponents now to be more specific i'm going to take the exponent in the numerator which is 3 and i'm going to subtract away the exponent denominator which is 4. now 3 minus 4 is negative 1 so this will give me 6 to the power of negative 1 which we just learned is 1 over 6. all right for the next one i have 7x cubed over x squared so i can write this as 7 times x cubed over x squared and what we can do here again 7 just stays the same that's unaffected we have the same base so x stays the same and we subtract 3 the exponent of the numerator minus 2 the exponent of the denominator and i end up with x to the power of 1 or just x so this would be 7 x all right for the next problem we have 6 x to the fourth power y to the seventh power over 2x squared y to the 8th power so there's more things i can do here i can cancel this 6 with this 2 and get a 3. that's how i'm going to start then i have x to the fourth power over x squared so that's x to the power of 4 minus 2. again base stays the same four is the exponent of the numerator subtract away the exponent the denominator which is two then i have y to the power of seven minus eight y and y so y stays the same the exponent the numerator is a 7 subtract away 8 which is the exponent of the denominator so this gives me 3 times x to the power of what 4 minus 2 is 2. so 3x squared then times y to the power of 7 minus 8 is negative 1. so i could rewrite this as 3x squared i can drag y to the power of negative 1 into the denominator and i would have y to the power of 1. right the y would stay the same but the exponent will go from negative to positive because it crosses the fraction bar all right the next thing we want to think about would be some power rules so we see something like a raised to the power of m this is inside of parentheses and it's raised to the power of n this is equal to a to the power of m times n so in other words if i have a power already and i raise that to another power it's called the power to power rule i keep the base the same and i multiply exponents so if i saw something like 2 cubed and i squared this 2 would stay the same and i would multiply 3 times 2 to give me 6. so this would be 2 to the 6th power you could prove that to yourself 2 cubed is 8. if i had 8 and i squared it i would get 64. now 2 to the 6th power we know is 64. so either way you can see that would be 64. now as the next example if we had a times b here and this whole thing was raised to the power of m i could say this is equal to a to the power of m times b to the power of m so as an example of this let's say we saw 2 times 3 and this was raised to the third power let's say well i could say this is 2 cubed times 3 cubed and you can prove this to yourself as well 2 times 3 is 6 6 cubed is 216. if i do it this way 2 cubed is going to give me 8. 3 cubed is going to give me 27 if i multiply 8 times 27 i get 216 as well right now we also want to look at we have a quotient here a over b this is raised to the power of m so we say this is equal to we have a over m over b over m so as an example of this let's say i had something like 10 over 2 and this is squared well this is 10 squared over 2 squared all right you can think about this as going here and also here so this would be equal to what 10 squared is 100 over 2 squared which is 4 so this is 25. now you think about this another way 10 over 2 is 5 5 squared is 25. now you might say why would you do it this way it seems like it's quicker with what you just did well it's important to just understand that you can go back and forth you might see something in this format and say hey it'd be pretty easy to do that calculation in this format so you can go back and forth between the two okay it doesn't matter all right so let's look at some examples if i have 6 cubed and then it's squared it's again that power to power rule so 6 stays the same and i would multiply exponents 3 times 2 is 6 so this would be 6 to the sixth power for this one i have 2 to the power of negative 1 times x to the 4th power and this whole thing is raised to the power of negative 2. so again the base in each case would stay the same so 2 stays the same and you'd have negative 1 times negative 2 which is 2. here i'd have x to the fourth power so x stays the same and we do 4 times negative 2 which is negative 8. so the base stays the same and you multiply the exponents now in this case since we know that x to the power of negative 8 is 1 over x to the 8th power i could just simply write this as 2 squared over x to the 8th power remember if i grab this right here and i bring it across the fraction line what i know is x stays the same the exponent goes from negative to positive all right the next one we're going to look at we have 4 times 3 and this is all raised to the power of 2 or you could say it's squared so this is the same as saying we have 4 squared times 3 squared now let's look at the next one all right for the next one we have 6xy inside of parentheses this is raised to the power of negative four so i could do this a few different ways the first way i could do it is i could just take the reciprocal of this guy right here and i could raise it to a positive four so i can say this is one over and then i'll have 6 x y this in parentheses raised to the power of 4 then i would use my power to power rule remember everything has an exponent of 1 right now so 4 times 1 would be 4 in each case so everywhere we'd have six to the fourth power x to the fourth power y to the fourth power so everything would have an exponent of four now another way you could have thought about that let's just erase this real quick we could say okay every exponent here is a 1 inside the parentheses so i'm just going to say okay 6 to the power of 1 is raised to the power of negative 4 so that would be 1 times negative 4 or negative 4. so i'd have 6 to the power of negative 4 same thing for x i'd have x to the power of negative 4 same thing for y i'd have y to the power of negative 4. then because these are all negative exponents i can drag everything across the fraction bar make the exponents positive so i can have 6 to the power of positive 4 x to the power of positive 4 then y to the power of positive 4. again i move these from the numerator to the denominator so i have 1 over 6 to the 4th power x to the 4th power y to the 4th power what about this one we have 10 over 5 and this amount is squared so i can write this as 10 squared over 5 squared and again you can prove that to yourself pretty easily 10 squared is 100 5 squared is 25 the result of this would be 100 over 25 which is 4. if i did it in here 10 divided by 5 would be 2 then if i square 2 i get 4. so either way it's the same all right then lastly i have x to the power of negative 1 times y this is over z cubed and then this is inside of brackets and it's raised to the power of negative 2. so you can think about each base here as staying the same and just multiplying the exponents so x stays the same negative one times negative two is two y stays the same remember it has an understood exponent of one so one times negative two would be negative two this is over we have z which stays the same 3 times negative 2 is negative 6. so now all i want to do x squared would stay the same this would go into the denominator and so i would change the sign of the exponent so it it'll be y squared this would go into the numerator so i change the sign of the exponent it would be z to the sixth power so x squared times z to the sixth power over y squared all right so let's look at a few practice problems before we kind of move on to our topic of polynomials so we have 2a to the power of negative 1 times b to the power of 0 and this is over we have 8 to the power of negative 5 and this is raised to the power of negative 5 times negative 2 a to the power of negative 4 times b squared times a squared so if i simplify in the numerator first i know b to the power of 0 is 1. so i can kind of cross this out and just put a 1 there 1 times anything is just itself so the numerator i'm going to write as 2 a to the power of negative 1. for my denominator i got a lot of stuff going on so i've got a to the power of negative 5 raised to the power of negative 5. so a stays the same negative 5 times negative 5 is 25 then times you have a negative 2 a to the power of negative 4 b squared a squared so let me write 2 a to the power of negative 1 over i have a to the power of 25 i have a to the power of negative 4 and i have a squared so keep a the same and just add those exponents 25 plus negative 4 is 21 21 plus 2 is 23 so i'll have a to the power of 23 there and i have that negative 2. now what i can do is i can cancel this 2 with a negative 2 and i can put a negative 1 there so instead of having a 2 there let me just put a negative in front of that a to the power of negative 1. all right so now the next thing i have is a b squared nothing i can really do with that so i'll just rewrite b squared so the last thing is let me write equals i'm going to put a negative 1 times i have a to the power of negative 1. now i have an 8 to the power of 23 here so what i can do is keep a the same i would subtract negative 1 minus we have 23 in the denominator so minus 23 and this would be over b squared so again i'll have negative 1 times we'd have a to the power of negative 1 minus 23 which would be negative 24. this is over b squared the last thing i'm going to do is i'm going to put this in the denominator and when i do that i'll have negative 1 over b squared times this a will be the same but the exponent because i'm dragging it into the denominator will be positive so a to the power of 24. so negative 1 over b squared times a to the power of 24. all right for the next one we have negative 2x to the power of negative 3 y to the power of 6. this is inside of parentheses and it's all squared then this is over we have negative 2y squared times negative x to the power of negative 2 times y to the sixth power so let me start out in the numerator here just think about this is going to be applied to everything right that's your power to power rule now i want you to be very very careful here this negative 2 is inside of parentheses it's inside of these parentheses here so that negative and that 2 are both squared so i could do this two ways i could say i have negative 2 like this squared or i could treat this negative as a negative 1 and say that's squared then times 2 squared then times x to the power of negative 3 squared and times y to the sixth power squared and if i go through on each thing negative 1 squared is 1. so this is 1. 2 squared i'm just going to leave it like that for now x to the power of negative 3 squared would be x to the power of negative 6 and then y to the 6th power raised to the second power y to the sixth power squared would be y to the twelfth power then this is over i have a negative two which i'm going to write as negative one times two then i have y squared then i have a negative x so i'm going to write negative 1 times x this is raised to the power of negative 2 then times y to the 6th power so a lot of things we can do and i want to go slowly i know a lot of you can do this much quicker but i want to go slowly so everybody can follow along so let me just start by simplifying in the denominator real quick so in the numerator i'm going to leave this as 2 squared times x to the power of negative 6 times y to the 12th power in the denominator i'm going to leave negative 1 times 2. what i'm going to do is i'm going to say i have y squared times y to the sixth power y stays the same two plus six is eight then i have times another negative one so negative one times negative one is one so we can erase both of these okay and i'll just slide this down and the next thing i see is times x to the power of negative 2. so what i know is if i have the same base here i have 2 squared and i have 2 down here you think about this as 2 to the first power this is going to tell me i have 2 to the power of what subtract exponents you'd have 2 minus 1 or 2 to the power of 1. then next i have x to the power of negative 6 and i'd subtract away a negative 2 because i have x to the power of negative 2 down here again this seems a little weird but if i have a negative 6 here that's what i start out with i'm subtracting away a negative 2 minus a negative is plus a positive so negative 6 plus 2. so negative 6 plus 2 is negative 4. so this would be x to the power of negative 4. then lastly i have y to the power of 12 here y to the power of 8 here so it'd be y to the power of 12 minus 8 or y to the 4th power so if i wrote this over 1 i could drag this down here and say this is what this is x to the power of 4. x stays the same the exponent's sign is going to change so i can just erase this and say we have 2y to the 4th power over x to the 4th power all right so for the last example we have 2x to the power of zero y to the fourth z to the seventh times two z y to the fifth over two x to the eighth power y to the seventh z to the eighth everything's inside of brackets and raised to the sixth power so that means that every item inside here is raised to the sixth power so two is raised to the sixth power x to the power of zero is raised to the sixth power y to the fourth raised to the sixth power so on and so forth now you could go through and do that now but it's gonna be easier if you simplify what's inside the brackets first so let's do that and let's do it in a quick manner so i'm gonna start out with two i have a 2 here and a 2 here so same base i would add the exponents so this is a 1 and a 1. right the exponents understood to be 1 there so that would be 2 to the power of 1 plus 1 which is 2 squared now i also have a 2 in the denominator so if i had 2 squared over 2 the exponent down here is a 1 as well that would be 2 to the power of 2 minus 1 and so this would be 2 to the power of 1. the next thing i come across is x so here i have x to the power of 0 which is 1 and down here i have x to the power of 8. so i'm just going to leave that x to the power of 8 down there then next i come across a y so y to the 4th power i have a y to the fifth power here so again if i did my product rule for exponents y stays the same i'd add 4 plus 5 that would give me 9. but then i have a y to the 7th power of the denominator so for that use the quotient rule for exponents y stays the same i would do 9 minus 7 which would give me 2. then the next thing i come across is z so i have z to the seventh power here i have a z here so the product rule for exponents tells me that z stays the same seven plus one would be eight then down here i have z to the eighth power so what would happen is i would have z to the eighth power over z to the eighth power which is one right or you can think about it this way i would take the exponent in the numerator which is eight subtract away the exponent of the denominator for z which is eight that would be z to the power of zero which is one you can get rid of z all right so we've dealt with all of that so don't forget your brackets here and this is all raised to the sixth power so two to the first power is raised to the sixth power two stays the same multiply one times six you get six y squared is raised to the sixth power y stays the same two times 6 is 12 so that's y to the 12th over x to the eighth is raised to the sixth power x stays the same 8 times 6 is 48 so we get 2 to the sixth power times y to the 12th power over x to the 48th power hello and welcome to algebra 2 lesson 34 in this video we're going to learn about adding and subtracting polynomials so at this point we should all know the basic definition of a term so a term is nothing more than a number a variable or the product of a number and one or more variables raised to powers so here to give an example of a term i have the term 3x to the fourth power plus the term 7x squared plus the term 2x minus the term x and then plus the term 7. now we separate terms with plus or minus symbols so in other words everywhere i see a plus or a minus i'm separating my terms so this 3x to the fourth power is a term 7x squared is a term 2x is a term x is a term and 7 is a term and if you go through those individually they fit the criteria given in this definition we have a number so a single number like 7 a variable so just a variable like x or the product of a number and one or more variables raised to powers so i have 2 times the variable x i have 7 times the variable x that's squared and i have 3 times the variable x that's raised to the 4th power so each of these here again so 3x to the 4th power 7x squared 2x x and 7 those are all examples of a term all right so next let's give a definition for a polynomial so a polynomial is either a single term okay so it can be one single term or a finite sum of terms where all variables have whole number exponents okay so we know what a whole number is that starts with zero and increases in increments of one forever and ever and ever and there are no variables let me highlight this there are no variables in any denominator so you want to look out for two things whole number exponents and no variables in any denominator okay so that's what's going to trip you up so i'm going to give you two examples of a polynomial so 3x to the ninth power minus 12x to the fifth power plus two this is a polynomial and all you're looking for is on every term do i have a variable where we don't have a whole number as the exponent or where we have a variable in the denominator if i look here i have x to the ninth power and i have x to the fifth power there's no variable in any denominator so this is a polynomial so we're good to go there this one is another example of a polynomial we have 4x to the 16th power minus 7x cubed minus 5x squared plus 8. so i look at my exponents here i've got a 16 a 3 and a 2. i'm good to go right nothing is in the denominator so we know that one is a polynomial now if i look at these two they're not a polynomial so 3x squared minus 1 over 2x minus 5 will have a variable in my denominator and that violates the definition of a polynomial this is in fact called a rational expression right we studied those in algebra 1 and we'll get to those pretty soon in algebra 2. but this is not a polynomial this other example here 4 x to the power of negative 3 plus 2 is also not a polynomial because we have this exponent of negative 3. so negative 3 is not a whole number okay if you see something like negative something or you see a fractional exponent something like that you don't have a polynomial again your exponents have to be whole numbers only all right so we have three polynomials that occur so often that we give them special names the first one is referred to as a monomial okay so a monomial so this is simply a polynomial with one term only so three examples we have 4x squared so that's a monomial negative 2x to the seventh power that's a monomial and negative 14x to the 15th power that's a monomial as well so in each case we just have one single term and that is the definition for a monomial all right so the next one would be a binomial so this is a polynomial with two terms only now again notice how the terms are separated by either a plus or a minus sign so when you see one of those signs you know you've separated two terms so this is a term and this is a term so we've got two terms in this polynomial i've got 4x squared plus 3x so that is a binomial then the same thing here i've got negative 2x to the seventh power minus five so i've got a term here and a term here the minus sign separates the two so that's a binomial as well and then as a final example i have negative 17x cubed plus 9x again this is a term here this is a term here the plus sign separates the two and so we know we have two terms and that fits the definition of a binomial all right the next one we run across is called a trinomial okay we think about a tricycle has three wheels a trinomial has three terms so a polynomial with only three terms is a trinomial so we have an example here four x squared plus three x plus two again everywhere you see a plus or a minus sign you're separating terms so you have a term here four x squared you have a term here three x and a term here of two so four x squared plus three x plus two has three terms and so it's a trinomial same thing here i have negative two x to the seventh power minus five x to the fourth power plus thirteen the minus sign and the plus sign separate the terms so i've got a term here of negative 2x to the seventh power a term here of 5x to the fourth power and a term here of 13. so we have a polynomial with three terms and so that is a trinomial all right the last example is negative 12x to the ninth power plus 9x minus 11. again this and this the plus and the minus separate the terms so i've got negative 12x to the ninth power i've got 9x and i've got 11. so three terms here in this polynomial so this is a trinomial all right so pretty easy stuff overall stuff that you would have learned in algebra one here's another kind of common thing you need to know and it's called standard form so a polynomial is written in standard form let me just highlight that so standard form when the powers are in descending order so this is starting all the way to the left and working your way to the right so i'm going to take the term that has the highest exponent on the variable and i'm going to put it all the way to the left then i'm going to go to the next highest then the next highest you know so on and so forth so if i look at this 12x minus 5x to the 7th power plus 9x squared minus 5 plus 10x cubed our variable is x what is the highest exponent on this variable well it's this 7 here okay so i want to write this term right here and i've got to include that negative the negative 5x to the seventh power i want to put that all the way to the left okay so if i put equals negative 5x to the seventh power then i would say what is the next highest exponent well this one's a two this one's a three and this one's understood to be one so i would go with this 3 here so this term 10x cubed would come next so plus 10 x cubed then i'd put plus 9x squared that would be next then i put plus 12x that would be next and then this constant here negative 5 would be all the way at the end so this would be minus 5. so this we could think of as being multiplied by x to the power of 0. you might see that in a textbook and be very confused you might say why would they do that so x to the power of 0 is understood to be 1 again as long as x is not 0. so negative 5 times 1 is just negative 5. so really i haven't changed anything there it's just a neat little trick to put a variable next to it so in standard form this polynomial is read as negative 5x to the seventh power plus 10x cubed plus 9x squared plus 12x minus 5. and again you can notice how your exponents go in descending order you've got 7 3 2 this would be a 1. and again if you wanted to you could write this as having an exponent on x of 0. so 7 3 2 1 and then 0. so in descending order all right so let's take this one and write it in standard form we have 9x minus 3 minus 7x to the 4th power plus 2x cubed so the highest exponent on the variable x occurs right here so i've got to take that negative and bring it with me okay so i'm going to put negative 7x to the fourth power and you might ask yourself if you're not familiar with a lot of things why am i taking that negative with me remember you can rearrange things with addition but not with subtraction okay so if i put this as plus negative 7x to the fourth power i just bring the negative with me and i can rearrange things as much as i want if i just took things as they were and i just move this over here and i didn't put my subtraction over there and i just left the signs where they were you'd have a problem there right i can't just move this over here and leave the sign where it is so i change this to plus negative and you can do it without doing it right you could just mentally know that and just put a minus 7x to the fourth power so there's no problems there now the next thing we want to think about is 2x cubed right because the exponent there is a 3. so then plus 2x cubed and then next i want plus 9x because the exponent there would be a 1. and lastly i just want that constant negative 3. or you can think about it as just minus 3. all right so i've rearranged everything to where the exponents are in descending order you've got a 4 a 3 a 1 and again you could write this as negative 3 times x to the power of 0 so then you could say you have a 0 here so negative 7 x to the 4th power plus two x cubed plus nine x minus three x to the power of zero or you can just put minus three right same thing so the last thing we're gonna review is called the degree so the degree of a term in a polynomial is the sum of the exponents on the variables of the term so all i'm going to do to find the degree is look at the exponent or it could be exponents if you have more than one variable and whatever the sum is that's the degree of the term so in other words with 5x we know the exponent here is understood to be 1 so the degree is 1. for this example 3x to the 7th power y to the fifth power we have two variables involved so what i want to do is some addition i have an exponent of 7 and an exponent of 5. 7 plus 5 is 12 so the degree is 12. so this isn't very difficult at all it's just something you just kind of need to know right once you know it and you understand the definition it's very very simple right just some simple addition for this one negative 15 x to the fourth power y to the ninth power z cubed again i'm just going to add the exponents so 4 plus 9 is 13 13 plus 3 is 16. so this plus this plus this your degree is 16. all right so let's extend this a little bit further now we're going to talk about the degree of the polynomial itself so the greatest degree of any term in the polynomial is known as the degree of the polynomial so as an example here i have 4x to the fifth power plus 2x cubed minus 7. so i would look at each term individually so for this term here the exponent is 5. so the degree of that term is 5. then this term here 2x cubed the exponent is a 3. so the degree of the term is a 3. and then for this negative 7 again a little trick here i can write x to the power of 0 next to it i showed you that earlier so the degree for this would be 0. now which term has the highest degree well this one does the 4x to the 5th power so since the degree for that term is a 5 that means the degree of the polynomial is a 5. so this polynomial has a degree of 5. all right let's take a look at another one so we have negative x to the seventh power y cubed plus seven x to the ninth power y to the fifth minus three x y plus eight so again i wanna look at each term individually to start if i look at this term here i have a seven as an exponent and a three i've gotta sum those two so seven plus three is ten for this one i have a nine and a five nine plus five is fourteen for this one i have a one and a one one plus one is two and then for this eight here i can write this as times x to the power of zero so this is zero in general when you look at a constant the degree is zero now the only exclusion to that is the number zero itself okay but anything other than zero if it's a constant it's going to have a degree of zero so if i think about this polynomial here we see that this is the big winner here right this term right here so the degree for this is 14 so the degree for the polynomial is 14. so the degree is 14. all right so now that we've kind of reviewed some of the things that we need to know let's jump in and start adding and subtracting polynomials so for the majority of you you've already done this this is a review and you remember how easy this is the only thing you have to really get down here is the concept of like terms so like terms have the same variable or it could be variables if we have a harder problem and they're raised to the same power or could be powers so as an example if i have something like x to the third power and i have 3 x to the third power those are like terms the coefficients or the number parts here don't matter okay don't pay attention that when you're trying to find like terms you just want to make sure that the variable is the same so i have an x here and an x here so that's good to go then i want to make sure my exponent's the same i have a cubed and i have a cubed so that's the same so these are like terms and how do we go about adding like terms well if i had something like 2x cubed plus 3x cubed all i need to do is add the numerical coefficients so two plus three which is five and then the x cubed part stays the same so times x cubed you can think about as having two apples plus three apples gives you five apples but if i wasn't adding like terms let's say i was adding oranges and apples well two apples plus three oranges is just two oranges and three apples okay i can't i can't really make that simpler so as an example of something that's not like terms let's say that we had 2x cubed and let's say 3 y to the fourth power well these are not like terms these are not like terms why is that the case well i have an x here and i have a y here so the variables are different the exponents are also different i have a 3 here and a 4 here but even if the exponents were the same it doesn't matter it needs to be the same variable raised to the same power okay so that's how you have like terms so if i added 2x cubed and 3 y cubed you can't combine that or make that any simpler there's nothing you can do and a lot of students make the mistake and say okay this is 5 x cubed y cubed you can't do that that's wrong okay this is wrong so very important that you understand that you can only combine like terms all right so let's erase all this i think all of you pretty much understand this at this point if you don't as we're working through the examples i think you will easily pick this up so i start out with adding this polynomial here 7x cubed plus 6x to the fourth power plus 2x squared and we're adding to this the other polynomial 7x to the fourth power plus 6x squared the first thing is i can just drop these parentheses i don't really need them so i can write this problem as seven x cubed plus six x to the fourth power plus two x squared plus seven x to the fourth power plus 6x squared now the next thing that i like to do and i don't do this anymore but when i started my teacher told me this and i loved doing it when i was struggling i want to look for like terms right away and i want to write them next to each other so it's obvious what i need to do so in other words if i start out with 7x cubed do i have another term with an x cubed in it no i do not so i don't need to do anything there i have 6x to the fourth power and x do i have another term where i have x to the fourth power yeah i have this one and this one then i have 2x squared do i have another term with x squared yeah this one and this one so let's reorder this i'm going to put 6x to the fourth power in the very front position then plus 7x to the fourth power then plus 7x cubed then plus 2x squared then plus 6x squared now i did two things there the first thing is i took the exponents that were highest and move them to the left and i went in descending order so this is the exponent with a 4 a 4 a 3 then a 2 then a 2. so it's in descending order so when i get my answer it's going to be in standard form okay so that's the first thing i did the second thing i did was i arranged it to where like terms were next to each other so these are like terms and these are like terms so now all i got to do is add the coefficients keep the variable part the same so 6x to the fourth power plus 7x to the fourth power add 6 plus 7 that's 13 and x to the fourth power just stays the same if i had 6 apples and 7 apples if i added them i would have 13 apples so then plus we have 7x cubed and then plus now we have like terms again we have 2x squared plus 6x squared 2 plus 6 is 8 and then x squared comes along for the ride now there's no way to make this any simpler we simply don't have any more like terms our variable is the same in each case but the exponents are different this is like adding apples oranges and nectarines okay if i had 13 apples plus 7 oranges plus 8 nectarines i can't really make that simpler if i had 13 apples and 7 apples i could just say i have 20 apples okay so that's kind of a real world example of what we're going through here so our answer is just 13x to the fourth power plus 7x cubed plus 8x squared and again notice how it's written in standard form you have x to the fourth x cubed and then x squared all right as a next example we have 6r plus r squared minus eight r to the fourth power then plus eight r squared minus six r to the fourth power plus eight and plus we have seven r minus two r cubed all right so the first thing i wanna do again is reorder these things i'm going to start out with a negative 8r to the fourth power so negative 8r to the fourth power remember i'm taking this sign with me i can think about this as plus negative 8r to the fourth power and then over here i see i have negative six r to the fourth power so i write those two next to each other again why did i start out with that because as i scan through here the highest exponent on r is a 4. so that's why i chose to put this all the way to the left now i'm going to scan through and see do i have r cubed yeah i've got one right here so again the sign comes with it i'm going to write negative 2r cubed and the next thing i'm going to look for is r squared so i've got that here and here so i'm going to put plus r squared then plus 8 r squared now i'm looking for r to the first power and so that's 6r and 7r so plus 6r plus 7r and then lastly i have this plus 8. so plus 8. now one of the things you might want to do especially when you first start out you might write everything and you might miss a term oh i forgot that one it's very very common so you can kind of go through especially if you're on an important test and check okay i've got negative 8r to the fourth power so i've got that negative 6r to the 4th power so i've got that negative 2r cubes i've got that r squared i've got that 8r squared i've got that 6r i've got that 7r i've got that and then 8 i've got that you can go through and say okay i've got a dot there there there there there there there and there so i've got all my terms so very important that you put everything in there otherwise you're not going to get the right answer so once we've set this up we're just going to combine like terms we've got like terms here and here and here so negative 8r to the fourth power minus 6r to the 4th power is going to be negative 14r to the fourth power then we've got minus two r cubed then plus you've got r squared plus eight r squared which is nine r squared then plus six r plus 7r is 13r then plus 8. so it's just as easy as that and a lot of times moving forward you're not even going to write this stuff out once you get good at this you're just going to say okay well negative 8r to the 4th power minus 6r to the fourth power write my answer negative 14 r to the fourth power and just kind of go from there i add these now just kind of mentally and just write down the answer i don't advise that until you get really good at this because it's very easy to make a mistake okay so just start out with it slow progress until you fully understand it and then you try to do it just a little bit faster so our answer here is negative 14r to the fourth power minus 2r cubed plus 9r squared plus 13r plus 8. all right so the next one is going to be 6n squared minus 5n minus 7n to the fourth power then plus n to the fourth power plus 6n squared plus 7n then plus 4n to the fourth power plus 4n so i scan through this and i see that 4 is the highest exponent so what has a 4 as an exponent so this term this term and this term so let's put equals i'll start out with negative seven n to the fourth power again take the negative sign with you plus n to the fourth power i'll do that next then plus four n to the fourth power now the next highest exponent would be a 2. i've got 6n squared i've got 6n squared again and that's all i've got let's write those so plus 6n squared plus 6n squared again then i've got my n to the first power so i've got minus 5n and i've got plus 7 in and i've got plus 4n so let's go ahead and write those i've got minus 5n plus 7n plus 4n okay so now we're just gonna combine like terms and so i have like terms here i have let me do this in a different color like terms here and then i have like terms i'm gonna switch colors again here so if i look at it what am i gonna have negative seven n to the fourth power plus n to the fourth power plus four n to the fourth power just add the coefficients so negative seven plus this coefficient is understood to be one so negative seven plus 1 is negative 6 negative 6 plus 4 is negative 2. this would be negative 2 times n to the fourth power just stays the same then plus you have 6n squared plus 6n squared you 6 plus 6 that's 12 times we have n squared that stays the same then you have negative 5n plus 7n plus 4n so you have negative 5 plus 7 which is 2 plus 4 which is 6 so plus 6 times n so i end up with negative two n to the fourth power plus 12 n squared plus six n all right now let's get into some subtraction problems and subtraction is not any more difficult than addition you just have to know one thing whatever polynomial is being subtracted away what you want to do is you want to change the sign of each term of the polynomial being subtracted away and then you just add so as an example we have five m cubed minus six m squared plus eight m then minus we have three m cubed plus five m squared now a lot of students get confused and they're like well what's the difference between having this minus six m squared and minus this whole thing here the difference is i have a minus sign outside of this set of parentheses so this whole thing is being subtracted away so if i'm subtracting the whole thing away what i can do and you'll see a lot of professors do this you'll see plus negative 1 there and they'll distribute the negative 1 to each term just as a reminder to the students to change the sign of each term and then just add so in other words if i do it this way i'd rewrite this problem as five m cubed minus six m squared plus eight m negative 1 times 3m cubed is negative 3m cubed and then negative 1 times 5m squared is minus 5m squared so again all i'm doing is i'm changing the sign of every term in the polynomial that's being subtracted away i could have just as easily said okay i've got minus 3m cubed and then minus 5m squared where students make another mistake is they'll drop the parentheses and let's say i have minus 3m cubed they got the first term right but then the second term they forget to change they just leave that as plus 5m squared and they get the wrong answer you need a minus here and here okay very important so once you have that you just combine like terms like we've been doing with addition okay so it's just one additional step and to kind of look at this i would write 5m cubed all the way to the left and then minus 3m cubed those are like terms and then minus 6m squared and then minus 5m squared those are like terms and then plus 8m so 5m cubed minus 3m cubed is 2m cubed negative 6m squared minus 5m squared is negative 11m squared and then i have plus 8m so again my answer here is 2m cubed minus 11m squared plus 8m all right let's take a look at another one so we have three b to the fifth power minus three b squared minus three plus eight b cubed then minus we're subtracting away this whole polynomial here six b to the fifth power minus five b cubed plus three b squared minus four then minus i'm subtracting by this whole thing b cubed plus two plus six b squared minus five b to the fifth power now the quickest thing you can do again you could just make this plus and change every sign here so this would be negative this would be positive positive positive change the sign here plus negative this would be negative negative positive okay change the sign of every term that's being subtracted away this first polynomial is not being subtracted away so it just stays as it is every polynomial that comes after it has a subtraction sign out in front of the parentheses that enclose it so you've got to put a plus sign and then change every sign for the terms inside the parentheses okay very important you understand that and again if you struggle with it put plus negative 1 and distribute the negative 1 to each term it's going to remind you to do that all right so what i'm going to do here if i scan through i see 5 is the highest exponent so i've got b to the fifth power i've got negative six b to the fifth power and i've got five b to the fifth power so that takes care of this one this one and this one all right the next highest i see is a three so i've got eight b cubed i've got five b cubed and i've got negative b cubed okay so that takes care of this one this one and this one all right so the next thing i see is b squared so i've got a negative three b squared i've got positive 3b squared and i've got negative 6b squared so that's going to take care of this one this one and this one and then lastly i have minus 3 i have plus 4 and i have minus 2. so that takes care of this one this one and this one and because i've highlighted everything i know i didn't miss anything and again it gets very very tedious with these kind of longer problems so you need some kind of process to tell you hey i've used all of these so if i start with three b to the fifth power minus six b to the fifth power plus five b to the fifth power again look at your coefficients three minus six is negative three plus five is two so i've got two and then times b to the fifth power here i've got eight b cubed plus five b cubed minus b cubed again i've got an eight a five and this would be a negative 1. understand that if you have a variable by itself the coefficient is 1. if you have a negative in front of it the coefficient is negative 1. you can't just not put anything so for this one 8 plus 5 is 13 13 minus 1 is 12. so this would be plus 12 b cubed then next i have negative 3b squared plus 3b squared minus 6b squared so i think about negative 3 plus 3 is 0 those are opposites and then i just have negative 6 b squared so minus 6 b squared and then lastly i have negative 3 plus 4 that's 1 then minus 2 is negative 1 so minus 1. so 2b to the fifth power plus 12 b cubed minus 6b squared minus 1. all right the next thing i want to look at is 3x squared plus 2x to the fourth power minus 10x cubed plus 4 then minus this polynomial here we have 4x squared minus 9 minus 6x to the fourth power plus 4x cubed then minus i have 3x cubed minus 9x squared plus 10 minus 2x again this first polynomial is just going to stay the same so just my 3x squared plus 2x to the fourth power minus 10x cubed plus 4. i'm subtracting away this polynomial here so all i want to do is put plus and then change the sign of each term so this becomes negative positive positive and then this would be negative for the next one it's the same i'm subtracting away this polynomial here so plus this is negative positive you put negative and then positive all right so once we've done that i'm going to scan through i see 4 is the highest exponent so i've got 2x to the fourth power i've got 6x to the fourth power and i don't have an x to the fourth power here so that just takes care of this and this now next i scan through and i see x cubed so i see i have negative 10 x cubed so minus 10 x cubed and then i have minus 4 x cubed and then i have minus 3 x cubed so that takes care of this this and this the next i see i have x squared so i have plus 3x squared i have minus 4x squared and i have plus 9x squared so that takes care of this this and this all right so lastly i have this 2x so that takes care of this no more x to the first power anywhere and then i have 4 9 and negative 10. so plus 4 plus 9 plus negative 10 or minus 10 whatever you want to do all right so that takes care of this this and this and again notice how i've highlighted everything so i know i didn't miss anything okay so i go through now and i say 2x to the fourth power plus 6x to the fourth power 2 plus 6 is 8 times x to the fourth power then next i have minus 10 x cubed minus 4 x cubed minus 3 x cubed so we've got negative 10 minus 4 that's negative 14 minus 3 that's negative 17. so minus 17 then times x cubed then i continue i have 3x squared minus 4x squared plus 9x squared so i've got 3 minus 4 that's negative 1 plus 9 is going to give me 8. so this is plus 8 x squared then i have my plus 2x nothing to combine with that and then i have 4 plus 9 which is 13 minus 10 which is 3. so plus 3. so we end up with 8x to the fourth power minus 17x cubed plus 8x squared plus 2x plus 3. hello and welcome to algebra 2 lesson 35. in this video we're going to learn about multiplying polynomials so as i've said in a lot of videos in algebra 2 most of you will find multiplying polynomials as just a review right we learned this back in algebra 1 and it's really quite an easy process so i'm going to start at the very beginning with just multiplying two monomials together right two single term polynomials together so i have here that we can multiply any two terms together using our commutative and associative properties remember the commutative property tells us we can multiply in any order and the associative property tells us that the grouping doesn't matter when we multiply so in other words if i look at 6x squared times negative 3x to the fourth power what i can do is i can rearrange things to where i put the 6 next to the negative 3 and i could group those together and then i could multiply this by the x squared times the x to the fourth power now once i do this i make it completely obvious how to find the answer here i know that 6 times negative 3 is negative 18. so i could just write that out in front and i know that x squared times x to the fourth power from the rules of exponents x stays the same and then we would add the exponents 2 plus 4 is 6. so the answer here is negative 18 x to the sixth power let's take a look at another one just like that so we have negative 4x squared y this is multiplied by 9x to the fourth power y cubed so again i can just rearrange things to make it easy to see what i need to do let's put the coefficients in one group so let's say we have negative four multiplied by nine and then another group let's do the x variable so x squared times x to the fourth power and then in another group let's do y so we have y times y cubed so again we set it up like this we make it completely obvious what we need to do and once we kind of get rolling with this we're not going to take the time to break it down this way we'll be able to mentally multiply two terms together and write our answer but for now we do negative 4 times 9 that's negative 36 we have x squared times x to the fourth power that's x raised to the power of 2 plus 4 which is 6. then we have y times y cubed remember this exponent is understood to be one so the base y stays the same the exponent one plus the exponent three gives me an exponent of four so my answer here is negative 36 x to the sixth power y to the fourth power all right so let's kind of move on now so when we multiply two polynomials together we're basically using our distributive property so if i have a polynomial 8x so this is just a monomial it's a one-term polynomial and i multiply by this binomial okay that's inside of parenthesis 8x plus 2y what would i do well again i have something outside of parentheses so we just use our distributive property so i'd multiply 8x times 8x so 8x times 8x then plus we'd have 8x multiplied by 2y so 8x times 2y and again as i just told you we're not going to take the time to kind of regroup things every time we know that if i'm multiplying 8x times 8x the first thing i'm going to do is multiply the coefficients together 8 times 8 would be 64. then next i'd multiply my variables i have x times x again this is an understood exponent of one this is an understood exponent of one so x stays the same we add one plus one and we get two so that's 64x squared then plus now i have 8x times 2y 8 times 2 is 16 x times y is what those aren't the same variable so it's just x y now because these are not like terms i cannot simplify this any further okay so i leave the answer as 64x squared plus 16xy all right let's take a look at another one so if we had 2x squared times the quantity 3x plus 4y again you've got a monomial times a binomial so i'm going to take this guy right here this 2x squared and i'm going to distribute it to each term inside of the parentheses so i would have 2x squared times 3x then plus 2x squared times 4y so in this first part right here 2 times 3 would be 6 and then x squared times x would be x cubed again you've got to understand that this exponent is 1. then we'd have plus we'd have 2x squared times 4y so 2 times 4 is 8 and then x squared times y all i can really do there is write x squared y so this is 6x cubed plus 8x squared y all right so now we come to another section we're going to multiply two binomials together and most of you again who have taken algebra 1 know that we can use something called foil okay foil to do this but i'm not going to use foil just yet we'll get to that at the later part in the lesson i'm going to show you a different way to do this so we use our distributive property and what we basically want to think about is each term of the first polynomial is going to be multiplied by each term of the second polynomial so in other words i would take 8b the first term of the first polynomial and multiply it by the second polynomial so we'd have 8b multiplied by the quantity 2b plus 8. then i'm going to add to that i'm going to take this second term of the first polynomial and i'm going to do the same thing i'm going to multiply by each term of the second polynomial so plus we'd have negative 7 multiplied by the quantity 2b plus 8. so again i'm going to use my distributive property 8b times 2b would be 16 b squared 8b times 8 would be 64b then plus you have negative 7 times 2b that's negative 14b then you have negative 7 times 8 which would be negative 56. now in this particular case we do have like terms the two terms in the middle here can be combined because they have a common b to the first power so i'm going to rewrite 16 b squared that's not going to change 64b plus negative 14b would be 50b so plus 50b and then minus 56. now i don't have any more like terms so that's my answer 16b squared plus 50b minus 56. all right let's take a look at another one like that again we have the product of two binomials and again if you wanted to use foil at this point you could if you know it you could use it but i want to do it again a different way so i'm going to take each term of the first polynomial multiply by each term of the second so i'm going to start out with v being multiplied by the quantity six v minus six and then i'm going to add to that six multiplied by the quantity six v minus six and again using foil is just a way to keep track of things it's a way to remember or easily accomplish the same task so v times six v would be six v squared v times negative six would be minus six v then you'd have plus six times six v is 36 v and then 6 times negative 6 would be minus 36 again you have like terms in the middle here so i'm going to rewrite 6 v squared so 6 v squared negative 6 v plus 36v is plus 30v and then minus 36. all right so let's talk about one that's a little bit more tedious what if you get a trinomial times a trinomial as we get more terms involved in the multiplication it becomes much more tedious not any harder the operation is the same it's just more tedious more things to keep track of so we don't have anything like foil for this what we have to do is we have to take again each term from the first polynomial multiply it by each term of the second polynomial so in other words i'd start out with 7x squared and that's going to be multiplied by this guy right here this 8x squared plus 6x plus 5. then i move to the next term in the first polynomial so i have plus this guy right here so 8x multiplied again by this second polynomial 8x squared plus 6x plus 5. then we'd have plus two that's the final term of the first polynomial and that's going to be multiplied by the second polynomial so times eight x squared plus six x plus five and this equals so now i'm just going to go through and use again my distributive property everywhere so 7x squared times 8x squared would be 1. well we've got to think about 7 times 8 is 56 we know that x squared times x squared is x to the fourth power 7x squared times 6x 7 times 6 is 42 and then x squared times x is x cubed then we have 7x squared times 5 so we know that 7 times 5 is 35 and then x squared would come along for the ride all right so let's move down to this one now so 8x times 8x squared would be what 8 times 8 is 64 x times x squared is x cubed then 8x times 6x would be what you know that 8 times 6 is 48 x times x is x squared then we have 8x times 5 we know that 8 times 5 is 40. so plus 40 and x comes along for the ride all right we've got one more of these to do so now we've got 2 times 8x squared that's 16x squared then 2 times 6x that's going to be 12x and then 2 times 5 would be 10. so all that work we're still not done though because we have to combine like terms if you look we have what we have 56 x to the fourth power nothing to combine with that but if i look at 42x cubed i've got like terms here so i've got 42x cubed and 64x cubed so we can combine those into 106 right because 42 plus 64 is 106 and then x cubed now the next thing i'm looking at is x squared we've got 35x squared here i've got 48x squared here and i've got 16x squared there so if i added 35 48 and 16 i would get 90 not so next i'd have plus 99 x squared and let me make that 2 a little bit better and then i have x to the first power i've got 40x here i've got 12x here so 40 plus 12 we know is 52 so this would be plus 52 and then x and lastly you just have plus 10. so plus 10 nothing to combine with that so you can see how tedious a problem like this is it just takes a lot of time to just set it up get going combine like terms and the more terms you have involved the easier it is just to make a mistake so you want to just be very very careful when you're working problems with a lot of terms all right let's take a look at one more like this so if i had the quantity 8x squared minus 4xy plus 6y squared and i multiply this by the quantity 2x squared plus 6xy minus 7y squared again what i want to do is take each term from the first polynomial multiplied by each term of the second polynomial now you don't have to go through and set it up like i did the last time you can do this more mentally so i could start out with this term right here 8x squared multiplied by this polynomial here let me just highlight this real quick so we know what we're working with so this times this okay so 8x squared times 2x squared 8 times 2 is 16 x squared times x squared is x to the fourth power then next let's say 8x squared times 6xy so 8 times 6 is 48 then we'd have x squared times xy now x squared times x would be x cubed and then y would just come along for the ride now lastly i do 8x squared times negative 7y squared 8 times negative 7 is negative 56 and then x squared times y squared would be x squared y squared okay so good to go there so let me erase this we're moving down to this term right here this negative 4xy so if i did negative 4xy times 2x squared negative 4 times 2 is negative 8. x times x squared is x cubed and then y would come along for the ride then negative 4xy times 6xy we'd have negative 4 times 6 which would be negative 24. x times x is x squared y times y would be y squared let me make this 2 a little better then the last one i'm going to do here is negative 4xy times negative 7y squared negative 4 times negative 7 is positive 28. x would come along for the ride and then y times y squared would be y cubed all right let's erase this now we're on the final term of the first polynomial so 6y squared times 2x squared 6 times 2 is 12 so plus 12 and then you'd have x squared y squared then we have 6y squared times 6xy 6 times 6 is 36 then you'd have x y squared times y would be y cubed alright then lastly we have 6y squared times negative 7y squared we know that's negative 6 times 7 is 42 y squared times y squared is y to the fourth power all right so again we see how tedious this is and even if you get into something with four terms five terms six terms it just gets worse so now i'm looking for like terms and i'm just gonna start out with looking at x so i have 16x to the fourth power so let's write that first then i have x cubed here so do i find x cubed anywhere else yeah i've got x cubed and let me erase this i've got the same thing here and here so i have that anywhere else no i do not so let's go with that 48 minus 8 is 40. so let's put plus 40 i have x cubed and then y so let's look at x again and see what we have so we've got x squared y squared x squared y squared so let's go with these and then we've got x squared y squared here and that's all that it appears so negative 56 minus 24 plus 12. well that's going to give me negative 68 and then of course times x squared y squared and then again in terms of x i've got x to the first power y cubed x to the first power y cubed again let's highlight these and we want to combine those so 28 plus 36 would be 64. so i put plus 64 that's x to the first power y cubed and then lastly i have this minus 42 y to the fourth power nothing to combine with that so i've got 16x to the fourth power plus 40x cubed y minus 68x squared y squared plus 64xy cubed minus 42y to the fourth power all right so now let's talk a little bit about foil now for any of you who took an algebra 1 course you're very familiar with foil you know that this is a method to multiply two binomials together we can't use foil in any other situation so it's only the product of two binomials okay now foil just makes it very easy to keep track of what's been done because you go in an order so foil is an acronym that stands for first terms outer terms inner terms or inside terms and then last terms so as an example here we have two binomials being multiplied together again i've got two terms here i've got two terms here if you had a two term polynomial multiplied by a three term polynomial you can't use foil or if you had a monomial times a trinomial you can't use foil it's got to be two binomials okay make sure you understand that now with this i'm going to use foil so let's just write this out as f o i l we'll just take it slow so the first terms means the first term that appears in each binomial so this one times this one so 3n times n is 3n squared then you go to your outer terms so what's outside well this is outside here this is outside here so 3n times negative 4 would be negative 12n then the inside terms so what's inside so you have 1 times n or just n then your last term so this is last and this is last 1 times negative 4 is negative 4. now you would just sum these so in other words i would get 3n squared plus you have negative 12 n so negative 12 n plus n and then plus negative 4 minus 4. now let me erase this and if i sum these all i'm going to do is combine the like terms that we have in our middle here so these middle two terms can be combined so i'd have 3n squared negative 12n plus n is negative 11n and then minus 4. and it's just that simple and a lot of times people can get really good at this and do foil in their head and they can just simply observe this and write the answer down especially for something that's pretty simple all right let's take a look at another one so we have 5m minus 2 that quantity times m minus 4. so again if i want to use foil i would do my first terms so 5m times m that's 5m squared then my outside terms 5m times negative 4 is negative m then my inner terms negative 2 times m is negative 2 m then my last terms negative 2 times negative 4 is positive 8. i would again combine the like terms in the middle so i'd start with 5 5m squared in the middle negative 20m minus 2m would be negative 22m and then plus 8. all right as another example let's say we have the quantity 2x minus 5y times the quantity 3x minus 4y so again i'm going to use foil so my first terms 2x times 3x would be 6x squared my outer 2x times negative 4y would be negative 8xy my inner negative 5y times 3x would be negative 15xy and then my last negative 5y times negative 4y would be positive 20y squared so again i'm going to combine the like terms in the middle here and so i'd have 6x squared negative 8 minus 15 is going to give me negative 23. so you'd have negative 23xy and then plus 20y squared all right let's take a look at one more of these so we have the quantity 8a plus 3b times the quantity 6a minus 5b again to use foil to start with the first terms 8a times 6a is 48 a squared then the outer 8a times negative five b would be negative forty a b then the inner three b times six a would be positive eighteen a b then the last three b times negative five b would be negative 15 b squared so what i want to do is i want to combine like terms in the middle in the middle right there so i'd have 48 a squared negative 40 a b plus 18 a b would end up giving me negative 22 a b and then minus 15 b squared so i'd have 48 a squared minus 22 a b minus 15 b squared all right so the last thing i want to talk about is what to do when you have more than two polynomials that are being multiplied together so a half here we can multiply more than two polynomials by multiplying pairs of polynomials until we have our product so this is the same thing as if i had let's say three times seven times two i can multiply any two together to start and then multiply that result by the third i could say 3 times 7 is 21 21 times 2 is 42 or i could say 7 times 2 is 14 14 times 3 is 42 or i could say 3 times 2 is 6 6 times 7 is 42. no matter how you do this you end up with 42. it's the same thing when we're talking about multiplying three polynomials together i could take 7b and multiply by the quantity 6b plus 3 then take that result and multiply it by the quantity 3 b minus 3 or i could do something different i can multiply these two together first multiply that by 7b again it does not matter right as long as everything gets multiplied together so what i'm going to do is i'm going to start out just going straight across i'm going to use my distributive property and say 7b times 6b is going to give me 42 b squared then 7b times 3 would be plus 21b so you'd have this now this quantity multiplied by this quantity 3b minus 3. now i've got two binomials here so i can use foil so to begin my first terms 42b squared times 3b would be 126 and then b cubed my outer 42b squared times negative 3 we know that's negative again 42 times 3 is 126 so negative 126 and now we just have b squared for the inside terms we have 21b times 3b 21 times 3 is 63. so plus 63 b times b is b squared and then for the last terms 21 b times negative 3 you'd have negative 63 b so in the middle here we can combine like terms we have negative 126 b squared we have positive 63 b squared so what we have is a 126 b cubed negative 126 plus 63 is negative 63 and then b squared and then minus 63 b all right let's take a look at one final problem so we're going to multiply three binomials together so again not a hard process just something that's tedious you're going to find the product of two of them first doesn't matter which two and then multiply that result by the third so let's go ahead and start out by just saying that i'm going to multiply the first one by the second one so my first terms 3x times 4x would be 12x squared the outer 3x times negative 4y would be negative 12xy the inner 2y times 4x would be positive 8xy and then for the last 2y times negative 4y that would be negative 8y squared now i can combine like terms in the middle negative 12xy plus 8xy would be negative 4xy so this would be multiplied by this final binomial here which is 6x minus 1. now i can no longer use foil because i have a trinomial times a binomial so i can't go okay first outer inner it doesn't work that way so what i'm going to do is i'm going to take each term from this binomial and multiply it by each term of the trinomial so in other words i'm going to take 6x to start and multiply it by this trinomial here so 6x times 12x squared 6 times 12 is 72 x times x squared is x cubed then next 6x times negative 4xy that's negative 24 x squared y 6x times negative 8y squared that's negative 48 x y squared now the next one's pretty easy i've got a negative 1 that's multiplying this trinomial here if i multiply everything by negative 1 i'm just changing the sign so i would go through here and just say okay well 12x squared will become negative 12x squared negative 4xy would become positive 4xy and then negative 8y squared would be positive 8y squared so that's all it does multiply by negative one you're just changing the sign so now i would just look to see if i can simplify this any further and i actually can't there are no like terms here we end up with 72x cubed minus 24x squared y minus 48xy squared minus 12x squared plus 4xy plus 8y squared hello and welcome to algebra 2 lesson 36. in this video we're going to learn about special products so again we've come to another scenario where if you've taken algebra 1 this lesson is nothing more than a refresher for you but it's an important refresher because we're going to look at polynomial products that occur so frequently that it's extremely useful to memorize the pattern associated with these products and you might do it just to be faster in your homework or on your tests but i can tell you when you start taking these standardized tests like the sat the act or even if you're going to grad school even if you've graduated from college already you want to take your gre or gmat or whatever you're taking those tests are designed with a time limit that in some cases requires you to be able to spit out these products very very quickly okay so this is just something you can use to do that so i want to start out with something that's very very basic something that most of you know so if i have something like the quantity x plus y times the quantity x minus y we can say this is equal to x squared minus y squared now first i want to do this with foil and then i want to show you why it ends up this way and how we can translate something generic like this to our actual problem so if i did foil here if i did x times x that would be x squared then if i did x times negative y that'd be minus xy then if i did y times x it'd be plus xy and then if i did y times negative y i'd have minus y squared so what happens with this the middle two terms are going to cancel and i'm left with x squared minus y squared so this is x squared minus y squared so you're looking to make sure that you have the same term in the first position of each binomial so i have an x and an x i've got to have different signs i have a plus and a minus and then i've got to have the same term in the second or last position of each binomial as well i have a y and a y so if it fits that pattern you can go ahead and say that you have the first guy or the first item that's going to be squared then we're subtracting away the second or the last guy squared and this is referred to as the difference of two squares i have something squared here minus something squared here so as another example let's say you saw the quantity v plus one times the quantity v minus one well again i think about this is v this is v so same item in the first position of each then this is one this is one same thing in the second position or last position of each then i have a positive and a negative so i have different sides so without going through and doing foil i can say i have the first guy which is v squared that's what i have there subtract away the second guy which is 1 squared 1 squared is 1. so i get v squared minus 1. and again if you want to prove this to yourself you can use foil v times v is v squared the outer would be minus v the inner would be plus v so we know those would cancel and the last would be negative 1 right 1 times negative 1 is negative 1. so again these cancel i end up with v squared minus 1. all right let's take a look at some more examples all right so let's say we had something like the quantity 8a minus 6 times the quantity 8a plus 6. so it's the same thing here i have 8a and 8a then i have 6 and 6 and your signs are different i have minus and i have plus so i don't have to go through and use foil i could do this very very quickly by just squaring 8a so 8a squared okay you've got to be careful that you're squaring this whole thing this whole term here i can't just do 8a squared like that i see that all the time that's not going to work i've got to square that whole term so i would square 8 8 squared is 64. i would square a that's just a squared then minus i'm going to look at my 6 and that's going to be squared 6 squared is 36. so just that quickly without doing foil or anything else i know my answer is 64a squared minus 36. and again we can prove it to ourselves 8a times 8a is 64 a squared the outer 8a times 6 would be positive 48a the inner negative 6 times 8a would be minus 48a and then your last negative 6 times 6 would be minus 36. so again you can see that these middle two terms are going to cancel positive 48a and negative 48a would cancel so 64a squared minus 36 is what you are left with so just a much quicker way to do this versus using foil let's take a look at another one so we have the quantity negative 6x minus 3y being multiplied by the quantity negative 6x plus 3y so again i'm looking at a negative 6x here and a negative 6x here i'm looking at a 3y here and a 3y here now the signs are different i have a negative and i have a positive when i do my pattern again i'm thinking about the first thing that occurs so this right here normally i'm not including the sign when i'm thinking about the second one in this case i need to include that negative when i think about negative 6x i can't just leave it off so i would square that so if i squared negative 6x it would look like this so that negative is inside the parentheses so it's basically negative 1 squared which is 1. 6 is squared 6 squared is 36 and x is squared x squared is just x squared so this is 36x squared then i'm subtracting away what is in the second position here it's 3y again when we think about what's in the second position i never think about the sign i just think about this right here so 3y if i squared 3 i'd get 9. if i squared y i'd have y squared and boom there's my answer 36x squared minus 9y squared just that quick but again on this one kind of the trap is to make sure that you include the sign when you think about that first term all right let's take a look at one more of these so suppose you have the quantity 4u minus 7v and you multiply it by the quantity 4u plus 7v so i would take this guy right here this 4u and i would square it so 4 squared is 16 u squared is just u squared then minus take this guy right here and square 7 squared is 49 and v squared is just v squared and then again boom just that quick i have my answer 16 u squared minus 49 v squared so again we're kind of done with this section i just want to recap and talk a little bit about this for a second you're looking for this pattern where you have the same thing in the first position of each in this particular case it's 4u and 4u then the second position of each has the same thing so 7v and 7v and then your signs are different you have a negative and you have a positive so that leads to this the first guy which is 4u squared minus the second guy which is 7v squared so you just memorize that pattern you can put it on a flashcard or something where you can flip through memorize it as you do your homework as you do your tests as you get into your standardized exams this is going to be very beneficial for you all right let's look at a different section now all right so the next thing i want to talk about is the square of a binomial and we have two different scenarios so we have the quantity x plus y squared this leads to x squared plus 2xy plus y squared so notice that when you have a plus sign here you have a plus sign here and here okay so i have this scenario i don't need to really think about the signs everything is going to be a plus now in the other scenario we have the quantity x minus y squared i have x squared minus 2xy plus y squared here i have a minus in the middle and so that leads to a minus in front of the second term so everything else is the same it's x squared minus 2xy plus y squared the terms are the same it's just this one sign that's different okay so it's very easy to memorize this one and it's very easy to apply it let's take a look at an example all right so we start out with the quantity 3x minus 8 and this is squared so this is a binomial squared so what would i do here well let me just write the pattern that we should know so if we do x minus y that quantity squared it's the first chi squared so it's x squared then remember we have a minus in front of the second term because we have a minus here then it's 2 times x times y then it's plus this guy squared so y squared so just match that pattern okay so if i take the first guy and i square so 3x and that's squared again you've got to make sure you square the 3 and the x then we would subtract away you have 2 multiplied by x times y so x and y just represent the first term and the second term here so 2 multiplied by 3x multiplied by 8. okay then the next thing we have is plus y squared so i'm going to add to this whatever is in the last position which is 8 squared so 8 squared and this may take you a few times to kind of get it down what represents what but x in this case is the first term so that's representing 3x here the first term y represents the second term here that represents 8 because 8 is the second term okay so once we've set that up let me erase this we don't need this anymore we will say 3x squared with the parentheses around the 3 and the x 3 squared would be 9 then x squared is just x squared and then we're subtracting away we have 2 times 3 times 8 2 times 3 is 6 6 times 8 is 48 and then times x and then lastly plus 8 squared is 64. so we get 9x squared minus 48x plus 64. and just really quickly i'm going to prove to you that this works and i'm going to show you how much longer it takes so if i had 3x minus 8 that quantity if it's squared i'm multiplying it by 3x minus 8 again don't make the mistake of saying the quantity 3x minus 8 squared is equal to 3x squared minus 8 squared that's wrong okay please don't make that mistake this has to be expanded out all right so once we have in this format we can use foil 3x times 3x is 9x squared the outer 3x times negative 8 would be negative 24x the inner negative 8 times 3x would be negative 24x and then the last negative 8 times negative 8 is positive 64. you can see you can combine like terms in the middle so you would end up with 9x squared minus 48x right negative 24x minus 24x is negative 48x then plus 64. now you can see that this answer is the same as this answer we just got this answer more quickly because we used a pattern right we recognized how we could get the answer we applied it we got it very quickly all right let's take a look at another one so we have 7n squared plus 8 that quantity and it's squared so again if i think about x plus y that quantity squared this would be what the first guy squared so x squared plus 2 times x times y plus the last guy squared and so the way i actually remember this i take this guy and say okay this is squared and i know this is squared so that goes at each end then i remember my exponent is 2 that's involved in a multiplication in the middle so i just say okay well i have 2 times x times y that's how i get the middle part 2xy that's what works for me i'm not sure what will work for you but you can certainly try that now if i go through and try to apply this again i have to match things up so in this case x which is the first term here represents 7n squared which is the first term here so those match up then y which is the second term here matches up with 8 which is the second term here so those match up so what i'm going to do is just say okay in this one i have the first term squared so 7 n squared will be squared then plus i have 2 times the first term times the second term so 2 times 7n squared times the second term which is 8 then plus y squared so it's the last term squared in this case that's 8. so 8 squared all right so let's erase this and let's just crank this out now so if we have 7n squared squared 7 squared is 49 n squared squared power of the power rule so n stays the same multiply 2 times 2 that's 4 then plus you'd have 2 times 7 times 8. i can do 2 times 7 first that's 14 14 times 8 is 112 so this would be 112 and then times n squared then lastly plus if we have 8 squared that's 64. so we end up with 49 n to the fourth power plus 112 n squared plus 64. let's take a look at another one let's suppose we had x minus 4y and this quantity is squared so i'm going to write out my pattern and i'm going to go with x minus y that quantity squared so we should know at this point as what it's x the first term squared then minus remember if you have a minus here you have a minus in front of the second term 2 times x times y 2 times x times y and then plus the last term squared y squared so just apply that here the first term squared so x is squared minus 2 times the first term times the second term so 2 times x times 4y and then plus the last term squared so 4y squared okay we can erase this now we don't need it and so we'd have x squared minus 2 times 4 is 8 so you'd have 8xy and then plus 4 squared is 16 y squared is y squared so you get x squared minus 8xy plus 16y squared all right let's take a look at one more of these so i have the quantity negative a squared minus 9b and this is squared now this one's a little tricky and i'm going to get to that in a second again if i have a subtraction in the middle i use this formula so x minus y that quantity squared this is x squared minus 2xy plus y squared now what makes this tricky is this negative right here this negative a normally we have a positive there right we don't think about our signs we just use whatever's there if you have a negative here you've got to account for that by putting it in okay so that's what makes it tricky a lot of students say okay i'll just put an a squared in there well a lot of students will throw this negative in there when they think about 9b right this sign right here in the middle is already figured in for you so you never need to use it okay you never need to think about it if i have a negative that's thrown at me there though i do need to put it in okay so that's the difference so what i want to do here i want to say i have the first guy squared so the negative a squared the whole thing so that's going to be squared then i subtract away 2 times the first guy again i've got to put the negative a squared in times the second guy now for the second guy i never include that negative or that subtraction sign out there i just take whatever's in that position just like here i just take y right i don't put negative y in so you've got to follow that formula to the t i'm going to put 9b in there and then lastly i'm going to have plus this guy 9b squared so 9b squared all right let's erase this and let's crank this up so if i have negative a squared squared i know this is basically negative 1 squared that would be one so this would be positive and then a squared squared keep a the same and multiply exponents 2 times 2 is 4. then next we would look at we have negative 2 times negative a squared times 9b negative times negative is positive so i know this would be plus and we think about 2 times 9 that's 18 and we'd have a squared times b so a squared b then lastly we have 9b that amount squared so plus 9 squared is 81. b squared is just b squared so we end up with a to the fourth power plus 18 a squared b plus 81 b squared so just something you want to look out for is when they throw these signs at you if you put a negative out in front of that first term okay so pay close attention to that all right so the last thing i want to talk about is when you have a binomial cubed so this one's a little bit more tedious but this one will also save you more time because as you multiply more terms it gets more tedious it becomes more easy to make a mistake and it takes you longer to do the process so memorizing these will have the greatest value for you so if i have the quantity x plus y and it's cubed i get x cubed plus 3x squared y plus 3xy squared plus y cubed now if i have the quantity x minus y and that's cubed it's the same thing only the second position and the fourth position have minus signs in front so x cubed minus 3x squared y plus 3xy squared minus y cubed now in terms of memorizing this guy it's a little bit more complicated i always just remember that i have x cubed and y cubed at the beginning and at the end so that's easy then i know i'm multiplying with a three in each of the two middle terms just like when we looked at the square of a binomial i knew i was multiplying by a two and it was two times the first term times the second term that's pretty easy to remember here you have a square involved in each case so it's kind of goes back and forth but if you just remember that the first term is squared in one case the second term is squared in the other case so it's three times one of these squared times one of these to the first power in each case in the first case we have three times the first term squared times the second term then we have three times the first term which is not squared times the second term which is squared okay and then the only difference between this one where we have a plus and this one we have a minus is the minus sign in front of the second and fourth terms and again i know this seems daunting but you work 10 or 15 problems looking at this and representing it and going back and forth you're going to have this down and it's going to save you a tremendous amount of time all right let's look at an easy one to start so let's suppose we had x plus 3 that quantity cubed well what's going to happen again if i put the quantity x plus y cubed what is this again i know i have x cubed in the beginning and i know i have y cubed at the end i know that i have two terms in the middle and i know it involves multiplication with three right so three and then three now all you need to remember is that i'm multiplying by the two terms here one of them is squared in each case one of them is not so if i start out by just saying okay x would be squared y is not then if i say okay now x will not be squared y will be squared boom got my answer it's just that simple so i'm just going to apply that technique here so i have x plus 3 that quantity cubed so the first guy is cubed so x cubed the last guy is cubed so 3 cubed let me put that all the way down here now i'm going to have 3 multiplying by these two terms for the two middle terms so let me put that in there and then in each case one of them will be squared one of them will not so i'm going to square x 3 will not be squared then i'm going to square 3 x will not be squared and that's it i've got my answer it's really really that simple all right so let's erase this and you can see how much quicker this would be versus kind of expanding this out into x plus 3 that quantity times x plus 3 that quantity times x plus 3 that quantity this will take a while okay doing it this way is going to be much much quicker for you all right so we have x cubed to start then we have three x squared times three three times three is nine so plus nine x squared then we'd have plus you have three times three squared times x so 3 squared is 9 9 x 3 is 27 27 times x is just 27x and then lastly plus 3 cubed which is 27. so we get x cubed plus 9x squared plus 27x plus 27. all right let's look at another easy one so we have the quantity x minus 2 and that's cubed so can we do this without writing out the pattern i think that we can so again i know that i want this first thing cubed so x cubed i know that i want this last thing cubed so 2 cubed and i'm going to go ahead and write that as 8. we know 2 cubed is 8. so let me write that all the way down here now as far as signs go i know in front of the second term and the final term or the fourth term i have minus signs so there'll be something here there'll be a plus sign and then something here now what do we remember we remember that these two middle terms here have a 3 that's leading so let's put a 3 in each case and i'm multiplying by each term here it's just that one of them will be squared in each case 1 will not so we start out with x being squared and then 2 not being squared then we'll have x not being squared and 2 being squared boom and we're done it's just that easy okay so we have x cubed minus 3 times 2 is 6 6 times x squared is 6x squared and then plus 2 squared is 4 4 times 3 is 12 12 times x is 12x and then minus 8. we're done okay and again think about how much time that saved us and we had to go through and do the pattern i could just kind of do this mentally and have this answer in a few seconds okay and you'll be able to do that too so if you're on a standardized test and actually for the quantity x minus 2 cubed you can get that answer in a few seconds and kind of move on versus somebody's got to take out scratch paper you know expand this into the quantity x minus 2 times x minus 2 times x minus 2 they might be there for a full minute okay so you're ahead of somebody by a full minute now that's time you can apply to a more difficult problem right so let's look at one final one of these it looks a little bit more complicated but really it's not we have the quantity 6z minus 4x and this is cubed so again what am i going to do i know that i want this first guy cubed so 6z i know i want that cubed i know i want this last guy cube so let me throw this all the way down here and say we have 4x and that's cubed i know that i'll have a minus sign in front of the second guy and the fourth guy so i'll have something here i'll have a plus sign and something here let me kind of move that down a little bit so it's more even and i know that i'm going to have a 3 that's multiplying the two middle terms now in each case one of the terms when we talk about the middle terms will be squared one of them will not so i'm going to multiply this by i'll have 6z squared times 4x in this case i'll have 6z not squared and 4x as squared and that's it i always have to go through and do some calculations so 6z cubed the 6 is cubed 6 times 6 is 36 36 times 6 is 216. z cubed is just z cubed then we'd have minus so then we'd have 6z squared which is basically going to be 36z squared and 36 times 3 is 108 then if i multiply by 4 i get 432. so minus 432 you would have x and then z squared so x z squared and then plus you have 3 times 6z times 4x that amount squared so this right here would be 4 squared times x squared we know 4 squared is 16. so let's write 16 x squared for that so 3 times 6 is 18 then 18 times 16 would be 288 and you'd have x squared z then lastly you have minus you have 4x cubed 4 cubed is 64 and then x cubed is just x cubed so we have 216 z cubed minus 432 x z squared plus 288 x squared z minus 64 x cubed hello and welcome to algebra 2 lesson 37 in this video we're going to learn about dividing polynomials so again for most of you if you've taken an algebra 1 course you already know how to divide polynomials and this lesson is nothing more than a review for you if you haven't taken algebra 1 or if you kind of struggle through algebra 1 you can pick this topic up from this lesson i'm going to kind of start at the very beginning i'm going to go a little quicker than i did in algebra 1 but still you can pick it up from this lesson the first thing i want to talk about is the easiest scenario if we divide a polynomial which is not a monomial meaning it's not a single term only by a monomial meaning it is a single term only so that's what we have here in this first example let's say i had 2p to the fourth power plus 4p cubed plus 12p squared and it's divided by 4p the very first thing you want to do is write this as a fraction so let's take this as the numerator of the fraction so 2 p to the fourth power plus 4 p cubed plus 12 p squared so that's going to be my numerator that's my dividend then i'm going to put this over my denominator or my divisor of 4p now we know from our work with fractions that we could split this up and write this as each term of the polynomial in the numerator over that monomial from the denominator in other words i can say this is 2p to the fourth power over 4p plus 4p cubed over 4p plus 12p squared over 4p and you can just think about taking this format right there's a common denominator and going to this format it's no different than if i had something like six over two plus let's say four over two right i could say this is six plus four over two six plus four is of course ten ten over two is five right so it's the same thing all right so let's erase this and let's focus on this so what i want to do is attack this problem piece by piece and what i mean by that is i'm going to start right here all the way on the left and i'm going to say 2 over 4 well i know 4 divided by 2 is 2. so i can cancel this with this i'll put a 1 here and a 2 here p to the 4th power over p to the first power i can cancel one factor of p so this would cancel and this will become a three then over here i go to this part next i have four p cubed over four p the fours will cancel p cubed over p to the first power is p squared then i go to this part i have 12 p squared over 4p 12 divided by 4 is 3 p squared over p is p to the first power so just look at what is remaining to write our answer we have 1p cubed or just p cubed this is over 2. then plus we have p squared then plus next we have 3p so we have 3p all right so p cubed over 2 plus p squared plus 3p so that's your answer now we know with division we can check our answer whatever our quotient is in this case this is the quotient this is the quotient we could multiply this by the divisor and we would get the dividend back and i'm not going to do that in the interest of time but you could pause the video and you could multiply 4p by each term here and you would come back to 2p to the fourth power plus 4p cubed plus 12p squared all right let's take a look at one more like this so let's say we have 24k to the fifth power plus 24k to the fourth power plus 12k cubed and this is divided by 4k cubed again the first thing i would do in this scenario is write this as a fraction so my dividend goes in the numerator so 24k to the fifth power plus 24k to the fourth power plus 12 k cubed then this is over 4k cubed and then i'm going to split this up so i'm going to say okay i have 24 k to the fifth power over 4k cubed then plus 24k to the fourth power over 4k cubed then plus 12 k cubed over 4k cubed all right so now what i want to do is just go through and simplify let me put equals here 24 divided by 4 is 6. k to the fifth power over k cubed would be k squared so the first result here would be six k squared then i'm gonna put plus for the next scenario again i have 24 over four that's six i have k to the fourth power over k cubed so this cancels and i have k to the first power so i have plus six k then lastly i have 12 over four which we know is three and then k cubed over k cubed is one so basically that's just plus 3 there so your answer or your quotient here is 6 k squared plus 6k plus 3. now again this is division so if you wanted to check it you could take your quotient right here multiply it by your divisor which is 4k cubed and you will get your dividend back which is 24k to the fifth power plus 24k to the fourth power plus 12k cubed okay so that's the very easy scenario that we first deal with when we're dividing polynomials in an algebra 1 course you usually have a section that's just devoted to that learning how to start out with dividing a polynomial by a monomial then you kind of move on and you look at two different other sections the first thing is dividing polynomials with long division so you divide a polynomial by another polynomial that is not a monomial and then you also hit another section where you do this and you have missing terms involved so the process is similar to when we divided whole numbers so if you are good at long division you might want to do a few problems just kind of get your brain reacclimated to that we're pretty much going to do the same thing here the first step to this is to write each polynomial in standard form and if you have a polynomial that's missing any powers for your variable you want to put 0 as a placeholder so i would start out with this example here 45n plus 5n cubed minus 18 plus 38n squared over n plus 6. so for right now i'm just going to keep it in this format but i'm going to write everything in standard form so remember i want the highest power on n all the way to the left so that's going to be 5 n cubed so 5 n cubed then next i'm looking for squared so plus 38 n squared then i'm looking for first power so plus 45 n then i'm looking for a constant or you could say n to the power of zero so minus 18 okay so that's in standard form now i'm not missing any powers for my variable i have a third power second power first power and basically power of 0. so i don't need to write 0 as the coefficient for any missing powers but we'll see an example of that in a little while so then this is over we have n plus 6. so that's already in standard form i have n to the first power plus my constant which is 6. all right let me erase this we don't need that now what we want to do is we want to write our problem in long division format so you'll recall if you had something like let's say i'm going to do this on a different page 651 divided by let's say 31 something like that well we could write this as 651 divided by 31 like that or we could also write this as 651 divided by 31 like that this is just long division format the 651 is the dividend this is the dividend and the 31 is the divisor okay the divisor so the divisor goes to the left of our symbol and the dividend goes underneath so we're going to do the same thing and what i'm going to do let me just kind of scroll down a little bit so we have some room i'm going to take the dividend here and let me label that this is the dividend and i'm going to put it under the long division symbol so i have my long division symbol and i'm going to write my dividend underneath so 5 n cubed plus 38 n squared plus 45 n minus 18. and then my divisor my divisor goes out to the left so n plus six okay so the process for long division with whole numbers as you'll recall let me kind of go back down to this example i'm going to erase this real fast it's going to give us some room and let me just scooch this to the left we use this d m s b r to remember this this is divide multiply subtract bring down and then repeat or remainder a lot of teachers will remind you of the family unit and use a little trick to remember this they'll say dad mom sister brother and then rover right so the family unit a mom and a dad a brother and a sister and then a family dog so i would start out with this first step of division i would say how many times does 31 which is your divisor go into and i'd start out with this piece by piece so say how many times does it go into 6 well it doesn't so you could put a 0 here and you could go through kind of the steps there or you could just make it a little quicker and you could expand this and say how many times does 31 go into 65 well 31 goes into 65 twice so i would put a 2 over the 5 right because that 5 is in the tens place that 2 stands for 20 and then i would multiply okay that mom step so 2 times 31 is 62 then i would subtract the sister step 65 minus 62 is 3 then i would bring down the brother step so this one would come down then i would repeat or remainder which is the rover step in this case we're going to repeat the process so we're coming all the way back up here and we're going to divide again so now 31 goes into 31 one time we're going to multiply 1 times 31 is 31 we're going to subtract 31 minus 31 is zero there's nothing to bring down we're at the step where we would repeat which we're not going to do or we'd have a remainder and there is no remainder so we saw that 651 divided by 31 is 21 okay so it's basically the same thing when we start doing our division here the only real difference here is i'm only going to think about leading terms so when i look at my divisor here i'm going to start by saying leading term into leading term so the question would be how many times does n go into 5n cubed that's my first step or my division step so i would say 5n cubed over n equals what well i know 5 is not going anywhere it's just n cubed over n so this would cancel with this and be this right n cubed over n would be n squared so you'd have 5 times n squared now a lot of students make this mistake and they'll just say okay well i'll just put 5n squared here and we'll just keep going you don't want to do that this has place value just like what we just saw this is the place to put n cubed right here in this call this is the place for n squared this is the place for n this is the place for your constants so my answer is 5n squared so i want to put it here over the 38 n squared once we've kind of done that first step the next thing is to multiply now when i multiply i multiplied by everything so i multiply this result here by this whole thing here so 5n squared times n would be 5n cubed then 5n squared times 6 would be 30 n squared now i want to subtract away and the main mistake here is that people just put a subtraction symbol out in front and they start subtracting so really i want parentheses around this and what i can do is i can just change the sign of each term that's being subtracted away and add so a quick trick to this would just be okay i list everything and then i just change the size so this would be negative and then this would be negative now i can just add so i put a little bar here and i'll add 5n cubed minus 5n cubed that's 0. 38n squared minus 30n squared is 8n squared now my next step is to bring it down so the next term is here i'll bring that down so plus 45n and then i just repeat the process again i'm going to go leading term into leading term so n goes into 8n squared and again if you want to write it out on the side 8n squared over n to figure that out that's fine most of you at this point can kind of do that in your head okay we know that 8n squared divided by n would simply be 8n so we would write plus 8n and look how it lines up with the place value this is where 45 n to the first power is this is where 8n to the first power so again we're maintaining our place value there now the next thing is again to multiply so 8n times n is 8n squared and then 8n times 6 is 48n now i can put a minus sign out in front and put my parentheses again the shortcut to this is just to remember i'm going to change the sign of everything there and i'm going to add so 8n squared minus 8n squared would be 0 and then 45n minus 48n would be negative 3n again i want to bring down as my next step so i'm going to bring down that negative 18. and then once again for the last time i'm going to go leading term into leading term okay so n goes into negative 3n how many times you're asking yourself the question what is negative 3n divided by n so n cancels within and i'm just going to get negative 3. so now i multiply so what is negative 3 times n that's negative 3n what is negative 3 times 6 that is negative 18. now you have the same thing here and here if i subtract the same thing away i'm going to get 0 right and you prove that to yourself again by changing the sign of everything and adding so this becomes plus and this becomes plus and we add negative 3n plus 3n is 0. negative 18 plus 18 is 0. so there's no remainder okay and sometimes you have a remainder sometimes you don't okay in this case we don't so our quotient is 5n squared plus 8n minus 3. so i just want to show you something let me erase kind of everything except for the answer and i'm going to prove to you that this works let me kind of scroll up a little bit i'll just kind of drag this up here all right so what i want to do in order to prove this i want to multiply the divisor by the quotient that should give me the dividend back if i got the correct answer this is how you can always guard against mistakes it's just like if i had 6 divided by 3 and this equals 2 i could say 2 times 3 gives me 6. it's the same thing i'm saying 5n squared plus 8n minus 3 times the quantity n plus 6 is going to give me 5n cubed plus 38n squared plus 45n minus 18. it's a more complicated example of that but it's the same concept all right so let's go ahead and do this if we had n plus 6 times 5 n squared plus 8 n minus three scroll down a little bit so n times five n squared is five n cubed then n times eight n would be plus eight n squared then n times negative 3 would be minus 3n then 6 times 5n squared would be plus 30n squared then 6 times 8n would be plus 48n then 6 times negative 3 would be minus 18. all right so if i combine like terms here i'll have 5n cubed and then 8n squared plus 30n squared would be plus 38n squared and then negative 3n plus 48n would be plus 45n and then minus 18. now this is exactly what we had as our dividend 5n cubed plus 38n squared plus 45n minus 18. so you can see that your division here is correct when i take the dividend and divide by the divisor i get the quotient so my quotient which is here times my divisor which is here gave me back my dividend which is here so you're always able to check your division with multiplication all right let's take a look at one where you have some missing terms involved so we have 32x to the fourth power minus 15x minus 50 plus 12x cubed this is over 4x squared minus 5. so again i want everything to be in standard form and anytime i have some missing terms i want to put 0 as a placeholder so this is a perfect example of that so in the numerator i've got 32 x to the fourth power i've got plus 12 x cubed now in order i have x to the fourth x cubed there is no x squared here so i can't just leave it off i've got to have it in there to do my division so what i want to do is i want to put plus 0x squared it's just a placeholder okay that's all it's doing okay so we need to have that then i put minus 15x and then minus 50. so now this is in standard form and we accounted for our missing term all right so for the divisor here we have 4x squared minus 5. well i have 4x squared i'm missing the x to the first power so i'm going to put plus 0x and then minus 5. all right so let's go ahead and set this up and we're going to do 32 x to the fourth power plus 12 x cubed plus zero x squared minus 15 x minus 50. we're going to put that under our long division symbol and then we're going to put our divisor out to the side so 4x squared plus 0x minus 5. so i'm going to go leading term into leading term so what is 32 x to the fourth power over 4 x squared well i know that 32 divided by 4 would give me 8. x to the fourth power over x squared would be x squared so i would get 8x squared as an answer now where am i going to put it i'm going to put it over the x to the fourth power now i'm going to put it over the x cubed i'm going to put it over the x squared so i'm going to have 8x squared right there let me erase all this and we're going to multiply let me kind of scroll down so we have enough room remember i'm going to multiply this by this whole thing so 8x squared times 4x squared would give me 32 x to the fourth power 8x squared times 0x would be plus 0x cubed and then 8x squared times negative 5 would be minus 40 x squared now i'm subtracting this whole thing away so again i want to just change the sign of everything so this would be minus this would be minus and this would be plus so let's put our little bar there 32x to the fourth power minus 32x to the fourth power is 0. 12x cubed minus 0x cubed is 12x cubed 0x squared plus 40x squared is 40x squared and then we want to bring down the next term so bring down negative 15x let's go ahead and go leading term into leading term so what is 12x cubed over 4x squared well we know that 12 divided by 4 is 3 x cubed over x squared is x to the first power so this would be 3x to the first power or just 3x and again i want to do my multiplication so if i do 3x times 4x squared i'm going to get 12 x cubed if i do 3x times 0x i'm going to get plus 0x squared then 3x times negative 5 is going to give me minus 15x okay so we want to change the sign of everything and then add so this would be negative this will be negative and this will be positive put our bar so 12x cubed minus 12x cubed would give me 0. 40x squared minus 0x squared would be 40x squared and then negative 15x plus 15x is 0. so i would bring down my next term here of negative 50 and i would go leading term into leading term so i would have 40x squared over 4x squared and so what's going to happen is this would cancel with this and i'll have 10. x squared over x squared is 1. so this would be plus 10. now i'd multiply let me erase all this 10 multiplied by 4x squared is 40x squared then if i had 10 times 0x remember 0x is just 0. so if i multiply 10 times 0 i get 0. so i could put a 0 here i could put plus 0 here and plus 0 here it doesn't matter you can even put plus 0x and plus 0x just as placeholders so things make sense then lastly i could do 10 times negative 5 that's negative 50 and you can see that this is going to result in zero right so if i change the sign of everything make this negative this negative and this positive 40x squared minus 40x squared is 0. 0x minus 0x is 0 negative 50 plus 50 is 0. so there's no remainder here we end up with 8x squared plus 3x plus 10. all right let's just kind of wrap up the lesson now just look at one with a remainder so we have m cubed plus 10 m squared plus 27 m plus 12 over m plus 5. so everything's already in standard form for you and everything is there right we're not missing any terms so i can go straight into my long division here i can just say i have m cubed plus 10 m squared plus 27 m plus 12. okay that's my dividend putting that under my long division symbol and then out to the left i'm going to have my m plus 5 my divisor so i'm going to go leading term into leading terms so m goes into m cubed how many times well we know m cubed over m would be m squared so that goes right here we have m squared m squared times m is m cubed m squared times five is plus five m again if i'm subtracting this whole thing away change the sign of everything and then add m cubed minus m cubed is 0. 10m squared minus 5m squared is going to give me 5 m squared the next thing i want to do is bring down this 27 m so plus 27 m and i just want to go leading term into leading term so m goes into 5m squared how many times well we know m squared over m is m so i would basically have that 5 times m so plus 5 m then i multiply 5 m times m is 5 m squared 5 m times five would be plus 25 m and scroll down a little bit get some room going now i want to subtract this away so this becomes minus this becomes minus change the sign of each term and then you can just add so five m squared minus five m squared is zero 27 m minus 25 m is 2m bring down your last term there so plus 12. and we're going to do one final time leading term into leading term so m into 2m you think about 2m divided by m the m's would cancel you'd be left with 2. so then 2 times m would be 2m and then 2 times 5 would be 10. so we subtract this away so this becomes minus and this becomes minus and we can just add 2m minus 2m is 0 12 minus 10 is 2. nothing else to bring down this is a remainder remainder okay we'll remember those from long division right basic long division now all we need to do with the remainder is write it over the divisor what that means is if i write my answer i have m squared plus five m plus two then plus here's my remainder it's two over the divisor which is m plus 5. so this is the remainder and it's over the divisor okay that's how you want to write this let's say you wanted to check the result what would you do well you could multiply this whole thing here by the divisor or you could multiply the quotient part by the divisor and then just add the remainder so just say okay plus 2. and it's the same thing it's a little faster if you just multiply everything as it's presented here so let me copy this so let's say i just multiply this by the divisor m plus five so i could take and distribute this m plus five to each term right i could write this as m squared times the quantity m plus five then plus five m times the quantity m plus five then plus two times the quantity m plus five then plus two over m plus five times the quantity m plus five now the reason i do that is so that when we get down here this cancels with this and i'm just adding 2 at the end so that's why it's the same either way if i just didn't have this in here and i did this multiplication and then i just added 2 i'm getting the same thing okay so that's why i'm saying it's the same either way all right so m squared times m is m cubed plus m squared times 5 is 5m squared then plus 5m times m is 5m squared then plus 5m times 5 is 25m then plus 2 times m is 2m then plus 2 times 5 is 10 then plus again this cancels i'm just adding 2. so i'm going to end up with m cubed plus 5m squared plus 5m squared is 10 m squared plus 25 m plus 2 m is 27 m plus 10 plus 2 is 12. so i get m cubed plus 10 m squared plus 27 m plus 12. that's exactly what we had as our dividend when we began m cubed plus 10 m squared plus 27 m plus 12. hello and welcome to algebra 2 lesson 38 in this video we're going to learn about synthetic division so for the majority of you who have only taken algebra 1 and this is your first time around in algebra 2 you've probably never heard or seen anything about synthetic division before what i can tell you is it's just a shortcut for dividing polynomials in a certain scenario so when we divide a polynomial by a binomial of the form i'm going to highlight this x minus k we can use a shortcut known as synthetic division okay this is a big time saver for so it's something you want to kind of memorize so the procedure only works again if i'm dividing by a polynomial that is a binomial of the form x minus k so some examples something like x minus 3 something like x minus 12. it could be x minus 2 billion okay but it's got to be in that format now you might say what if i have x plus something that's okay too if i had x plus 10 for example i could rewrite this mathematically as x minus a negative 10. and you're going to see where we're going to do that when we get to our examples what we can't have is an exponent on the variable that is not a one so in other words i can't say something like x squared plus seven or something like x cubed minus three we also don't want to see a coefficient that is not 1 on the variable either okay so this is what we're looking for something in the format of x or some variable could be y z q whatever you want to choose minus some value and again we could have plus some value by using this trick minus a negative of that value all right so we all know how to divide x to the fourth power minus 3x cubed plus 7x minus 26 divided by x minus 3 using long division right we covered that in algebra 1 we just reviewed it in algebra 2. so you could pause the video now and you could do that and see what answer you get then come back and see how we do it with synthetic division or if you just want to follow along with the synthetic division that's fine as well the first thing you want to do just like if i was doing polynomial long division you want to make sure everything is in standard form and you want to make sure that if you have any missing powers for your variable that you put 0 times that variable raised to that power so for example here it's in standard form but i'm missing x squared so i want to rewrite this and say i have x to the fourth power minus 3x cubed plus 0x squared okay 0 is the coefficient for x squared then plus 7x minus 26 and this is divided by we have x minus 3. so this is in standard form and it's in the format of x minus k in this case 3 represents our value k what we want to do now we just want to do the numerical information from the dividend i think about this kind of the way that i think about matrices when we solved a system of linear equations using matrices we just took the numerical information out we didn't really deal with the variables so it's going to be the same thing so let's set up a long division like we would but instead of putting everything underneath let's just take the numerical information so what's the coefficient for x to the fourth power well it's a 1. so let's put that in there then we'd have a negative 3 as the coefficient for x cubed then we'd have a 0 as the coefficient for x squared and we'd have a 7 as the coefficient for x to the first power then our constant is negative 26. so just the numerical information now out here where we normally put our divisor all we want to do again if this is in the format of x minus k we just want to plug in whatever is in the place of k alright so in this case that's the value 3. now here's where students get confused if i have something that's already x minus some number i just throw in this value so this is going to be positive if i had let's just say for example my divisor is x plus three i start out by rewriting that as x minus the negative of three okay in this case my value for k by line this up x minus k is now negative 3 and that's what i use over here that's a big source of confusion for students and that might not make sense for you right now but again as we work more examples it's going to make sense so right now again this is x minus 3. it perfectly fits the format of x minus k so i just plug in this 3 right here that's all i'm looking to do now the other thing i want to tell you is there's many ways to write this in your textbook or in your class or in your tutoring session or wherever you get this information you might see this drawn differently it's the same procedure either way do it however you want to it's just a matter of just getting the right answer okay now once you've set this up there's going to be three rows involved so this is kind of row one you're going to need to leave a little space for row two then you draw a little line down here and this is going to be row three below it okay so you've got row one row two row three now the very first thing you're going to do is take your leftmost number that's under this long division symbol and you're going to drop it all the way down to row three so this is a one coming all the way down now the procedure is very simple it kind of repeats itself over and over again once i've dropped this down i'm just going to multiply 3 times 1 is going to give me 3. so i'm going to write it in the next position here so in this next column in row 2. okay so now i have negative 3 and 3 there all i'm going to do is i'm just going to add so negative 3 plus 3 is 0. now i'm going to multiply again 3 times 0 is 0. put the result here i'm just going 1 column to the right and i'm doing the same thing so now i'm going to add 0 plus 0 is 0. you guessed it now i'm going to multiply 3 times 0 is 0. put that right here 1 column to the right and at 7 plus 0 is 7 and i'm going to multiply 3 times 7 is 21 put that right here one column to the right then add negative 26 plus 21 is negative 5. so you see how quickly we went through this and i can tell you we already have our answer okay you don't see that yet but we already have our answer these are the coefficients and the remainder so if i start all the way to the right this is going to represent my remainder then if i move to the left this represents my constant or my coefficient for x to the power of 0. as i go to the left the exponent on x is going to increase by 1. so this is the coefficient for x to the first power this is the coefficient for x squared and this is the coefficient for x cubed now i can also start from the left and go to the right again everything's in standard form so we know that this would be the highest exponent and it would decrease by 1 as i move to the right so i could have figured out the same thing just by knowing that when we get our results it's going to be one degree less than the dividend now in the dividend the highest exponent was a four so the degree was a four here the highest exponent is going to be one less so a three so i would begin with x cubed now i have the numerical information so i can just write my answer so what i would do is i would say okay i have 1 as the coefficient for x cubed so that would be 1x cubed or just x cubed i have 0 as the coefficient for x squared 0x squared is just 0. so i don't need to put that i have 0 as the coefficient for x to the first power 0 times x is just 0. i don't need to put that i have 7 as the coefficient for x to the power of 0. x to the power of 0 is just 1. it's like 7 times 1 is just 7. 7 is my constant i put x cubed plus 7 and then i have my remainder so plus you have negative 5 and you place this over your divisor which is x minus three so this is your answer right here it's x cubed plus seven plus negative five over the divisor x minus three and you can see how much quicker that would be versus setting up a long division and going through and having to involve the variables just working with the numbers made it significantly quicker all right let's go ahead and take a look at another one i know some of you will kind of be like well i'm a little little fuzzy on that it just takes practice so we're going to work a lot of problems here so we have n to the fifth power plus n to the fourth power minus n cubed plus 18 n squared plus eight n minus three and this is over n plus three so again i'm looking for standard form and i'm looking for no missing powers on the variable n so i have the fifth power four thirds squared and first power so i'm good there and everything is in standard form so i'm good there so now i'm just going to set up my synthetic division so underneath this long division symbol i'm going to put all the numerical information so the coefficient for n to the fifth power is 1 then the coefficient for n to the 4th power is 1. the coefficient for n cubed would be negative 1. so let's put a negative 1 there then the coefficient for n squared is 18 and the coefficient for n to the first power is 8 then my constant is a negative 3. so out here what am i going to put again this is where students make a mistake they go okay i'm just going to throw a 3 over there what's wrong we want in the format of x minus k and we want k to go right there now in this case i have let's just say for example this is an n let's just say this was x okay if i have x plus 3 i need to rewrite it to match this format x minus a negative 3. and so this right here is representing k and so k has to be negative 3 in this scenario take the time to understand that because that's a common source of confusion and it leads to a lot of wrong answers this number right here has to be based on what is given to you there if you have some variable plus a positive number this is going to be the negative of that number if you have some variable minus a positive number this is going to be just the number okay that's being subtracted away all right so once we have that again we have three rows going on so let's let's kind of make a little bar here and let's drop this guy into row three to begin and i can number the rows you don't have to it's not traditional two but i just do it so that you understand what i'm saying so i'm dropping this first one all the way to the left into row three that's the first thing you do and then you start a series of multiplication and addition so i multiply negative three times one to kind of get the ball rolling that's going to give me negative three now i'm going to add 1 plus negative 3 is negative 2 now i'm going to multiply negative 3 times negative 2 is 6. now i'm going to add negative 1 plus 6 is going to be 5. now i multiply negative 3 times 5 is negative 15. now i add 18 plus negative 15 or 18 minus 15 is going to be 3. now i multiply negative 3 times 3 is negative 9. now i add 8 plus negative 9 is negative 1. now i multiply negative 3 times negative 1 is 3 and then negative 3 plus 3 is 0. since we have a 0 in the final position that means we have no remainder okay no remainder we know already that this will be 1 degree less than the dividend the dividend has a degree of five this will have a degree of four so that means the highest exponent would be a four so one is the coefficient for n to the fourth power so if this is the coefficient for n to the fourth power we just have n to the fourth power then i would have minus 2 n cubed right this is the coefficient for the n cubed this is the coefficient for n squared so then plus 5n squared and then this is the coefficient for just n to the first power so plus 3n and then that's my constant so minus 1. again that's 0 there so there's no remainder so this would be your answer for the problem all right for the next one we're looking at negative 18r cubed minus 9 plus 7r to the fourth power plus 14r squared minus 4r this is over negative 2 plus r so the first thing i want to do is write each one of these the numerator and denominator in standard form so i'll have 7r to the 4th power to begin right 4 is the largest exponent then next i'll have minus 18r cubed then i'll have plus 14r squared then after that i'll have negative 4r and lastly i have minus 9. okay then this is over we have negative 2 plus r we need to switch that we need to write this as r minus 2. so now this is in standard form and i'm going to set up my little long division symbol here and i'm going to take the numerical information out of the numerator or the dividend so i want a 7 i want a negative 18 i want a 14 i want a negative 4 and i want a negative 9. then over here remember if this follows the format of a variable minus a positive value we just take this value right x minus k k gets plugged in right there so we just want a 2 right there and one more time i just want to cover this if i had something like r plus 2 i would rewrite this as r minus a negative 2 and i'd be plugging in a negative 2 in that scenario very important you understand what to plug in right here okay big source of confusion all right so let's go ahead and put a little bar here we'll bring this down and we'll begin so i'm not going to label row one two and three you should know this at this point this is the third example so after i bring 7 down to start it's multiply then add then multiply then add it just keeps going like that so 2 times 7 i multiply that's 14. then i add negative 18 plus 14 is negative 4. then i multiply 2 times negative 4 is negative 8 then i add 14 plus negative 8 is 6 then i multiply 2 times 6 is 12 then i add negative 4 plus 12 is 8 then i multiply 2 times 8 is sixteen now i add negative nine plus sixteen is seven so again i can work left to right or right to left i know this is my remainder and i know this would be the constant then this would be the coefficient for r to the first power this would be the coefficient for r squared and this would be the coefficient for r cubed and again it's always going to be one degree less than the degree of the dividend you look at the dividend the highest exponential power is 4 so the degree of this polynomial here in the dividend is 4. the degree of the polynomial down here is a 3 right it's 1 degree less so to write our answer again if this is the coefficient of r cubed we would have seven r cubed then minus four r squared then plus six r then plus eight and then plus seven over our divisor is going to be r minus two so this is your answer here you end up with seven r cubed minus four r squared plus six r plus eight plus seven over r minus two all right let's take a look at a couple more so i have v to the fourth power plus six v minus forty minus eight v cubed this is over v minus eight so again let's put things in standard form and let's account for missing powers of the variable v so if i start out with v to the fourth power then minus eight v cubed i'm missing v squared so let's put plus zero v squared then plus 6v then minus 40. all right then down here i have v minus 8 don't need to change anything there and again this follows the format of x minus k and here x is just v and k is just 8. all right so let's set everything up and again i'm going to make a little long division symbol and underneath that long division symbol i'm going to take the numerical information here so the coefficient for v to the fourth power is a 1 then i would go to the next one i have negative 8 then i'd have a 0 then a 6 and then finally a negative 40. now over here and again if it's in the format of x minus k this fits this perfectly i just want this whatever k is so in this case this is going to be an 8. again i'm going to say this one more time if i had something like v plus 8 i would rewrite it as v minus a negative 8 and we'd have a negative 8 out here okay again most common mistake that i see so i want to just keep saying it over and over again so that if you do make the mistake you're like oh he warned me about that and then you'll catch yourself and it'll stop you from making the mistake in the future all right so let me put a bar down here and again it's a simple process once you have everything set up i bring this down so bring down the 1 and i just multiply then add multiply then add multiply then add very very simple so 8 times 1 is eight negative eight plus eight is zero a times zero is zero zero plus zero is zero a times zero is zero six plus zero is six eight times six is 48 negative 40 plus 48 is eight so again we know that this is the remainder this is the constant this would be the coefficient for v to the first power this would be the coefficient for v squared and this would be the coefficient for v cubed and again if i look at my dividend here it's got a degree of four here i have a degree of three so again i could have started to the left and said well i know it's one degree left so this would be the coefficient for v cubed then the coefficient of v squared you know so on and so forth so to write our answer i'd have 1 times v cubed or just v cubed i'd have 0 times v squared and 0 times v i need to write that so then plus my constant 6 and then plus 8 my remainder okay let me label that this is the remainder over my divisor which was v minus eight okay v minus eight so v cubed plus six plus eight over v minus all right for the last problem we have 11x plus 2x squared plus x to the fifth power plus 10 minus x to the fourth power minus 3x cubed and this is over x plus 1. so again let's write it in standard form and let's account for any missing powers of x so let's say that this is x to the fifth power and we're going to have minus x to the fourth power then minus 3x cubed and then we would have plus 2x squared and then we'd have plus 11x and then lastly plus 10 and this is over we have x plus 1. now i'm going to go ahead and write x plus 1 as x minus a negative 1. again i want this to be in the format of x minus k here k is going to be negative 1. i'm going to make my little long division symbol we're going to take the numerical information from the dividend so i have a coefficient of x to the fifth power of one then i would have negative one then negative three then two then eleven then ten then when i think about k again it's negative one here because i wrote x minus a negative one so this over here is going to be a negative one make sure you do that or you will get the wrong answer okay very important all right let's put our bar down here let's bring this down to begin and again we're going to multiply then add multiply then add it's very very simple all right so we start out with multiplication negative 1 times 1 is negative 1. now we add negative 1 plus negative 1 is negative 2. now we multiply negative one times negative two is two now we add negative three plus two is negative one now we multiply negative one times negative one is one now we add two plus one is three now we multiply negative one times three is negative three now we add eleven plus negative 3 is 8 now we multiply negative 1 times 8 is negative 8. now we add 10 plus negative 8 is 2. so we're done we know that this is our remainder this is our remainder this is our constant this is our coefficient for x to the first power our coefficient for x squared our coefficient for x cubed and our coefficient for x to the fourth power so what we would have as our answer is one x to the fourth power which is x to the fourth power minus two x cubed minus x squared plus three x plus eight then plus you'd have two over remember your divisor here was x plus one okay we wrote it as x minus a negative one but originally it was x plus one so two over x plus one so this right here will be your answer x to the fourth power minus two x cubed minus x squared plus three x plus eight plus two over x plus one hello and welcome to algebra 2 lesson 39 in this video we're going to learn about polynomial functions so essentially what we're going to look at in this lesson we're going to look at polynomial functions and operations with polynomial functions so i want to just start out with a polynomial function we have f of x is equal to we have negative x cubed plus 5x minus 10. now in a previous lesson in algebra 2 i taught you about function notation so if i ask you to find f of 0 f of 2 or f of negative 2 at this point you should be able to do that essentially what i'm doing if i ask for something like f of 0 i'm saying what is the function's value if the independent variable x is 0. so i would have negative i'd plug in a 0 for x and that would be cubed plus 5 times i'd plug in a 0 for x then minus 10. so we can see that these two would be zero right so i'd have zero plus zero which is zero then minus ten which is just negative ten so f of zero is negative ten or the function's value when the independent variable x is zero is negative 10. all right what if we did f of 2 okay f of 2. same concept what i'm going to do is i'm going to replace every x with a 2. so negative and then i have a 2 being plugged in for x this is cubed then plus 5 multiplied by again 2 is plugged in for x then minus 10. so 2 cubed is 8 i'd want the opposite of that notice how the negative is not inside of parentheses it's outside here okay so the opposite of 8 would be negative 8. then plus we have 5 times 2 which is 10 then we have minus 10. so we know 10 minus 10 is 0 so that's basically canceled we're left with just negative 8 there so f of 2 is negative 8. all right then the last one we wanted to look at is f of negative 2. so again i'm just going to plug in for my variable x so negative then we plug in a negative 2 and that's cubed then plus we have 5 times plug in a negative 2 for x then minus 10. so negative 2 cubed is negative 8. again this negative here is inside the parentheses so it's included negative 2 times negative 2 is 4 4 times negative 2 is negative 8. so i would have the opposite of negative 8 which is positive 8. then we have 5 times negative 2 which is negative 10 then minus 10. so we would have 8 minus 10 which is going to give me negative 2 and then negative 2 minus 10 which is going to give me negative 12. so f of negative 2 or the function's value when x is negative 2 is going to be negative 12. all right so let's look at one more just as a review if i had something like g of x equals 2x squared minus 7x plus 5. and i wanted to find g of 3 g of negative 5 and g of negative 2. what would i do again i'm just going to be plugging in for the variable it's not very hard if i want g of 3 that's telling me just to plug in a 3 everywhere i see an x so 2 times we'd have a 3 plugged in there that's squared minus 7 times plug in a 3 for x here then plus 5. 3 squared is 9. 9 times 2 is 18 so this would be 18 then minus 7 times 3 is 21 then plus 5. if i take 18 and i subtract away 21 i get negative 3 negative 3 plus 5 is positive 2. all right the next thing i want to look at is g of negative 5. again all i'm doing is i'm plugging in a negative 5 for my variable x so 2 times plug in a negative 5 for x that's squared minus 7 times plug in a negative 5 for x plus 5. negative 5 squared is 25 25 times 2 is 50. so this would be 50. negative 7 times negative 5 is positive 35 so plus 35 and then plus 5. so 50 plus 35 is 85 85 plus 5 is 90. all right then one final one we want to find g of negative 2 okay g of negative 2. so all i'm going to do once again super super simple if i have g of negative 2 i plug in a negative 2 everywhere i see an x so i have 2 times negative 2 plug that in for x that's squared minus 7 times plug in a negative 2 for x plus 5. very very easy negative 2 squared again the negatives involved in the parentheses so that would be four four times two would be eight so you'd have eight negative seven times negative two is positive fourteen so plus fourteen and then plus five so eight plus fourteen is going to give me 22 then 22 plus 5 is 27. so pretty easy overall but something i wanted to do a refresher on before we started kind of going to the next level in a moment we're going to look at operations with polynomial functions all right so the very first thing we're going to learn how to do is adding and subtracting functions okay so if you see something like f plus g of x with this notation here all this is telling you to do is take f of x and add g of x okay so all you're going to be doing is basically combining like terms it's just like if i added two polynomials together it's no more difficult okay you've just got to get used to the notation it's the same thing if i see something like f minus g of x this is f of x minus g of x now it's going to get more complex because they're going to ask you for stuff like this they're going to say f plus g of let's say 2. all i'm doing is i'm finding the function f of x plus the function g of x and i'm going to take that result and i'm going to plug in a 2 everywhere there's an x and i'm going to find out what that value is okay that's all this is asking for okay so let's take a look at an example so suppose you see something like f of x equals negative 11 x to the fourth power plus 10x cubed plus 14x squared and you see g of x is equal to negative 2x to the fourth power minus 11x cubed minus 9x squared now the problem we get we're told to do f minus g of x so all i'm going to do this is super super easy the order matters f comes first so i want f of x then minus g of x okay so what's that going to be equal to well if i have my function f of x it's what it's negative 11x to the fourth power plus 10 x cubed plus 14x squared now if i subtract away my function g of x remember if i'm subtracting something away we've got to put it inside of parentheses so this whole thing needs to be subtracted away don't make the mistake of just subtracting away the first kind of term there you want minus and then inside of parentheses you want negative 2x to the fourth power minus 11x cubed minus 9x squared with subtraction remember the order always matters so i have f of x here and then i have g of x here okay very important if this was g minus f of x the order would be flipped all right so once we've kind of set this up it's no different than just subtracting polynomials the only thing you really have to get used to is just the notation involved once you get used to what it's telling you to do the actual operation itself is quite simple so i would rewrite this as negative 11x to the fourth power plus 10x cubed plus 14x squared i would distribute this kind of negative to each term so minus a negative 2x to the fourth power would be plus 2x to the fourth power minus a negative 11x cubed would be plus 11x cubed and then lastly minus a negative 9x squared would be plus 9x squared so all i need to do now is just combine like terms and i have my answer negative 11x to the fourth power plus 2x to the fourth power would give me negative 9x to the fourth power then i would have 10x cubed plus 11x cubed that would be plus 21x cubed then i would have 14x squared plus 9x squared that would give me plus 23x squared so this is my answer here this is f minus g of x okay that's the result that you get it's negative 9x to the fourth power plus 21x cubed plus 23x squared all right so let's say we had the same function for f of x the same function for g of x but now we've changed the problem to f plus g of x so we want to find this so now we're just going to add f of x plus g of x okay and with addition the order does not matter so i could put f of x first or i could put g of x first i would get the same answer right it would not matter with subtraction the order does matter okay so that's where you've got to pay close attention so i'm going to start out by writing my f of x so this is negative 11x to the fourth power plus 10x cubed plus 14x squared then i'm going to add to this i'm going to have negative 2 x to the fourth power and then minus 11x cubed and then minus 9x squared so what would the result here be well again i'm just going to combine like terms just like i would if i was adding polynomials negative 11x to the fourth power plus negative 2x to the fourth power is negative 13x to the fourth power if i have 10x cubed minus 11x cubed that's going to give me negative x cubed and then lastly if i have 14x squared minus 9x squared that's going to give me plus 5x squared so i'm going to end up with negative 13x to the fourth power minus x cubed plus 5x squared all right so let's kind of take things up just a notch and let's say we had f of x equals 14x to the fourth power minus 8x cubed plus 11x and we have g of x is equal to negative 3x to the fourth power minus 5x minus 3. let's suppose you see something like this f plus g of 2. so again this notation is just something you have to get used to let me think of what in the world does that mean what it means is i'm going to find f of x plus g of x and when i get that result i'm just going to plug in a 2 for x another way to explain is i could find f of 2 and g of 2 and then i could just sum those together it's just two different ways that you could achieve this and i'm going to do both for you so the first way i'm going to do this is i'm going to find f of 2 plus g of 2 okay so to do that f of 2 would be what it would be 14 times i'd plug in a 2 for x so 2 to the 4th power minus 8 times i'd plug in a 2 for x so that's cubed plus we'd have 11 times plug in a 2 for x then plus now i want g of 2 so g of 2 i'm plugging in at 2 for x so negative 3 times you'd plug in a 2 and that would be to the 4th power minus 5 times plug in a 2 and then minus 3. all right so what does this give us 2 to the 4th power we know is 16 then 16 times 14 is 224 then we have minus 2 cubed is 8 8 times 8 is 64. so 64. then plus 11 times 2 is 22. then we have 2 to the fourth power we know that's 16. you multiply that by negative 3 that's going to give you negative 48 so let's put minus 48 then here we have negative 5 times 2 that's negative 10 so negative 10 and then minus 3. so now we're just going to work left to right 224 minus 64 would give us 160 plus 22 would be 182 minus 48 would be 134 minus 10 would be 124 then minus 3 would be 121. so this is f plus g of 2. now the other way we could have got this just remember you got 121. let's erase this the other thing we could have done is we could have first added f of x plus g of x if i have f of x plus g of x this is going to be equal to we'd have 14x to the fourth power minus 8x cubed plus 11x and then plus g of x which is negative 3x to the fourth power minus 5x minus 3. again all i'm looking to do is just combine like terms here so i put equals 14x to the fourth power minus 3x to the fourth power is 11x to the fourth power negative 8x cubed nothing to combine with that i have 11x and minus 5x that's positive 6x and then i have minus 3. so once i have this result what i'm looking for is to substitute a 2 in everywhere i see an x so i'd have 11 times you'd put a 2 in 4x so 2 to the fourth power minus 8 times plugging a 2 in for x so 2 raised to the third power plus 6 times 2 minus 3. so the result of this we know is going to be 121. 2 to the fourth power is 16 16 times 11 is 176. we have 2 cubed that's 8 8 times 8 is 64. so we're subtracting away 64. then we're adding 6 times 2 is 12 then we're subtracting 3. so 176 minus 64 is 112. if i then add 12 i get 124 if i subtract 3 i get 121. again f plus g of 2 is 121. all right so let's take things up just a notch so suppose you had f of x equals 5x squared plus 14x plus 12 and g of x equals 14x cubed minus 12x squared and then h of x a third function is equal to 7x squared minus 11x if we wanted to find f plus g plus h of negative 1 what would we do well one thing we could do is we could find f of x plus g of x plus h of x first and then we could take that result and we could plug in a negative one for x so let's just go ahead and crank this out so we would have f of x is 5x squared plus 14x plus 12 and let me scroll down plus g of x which is 14x cubed minus 12x squared plus h of x which is 7x squared minus 11x again once i have this notation down it's pretty easy because all i'm really doing is just adding polynomials so my highest exponent is a 3. so let's start with 14x cubed nothing to combine with that i've got 5x squared i've got negative 12x squared and i've got 7x squared well i know that 5 and 7 make 12 so 5x squared plus 7x squared would be 12x squared then i've got a negative 12x squared those would cancel out then i'd have 14x and negative 11x that would give me positive 3x and then lastly i'd have plus 12. so what i have now is i have f plus g plus h of x and this equals 14x cubed plus 3x plus 12. so to find f plus g plus h of negative 1 all i would do is take this and i would plug a negative 1 in everywhere i see an x so i would have 14 times you plug in a negative 1 for x and that's cubed plus 3 times negative 1 plug that in for x plus 12. let me kind of slide this stuff down just a little bit myself a little room so this is equal to negative 1 cubed is negative 1. so negative 1 times 14 is negative 14. 3 times negative 1 is negative 3 so we'd have minus 3 and then plus 12. so negative 14 minus 3 would be negative 17 negative 17 plus 12 is equal to negative 5. so f plus g plus h of negative 1 is going to be negative 5. all right so now that we've kind of looked at adding and subtracting functions let's talk about multiplying and dividing functions so we also encounter problems that ask us to multiply or divide functions so if i see something like f times g of x this is asking me for f of x times g of x again after you see this enough it becomes very very simple it's just a matter of understanding the notation the actual operation itself is quite simple all right so let's say we saw f of x which is equal to 2x plus 5 and we see g of x which is equal to 3x squared plus 4x minus 6 and we have f times g of negative 1. so what this is telling me to do is multiply f of x times g of x and i could take that result and plug in a negative 1 for x the other thing i would do which might be a little quicker in this scenario is find f of negative 1 and multiply that by g of negative 1. either way i would get the same answer and again i'll show that to you so let's start out by doing it this way so f of negative one if i plug in a negative one for x i'd have two times negative one which would be what this would be negative two negative two would then be added to five that would give me 3. so you basically have 3 here times if i plug in a negative 1 here negative 1 squared is 1 so i'd have 1 times 3 or 3. so this is 3 i'd have 4 times negative one which is negative four so this is negative four and this is minus six so three minus four is negative one and then negative one minus six is negative seven so this ends up being three times negative seven which is negative 21. now that went by very very quickly because of the way i did it if we did it the other way we're going to take a little bit longer for the simple fact that i've got to multiply these two polynomials together first okay and that can be quite tedious sometimes so let's do it the other way so let's say that we said that we wanted to do f of x times g of x okay what would that be equal to well we would have 2x plus 5 multiplied by we have that other polynomial 3x squared plus 4x minus 6. so let's go ahead and crank this out so let me take 2x and multiply it by 3x squared that would give me 6x cubed then 2x times 4x would give me positive 8x squared then 2x times negative 6 would be minus 12x now i have 5 times 3x squared that's plus 15x squared i have 5 times 4x that's plus 20x and i have 5 times negative 6 which is minus 30. all right so looking at this now i just need to combine like terms so i have my 6x cubed okay my 6x cubed nothing to combine with that then i have 8x squared and 15x squared so what that would give me is 23 okay positive 23 x squared then next i have negative 12x plus 20x that's going to give me 8x okay so plus 8x and lastly i have minus 30. so then once we have it in this format i would just plug in a negative 1 for x so f times g of negative 1 would be 6 times plug in a negative 1 for x that's cubed plus 23 times plug in a negative 1 for x that's squared plus 8 times plug in a negative 1 for x and then minus 30. so let's scooch this down kind of get a little room going so we know that negative 1 cubed is negative 1 6 times negative 1 is negative 6. i know that negative 1 squared is 1 23 times 1 is 23 so plus 23. 8 times negative 1 is negative 8 and then minus 30. so negative 6 plus 23 is going to be 17 17 minus 8 is 9 9 minus 30 is negative 21. and look how much longer it took to do it this way when you're working with multiplication or division it's generally going to be quicker to substitute in first then do the multiplication okay versus going through and finding you know for example here f of x times g of x then plugging in all right so we have f of x is equal to three x minus five we have h of x is equal to eight x minus eight we have g of x is equal to four x minus 1 and we're asked to find g times h times f of 0. again i could do this the long way and i could say 3x minus 5 times 8x minus 8 times 4x minus 1. and then once i get that result i could plug in a 0 for x or i could do it very quickly by finding f of 0 h of 0 and g of 0 and then just multiplying those together so f of 0 would be what plug in a 0 there i'd have 3 times 0 which is 0 minus 5 so this would be negative 5. h of 0 8 times 0 is 0 you'd have minus 8 or negative 8. g of 0 you'd have 4 times 0 which is 0 then minus 1 so this is equal to negative 1. so basically what you have is negative 5 negative 5 times negative 8 times negative 1. and what's that going to give me well if i do negative 5 times negative 1 i get 5 5 times negative 8 is negative 40. all right so let's take a look at one more and this one's going to involve division so let's suppose i have f of x is equal to x cubed plus 7x squared minus 9x minus 4 and then g of x is equal to x plus 8. so what i'm asked for and i'm going to show you this on the next page i'm asked to find f over g of negative 4. so what this is saying is i want f of x over or divided by g of x when i get the result i want to plug in a negative 4 for x so again you can do this the fast way plug in a negative 4 for x in each function and then do the division or the slow way which would be to crank out f of x over g of x and then plug in a negative 4. so let's do it each way and i'm going to start out with the quick way so if i plug in a negative 4 in other words if i had f of negative 4 here and if i had g of negative 4 here what would i have well this would be negative 4 cubed plus 7 times negative 4 squared minus 9 times negative 4 minus 4. negative 4 cubed would be negative 64. so that's negative 64. negative 4 squared is 16 16 times 7 is 112 so plus 112 negative 9 times negative 4 is positive 36 and then minus 4. negative 64 plus 112 is 48 48 plus 36 is 84 and then 84 minus 4 is 80. so i could say f of negative 4 is 80. so what's g of negative 4 if i just plug in a negative 4 there i know that negative 4 plus 8 would be 4. so now all i want to do if i want f over g of negative 4 i know it's 80 divided by 4 which is just 20. okay so that's my answer to kind of do this the long way i would take f of x and divide it by g of x and again when i'm done with that i would plug in a negative 4 for x and again you'd find out that the value you get is 20. so let's go ahead and do this the long way so i would have x cubed plus 7x squared minus nine x minus four over x plus eight now what's the quick way to do this we just in the last lesson learned about synthetic division and we see that we have a perfect scenario for synthetic division and all i need to do really is convert this x plus 8 into x minus a negative 8. so i could divide this very quickly with synthetic division i'll use my coefficients here 1 7 negative 9 and my constant negative 4 and i'm going to use my value negative 8 out here put my little line there drop this 1 down and i'll get going negative eight times one is negative eight seven plus negative eight is negative one negative eight times negative one is positive eight negative nine plus eight is negative one negative eight times negative one is 8 then i would want to do negative 4 plus 8 that would give me positive 4. so i already have my answer now remember the degree of this polynomial and the answer is one less than the degree of the polynomial here since this is x cubed as the highest power this polynomial has a degree of three the highest exponent is a three here the highest exponent will be a two it's of degree two so this would be the coefficient for x squared this would be the coefficient for x to the first power this would be your constant and this would be your remainder okay all right so if we wanted to write this you would have 1x squared or just x squared minus 1 x to the first power or just minus x and then minus 1 and then plus 4 over your divisor was x plus 8. x plus eight so let's copy this and we'll come down here and we'll put this is equal to this f over g of negative four is equal to this if we substitute in a negative 4 everywhere we see an x okay so if i had negative 4 squared to begin negative 4 squared is 16. then minus a negative four is the same thing as plus four then you'd have minus one then plus here you have four over you have negative four plus eight negative four plus eight is four four over 4 is 1. so essentially you can see that these two would cancel that's zero you're left with 16 plus 4 which is 20. and again that's what we got at the very beginning of this very easily just by plugging in a negative 4 for x in each function and dividing right dividing the resulting numbers you had 80 divided by 4 which gave you 20 as well so much much quicker to do it that way but i wanted to go through it the long way just to prove to you that it works either way hello and welcome to algebra 2 lesson 40. in this video we're going to learn about factoring out the gcf otherwise known as the greatest common factor now before we actually get into factoring i need to review something known as the greatest common factor or the gcf this is how we abbreviate that so this is the largest term that each term of the polynomial is divisible by so i want to start out with a brief example so we want to find the gcf for 6x squared 12x and then 18x to the fifth power so for the number parts we've been doing this since pre-algebra okay if i wanted to find the gcf for 6 12 and 18 a lot of you at this point can eyeball that and say well i know it's 6 right because 18 divided by 6 would be 3 12 divided by 6 would be 2 6 divided by 6 would be 1 right so that's easy to do at this point for us but if you have no clue on how to do it the official procedure would be you take each number and you find the prime factorization so 6 would factor into 2 times 3. 2 is prime and so is 3. 12 we could factor as 4 times 3 3 is prime 4 is going to be 2 times 2 and 2 is prime then next we have 18. so 18 i could write as 9 times 2 2 is prime then 9 is 3 times 3 3 is prime now once we have the prime factorization for each number what we're looking to see is what prime factors are common to everything okay common to everything so to make that crystal clear i could write six right here as two times and put a big space out here times three i could write twelve as 2 times 2 times 3 and then i could write 18 as 2 times leave my space 3 times 3. now the way i've written it here makes it easy to see what's common to everything everything has one factor of 2 everything has one factor of 3. so to get my gcf for 6 12 and 18 i just take the prime factors that are common to everything and i multiply them together 2 times 3 is 6 okay so 6 would be your gcf and the reason for that is because everything involved has those prime factors when the number is built so in other words 18 has a prime factor of 2 and a prime factor of 3. so i know 18 would be divisible by 6. same thing with 12. i know it'll be divisible by 6 and then same thing with 6 i know it would be divisible by itself 6. so again the gcf is 6. so let me let me erase all that all right so when we think about our gcf i'm just going to put 6 right here and then i'm going to put a question mark for the variable part the variable part is pretty simple overall you have x squared you have x and you have x to the fifth power now you could go through and do the same thing you could factor x squared and say x squared is x times x then x is just what it's just x and then x to the fifth power is x times x times x times x times x you can go through and say what's common to everything well just one factor of x so the variable part is just one x one x but what's the quicker way to do that well when you have variables involved the first thing you want to check to make sure of is that everything has that variable in other words this has an x this has an x and this has an x then you're just going to look for the smallest exponent in this case the smallest exponent would be this one right the understood exponent of 1. the reason you look for the smallest exponent that's going to tell you the fewest number of factors of that variable that you're going to have in this case i only have one x here everything else if this is the smallest we'll have at least that and so that's going to be common to everything so my gcf here would be 6x all right let's take a look at another one let's say i had 14y squared x 21x cubed y squared and 42 x to the ninth y cubed now again a lot of you can kind of eyeball this and look at the numbers and say well i know that 42 is divisible by 14 it would be 3. i know that 21 is not divisible by 14 but i know that 21 and 14 would share a common factor of 7. so my gcf for the number part would be a 7 because 14 is divisible by 7 21 is divisible by 7 and 42 is divisible by 7. right so once you have enough experience you can kind of eyeball things and say i know what my gcf is but for right now i want to just go through it the long way for those of you who are a little confused about what i just did if i think about 14 it's what i could factor it into 7 times 2 7 and 2 are both prime numbers if i think about 21 i can factor it into 7 times 3 7 and 3 are both prime numbers if i think about 42 i could factor that into 7 times 6 7 is prime 6 i could factor into 2 times 3 2 and 3 are both prime so look at what's common to everything everything has one seven and then this does not have a three this does not have a two so a two would not be common to everything and a three would not be common in everything so i can only take what's common everything and that's a 7. so for the number part the gcf would be 7 and then you have some variable there all right so let's erase this now for the variable parts i see that i have the variable y and i have the variable x involved now i have an x and a y in every term here so i'm just going to pick the smallest exponent on each so i'm just going to write x and i'm going to write y and then i'm going to write the smallest exponent that appears on each so here for x i have x to the first power i have x cubed and i have x to the ninth power so the smallest exponent is a one so this just stays as x or you could write x to the first power like that if you wanted to for y i have y squared y squared and y cubed the smallest exponent that appears on y is a 2 so this would be y squared here so my gcf here is going to be 7x y squared well let's do one more of these and then we'll get to some factoring so we have 25z to the fourth power y squared we have 10 z and we have 50 zy so for the number part again try not to get into a factor tree just notice that everything here is divisible by five right we think about our divisibility rules i know everything ends in a five or a zero here so 25 divided by five would be what that would be five so we know this is five times five ten we know is what five times two fifty we know is five times 10. now 10 is what it's 5 times 2 so let me write 5 times 5 times 2. so i've factored each number here but i can kind of do that in my head and i know you can as well especially with these smaller numbers everything has one five everything does not have a second five and then everything does not have a two so really the only prime factor that's common everything is five so for my gcf the number part would be a 5 okay and a 5 only all right for the variable parts we see we have z and we have y involved now i have a z here a z here and a z here the smallest exponent that appears on z is a one right you have a one here and a one here so i just put z now for y i have y squared here and i have y to the first power here but i don't have a y here so it's not common to everything okay and if it's not common to everything it can't go into the gcf the gcf is the largest term that each term here would be divisible by so each term here would not be divisible by something with a y in it because there's no y here okay so the gcf is just going to be 5 z so now you know how to find the gcf for a group of terms let's talk about how to factor out the gcf from a polynomial to do this i'm going to start out with a very simple example let's say i had 6 plus 15 and what i would do is i would write 6 as 2 times 3 and i would put plus 15 i'd write as 5 times 3. all we're doing is we're just kind of reversing the distributive property notice that i have a 3 that's common to each right 3 is the greatest common factor between 6 and 15. all i'm going to do is pull that 3 outside of a set of parentheses so 3 goes here and then in these spaces i'm just going to have what's left i pulled out a 3 here so i'm just going to have a 2 that's left i pulled out a 3 here so i'm just going to have a 5 that's left and you can think about reversing this if i did 3 times 2 i'd be right there and that would give me 6. then plus if i did 3 times 5 or 5 times 3 it's the same thing i'd have that which would give me 15. so instead of kind of going this way like i'm normally doing i'm just kind of going backwards right i'm reversing the distributive property all right so let's look at the first example so we have 4x plus 12. the first thing i'd want to do is figure out what is the gcf here what is the greatest common factor well i know there's going to be no variable involved because i have an x here but no x here remember it's got to be common everything so i'm just looking at a number and this one's very easy because i think about 4 and i think about 12. i know that 12 is divisible by 4. 12 divided by 4 is 3. so the greatest common factor here would just be 4 right 4 divided by 4 is 1 12 divided by 4 is 3. and again if you needed to see that you could break that down the long way you could go okay well 4 is 2 times 2 and 12 is what it's 4 times 3 4 is 2 times 2 so really i have two factors of 2 here and 2 factors of 2 here so each one here is going to be divisible by 4 so the greatest common factor is going to be a 4. all right so once we've figured that out all i need to do for this example is pull a 4 out so i could write this as 4 times x like this is to completely show what's going on plus if i divide 12 by 4 i get 3. so 4 times 3 and this would be equal to just pull the 4 out place it outside of a set of parentheses and in the parentheses would have what's left here would just be an x plus here it would just be a 3. the other way you could think about that remember multiplication and division are opposites so i know if i do 4 times x i get 4x i know if i do 4 times 3 i get 12. so i can reverse that and say if i'm pulling a 4 out from here well then 4x divided by 4 would be what that would be x so that's going to go right there i could do 12 divided by 4 and that would give me 3 so that's going to go right there and i'm just thinking about multiplication and division as their opposite operations so that's another way you can do that and that's in fact what you're going to do when you do this mentally when i factor this now i just look at it and i say okay well i know that 12 is divisible by 4 so 4 is the gcf that's the first thing i come across then after i put 4 outside of a set of parentheses i would say okay well 4x divided by 4 would be x that's how i get my first term then i put my plus and x i say 12 divided by 4 would be 3. that's how i get my second term there all right let's take a look at another one so i have negative 2v to the fifth power plus 5v cubed so if i think about negative 2 it's just negative 1 times 2 2 is a prime number 5 is a prime number there's nothing common there number wise so in my gcf i'm not going to have a number involved other than i could use 1 and we'll talk about this later i could also use negative 1 if i wanted to okay i'm allowed to factor that out but for right now i'm not going to think about any numbers i'm just going to look at the variable i have v to the fifth power and i have v cubed so everything has a v the smallest exponent is a 3. so i know that's what i can factor out of v cubed so let's put equals we'll put v cubed and then inside i'll have two spaces here again i could write this the long way and say i have negative two times v cubed times v squared plus 5 times v cubed i could circle v cubed okay a lot of students like to do this and say okay if i pull that out what's left well here it's just negative 2 times v squared plus here it would be 5. okay you could do it that way now another way you can do this again if i don't know what this is let me just erase this i can just use division okay i could say well if i have negative 2v to the fifth power divided by v cubed what's that going to give me it's going to give me negative 2 v to the fifth power over v cubed is v squared that's what goes here negative 2v squared and i will do the same thing with this term i would say 5 v cubed divided by v cubed is equal to what well again i have 5 v cubed over v cubed is 1. so i get a 5 so you could use division or you can kind of separate things with multiplication where it's obvious what's going to be placed in here for these terms whatever is more comfortable for you i can tell you that you're probably as you get higher in math going to use division okay it's just easier to do that it's more efficient and in most cases you're going to use division in your head we know if we wanted to check this we could use our distributive property v cubed times negative 2 v squared would give me back to negative 2 v to the fifth power then plus v cubed times 5 would be 5 v cubed so good to go there all right let's take a look at the next example so suppose we have negative 8 x cubed plus 12 x so we think about the numbers here i have negative 8 which i'm going to write as negative 1 times 8 8 is what i think about that as let's write negative 1 times 2 times 2 times 2. we should know at this point that 8 is 2 cubed for 12 we have 1. we have 4 times 3 right 4 times 3. so 4 is 2 times 2 then times 3. so kind of a quicker way to approach this would say okay well i have negative 8 here so just think about 8 for a second and then i have 12. so both of those are even numbers and i would say well i know that 12 breaks down into 4 times 3. i know 8 is not divisible by 3 but it is divisible by 4 and so right away i figured out that 4 is my gcf but if you have to do this the long way for a little while that's fine you see that here i have 2 factors of 2 there 2 factors of 2 there that's what's common and 2 times 2 is 4. so my gcf here the number part is a 4. all right for the variable part we have an x cubed and an x to the first power again it's common to everything we go with the smallest exponent which is 1 so this is 4 and then x all right so if we want to pull this out we would say this is equal to and let me kind of scooch this down we'll put our 4x outside of a set of parentheses put my two spaces in there what goes here and here let's just use division okay let's just use division so if i had negative eight x cubed divided by four x negative eight x cubed divided by four x negative eight divided by four is negative two and then x cubed over x is x squared okay and with that next let's use 12x and let's divide that by 4x we know that 12 over 4 is 3 x over x is 1 so this is a 3 here okay and again if you use multiplication you'll get back to this 4 times negative 2 is negative 8 x times x squared is x cubed then plus we have 4x times 3 4 times 3 is 12 x comes along for the ride you get 12x back so our answer here is correct 4x times the quantity negative 2x squared plus 3. let's take a look at another one so we have 8x to the sixth power plus 80x to the sixth power y to the fourth power plus 40x to the ninth power so i want you to kind of eyeball this and i want you to without doing a prime factorization pause the video and see if you can figure out what the gcf is i think that you can do that okay so i'm going to assume that you tried that and really the method that i would use i'll see that i have an 8 here and 80 here and a 40 here everything you should notice here is divisible by 8. 8 divided by 8 is 1. 80 divided by 8 is 10 40 divided by 8 is 5. so that tells me right away that the gcf the number part at least will be an 8. now for the variable part i have an x an x and an x i have a y here but i don't have a y anywhere else so the gcf will only contain an x and then the smallest exponent is a 6 right i have a 6 there a 6 there and a 9. so 8x to the sixth power is my gcf so to factor this let's put equals we're going to put 8x to the sixth power out in front and inside the parentheses i'm going to have three spaces so what's going to go here here and here well all i'm going to do is i'm going to take that gcf and i'm going to use it as the divisor of each term here so in other words i would start out with 8x to the sixth power and i would divide it by 8x to the sixth power of course 8x to six power over eight x to the sixth power is one and next we'd have plus so i'd move to this next term i'd have 80 x to the sixth power y to the fourth power divided by eight x to the sixth power so i know that eighty divided by eight would be ten so a ten would go there x to the sixth power x to the x power is one and then you'd have y to the fourth power all right let's do one more here we have 40x to the ninth power so 40 x to the ninth power we're going to divide that by again the gcf of 8x to the sixth power we know that 40 divided by 8 is 5 x to the ninth power of x to the sixth power is x cubed so this would be 5 x cubed now you can check this if you want to again you would use your distributive property and just kind of reverse this and you'll get back to this right so our answer here we have 8x to the sixth power times the quantity we have 1 plus 10y to the fourth power plus 5x cubed all right let's take a look at another one let's say you have negative 28 x to the fourth power y to the fourth power plus 24 x y cubed plus 32 x y squared so again if i think about the number part here without making a factor tree without going through all that just just think about some things that we notice the first thing i notice is that everything is an even number so i know everything would be at least divisible by 2 okay now the next thing i would check and say is everything divisible by 4 yeah it is forget about this negative 4 seconds pretend it doesn't exist 28 divided by 4 is 7 24 divided by 4 is 6 32 divided by 4 is 8. the fact that this divided by 4 is 7 is a big thing 7 is a prime number okay so i know that i can't further break this down it's just going to be 2 times 2 times 7 okay so let me let me just write that right here real quick this is this is 2 times 2 times 7 and if i consider that negative it's really negative 1 times 2 times 2 times 7. now if this is not divisible by 7 which it's not and this is not divisible by seven which it's not i'm done all right i know that everything will only have a four in common okay so i can stop and say my gcf the number part at least would be a four now don't get me wrong there's times where you get really big numbers and you have to make a factor tree you can't get around that but in simple cases like this you should be able to eyeball these things and come up with your gcf okay and you're going to be expected to do that especially as you move up and you start taking standardized tests right it's going to be built into the time limit all right so now i think about my variable part i have an x everywhere and i have a y everywhere so x the smallest exponent would be a 1 so i'd put an x there and then for y the smallest exponent would be a 2. so i'd put y squared so my gcf is 4 x y squared all right so if i want to pull that out let's kind of scroll this down let's put equals which we already did and then 4 x y squared outside of a set of parentheses and we'll have three terms in here and again to get the terms i just take each term and i divide it by the gcf and i get the term right because again division is the opposite of multiplication so if i multiplied i'd go back to what i had okay so negative 28 okay negative 28 x to the fourth power y to the fourth power divided by 4xy squared or x y squared we think about negative 28 divided by four is negative seven so this is negative seven we think about x to the fourth power over x this is x cubed when we think about y to the fourth power over y squared this is y squared so this would be negative seven x cubed y squared all right for the next one we have 24 x y cubed over we have 4 x y squared okay so let's look at this 24 divided by 4 is 6 x over x is 1 y cubed over y squared is y to the first power so this is 6 y [Music] and let's erase this and we'll do our last one now so we have 32 x y squared over 4 x y squared so 32 divided by 4 is 8 x over x is 1 y squared over y squared is 1. so i'm going to put an 8 there all right so let's erase this real quick and we see our answer is 4xy squared times the quantity we have negative 7x cubed y squared plus 6y plus 8. now if you wanted to check to make sure that you factored this correctly you could use your distributive property and distribute this 4xy squared to each term and you can see that you get this polynomial up here back all right so the next thing i want to talk about we have here that in some cases we may prefer to factor the negative of the gcf so this is something that might be very useful for you in fact in the next section when we start reviewing factoring by grouping a lot of times we'll use the negative of the gcf to end up with a common binomial factor so when we talk about pulling out the gcf we could do the gcf or the negative gcf and it's still the same thing in terms of getting the correct answer okay so if i look at something like this we have negative 25 x squared y cubed negative 15 x squared y to the fifth and negative 50x cubed y cubed i could pull that negative out to start okay if i want to when i do that i'm going to have the negative gcf not the gcf itself right we think about the greatest common factor we know that the positive version is larger than the negative versions that's why we say the negative of the gcf when we do something like this now if i think about what the gcf is here think about 25 15 and 50. they're all divisible by 5 right 25 divided by 5 is 5. 15 divided by 5 is 3. so i can stop there i don't even need to go there i know that these two only share a common factor of five and this one also has a factor of five so i can just put five down again if i'm doing the gcf i want positive five well because it's convenient if i want to do the negative of it i just do negative five right and both answers are acceptable now when i think about the variable part i've got x squared x squared and x cubed so the smallest exponent is a 2 so i go with x squared then i've got y cubed y to the fifth and y cubed the smallest exponent is a 3 so i go with y cubed so inside the parentheses now these won't be minuses anymore they'll be pluses okay and why is that the case because i pulled the negative out right so everything inside will now be positive because negative times positive would bring me back to negative negative times positive bring me back to negative negative times positive bring me back to negative very important you understand how to deal with signs when you're factoring okay so i'm going to do some division negative 25 x squared y cubed divided by negative 5 x squared y cubed what do i get negative over negative again is positive 25 over 5 is 5 and then these cancel themselves out so this first position here is a 5. for the second position i would have negative 15 x squared y to the fifth power over negative 5 x squared y cubed so negative 15 over negative 5 is going to give me positive 3 x squared over x squared is 1 y to the fifth pi over y cubed is y squared so this would be 3y squared and then lastly we would have negative 50 x cubed y cubed over negative 5 x squared y cubed and so negative 50 over negative 5 is 10 x cubed over x squared is x to the first power y cubed over y cubed is 1. so this would be 10x so 10x now what you could have done again this is what i'm trying to do here is explain to you that i didn't have to factor out this negative i could have said the gcf is 5x squared y cubed so i could have just factored out a positive the only difference is all the signs in here would change this would be negative this would be negative and this would be negative so it's just a matter of preference okay what do you want it to be if i wanted to pull the negative out i'm allowed to do that that's all i'm telling you you could factor out the gcf or the negative of the gcf and it's still going to be an acceptable answer either way hello and welcome to algebra 2 lesson 41 in this video we're going to learn about factoring by grouping all right so again for most of you that took an algebra 1 course you've had experience with factoring by grouping it's not a very difficult concept it's something you just have to do enough practice with so for factoring by grouping the very first thing you want to do you want to look for the gcf again the greatest common factor of all terms now when this is not a one or we could also say when it's not a negative one you want to factor this term out to begin so what are we saying here well let's say if we had a four term polynomial and everything had an x as an example well you'd want to pull that out before you even start doing anything because that's common to everything when you don't do that you're going to end up with additional factoring after you do the process and you might miss it okay and that's going to give you only partial credit on your tests your homework whatever you're working on so you want to always make sure you pull things that are common out first before you go into the steps all right now here's the next part and this is where it gets a little confusing you want to arrange the terms into two groups of two terms each so this is for a four term polynomial so you'll have group one and group two now i have such that each group has a common factor now in some cases that common factor might be one or negative one the whole goal is to end up with a common binomial factor okay so you're going to have to play around with this a little bit this isn't something where you can just kind of read a procedure and it's always going to be the same so i have here that you'd factor the gcf out of each group again that might be factoring out something that's not one from one group it could be factoring out one from another group and i have here we should be left with a common binomial factor now if you try a certain configuration and it doesn't work i have here in some cases we must try a different grouping so don't get discouraged right away or don't say oh this isn't factorable right away a lot of the problems you're going to see especially if you're intersection on factoring by grouping are going to be factorable you just have to change around the terms let's look at the first example and this one's pretty easy i'm going to start out by just going with the way things are presented and i'm going to show you that you can do this multiple ways so we have 18 n cubed plus 21 n squared minus 6n minus 7. so again the first thing i want to think about is is there something other than 1 or negative 1 that i could pull out from every term here set it outside of some parentheses the answer to that is no and what i look for right away i see this 7 here the 7 is a prime number so i know that it's only divisible by itself or 1. could i pull a 7 out from this 6 no 6 is 2 times 3. so i can just stop there i know i'm not going to be able to pull a number again other than 1 or negative 1 out from every single term of this polynomial now variable wise you see that you have an n an n and an n but i don't have an n here so that's not going to work either so we're not going to be able to pull anything common out to start all right so then we kind of go into our grouping so i think about the first two terms as a group so we could say this is you know group one and then i think about the next two terms as a group so let's say this is group two so in the first group or group one what could we pull out let's just say this is a separate problem and we go back to what we're learning in the last video where we factored out the gcf well if i think about 18 and 21 they have a common factor of three right 21 divided by 3 is 7 18 divided by 3 is 6 and 6 is 2 times 3 so i can't go any further so if i pulled out a 3 and put that out in front and then variable wise i have n cubed and n squared the variable is the same so i know i'm going to have an n and the smallest exponent is a 2 so it would be squared so let's say i pulled that out from that first group what's left what are the two terms inside now remember we can use division eighteen divided by three would be six n cubed divided by n squared would be n then 21 divided by three would be seven and square root of n squared is one so i've factored the gcf out from group 1. i've written it like this now i'm going to put plus from group 2 negative 6n minus 7. what could i factor out what's the gcf there well 6 and 7 you're not going to get anything out of there other than 1 or you could do negative 1. you can't do anything 7 is prime 6 is 2 times 3. the n is only part of this negative 6 n there's no n over here so in this particular case i'm going to try 1 and then i'm going to try negative 1 to see if i end up with a common binomial factor it's obvious because the signs here are positive and positive that i need to pull out a negative 1 right because this is a negative and this is a negative so if i pull out a negative 1 what's going to happen is the sign of each term in here is going to change negative 6n divided by negative 1 is 6n negative 7 divided by negative 1 is positive 7. so now if you look we have the quantity 6n plus 7 and the quantity 6n plus 7. so we have a common binomial factor and just like when we factor anything else out we could just pull it out so let's pull that outside so we'd have 6 n plus 7. and if i've pulled that out well then this is left here so i'm going to start another set of parentheses and say this is left so 3n squared is left then i have my plus and i'm left with if this is gone i just have the negative 1. so i can put plus negative 1 or most of you would just rather write minus one doesn't really matter and i'm left with the quantity six n plus seven times the quantity three n squared minus one okay very very easy now i wanna show you a different grouping and show you that there's alternative ways to kind of get the same answer and another thing you might want to do you might want to pause the video and you might want to foil this and prove to yourself that it goes back to this a lot of students have trouble believing that so let me erase this as well for right now we don't need that anymore and let's say you started this problem out by saying okay well i have an 18 here and i have a negative 6 here well each of those would be divisible by 6 so those should be in a group together okay those those could be group one for example and then you might say okay well this has 21 and this has a negative seven those could be a group because each is divisible by seven okay so let's try it that way let's make group 1 18 n cubed then you'd have minus 6 n let's make group 2 we'll have plus 21n squared minus 7. so this would be your group 1. and now this would be your group 2. so from group 1 now i can pull out a 6n so if i pulled out a 6n what would i have again 18n cubed divided by 6n would be what that would be 3 n squared and then negative 6n divided by 6n would be negative 1. i'm going to have plus from group 2 i can pull out a negative seven or i can pull out just a seven and it's just a matter of the signs now if you look at group one and what we pulled out the binomial inside the parentheses is positive then negative if i pull something out here i'm going to want that same configuration i want positive then negative so seeing that i know i want to just pull out a positive 7. so pull out a 7 and what would happen 21 n squared divided by 7 is 3n squared negative seven divided by seven is negative one so minus one and again you get your common binomial factor of three n squared minus one again that's what you have right here if i factored it out i would get the same thing as i got here right i'd pull out the 3n squared minus 1 it would be there it'd be in a different position to be in the first position here but that doesn't matter the order of multiplication is irrelevant then i would have the quantity 6n plus 7 as my other binomial and that's what i have here so either way you do that you end up with the same answer so a lot of different ways to kind of rearrange things and make things work all right let's take a look at another one now so let's suppose we had 49 x cubed plus 35 x squared minus 42 x minus 30. so again the first thing i want to think about is there anything that's common to everything i know 49 is 7 times 7 so again 7 is a prime number so i go through and look at everything 35 is divisible by 7. 42 is divisible by 7 but 30 is not so i can't pull anything out number wise variable wise is the same i've got x x x but no x here so i'm not going to be able to pull anything out other than a 1 or a negative 1 which again it's a waste of time at this point so now i would think about okay i want two groups of two terms each where each group has a common factor now again that common factor might have to be one or negative one so i would start by thinking about okay if this was a group let's say this was group 1. what could i pull out let's just say i pulled out a 7 because both of these are divisible by 7 and then an x squared right they each have an x the smallest exponent is a 2. so 7 x squared what would that leave me with 49 divided by 7 is 7 x cubed divided by x squared is x and plus 35 divided by 7 is 5 x squared divided by x squared is 1. so let's think about group 2 now so this is going to be group two and our goal is to pull something out to where we end up with a seven x plus five that quantity so what could we do we think about what's common the first thing is if i have 42 and a half 30 number wise the greatest common factor is a 6 right because 42 divided by 6 is 7 that's a prime number 30 divided by 6 is 5 that's a prime number so the best i could do is pull out a 6. now one thing you want to notice is if i pulled out a positive 6 the two terms inside the parentheses would be negative the two terms inside the parentheses here are positive so i could just do the opposite of that just pull out a negative 6 and what's going to happen is the two terms inside the parentheses would be positive negative 42 x divided by negative 6 would be positive 7x negative 30 divided by negative 6 would be positive 5. and voila i get my common binomial factor of 7x plus 5. so now i just want to pull that out again just pull that out if i have the quantity 7x plus 5 outside and then inside the parentheses what's left okay you think about what's left we'd have a 7x squared and then you'd have a minus 6 minus 6. so this factors into the quantity 7x plus 5 times the quantity 7x squared minus 6. and again this isn't the only way you could have done that i could have rearranged this into two different groups and gotten the same result so let's say i did 49 x cubed and then minus 42x that's a group each would be divisible by 7x let's say i did another group where we have 35x squared minus 30 each is divisible by 5. so in the first group i would pull out a 7x and i'd have 7x squared minus 6. in the second group i'd pull out a 5 and i'd have 7x squared minus 6. so what do i get again i have that common binomial factor of 7x squared minus 6 which matches this and then if i pull that out i'd have a 7x plus 5 which matches this so again i'm just showing you in a lot of cases there's more than one configuration that's going to work all right let's take a look at another one so suppose we have 75 a b minus 30 plus 45 a minus 50 b so the first thing again is there anything i can pull out before i get started well in this case there's going to be every number here involved ends in a 5 or a 0. so i know everything's divisible by 5 so i could pull that out so if i pull out a 5 to start 75 divided by 5 is 15 so this would be 15 a b minus 30 divided by 5 is 6 plus 45 a divided by 5 would be 9a minus 50 divided by 5 would be 10 then times b would be 10b so we pull that out before we even begin continuing i do my normal process with the inside part forget about the 5 outside that's just going to multiply whatever the result of this is so inside here let me just write the 5 again i'm looking for two groups of 2 where each group has a common factor and again that common factor might end up being 1 or negative 1. now one thing that you would see right away if i look at 15 a b and i look at negative 6 they don't share a b okay only a b in this term if i look at 9 a and negative 10 b again they don't share a b i'm going to want to put the terms with b's next to each other let me show you what happens if i don't so 15 a b minus 6 plus 9 a minus 10 b there's no way to factor this the way it is if i look at this group of 2 i could pull out a 3 okay i could pull out a 3. and what would be left is five a b minus two but over here i can't pull out anything so plus i could pull out a one if i want or negative one let's just try one so 9 a minus 10 b i don't have a common binomial factor here so i'm kind of stuck there what i need to do in this situation is just rearrange things okay so i'm going to put 5 times inside of parentheses i'm going to put the terms with b's next to each other so 15 a b minus 10 b and then i can put minus 6 plus 9 a so now let me kind of scroll down get some room going if i look at this first group what can i pull out well i can pull out a 5b so 5 times i pull out a 5b from there and inside the parentheses 15 divided by 5 is 3 a b divided by b would just be a then if i did negative 10 b divided by 5 b i'd have minus 2. now from this next group i can pull out a positive 3 or negative 3. let's just start by pulling out a positive 3. so if i pulled out a 3 i'd have negative 6 divided by 3 which is negative 2 then 9a divided by 3 would be plus 3a now a lot of you at this point if you came to this would say up i don't have a common binomial factor but pay close attention here you have 3a minus 2 and you have negative 2 plus 3a those are the same remember if i'm adding i can switch the order if i just switch the order here okay and just change the terms around i've got the same thing so i can just put 3a minus 2 3 a minus 2 and now i have a common binomial factor so if i factor that out remember the 5 stays out in front i would have 3a minus 2 3 a minus 2 times again if i pulled this out what's left well i have 5b plus 3. so 5 times the quantity 3a minus 2 times the quantity 5b plus 3. and again there are scenarios where you have to change things around otherwise it will not work and that's a perfect example so i just wanted to quickly come back up to the original problem and i want to show you what happens if i don't pull a 5 out to start if i don't pull a 5 out to start we already know that i can put these two next to each other and these two next to each other and i will get a common binomial factor so let's do that let's do 75 a b minus 50 b and then let's switch the order here just put plus 45 a minus 30. so from the first group if i think about 75 and i think about 50 i know that the gcf is 25. so 25 and then i know that i have a b here and a b here so let's do 25b so 75 divided by 25 is 3 a b divided by b is a then minus 50 divided by 25 is 2 b over b is 1. so from these two 45 and 30 i know that i could pull out a 15. so if i pull out a 15 i would have 45 divided by 15 which is 3 then times a and then negative 30 divided by 15 is negative 2. so let's continue here i've got a common binomial factor of 3a minus 2. all right so that would come outside so 3a minus 2 and then what's left when i pull that out what i have 25b plus i'd have 15. now you might say well okay that's fine it's not whenever you factor you must factor completely okay so is there anything else that i could pull out from 3a minus 2 no there isn't is there anything else i could pull out from 25b plus 15 yes there is 25 is divisible by 5 and so is 15. so i would actually have to pull a 5 out from this and say that and i'm just going to switch the order to make this convenient i'd have 5 times 25 divided by 5 would be 5 then times b plus 15 divided by 5 is 3 then times this other 3 a minus 2. so if i reported this as my answer which is a common mistake i'm not going to get full credit because i didn't factor completely when i look at this answer 5 times the quantity 5b plus 3 times the quantity 3a minus 2 i've factored completely nothing else i can take out of here nothing else i can take out of here right other than 1 or negative 1. and again that's why i preach to pull something out if it's common when you begin because then you don't have to worry about this happening at the very end so let's look at one more and we're just going to wrap this lesson up so we have 56 n squared a b minus 40 n squared m a plus 70 n squared a squared minus 32 n squared mb so let's see if we can pull anything out 56 i know is 8 times 7 and i know 8 is 2 cubed but let's not go that far 40 is 8 times 5 seventy i would think about ten times seven and thirty two would be what eight times four now forget about the seven and the five that's not going to work everything is divisible by 2. i can pull out a 2 to begin now variable wise i start with n this has an n squared n squared n squared n squared i can pull that out then i think about a so i have an a here an a here and an a here but not an a here then i think about b a b here but not here then i think about m and there's no m here so really all i can pull out is 2n squared that's fine so inside of parentheses 56 divided by 2 would be 28 n squared over n squared is 1 so i'll just have 28 a b then we have minus you think about 40 divided by 2 that's 20 n squared over n squared is one and then you'd have m a or you could write a m doesn't matter then plus 70 divided by 2 is 35 n squared over n squared is 1 and then a squared then minus 32 divided by 2 is 16. n squared over n squared is 1 so you'd have mb or you could put b m okay so let's think about if we had a group here and a group here would that work out well i just think about the variables for a minute if i look at this one right here i pull out an a i would have a b and an m i can't pull any variables out here so i'd be left with a squared and bm so that configuration is not going to work i don't need to waste my time so configuration wise what i want to think about is putting these two next to each other because they each have an m so i'd be left with an a and a b i pulled out the m with these two they each have an a so if i pulled out the a i'd be left with a b and i'd be left with an a so that's what i want you can see that that would work out i might have to change the order around but again if you're adding things that doesn't matter all right so i want to flip the configuration let's do 2 n squared let's take 28 a b and let's stick that next to plus 35 a squared and just for the sake of completeness let's go ahead and just rearrange that let's say we have 35 a squared plus 28 a b let's do it that way okay then next i have this minus 20 am so let's put plus i'm just going to put this in parentheses negative 20 am and then minus you have this 16 bm so from the first group i'm just going to copy 2n squared so from the first group here i could pull out a 7a what's left 35 divided by 7 is 5 a squared over a is a then plus 28 divided by 7 is 4 a b over a is b then plus from the second group i can pull out a negative because both of these are positive so let's do negative and then 20 and 16 are each divided by four so let's do negative four and then there's an m that's common so negative 4m so inside i'd have negative 20 divided by negative 4 which is 5 am over m which would be a then negative 16 bm divided by negative 4m is positive for b so what do i have here i have a common binomial factor of 5 a plus 4 b so this will be 2n squared times if i pulled this out i'd have that 5a plus 4b and then what's left well i'd have 7a plus over here i would have negative 4m so i'm just going to write minus 4m so we end up with 2n squared times the quantity 5a plus 4b times the quantity 7a minus 4m hello and welcome to algebra 2 lesson 42. in this video we're going to learn about factoring trinomials with a leading coefficient of one so once again we're on a topic that if you took an algebra one course you're very familiar with this this is a very very easy scenario to deal with i just want to start by talking about what a trinomial is and how we're going to factor it into the product of two binomials so i want you to recall that a trinomial is a polynomial with three terms so as a generic example this is a trinomial we have ax squared plus bx plus c so you've got a term a term and a term three terms that gives you a trinomial now in this particular case that we're going to look at today we're going to factor this trinomial where the leading coefficient in this case that's represented with a is going to be a 1 and it's going to be factored into the product of two binomials okay so one binomial times another binomial now when we have a binomial times another binomial we already know a shortcut for that known as foil okay so if i just use foil here let's just observe a few things you do first times first x times x is x squared then the next thing i would do is the outer x times 9 is plus 9x then the next thing i would do is the inner 3 times x is plus 3x then the last thing i would do is the last 3 times 9 is plus 27. now if i have x squared plus 9x plus 3x plus 27 i can combine like terms and i can say this is x squared plus 9 plus 3 is 12. so the middle would be 12 and then that common variable of x and then plus 27. so the question for today is how to go from this to this okay we're going to be factoring remember that's the reverse of the distributive property which is what foil really is so if i'm going to go from this to this let's just do it in reverse i know that i would set up two sets of parentheses here and if the leading coefficient on this is a 1 well then i know that this times this gives me this so this is going to be x and this is going to be x so that's always going to be the case if i had y squared in this position this would be y and this would be y if i had z squared z and z if i had q squared q and q okay so the first position of each is very easy to get where you have to think a little bit is there's four steps to foil so you need the o the i and the l you've got to work that out and the way you work that out is you need two integers here that are going to sum to this middle coefficient the coefficient for x to the first power and that's going to give you a product of this final constant term so you would say you want a sum of 12 and a product of 27 and why is that the case well the outer and the inner if you remember this step the outer was 9x the inner was 3x we add those together that gives us 12x so this we're looking from the outer plus the inner and then this comes from the last 3 times 9 that gave us 27. so we're working out these three by just finding out two integers whose sum is 12 and whose product is 27. and we can see that 9 and 3 gave us the 12 in the middle and then 9 and 3 gave us the product of 27 in the end so if i started without knowing this information if i just scrolled past it and said this is what i have if i take 27 really the only factors would be what 1 times 27 or 3 times not i can't really do anything else so i'd have to work with 3 and 9 and that happens to work in this case x plus 3 and then x plus 9. so that's how you go back and forth you just find two integers whose sum is the middle term and whose product is the final term the first spot in each is going to be super simple okay so you just have to work out the final spot in each so let's take a look at an example and if you didn't understand anything that i just said that's okay because you're going to pick this up right away so the first thing that i tell people to do is just set up the parentheses if you just get that set up and say okay well i know if this is x squared for my first it would be x and then x all i have to do now is fill in this blank and this blank very easy to do find two integers whose sum is negative seven so i want this for the sum and whose product is 12. so i want a product of 12. now if i think about 12 if i factor it i've got 1 times 12 i've got 2 times 6 i've got 3 times 4. now let me stop for a second if i want a positive product but a negative sum i know that i've got to have a negative and another negative negative times negative is positive negative plus negative is negative so thinking about my two integers i know that negative three and negative four would work perfectly negative 3 times negative 4 would be positive 12 negative 3 plus negative 4 would be negative 7. so what i want here is x minus 3 that quantity times the quantity x minus 4. and if you want to you can check this with foil x times x is x squared the outer would be minus 4x the inner would be minus 3x you sum those you get negative 7x and then the last would be negative 3 times negative 4 which is 12. so you would get x squared minus 7x plus 12 back all right so for the next one we'll look at it's x squared plus 2x minus 35. so again i want to set up my parentheses here and i want to just say okay well i know that this times this will give me this so i need x and x the first position of each one is always a breeze to get it's this position here and here that we have to do a little work for again two integers that are going to sum to two so sum to the coefficient for that middle term and then whose product is that constant at the end so this is the product all right so two integers we think about negative 35 if i have negative 35 as a product one of these has to be positive one of these has to be negative so just think about 35 for a second you've got 1 times 35 can't play with the signs and make that work it's not divisible by 2 it's not divisible by 3 not divisible by 4 it is divisible by 5. i've got 5 times 7. now i can make 5 and 7 work but i got to play with the signs the sum is positive 2. so the larger absolute value needs to be positive so if i do positive 7 and i do negative 5 that will work positive 7 plus negative 5 is going to give me positive 2. positive 7 times negative 5 is going to give me negative 35. so what i want is the quantity x plus 7 times the quantity x minus 5. take a look at n squared plus 12 n plus 20. so again let me just set up my parentheses here and again if this is n squared then this is n and this is in if it was q squared this is q and this is q so very easy to get the first position in each again what we think is right here and right here so two integers whose sum is 12 okay the sum is 12 and the product is 20. now if i think about factors of 20 i think about what you have 1 times 20 that's not going to work 2 times 10 that would work 2 times 10 is 20 2 plus 10 is 12. so n plus 2 and plus 10 right those two quantities the quantity n plus 2 times the quantity n plus 10 would give you n squared plus 12n plus 20 back so in some cases we're going to have a common factor involved so 4r squared plus 20r plus 16 there's a common factor of 4 that we can pull out and i'd be left with r squared plus r plus four and this may trip you up because if you're in this section in your textbook and all of a sudden you see a coefficient that's not one you might say well i haven't gotten to that yet and that's a more difficult scenario but a lot of times they're going to give you these problems to where you have to factor something out first and then you can use your normal procedure so 4 just stays out there you don't need to do anything with it just put the parentheses like you normally would and then just think about what's inside here okay what's inside here so i know if i have r squared it's r and r that's easy then i need two integers whose sum is five so a sum of five and a product of four okay a product of 4. so that's really easy because if you think about 4 it's really 1 times 4 or 2 times 2. 2 plus 2 is 4 that doesn't work for the sum so i'd have to go with 1 and 4. if i do 1 times 4 that's 4 1 plus 4 is 5. so this would be plus 1 and this would be plus 4. so i end up with 4 times the quantity r plus 1 times the quantity r plus 4. let's take a look at another one like that so we have 6p to the fourth power minus 84 p cubed plus 240 p squared so obviously if this is supposed to be a one here the coefficient for the leading term we've got to do some additional work so i notice that i can pull out a 6p squared from everything and so what would happen is this would be p squared then minus this would be a 14p then plus this would be 40 and then we could close our parentheses now all i need to do is think about factoring this guy now and let's just set this up so we'd have 6p squared and set up your parentheses again if i'm factoring this and p squared is the leading term this is p and this is p for these positions here i just want two integers whose sum is negative 14 and whose product is 40. now that's really easy when you think because i know that again if i have a positive for my product and a negative for my sum i know i need a negative and a negative and just think about 40. you think about 4 times 10 if i did negative 4 times negative 10 that's positive 40. if i did negative 4 plus negative 10 that's going to give me negative 14. so p minus 4 and then p minus 10 there so 6p squared times the quantity p minus 4 times the quantity p minus 10. all right so let's talk about another scenario that confuses people a lot you might see two variables involved so let's say you come across something like x squared plus 8xy plus 7y squared and you go how in the world am i going to factor that it's actually no more difficult than what you've just been factoring if you just set up your parentheses and you just think for a minute we know that the first term here times the first term here will give me x squared and that's how i have x times x well if i know that this is the result of multiplying this times this i know that that's going to come from y times y now if this is set up i just need to figure out what the coefficient of y is there and there because the outer and the inner will have x y involved so that's going to give me this right here i just got to get the coefficient for y worked out in each case and i do that the same way forget about that second variable i don't even care about it i've already set it up for myself so all i need to really do is say give me two integers whose sum is eight and whose product product is seven well that's easy seven is prime so it can only be one and seven and that works out so put plus seven put plus one so you have the quantity x plus seven y times the quantity x plus 1y or just x plus y and check that through foil x times x is x squared your outer would be x times y or plus x y your inner would be 7y times x or plus 7xy and your last would be 7y times y which would be plus 7y squared if you combine like terms here so you'd end up with x squared plus 8xy plus 7y squared which is exactly what you started with right there so if you have two variables involved don't panic it's just as easy just make sure you figure this part out and this part out before you move on and figure out that sum and product part that we've been talking about for the whole lesson let's take a look at one last problem and then we're going to talk about prime polynomials and let's give you one example of that so we have x squared minus 12xy plus 27y squared so again if i see two variables involved i'm not going to panic set up your parentheses i know that this would be x and this would be x and if i have a y squared in the final position here i know that came from this times this so i'm just going to put a y here and a y here and then i just say okay well two integers whose sum is negative 12 this is my sum and whose product is 27 so this is my product super simple okay we think about 27 27 i could do 1 times 27 or i could do 3 times 9. now i've got to work out some things with the signs here because i've got a positive product and a negative sum so that means i want a negative and another negative negative times negative is positive negative plus negative is negative so what i can do here is say a negative 3 and a negative 9. negative 3 times negative 9 is positive 27 negative 3 plus negative 9 is negative 12. so i would have the quantity x minus 3y times the quantity x minus 9y okay super super simple now let me give you an example real quick of something that's prime you're going to come across these polynomials that you just can't factor okay so let's say i had something like x squared plus 7x plus 14 and i said go ahead and factor that for me well you go ahead and start out by saying okay well i can do that x and x that's going to be leading and then two integers who's what whose sum is 7 whose product is 14. so this is the sum and this is the product and this is a great example to do this because it's very easy because 14 really is what it's only 1 times 14 or 2 times 7. now if i add 1 and 14 i get 15. that doesn't work if i add 2 and 7 i get 9 so that doesn't work so there's no configuration that would work and because this is positive and this is positive i know the signs for the integers have to be positive so really i'm out of options here and i've got to stop and say well this polynomial is prime this is prime and using rational numbers i will not be able to factor this hello and welcome to algebra 2 lesson 43 in this video we're going to learn about factoring trinomials with a leading coefficient that is not one and we're going to be using the factoring by grouping method so in the last lesson we reviewed how to factor a trinomial with a leading coefficient that is one so this is by far the easiest scenario you're going to come across so i have an example of that so let's say you had something like x squared plus 5x minus 24 and here you can see the leading coefficient or the coefficient for the squared variable x is a 1 right if nothing's displayed there the coefficient is understood to be a 1. so in this particular scenario it's very easy to factor this into the product of two binomials i can just set up my parentheses and i know if i have x squared the first position here is x and the first position here is x as well x times x would give me x squared now the only thing that i really need to figure out is what goes here and here and to do that i think about two integers whose sum is 5 and whose product is negative 24. so i could think about the fact that i have a negative 24 a negative product comes from what comes from a positive times a negative so i could just write that one of these is positive and one of these is negative and i could just think about factors of 24. i've got 1 times 24 i've got 2 times 12 i've got 3 times 8. now when i see something that i know in my mind those two integers if i play around the signs could get me to 5 i just stop right 3 and 8 if i play with the signs will work if i had a negative 3 and a positive 8 well negative 3 times positive 8 would give me negative 24 negative 3 plus 8 would give me positive 5. so i know i want x plus 8 and i want x minus 3. so that's how you go about factoring these easy scenarios not a whole lot of work you just have to go through the factors and figure out what two integers again sum to this the coefficient for the variable that's raised to the first power and have a product of this your constant at the end now what we're going to look at today is something that's a little bit more challenging okay what if we have a scenario where the coefficient on the squared variable is not a 1. so in this example i have 8x squared plus 51x plus 18. so you notice that this is an 8 here so i can't just go ahead and set this up and say okay well i have you know x times x that's no longer going to be the case so now instead of just having to find two i have to find all four positions so how can we do this well there's two main methods that are taught in every textbook one is going to be reversing the foil process which i personally don't like and the other is going to be a method that relies on factoring by grouping now i like this method i think it's quicker and more efficient so i'm going to start by teaching you that in this lesson and the next lesson i'm going to cover reverse foil all right so the first thing you need to understand is that in your textbook when they're teaching this method they're always referring to a b and c and you need to know what those are so generically when we see a trinomial we generally see ax squared plus bx plus c so what does a represent a represents this it's the coefficient for the squared variable then what does b represent b represents this it's the coefficient for the variable raised to the first power then what does c represent c represents this this is the constant so if your book says something like we want to find two integers whose product is ac and whose sum is b well i know that a product of ac is what well a is 8 in this case and c in this case is 18. so i would need to multiply 8 times 18 and that would give me 144 and then a sum of b a sum of b well b is 51. so what i'm going to do here is i'm going to think about some factors of 144 and i could do that with a factor tree or i can kind of do that by just going through things doesn't really matter let's just make a factor for you 144 is what 12 times 12 most of us know that and then 12 is what it's 4 times 3 we know 3 is prime and then 4 is 2 times two so to go through this i can think about two integers whose product would be 144 and whose sum would be 51 okay well we could start with one times 144 that's obviously not going to work you can just mark that out then you could go on to let's say 2 times 72 again that's not going to work mark that out then is it divisible by 3 yeah of course it is so i could do 3 times 48. now let me stop for a second when you see something like this you say okay well 3 and 48 if i had 3 plus 48 that would in fact give me 51. so it's just a matter of going through the factors until you get the two integers that you need so once i know that i need 3 and 48 it's a little bit of a different process from before let me kind of erase this i'm going to say my two integers are positive 3 and positive 48. so for some of you you'll stop and you'll say okay well i've figured it out this is going to be x plus 3 that quantity times the quantity x plus 48 and i'm done well no not exactly okay you need to take this information right here and use it to rewrite this polynomial here okay remember when we use factoring by grouping and i'm just going to erase this real quick when we use factoring by grouping we are generally factoring a four term polynomial so what i want to do is i just want to make this trinomial into a four term polynomial and the way i'm going to do that is i'm going to rewrite this middle term this 51x using these two integers so i would have 8x squared i'm going to put plus 48x then i'm going to put plus 3x and then plus 18. so mathematically have i done anything illegal no because 51x is the same as 48x plus 3x so i've just rewritten this polynomial as a polynomial with four terms now once i've done that what you're going to see is that you can use factoring by grouping now to get this into the product of two binomials so from the first group i could pull out an 8x what would be left is what you'd have x plus six from the second group i could pull out a three what would be left is x plus 6. so i have a common binomial factor of x plus 6. so if i factored that out what would i have well i'd have x plus 6 that quantity times what's left you'd have 8x plus 3. so the quantity x plus 6 times the quantity 8x plus 3. now i'll leave it up to you to pause the video and check this using foil but you will in fact get this back 8x squared plus 51x plus 18. so this is really quite a simple process sometimes it will be a little tedious depending on what your integers are that you're trying to find but it is something that you can go step by step on all right let's take a look at another one so we have 10x squared plus 43x minus 9. again just for the first few i want to just identify things so this is ax squared plus bx plus c so a here is 10 b here is 43 and c here is negative 9. okay the negative and the 9. that's why i circled it so if i want two integers whose product again the product is a c so what is a it's 10 what is c it's negative 9 so ac would be 10 times negative 9 that's going to be negative 90 and we want a sum of this middle term b so a sum of 43. all right in order to figure this out i think about the factors of 90. forget about the fact that it's negative 90. just think about 90 for a second 90 is pretty easy i don't need to really make a factor tree you've got 1 times 90 and again you've got to flip the signs around and make this work because this is going to be a negative product but i already know that this won't work out because it's never going to sum the 43 no matter how much i change the signs around then the next thing i would say is 2 and 45 so 2 times 45 well immediately i found something that would work if i did a negative 2 and a positive 45 well yeah 45 minus 2 is 43 that's what i'm looking for and then negative 2 times positive 45 is negative 90 that's what i'm looking for there so remember i'm not done once i've figured this out i've got to use this information to rewrite this middle term so what i'm going to do is i'm going to say i have 10x squared i'm going to do minus 2x then plus 45x and then minus 9. all right so once i've rewritten this trinomial as a polynomial with four terms again i can use factoring by grouping so let me erase this and we'll set this up so from the first group i could pull out a 2x so pull out a 2x i'd be left with 5x minus 1. from the second group i could pull out a 9 i'd be left with 5x minus 1. so you see you have a common binomial factor of 5x minus 1 and we can just factor that out so we'd have 5x minus 1 that quantity times the quantity 2x plus 9. so this is our answer the quantity 5x minus 1 times the quantity 2x plus 9. all right let's take a look at another one so we have 6x squared minus 5x minus 25. so again let me just label things so this is ax squared plus bx plus c this is a this is b in this case the negative and the 5 so i'm going to circle it so that's clear and this is c negative 25 negative and the 25. so again if i want two integers whose product is ac a is 6 c is negative 25. so 6 times negative 25 is going to give me negative 150 and then we want a sum a sum of negative 5. so we think about factors of 150 forget about the negative 4 second just think about factors of 150. 150 so you've got 1 times 150 again no way you can make the signs work there then you'd have 2 times 75 can't make the signs work there you would have 3 times 50 that's not going to work you would have 5 times 30 that's not going to work you would have 6 times 25 again not going to work then the next one you come to would be 10 times 15. if you think about this i can make that work with the signs if i had a positive 10 and a negative 15 i can make that work so i'm always mentally kind of saying okay well i know it would work with the product but can i mess with the signs here and make it work well yeah if i think about 15 minus 10 in my head i get positive 5 i know i need a negative 5 but i can just play with the signs and make that work so that's that's kind of what i'm doing here okay that's what i'm thinking about so let me erase this we've found our two integers and again i just want to rewrite this middle term i want to rewrite negative 5x using these two integers so i would have 6x squared i can say plus 10x so i could say minus 15x wouldn't matter so let's just say plus 10x and then let's say minus 15x and then minus 25. so now that i've rewritten this as a four-term polynomial i can use factoring by grouping so from the first group i could pull out a 2x and i'd be left with 3x plus 5. from the second group i would pull out a negative 5 and i'd be left with 3x plus 5. so i have a common binomial factor of 3x plus 5 and of course i can factor that out so i'd have 3x plus 5 here that quantity and then times what's left we'd have the quantity 2x minus 5. so your answer for this would be the quantity 3x plus 5 times the quantity 2x minus 5. all right for the next problem i'm going to do something that's a little bit more challenging a lot of students see these types of problems and they just shut down and really you shouldn't because the procedure is the same it's no more difficult so let's say you see something like 8x squared minus 26xy plus 20y squared the first thing is that notice we have an even number here here and here so before i even start i can pull that out so just pull out a 2. so pull out a 2 you get 4x squared then minus 13xy and then plus 10y squared now once that's done you think about the middle part here and again a lot of students freak out because they only have an x normally or whatever the variable is now you have an x and you have a y to deal with and like what am i going to do there's two variables i'm going to tell you what you can do don't even think about the second variable just do your normal process and i'll show you how to work that variable in in a second so if i just ignore the fact that i have a y just kind of line it out for a second and just say okay well if i had this scenario 4x squared minus 13x plus 10 what would i do i would find two integers whose product is what a times c well a is 4 c is 10. 4 times 10 is 40. and then whose sum is b b is negative 13. now if i expand on this 4 second i bring my y back really all i'm going to do is say well b is now negative 13 times y it's the coefficient for the x to the first power so what's multiplying x to the first power well negative 13 is and also y is so i can just say this is negative 13y and then what's my term at the end here well it's 10 y squared so normally i do the coefficient for the squared variable times the term at the end we can do that again so 4 times 10y squared would be 40y squared no more difficult okay you just include the second variable but you could also just leave it off if you wanted to and just make sure that you put a y next to the two integers that you come up with okay so let me just say this find two integers whose sum is negative 13 and whose product is 40 and attach a y at the end of those two integers and i'll show you that that works so for 40 we know we have 1 times 40 that's obviously not going to work 2 times 20 won't work not divisible by 3 4 times 10 won't work 5 times 8 we can make that work if i had a negative 8 and a negative 5 negative 8 times negative 5 is positive 40. negative 8 plus negative 5 is negative 13. and what i do is just attach a y to the end of each so negative 8y and then negative 5y because negative 8y plus negative 5y is negative 13y negative 8y times negative 5y is 40y squared so it's not any more difficult when you have a second variable you just kind of throw it in there right it'll work itself out really the only challenging part is finding the two integers the variable is a breeze so knowing that this is what i have now let me kind of erase this i'm still going to rewrite my middle term so i'm going to say this is equal to we'll have 2 times the quantity you'll have 4x squared now the middle term i have negative 13xy so i'm going to use these two i'm going to put minus 8 and then x and then y and then minus 5 and then x and then y so basically i just said minus 8xy minus 5xy we know that would end up being negative 13xy then plus we have our 10y squared close the parentheses there and the process is the same i'm just factoring by grouping here so in my first group i could pull out what i could pull out a 4x that would leave me with x minus 2y in my second group i could pull out a negative 5y and that would leave me with x minus 2y so i have a common binomial factor of x minus 2y and i'm just going to pull that guy out so i'll have 2 times the quantity we'll have x minus 2y and then times the quantity 4x minus 5y now let's take a look at another example with two variables involved so let's say we saw something like 12x squared plus 33xy minus 9y squared the first thing you'd notice is you could pull a three out before you begin and this is very important to do this because it saves you work in the end right you don't want to end up with additional factoring because you might miss it right so if i pulled a 3 out i'd have 4x squared plus 11xy minus 3y squared another thing that benefits you to pulling this out early is you're going to end up working with smaller numbers right smaller numbers are just easier to work with so once i've done that again i want to use the same process it doesn't matter that there's two variables involved i can label this as a okay so i can say a is four i can say b is what b is a new role remember it's the coefficient for x so if x is being multiplied by 11 and by y i could really say that b is 11y and then c is what it's this whole thing here it's this negative 3y squared so if i want a times c i would do 4 times negative 3y squared so ac is negative 12 y squared and then b is 11y but forget about these variable parts right here this y squared and this y again if i work out the integers in other words two integers whose product is negative 12 and whose sum is 11 i can just tack a y on to the end of each and i'll have my answer so if i want two integers whose product is negative 12 i think about 12 i know i can do 1 times 12 and if i play with the signs if i do positive 12 and negative 1 that would work so positive 12 again put a y at the end and negative 1 again put a y at the end that's going to work positive 12 y minus 1y if i threw an x in the mix and i sum those two i get 11xy i'm going to use that to rewrite my middle term so i'm going to have 3 times the quantity 4x squared and i'll do plus 12 don't just put plus 12y remember this you have an x involved there so it has to be 12 12xy and then minus 1xy or minus xy and then minus 3y squared let's scroll down get a little room going so factoring the inside here by grouping we have three times inside of parentheses in the first group i can pull out a 4x so i pull that out i'd have an x plus 3y inside in the second group i could pull out a negative y so i'd be left with x plus 3y inside and again i get a common binomial factor of x plus 3y that i can pull out so i'd have my 3 outside of parentheses and then i'd have x plus 3y that quantity and then times i'd have 4x minus y that quantity so this is my answer 3 times the quantity x plus 3y times the quantity 4x minus y hello and welcome to algebra 2 lesson 44. in this video we're going to learn about factoring trinomials with a leading coefficient that is not 1 using the reverse foil process so again for those of you who took algebra 1 you've seen this content before so it's not too challenging for you but you might just need a refresher on it so how would we go about factoring this using reverse foil if i see something like 3x squared minus 19x minus 14 and i want to reverse the foil process so i'm going to factor this into the product of two binomials and let's just write foil out for a second so it's f for first terms o for outer terms i for inner terms and l for last terms so the first thing you would think about is the fact that this times this okay would give me this now this is a pretty easy scenario because the coefficient of x squared is a prime number right three is only one times three or if we're really thinking it could be negative one times negative three but you don't wanna put that there because that would complicate things okay and i'll show you that at the very end what would happen if you put negative 1 and negative 3 in so i just think about 3 and 1 okay 3 and 1. now i've got an x involved there because it's x squared so i would have 3x and then x okay 3x times x is 3x squared so we've got that down and the order that i put this in wouldn't matter if i put my x here and my 3x over here that's completely irrelevant the next thing okay the next thing that you would think about is the fact that the last terms so this times this will have to give me this okay it'll have to give me negative 14. so i would think about factors of negative 14 but it's got to be such that the outer and the inner are going to combine to give me that negative 19x so that's where this trial and error is going to come in i've got to go through all the possibilities so for negative 14 i've got to think about positive 1 times negative 14. and i've got to think about negative 1 times positive 14. i've also got to think about positive 2 and negative 7 and then i've got to think about negative 2 and positive 7. okay so more scenarios are involved when you have something that's negative because it's a negative times a positive you've got to go back and forth between the possibilities so the only thing we can really do is just go through and do trial and error so in other words i would set up some scenarios so i have 3x i have my x and let's just do another one down here and sometimes i'll write out a bunch at once and just quickly go through the possibilities so let's start with this positive 1 and negative 14. so plus 1 here minus 14 here and then you've got to reverse that so you're going to go minus 14 and plus 1. because you've got to check different outer and inner products in other words for this one i'd have 3x times negative 14 and 1 times x now 3x times negative 14 is negative 42x 1 times x is 1x negative 42x plus x is going to be negative 41x so that's not going to work this isn't a possibility so you go to this one the outer would be 3x times 1 which is 3x the inner would be negative 14 times x which is negative 14x so negative 14x plus 3x is negative 11x so that's not going to work either so you can line that out and say well positive 1 and negative 14 will not work so then you go to the other scenarios so let me let me erase this and so in this scenario we have a negative 1 so i'll put a negative 1 here and here and a positive 14. so i'll put that here and here so what am i checking so 3x times 14 would be 42x and the negative 1 times x would be negative x so 42x minus x is 41x so that won't work for this one 3x times negative 1 would be negative 3x 14 times x would be 14x so 14x minus 3x would be 11x so this is gone that's no longer a possibility we can move on to the other scenarios now when we get to this scenario here again we have positive 2 and negative 7. so plus 2 negative 7 negative 7 plus 2. all right so let's check this one so 3x times negative 7 is negative 21x and then 2 times x is positive 2x and that would work negative 21x plus 2x would give me negative 19x 2 times negative 7 is negative 14. so this is your correct factorization let me erase all this i don't need to try anything else and sometimes you hit it right away other times it takes you a while but this is our correct factorization and if you want you can use foil and check it 3x times x is 3x squared [Music] then the outer 3x times negative 7 is negative 21x then the inner 2 times x is positive 2x then the last 2 times negative 7 is minus 14. if i combine like terms here in the middle i'm going to end up with 3x squared minus 19x minus 14. that's exactly what i started with right there now let me give you a little insight into why we wouldn't want to say start out with negative 3x and negative x i can factor out a negative 1 from this if i wanted to i could say this is negative 1 times and i can make this negative 3x and this minus 2. i just pull the negative 1 out i'm legally allowed to do that then times i could pull a negative one out from here as well so negative one times you'd have negative x plus seven this is mathematically the same as this negative one times negative one is one so really i could write this as negative 3x minus 2 that quantity times negative x plus 7. so i could have began if i wanted to with negative 3x and negative x so when you see something positive that's leading you can do this if you want but working with negatives in the leading position makes it a little bit more complex you want to make things as easy as you can make them so it's just better to start out with a positive here and a positive here if this is positive you know you can make that work because of what i just showed you you can always factor out a negative one from each of those multiply the two negative ones together and get one one times anything is just itself and just to completely prove this to you because i know some of you are like no that doesn't seem right if you do foil on this negative 3x times negative x is what that's three x squared negative times negative is positive 3 times 1 is 3 x times x is x squared the outer negative 3x times positive 7 is negative 21x then my inner negative 2 times negative x is positive 2x then my last negative 2 times positive 7 is negative 14. so if you combine like terms in the middle again you would get 3x squared minus 19x minus 14. so again that's why i just started out with this because it's going to be the same as this okay and it's just easier to work with positives in the leading position all right let's look at another one now and we did our first one so it should go a little bit quicker so we have 2x squared plus 23x plus 63. all right so we're going to factor this into the product of two binomials now my coefficient for the leading term here is a 2 that's a prime number so again that's a very easy scenario because i know 2 is just 2 times 1. and again for the reasons that i just covered i don't want to think about negative 2 and negative 1. so i'm just going to go 2x times x and be done with that now i just need to think about 63 here what are the factors of 63 so what times what would give me 63 but then through the process of combining like terms with the outer and the inner we would want to get a 23x in the middle and for this everything's positive so i'm just going to think about positive numbers here so 1 times 63 it's not divisible by 2 it is divisible by 3 so 3 times 21 not divisible by 4 not divisible by five is divisible by seven seven times nine and that's all we're going to get so lots of possibilities here now if you think about this this right here is very far apart if you think about the numbers 1 and 63 are far apart if i was multiplying 2 times 63 i'd have a really big number or if i was just multiplying 63 times 1 i'd have 63 this one isn't going to work you can eyeball that and see that you can just eliminate that right away now 3 and 21 i might be able to get that to work i just have to check the different possibilities so 2x x so 3 and 21. so i'm only going to check 3 positive 3 here and positive 21 here and because i'm only using plus signs here it's much quicker the other scenario is positive 21 here and positive 3 here so check your outer and inner 2x times 21 would be 42x 3 times x will be positive 3x this would combine to be 45x so this isn't a possibility we can line that up for this one 2x times 3 would be 6x and then 21 times x would be 21x so that would be 27x if we added so that's not a possibility either so we can line this out so now our last possibility is to use 7 and 9. so i would do 7 here and 7 here 9 here and 9 here so the outer 2x times 9 would be 18x and then 7 times x would be 7x if you add those together you get 25x that's not 23x so this is eliminated so it has to be this possibility or it would be prime so 2x times 7 would be 14x and then 9 times x would be 9x and if we combine 14x and 9x we do get 23x so we do get that so this is the big winner so let me erase everything and again you can check this through foil 2x times x is 2x squared the outer 2x times 7 would be 14x the inner 9 times x would be 9x 14x plus 9x would be 23x so plus 23x and then 9 times 7 would be 63. so you can see you've got exactly what you started with 2x squared plus 23x plus 63. all right let's look at a few more i think for most of you kind of figured this out by now we're going to look at one where the leading coefficient is not a prime number it makes it more tedious not any more difficult so 40x squared minus 172x plus 112 is what we're going to try to factor obviously if you look everything's even right i've got a 0 a 2 and a 2. but if you further think about this you should know that 40 is divisible by 4 and 112 is divisible by 4. those are obvious for 172 forget about the negative per second for 172 well 72 is divisible by 4 so i know 172 is so that means i can pull a 4 out before i even start so if i pull a 4 out i'm going to have 10x squared minus 43x plus 28. all right so i want to factor the inside of this now so we'll have four times inside of my parentheses now if i have 10x squared it is not obvious what's going to go here and here 10 is not a prime number the factors of 10 are 1 times 10 and 5 times 2 and again i know you could do negative 1 times negative 10 or negative 5 times negative 2 but we don't need to think about that again for reasons that we already covered so i've got to go through and try as different scenarios 5x times 2x or the other scenario i would have is 4 times the quantity you'd have 10x and then times 1x or just x so these are what we're going to consider here it becomes more complex because the final term is positive and the middle term is negative so what does that tell me it tells me that to get this i had to have a negative times another negative so that tells me that these positions here are going to be negative values so what are the factors of 28 considering only the negative values well i know it's 1 times 28 so that would be let me kind of just make a line here negative 1 times negative 28 i know it's 2 times 14 so negative 2 times negative 14 i know it's 4 and 7 so we think about negative 4 and negative 7 and so that's all we're going to have so let's start out with a scenario where i have negative 1 here and i have negative 28 here now one thing that i haven't told you and i've talked about this in algebra one if there's not a common factor in what you're trying to factor so in other words we're trying to factor this forget about this because you say oh there's a common factor of 4. that doesn't matter we've already pulled that out if there's not a common factor here then none of the factors for this meaning neither of these binomials would have a common factor 2 and 28 have a common factor of 2. so this isn't a possibility i can try the other alternative for this which is if i have a 28 here and a 1 here so 4 times the quantity 5x minus 28 times the quantity 2x minus 1. that's a possibility but you'll see that we're going to rule this out right away because the outer 5x times negative 1 is negative 5x and the inner negative 28 times 2x is negative 56x those two are not going to combine to give me negative 43x so that is not a possibility here now let's erase that and i'm going to move on to the next scenario which is negative 2 and negative 14. now we can eliminate that right away and why can we eliminate that right away well if i put a negative 2 here and a negative 14 here i've got a common factor of 2. if i switch the order of that and i put a negative 14 here and a negative 2 here still got a common factor of 2. and it's going to be the same thing when i get down to this if i need to if i put 14 here and 2 here common factor of 2. if i put 2 here and 14 here common factor of 2. so either way you do that in any scenario this is eliminated you don't need to worry about that now i want to check this one right here in this and if that doesn't work we're going to move on to this scenario so if i put a negative 4 here and a negative 7 here common factor of 2. so that arrangement cannot work so then what i want to try is negative 4 here and negative 7 here let's see if that works so i would have 5x times negative 7 that would be negative 35x and then i would have negative 4 times 2x which would be negative 8x so it looks like that's the big winner negative 35x minus 8x is negative 43x right that's what we want right there all right so let's erase everything and we're lucky because we didn't have to go into this scenario at all all right so we found the correct factorization if you want to you can go ahead and check this you could do 5x times 2x that would give you 10x squared the outer 5x times negative 7 will be minus 35x the inner negative 4 times 2x would be minus 8x negative 35x minus 8x would be negative 43x and then the last negative 4 times negative 7 is positive 28. so that gives you this and you're multiplying this by 4 and when you multiply it by 4 4 times 10x squared will give you 40x squared 4 times negative 43x would be minus 172x and then 4 times 28 would be positive 112. so you get exactly back to this all right let's take a look at one final problem and i gave you one here that has two variables involved you've seen factoring with this before you know at this point that it's not any more challenging right you can almost just ignore the second variable work everything out and then come back to it so if i have 8x squared plus 34xy plus 36y squared the first thing you would notice is that what everything is divisible by 2. so i could pull that out before i even begin so if i pull out a 2 i'd have 4x squared plus 17 x y plus 18 y squared and so once that's done we want to factor this inside of parentheses into the product of two binomials so i think about 4x squared i know that can come from what it could come from 4x times x or it could come from 2x times 2x so we've got kind of two scenarios here and again we're not going to think about the negative version negative 2x times negative 2x or negative 4x times negative x but the reason we talked about earlier all right so let's do this one or you'd also have this one now once we've worked this part out i know that y times y gives me y squared you can just throw this in and forget about it okay we know that from working previous problems y times y is y squared when you do your outer you get x y as a variable when you do your inner you get x y as a variable so when you combine like terms you're going to have a term with x y so just put your y in there or whatever your second variable is and forget about it don't stress about having two variables super super simple all right so the next thing is to work things out you know that you're just going to focus on this 18 here and everything's positive so i'm just going to focus on positives and so i would have what for 18 1 and 18 2 and 9 and then 3 and 6. now i've got to go through many possibilities here so let's think about for a second what we can eliminate right away for 1 and 18 it would not work in here at all because i'd have to put an 18 in here somewhere and it would have a common factor of 2. so it wouldn't work in this one it might work in this one but i'd have to try this configuration here an 18 here and a 1 here so just think about 4 times 18 that would give us 72 and you'd have 1 times 1 which would be 1. so 72 plus 1 would be 73 so that's not going to work out in either one so you can go ahead and get rid of that all right so let's erase this next let's think about 2 and 9. so in the first one i can't use it at all again because i'd have to put a 2 somewhere i can't use a 2 here or here because i have a common factor of 2. there's no common factor in this right here that we're trying to factor i know there's one up here but again we've already pulled that out so there shouldn't be one in any of the binomials there so it's not going to work there in this one i can put a 2 here and put a 9 here i can check that scenario so 4 times 2 would be 8 9 times 1 would be 9 8 plus 9 is 17 so we have found what we need so let's erase everything let me just kind of slide this up so we can check it through foil so i'm going to put my 2 out in front 4x times x is 4x squared my outer 4x times 2y would be positive 8xy my inside 9y times x would be positive 9xy and my last 9y times 2y would be positive 18y squared again we think about those two middle terms here 8xy plus 9xy would be 17xy so let's just erase this and put plus 17 xy and if i distributed that 2 i'd be back to 8x squared plus 34xy plus 36y squared hello and welcome to algebra 2 lesson 45. in this video we're going to learn about factoring polynomials using substitution so in some cases we can factor a more complex polynomial by making a substitution so i want to jump right in and just look at an example here so let's suppose you see 7x to the fourth power plus 20x squared plus 12 and you're asked to factor this well you might look back at me and say well i don't know how to do that i'm used to seeing something that looks like this ax squared plus bx plus c now i've got something times x to the fourth power plus something times x squared plus some constant so the way this works if you look at this particular problem i could rewrite this and i just have to show this to you as 7x squared that's squared now if i take x squared and i square it what do i have i have x to the fourth power so i haven't changed anything i just rewrote it then plus 20x squared then plus 12. now the benefit or the reason to writing it like this you see that i have an x squared here and an x squared here so what if i made a substitution and i let a variable be equal to or the same as my x squared so that variable could be whatever you want you could use y or z or q or n or whatever you want to do let's just go ahead and use u so u is going to be equal to x squared so that means everywhere i see an x squared i'm going to plug in a u so let me erase this we don't need this anymore and so i'm going to rewrite this problem as 7 i have an x squared there so i'm going to write a u then that's squared so 7u squared plus we'd have 20 then i have an x squared so that's going to be a u then plus i have 12. so now if i look at this right here it matches a format that i'm familiar with okay very easy to factor this we just talked about it we can use reverse foil if we want or we can also use a method that relies on factoring by grouping i prefer to use factoring by grouping so let's go ahead and factor this guy i would want two integers whose product is 7 times 12 that's 84. so the product is going to be 84. and then the sum is what the sum is to that middle coefficient so a sum of 20. so think about factors of 84. you think about 84 it's what you got 1 times 84 that's obviously not going to work right because the two integers have to sum to 20. then you've got 2 and 42 that wouldn't work you've got 3 times 28 which wouldn't work 4 times 21 which wouldn't work it's not divisible by 5 but it is divisible by 6. it's 6 times 14. now 6 and 14 sum to 20 so that's your magical combination there so let's erase this we know what we need to erase this so we take those two integers and we rewrite the middle term so i would say that i have seven u squared plus i'll put 14 u plus i'll put 6u and then plus 12. now i'm going to factor using grouping so i'm going to take the gcf out of each group so from the first two i'm going to pull out a 7u and that's going to leave me with what i'd have a u plus a 2. from the second group i'm going to pull out a 6 and i'm going to have a u plus a 2. so i can pull out the common binomial factor of u plus 2. and that's going to give me what that's going to give me u plus 2 that quantity times what's left i'd have 7u there plus i'd have 6 there so u plus 2 that quantity times the quantity 7u plus 6. so if i'm taking a test or i'm doing my homework and i'm asked to factor this can i report this as my answer no i made a substitution to make it easy on myself so i've got to substitute back in the end if u is equal to x squared that means i can take every u that i have okay so there's a u here and here and i can replace it with x squared so i can say that my answer is what it's x squared plus two that quantity times the quantity seven again you've got a u there so x squared plus six so i have factored this seven x to the fourth power plus 20x squared plus 12 as the quantity x squared plus 2 times the quantity 7x squared plus 6. again just using a simple substitution let's take a look at another example so we have 27x to the sixth power minus 54x cubed plus 15. so before we do anything we should notice that everything here is divisible by 3. so i want to make sure i pull that out so i don't forget about it or have additional factoring in the end so if i pulled out a 3 to start this 27 would now be a 9 and still have my x to the sixth power then minus 54 divided by 3 would be 18 then x cubed then plus we know 15 divided by 3 is 5. so this is what i'm working on now i just want to factor what's inside the parentheses now again we can use a substitution if you look at what we normally try to factor it's ax squared plus bx plus c you think about this exponent as a 1 this exponent is a 2. so this exponent here is double this exponent here in the last example we looked at we had an exponent that was a 4 which was double the exponent of a 2. in this example here you have an exponent of a 6 that's double the exponent of a 3. so when this occurs you can use a substitution okay and you're going to take a variable let's say u or q or n or whatever you want to do let's just say z for this example so let's say z that's going to be equal to the variable raised to the smaller power so z is equal to x cubed in this case now since this exponent is double this one what i can do using my rules of exponents i can rewrite this i'll put three times inside of parentheses 9 x cubed so you want to match that but i'm doubling this exponent so i'm just going to raise it to the second power right or square it by the rules of exponents x cubed squared would be x to the sixth power then you'd have minus 18 x cubed then plus 5. and the reason you write this this way is so it's obvious what you're going to do when you substitute here's x cubed and here's x cubed so what i'm going to end up doing when i make my substitution i'm going to say that i have 3 times inside of parentheses 9 this right here will be replaced with a z z equals x cubed i have x cubed right there so that would be z and then squared then minus i have 18 times here comes x cubed again so i'm going to substitute with a z and then plus 5. so all i have to do now is just factor this as i normally would once i'm done i go back and i resubstitute and i have my answer so give me two integers whose sum is negative 18 so the sum is negative 18 and whose product okay whose product is 9 times 5 or 45. now we know from the rules of signs that this has to be a negative and another negative negative plus negative is negative negative times negative is positive so i'm thinking about negative values but in thinking about that i can just think about positives and then just switch over so think about the factors of 45 you've got 1 and 45 which obviously wouldn't work it's not even so it's not divisible by two it's three and fifteen three times fifteen would be forty-five now what's interesting here is three plus fifteen is eighteen but i just want negatives involved so i would think about negative 3 and negative 15. that's what i want negative 3 times negative 15 is positive 45 negative 3 plus negative 15 is negative 18. so i'm going to use those two integers to rewrite my middle term okay so let me erase this and i'm going to say that this is equal to we're going to have 3 times inside of parentheses 9z squared minus 15z minus 3z plus 5. and so i'm going to use factoring by grouping in the middle here i'm going to have my 3 out in front from the first group the first 2 i could pull out a 3z so pull out a 3z that would leave me with a 3z minus a 5. then in the second group i have negative 3z and i have 5. well if i look at trying to get a common binomial factor i just want opposite signs i want positive 3z and negative 5. so just pull out a negative 1. pull out a negative 1 you can change the signs this would be 3z this would be negative 5. and so now i have that common binomial factor that i'm looking for this 3z minus 5. so i can take that and factor it out i'd have a 3 out in front factor out the quantity 3z minus 5 and it would be multiplied by what's left you'd have a 3z here minus a 1 here so this factors into 3 times the quantity 3z minus 5 times the quantity 3z minus 1. now am i done with this problem no i substituted to get here so i've got a substitute to finish this up so let me erase this and i can kind of move things up all right so we know that we substituted a z for x cubed so everywhere there's a z i can just go back and put x cubed it's just that simple x cubed here and x cubed here and i'm going to have my answer for the original problem so it would be 3 times the quantity 3x cubed minus 5 times the quantity 3x cubed minus 1. and again if you don't believe me go ahead and foil this out you'll get a result multiply that result by 3 and you'll get back to 27x to the sixth power minus 54x cubed plus 15. all right so let's take a look at another one so i just want to show you another scenario where you can use substitution we have 4 times the quantity x minus 5 squared minus 4 times the quantity x minus 5 minus 15. so if someone said go ahead and factor this again you might have a little bewildered look on your face but again if you see something that's common like the quantity x minus 5 i could just replace that with a variable so i can say hey i can let u well let's just go ahead and do q i can let q be equal to the quantity x minus 5. so everywhere i see an x minus 5 i could replace it with a q so what would that give me well i'd say 4 q squared minus 4 q minus 15. can i factor this yes i can i can find two integers whose product is negative 60. so the product is negative 60 and whose sum is negative 4. so the sum is negative 4. so we're going to have to think about this a little bit more because we have a negative product so that means we have one positive integer and one negative integer so let's think about some combinations that we can do so for factors of 60 i'm going to throw out things like 1 times 60 or 2 times 30 they're too far away then next you'd have 3 times 20 also too far away you've got 4 times 15 that wouldn't work you've got 5 and 12 that wouldn't work then you come to 6 and 10. so if i had 10 minus 6 that would give me 4. but i want negative 4 so i'd have to put positive 6 and negative 10 and that would give me a negative 4. so i'm looking for a negative 10 and a positive 6. negative 10 times 6 is also negative 60. so let's use these integers to rewrite that middle term let's erase this and say that we have 4q squared minus 10q plus 6q minus 15. so if i use my factoring by grouping i can pull out a 2q from the first group that would leave me with 2q minus 5. from the second group i can pull out a 3. so plus 3 times you'd have 2q minus 5. so if i factor out the common binomial factor i'd have the common 2q minus 5 that's pulled out and then times what's left which is 2q plus 3. so this is my factorization but again i'm not done because q is representing x minus 5. so let me erase this up here and i'm going to substitute so inside of parentheses i have 2 remember q is the quantity x minus 5. make sure you use parentheses they're very very important because 2 is multiplying this whole thing then minus 5. then times you have 2q so 2 times the quantity x minus 5 and then plus 3. all right so let's erase this and we would think about this so if i go through and kind of simplify 2 times x is 2x 2 times negative 5 is minus 10 and then minus 5. well negative 10 minus 5 is negative 15. so let's write 2x minus 15. and then for this one 2 times x is 2x and then 2 times negative 5 is negative 10 and then you have plus 3 negative 10 plus 3 is negative 7. so i have factored this into 2x minus 15 that quantity times the quantity 2x minus 7. now you might sit there and say there's no way that this is the factorization of this but in fact it is and i'm just going to put this off to the side and i'm going to prove that to you and let's say you were to look at this problem and say okay i can't start out by just using the distributive property i've got to expand this so i would have 4 times we know how to do this from our special products formulas this would be x squared minus 2 times x times 5. 2 times 5 is 10. so we'd have 10x and then plus this guy squared 5 squared is 25. so let me just stop and do the 4 times this this would be 4x squared minus 40x plus 100. okay so let me just erase this now for this one i have a negative 4 times this it's not squared or anything so i can choose my distributive property negative 4 times x is negative 4x and then negative 4 times 5 is plus 20 and then you have minus 15. okay so we have 4x squared nothing to combine with that i've got negative 40x minus 4x that's minus 44x and then i've got 100 plus 20 which is 120 minus 15 which is positive 105. now let me erase this and you could do one of two things now you could factor this into this or you could foil this into this let's go ahead and foil this into this because it's a little bit quicker so 2x times 2x would be 4x squared the outer 2x times negative 7 would be negative 14x the inner negative 15 times 2x would be negative 30x and then the last negative 15 times negative 7 is positive 105. so if you combine like terms in the middle here you see that you would get 4x squared minus 44x plus 105 which is exactly what you have right there so just a much quicker way to factor think about having to factor this versus what we factored it was quicker in the end right just to make the substitution and then go back hello and welcome to algebra 2 lesson 46 in this video we're going to learn about special factoring so we did a lesson on this back in algebra 1 but again i think it's very important to review this because you're going to come across factoring scenarios that occur so frequently that it's super beneficial for you to memorize the generic formulas and this can be used on your homework your tests or something important like the sat or the acts where they're timed and you have a certain amount of things and you just have to be as fast as you can so i'm going to start off by talking about the difference of two squares so this is something that goes back and forth we learn when we talk about special products and then we learn when we talk about special factoring so if you see something squared minus something else squared it factors into you have the first thing that's squared goes in the first position here and here the second thing that squared goes in the position here and here the second position of each so you can remember first goes with first so this is first goes in the first position of each second goes with second so this is the second it goes in the second position of each and then the sine is going to be different each time so you have a plus here and then a minus here you could change that around and put the minus here and the plus here wouldn't affect the answer and this is based on the fact that if i do foil let's say i have the quantity x plus y times the quantity x minus y what's going to happen is those two middle terms are going to cancel each other x times x is x squared the outer x times negative y is minus xy the inner y times x is plus xy and the last y times negative y is minus y squared so you can see that this is going to cancel and you're left with x squared minus y squared so if i see something squared minus something else squared i can quickly factor it using this technique so let's let's just say i saw something like i don't know x squared minus four we know x squared is a perfect square but is four well of course force two squared so make this simple on yourself and just rewrite this as 2 squared let's say this is 2 squared so then following this format let me just set up the parentheses i know one of these is plus and one of these is minus the first thing that's squared in this case that's x appears again in the first position of each so whatever square is going to appear in the first position of each and then the second thing that squared in this case it's y is going to appear in the second position of each so we've got a y there and a y there here what squared is 2 so that would appear here and here and again from your special products formulas you know that the quantity x plus 2 times the quantity x minus 2 would be x squared minus 2 squared or x squared minus 4. okay you can go back and forth between the two so let's say i gave you something like 4x squared minus 1 and i told you to factor well the first thing is there's two terms only and there's a minus sign so i'm immediately thinking the difference of two squares so what i'm going to ask myself is is this a perfect square yeah i could write this as 2x that quantity squared and then is this a perfect square yeah 1 squared is 1. so i could write 2x that quantity squared minus 1 squared so if i transform this into this i can use my little formula set up your parentheses this is how i remember it put your plus and your minus whatever is squared first goes in the first position of each 2x is what squared so put a 2x here and a 2x here whatever is squared second goes in the second position of each and that's it i've got the quantity 2x plus 1 times the quantity 2x minus 1. if you want to use foil and check it 2x times 2x is 4x squared the outer 2x times negative 1 and the inner 1 times 2x would cancel you'd have positive 2x and negative 2x that's gone the last is 1 times negative 1 which is negative 1 or minus 1. so you have 4x squared minus 1 and you start out with 4x squared minus 1. so we've got a very very simple and fast approach to factor this all right let's take a look at another one so we have 25p squared minus 16. so again two terms only and a minus sign so immediately you should be thinking about the difference of two squares so is 25 a perfect square yeah it's 5 times 5 or 5 squared we know p squared is a perfect square so i could write this as 5p in parentheses squared then minus is 16 a perfect square yes it's 4 squared so what i can do is i can factor this again set up your parentheses put a plus and a minus and then whatever squared in the first position in this case that's 5p is going to go here and here whatever squared in the second position is going to go here and here first position goes in the first position of each second position goes in the second position of each and we get our answer this factors into the quantity 5p plus 4 times the quantity 5p minus 4. all right for the next one we're looking at 169x to the fourth power minus 100 so again two terms and you got a minus sign so we're looking for the difference of two squares so 169 is what it's 13 squared so let me write 13 there and x to the fourth power is what i can really think about this again using the rules of exponents as x squared squared okay so what i want to write is 13x squared and then this thing right here inside of parentheses the 13x squared would be squared this mathematically is the same as this right i just rewrote it then minus we know 100 is 10 squared so let's write 10 and that's squared so to factor this again set up your parentheses very very easy one is plus one is minus let me put minus and then plus the order doesn't matter and then whatever is squared in the first position here it's 13x squared is going to go first in each case so 13x squared and then 13x squared and then whatever squared in the second position in this case that's 10 is going to go in the second position of each so a 10 here and a 10 here so we factored this as the quantity 13x squared minus 10 times the quantity 13x squared plus 10. all right so let's look at the next one which is where we talk about perfect square trinomials so we all know from special products that the quantity x plus y squared turns into x squared plus 2xy plus y squared so obviously we can factor and go from x squared plus 2xy plus y squared to the quantity x plus y squared now if i have x squared minus 2xy plus y squared i can factor that into x minus y that quantity squared let's take a look at an example of this we have 50x squared plus 60x plus 18. so let's say i tell you to factor this now when you look at 50 you would say well that's not a perfect square when you look at 18 you'd say well that's not a perfect square so you might give up on this idea of it being a perfect square trinomial but before you do that just realize that everything here is divisible by 2. so if i started out by just pulling a 2 out inside the parentheses that have 25x squared plus 30x plus 9. so now if i think about what's going on i know 25 is a perfect square and i want to check this 25 is a perfect square it's 5 squared x squared is a perfect square so i can write this as 5x that amount squared now the next thing you would check is to say is 9 a perfect square well yeah 9 is 3 squared okay now what we check is this middle term 2 times this times this 2 times 3 is 6 6 times 5 is 30 times x is 30x so 2 times 5x times 3 does equal 30x so it all checks out so that means i can factor this into what two times it's going to be this guy right here 5x plus this guy right here 3 and this quantity is squared so whatever is squared here plus whatever is squared here that quantity is squared very easy to set up once you know that you have a perfect square trinomial so again we get 2 times the quantity 5x plus 3 squared now let's take a look at another one so we have 49 n to the fourth power minus 42 n squared plus 9. so the first thing i'm going to look at is the n's so 49n to the fourth power i know 49 is a perfect square it's 7 squared i know n to the fourth power is a perfect square it's n squared squared so i could write 7 n squared squared for this 9 i could write 3 squared so so far everything checks out now again i want to think about i have a negative here or a minus whatever you want to think about is 2 times this right here 7 n squared times this right here 3 equal to 42 n squared forget about the negative so 2 times 7 is 14 14 times 3 is 42 42 times n squared is 42n squared so that's good to go there you just need to remember the minus sign so with this formula when i factor i have something minus something squared and what it is is this first term here that was squared which is 7n squared minus because of the minus sign here the last term that's squared in this case that's three so you get the quantity seven n squared minus three and this is squared all right let's take a look at another one so we have nine x squared minus twenty four x plus sixteen minus y squared so you might look at this problem if i told you to factor it and try to do something like factoring by grouping because it's a four term polynomial but you can't factor it that way you need to use a little trick here so i have to show this to you because you probably won't see it right away if i was to enclose this part only inside of parentheses let's say we have the quantity 9x squared minus 24x plus 16 then minus y squared now i can factor this because this part right here is a perfect square trinomial so let's go ahead and factor this this is 9x squared so this i could write as 3x squared this is 16 so that's 4 squared and then for the middle i have 3x times 4 which is 12x times 2 which is 24x so that matches up so all i need to do is say that i have what 3x this guy right here i have a minus right there so minus and then 4 this guy right here so the quantity 3x minus 4 squared minus y squared now what do i have here i have the difference of two squares i have something squared minus something else squared so let's further factor this into the quantity 3x minus 4 plus y times the quantity 3x minus 4 minus y now if you go back and you multiply these together you will end up with 9x squared minus 24x plus 16 minus y squared all right for the final section we want to talk about the sum or difference of cubes so something like the difference of cubes you have x cubed minus y cubed so this is going to factor into the quantity x minus y times the quantity x squared plus xy plus y squared so the main thing to remember here is that if you have a negative here the first sign you see is going to be negative with the sum of cubes you have x cubed plus y cubed everything is the same except for the sine here and here the final sign is always plus and either so the way i always remember this is that the first sign is always going to match the second sign is not going to match and the last sign is always positive so in other words if this is minus this is minus flip the sign it's plus this is always plus if this is plus this is going to match so it's plus flip the sign because it's not going to match it's minus and then this is always plus so that's how i personally remember it now for this part right here the multiplication it's pretty easy you just have an x and a y and then you have squared and squared at the end and x y in the middle so not too hard to memorize that it's just something you have to practice over and over and over again and then you know 15 or 20 of these you pretty much have it down packed but i can tell you unless you do a lot of problems you know over time it's something that you will forget and have to revisit so let's take a look at some examples all right so let's say you had something like negative 250x cubed minus 128. so forget about the negative for a second is 250 a perfect cube no it is not okay and if you didn't know that off top your head you could factor it and figure out that it's not but a lot of you will see right away that what you could pull out a 2 and you would have 125 which is a perfect cube you could pull out a 2 from here and you'd have 64 which is a perfect cube so a lot of these problems are just going to set up nicely for you so instead of just pulling out a 2 let's pull out a negative 2 make it easier on ourselves we'll pull out a negative 2 and we'll be left with 125 x cubed plus because i'm pulling out a negative 64. let's rewrite this down here we're going to put negative 2 times the quantity 125x cubed 125 is what it's 5 cubed x cubed is x cubed so let's write 5 x and that is going to be cubed then plus you have 64. 64 is what it's 4 cubed 4 cubed all right so how do we take this part right here which is the sum of cubes and factor it let's put our negative 2 out in front and then we know we're going to have a binomial and a trinomial so hopefully you can remember that so remember it was the first sign is the same so this will be plus the second sign you see is flipped so that means this one's going to be minus and the last one is always plus now this position here and here is going to match this and this right so it's just 5x and it's 4. so that's easy to remember this position here and here it's going to be the first position squared and the last position squared so 5x squared would be 25x squared then 4 squared would be 16. so that's done the middle part is just this times this 5x times 4 would be 20x so what we end up with is negative 2 times the quantity 5x plus 4 times the quantity 25x squared minus 20x plus 16. all right for the next one i have 8x to the sixth power plus 1. so do i have a perfect cube here do i have a perfect cube here well yeah 8 is 2 cubed then x to the 6 power i could write as x squared cubed so let's put this all in parentheses and raises to the third power and then we have plus we have one which is a perfect cube right one to the third power is just one so now that i have this in this format i can factor it very easily i know that again let's set up our parentheses so i've got a binomial here and a trinomial here so i know this position and this position is very easy it's the first goes first last goes last so i have 2x squared and i have one very simple now what sign do i want remember if this is plus the first sign is the same the second sign is different so plus minus the last line is always plus very easy way to remember that now what goes here and here well it's this squared the first thing squared goes in the first position so if i square 2 i'd have 4 if i squared x squared x would stay the same 2 times 2 is 4. so this is 4x to the 4th power so then the same thing goes here i want 1 squared 1 squared is just 1. now what goes in the middle is this times this 2x squared times 1 is just 2x squared so i end up with a quantity 2x squared plus 1 times the quantity 4x to the fourth power minus 2x squared plus 1. let's take a look at one final problem so we have m to the ninth power minus 343. so m to the ninth power that's easy i could write that as m cubed cubed right all i have to do with exponents if i look at an exponent of nine and i want to have something cubed just divide 9 by 3. 9 divided by 3 is 3 because i know when i use my rules of exponents if i raise a power to another power i'm multiplying so i can just say well if i'm taking m cubed and i raise it to the third power m stays the same 3 times 3 is not then minus and then what about 343 well that's going to be 7 cubed so i've got the difference of 2 cubes so what i'm going to do again let's set this up you've got a binomial and then you've got a trinomial the first sign is the same the second sign is different and the last sign is always positive so the first position of each is easy i want the first position here to be the first position here i want the second position or the last position here to be the last position here that's done then i want to take the first thing and square m cubed squared is m to the sixth power then what's going to go here i'm going to take the last thing and square it 7 squared is 49. in the middle i multiply the 2 together so m cubed times 7 is 7 m cubed so i end up with a quantity m cubed minus 7 times the quantity m to the sixth power plus seven m cubed plus 49. hello and welcome to algebra 2 lesson 47 in this video we're going to learn about solving polynomial equations by factoring so in some cases polynomial equations can be solved using factoring this technique revolves around using the zero factor property let me highlight that this zero factor property now your textbook might refer to this as the zero product property but either way whether you know it is the zero factor property or the zero product property it's important for you to understand what it means so here's an explanation of the zero factor or again the zero product property let me highlight this you have if a times b equals zero so let's stop for a second a is just some unknown b is some unknown so this could be x times y could be z times q it could be presented in any different way two unknowns are multiplied together and the result is zero if we have that then one of the following scenarios has to be true based on the fact that if i multiply something by zero i get zero so it could be that a is zero and b is not zero so in other words zero times something that's not zero would give me zero it could be that b is equal to zero and a is not zero so something that's not zero times zero would give me zero or the third and final possibility could be that a and b are both equal to zero zero times zero would also give me zero so this is very important to understand and when we start solving these equations using this property it's not going to look like this it's going to look slightly different and i'm going to refer back to the zero factor property and you go oh yeah that makes sense now okay but it's very important that you understand this conceptually that way when we start working through the exercises you understand what's going on all right so the focus for today is going to be to solve quadratic equations by factoring we did this back in algebra one and then most of you will recall that some of these we couldn't factor and when we had something that we couldn't factor we learned later on in algebra one we could use something known as completing the square or we also can use the quadratic formula so a quadratic equation is one where the highest power is two so in other words if i see something like ax squared plus bx plus c equals zero you look at your exponents this one's a two this one's obviously a 1. and so the highest exponent is 2 so this is a quadratic equation now with your quadratic equation you have a as the coefficient for the squared variable b is the coefficient for the variable raised to the first power and c is the constant a b and c can be any real number with the one exception that a is not allowed to be 0 because you obviously wouldn't have a quadratic equation anymore if i had 0 times x squared that's just 0 and in that case i have bx plus c equals zero okay that would be a linear equation now so let's talk a little bit about the steps that we would use to solve a quadratic equation by factoring now if you get something that's more advanced than a quadratic equation and it's factorable you're basically going to use the same steps you might see something with the highest power being a 3 or 4 or 6 and there might be some tricks that you can use you might be able to factor using substitution you might be able to factor using grouping there's all kinds of scenarios where this will be beneficial for you so solve a quadratic equation by factoring the first thing you want to do is write the equation in standard form so standard form we've talked about a lot so the left side over here we have ax squared plus bx plus c and if i was writing a polynomial in standard form we know we want it in decreasing order of powers so i would start with the highest power which is a 2 then i would go to the next highest power which in this case is a 1 then i would go to kind of the next highest power which we have a constant so we could say that variable is raised to the power of 0. so you'll notice that this is set equal to zero and this is so we can use our zero factor property we want this to be zero all right so the next step is to factor the left side you've got a trinomial on the left side you've got 0 on the right so you factor the left side that's going to put it into the product of two binomials then we're going to use the zero product property and set each factor with a variable equal to zero and the reason we say with a variable sometimes you can pull something else out okay before you start factoring and you don't need to set that equal to zero because that would be irrelevant all right the last thing is obvious you just want to solve the equation and check all right so let's get started with the first example so we have x squared minus 12x equals negative 36. so the very first thing i want to do is put this in standard form so all i need to do is add 36 to both sides of the equation so i would have x squared minus 12x plus 36 is equal to negative 36 plus 36 is 0. so the left side if you look at this is a trinomial in standard form i've got x squared x to the first power and i've got my constant okay so that's what i want on the right side i have a zero so what i want to do now is i want to factor this left side into the product of two binomials very easy to do i know that i have x squared here and this is a 1 is the coefficient for x squared so this would be x and this would be x and i simply need to find two integers whose sum is negative 12 and whose product is 36 well i know i'm looking for a negative and another negative negative plus negative is negative negative times negative is positive okay so let's just think about factors of 36 i've got 1 and 36 i've got 2 and 18. i've got 3 and 12 i've got 4 and 9 i've got 6 and 6. now think about 6 and 6. 6 plus 6 gives me 12. so i just need to work out the signs remember we already figured out it's a negative and a negative so negative 6 plus negative 6 would give me negative 12 negative 6 times negative 6 would give me positive 36. so x minus 6 that quantity times x minus 6. now you'll notice that essentially we could have found this by using our special factoring this is a perfect square x squared and 36 is a perfect square and remember how to check the middle term you think about is this right here this 12x 2 times x times 6 well yeah 2 times 6x is 12x so we know this would have factored into the quantity x minus 6 squared now in this particular case because we have the same thing twice i don't need to set each one equal to zero i only need to say hey x minus six equals zero what's the solution for that just drop the parentheses and solve x minus six equals zero i add six to both sides of the equation and i get that x is equal to six now let me stop and explain what i did if we consider the zero factor property remember this is multiplication here this is multiplication so if the quantity x minus six is multiplied by the quantity x minus six well one of these and it happens to be the same thing has to be 0. so if i can make this equal to 0 then 0 times whatever this is it doesn't matter that's the same thing if it was something different if it's multiplied by this it would give me 0. so in other words if i plugged in a six here six minus six would be zero zero times this whatever it would be and i know it's the same thing in this case would give me zero so that's how the zero product property or zero factor property however you wanna say it allows us to solve these equations when they're able to be factored so if you wanted to check this there's only one solution in this case we'd plug in a 6 everywhere there's an x so you'd have 6 squared which is 36 minus 12 times 6 which is 72 and this should equal negative 36 and it does 36 minus 72 is negative 36 so you get negative 36 equals negative 36. so this checks out all right let's take a look at one that's a little bit more challenging so we have 7x squared plus 84 is equal to negative 7x squared minus 77x so again i want to put this in standard form so i'm going to take everything and just move it to the left the right should be 0. so i'm going to add 7x squared to both sides of the equation and i'm going to add 77x to both sides of the equation so this is going to cancel and become 0. so on the left side i'm going to have 7x squared plus 7x squared which is 14x squared then plus 77x then finally plus 84. remember the right side is 0. so this equals zero don't forget the equal zero it's very important that you put equal zero i see a lot of students forget to put this they factor it and then they've lost their way they don't know what to do next all right so i want to factor this now this is the harder scenario where the coefficient here is not a one so again you can use reverse foil you can use factoring by grouping you can basically do whatever you want to do it's just something that's going to work for you now the first thing you should notice here is that you could pull a 7 out so if i pull out a 7 this would be 2 x squared plus 11 x plus 12 inside the parentheses this is equal to 0. so keep my 7 outside and i want to factor this i'm going to use grouping because it's just a little bit easier for me to do so what i want to do is find two integers whose product is 2 times 12 or 24 so i want the product to be 24. so product is 24. the sum is this middle coefficient of 11. so you think about 24 and right away you think about 3 and 8 right 3 plus 8 is 11. 3 times 8 is 24. so i could write that middle term we'll have 2x squared we'll put plus 8x plus 3x plus 12. again for those of you who are unfamiliar with this process all i'm doing is i'm rewriting the middle term using those two integers that i found so instead of 11x i've expanded this to 8x plus 3x 8x plus 3x is 11x so it's legal so let me put equals 0. now i'm going to factor inside the parentheses using grouping so this is super easy we have 7 times from the first group i'd pull out a 2x what would be left is x plus 4 then plus from the second group i'd pull out a 3 what would be left is x plus 4 so you can see you have this common binomial factor here of x plus 4. so i can factor that out i would have 7 times the quantity x plus 4 then times the quantity you'd have 2x plus 3. now remember this equals zero okay so let's scroll down and get some room going now i have three factors here seven is multiplied by the quantity x plus four which is multiplied by the quantity two x plus three so three factors this is a factor so is this so is this now 7 does not have a variable involved do i need to set it equal to 0 no i'm not going to waste my time i'm only going to set a factor with a variable equal to 0. so i'm going to take this quantity x plus 4 and set it equal to 0. so x plus 4 could be equal to 0 or okay we use this connective word or 2x plus 3 could be equal to 0. so it could be true that this equals 0 or this equals 0. those are our two scenarios because remember if a times b equals zero then either a is zero b is zero or they're both zero okay so that's what we're just doing here it's just a little bit more complex all right so to solve the first one i just subtract four away from each side and i get x is equal to negative 4. for the second one i subtract 3 away from each side this cancels i get 2x is equal to negative 3 i divide both sides by 2 and we end up with x is equal to negative three halves so in solution set notation remember we'll put some curly braces let's list negative four comma negative three halves those are two solutions and that's it we're done if you want to pause the video and check i would welcome you to do that i've already checked this so i know it works out and in the interest of time i'm not going to check anymore but all you would need to do is just plug in a negative 4 for every x in the original equation you'll verify that the left and the right side are equal then you'll plug in a negative 3 halves and for each x in the original equation and again you'll verify that the left and the right side are equal all right let's take a look at another one so we have 25x squared minus 16x plus 3 and this is equal to negative x plus 1. so again move everything over to the left so i'm going to add x to both sides of the equation and i'm going to subtract 1 away from each side of the equation so the right side has cancelled and it is now 0. on the left side i still have my 25x squared and then negative 16x plus x you could say plus 1x is negative 15x and then 3 minus 1 is 2. so plus 2. all right so now this equals 0. don't forget that and the simple fact is there's nothing i can pull out before i start this is divisible by 5 this is divisible by 5 and 3. this is a 2 it's prime so it's only divisible by 2 and 1. so if i think about what i'm going to do i'm just going to factor this using grouping so we want to find two integers whose product is 25 times 2 or 50. so the product is 50 and then the sum is negative 15 that middle coefficient so we think about this we know that it's going to be a negative and another negative so go through the factors of 50. right away i think about 10 and 5. negative 10 and negative 5 would work right negative 10 plus negative 5 is negative 15. negative 10 times negative 5 is positive 50. so again i'm going to use that to rewrite this middle term i'm going to have 25x squared minus 5x minus 10x plus 2 and this equals 0. so now that i have a four term polynomial i'm going to just factor using grouping so from the first group i could pull out a 5x that would leave me with 5x minus 1. from the second group because i want to match the signs i'm going to pull out a negative and i'm going to be pulling out a negative 2. so minus 2 here and then we'll have 5x minus 1. so we see we have a common binomial factor of this five x minus one and we're going to pull that out so i'll set this equal to zero we'll pull that out we'll have five x minus one that quantity multiplied by five x minus 2 that quantity this equals 0. all right so again i have two factors here this is multiplication so this times this gives me 0. so it must be true that 5x minus 1 equals 0 or 5x minus 2 equals 0. so we just need to solve each equation so we can get our solution here so what we want to do let's just add 1 to both sides that'll cancel you'll have five x is equal to one divide both sides by five and you'll get x is equal to one fifth over here if i add two to both sides of the equation that'll cancel you'll have x is equal to two divide both sides by five and you'll get x is equal to two fifths so in solution set notation we'll say inside of curly braces we have one-fifth comma two-fifths so those are our two solutions x equals one-fifth or x equals two-fifths again you can pause the video check each solution and verify that we got the correct answer all right for the last one we're going to look at we have 8x cubed plus 22x squared plus 14x equals 0. so you might jump out and say whoa that's not a quadratic equation when we can factor something like we will be able to here you're going to end up being able to get the solution that way versus using some other method so for 8x cubed plus 22x squared plus 14x equals 0 on the left side it's already in standard form the first thing you should notice is that there's a common factor of x and there's also a common factor of 2. so i could pull out a 2x and what would be left is 4x squared plus 11x plus 7 this is equal to 0. now inside the parentheses we have a quadratic equation so again that's something that's familiar to us so we have our 2x that just stays out in front and i want to just factor the inside so if i think about two integers whose product would be what 4 times 7 is 28 so we think about a product of 28 and a sum of 11. well 4 and 7 give me that criteria 4 plus 7 is 11 4 times 7 is 28. so again use that to rewrite that middle term there so i would have 4x squared then plus we'll do 4x then plus 7x then plus 7 okay and this equals 0. all right so i'm going to have my 2x here from the first group i can pull out a 4x then inside i would have x plus 1. from the second group i can pull out a 7 and inside i'd have x plus 1. so you have your common binomial factor let me just kind of scroll down a little bit you have your common binomial factor of x plus 1 that quantity and we can factor that out so i'm going to have my 2x out in front i'll have that quantity x plus 1. and then what's left we'll have 4x plus 7. now this equals 0. so now i have three factors here with a variable 2x is multiplied by the quantity x plus 1 which is multiplied by the quantity 4x plus 7. so i've got to set each one equal to 0. again let me label this so it's crystal clear this is a factor this is a factor and this is a factor just like if i had a times b times c equals zero it's got to be true that a equals zero or b equals 0 or c equals 0 or it could actually be in this case because we have three of them any two of them equal zero and the other one doesn't or all three of them equal zero okay so lots of possibilities but the point is here that each one of these has to be set equal to zero because it's a possible solution for us so we would say two x is equal to zero or x plus one is equal to zero or four x plus seven is equal to zero so let's solve these three equations all right 2x equals 0 is very simple you divide both sides by 2 you get x equals 0. then for this one x plus 1 equals 0 we subtract 1 away from each side we get x is equal to negative 1. then for 4x plus 7 equals 0 we subtract 7 away from each side we get 4x is equal to negative 7. we divide both sides by 4 and we get x is equal to negative seven fourths so i'll put my or kind of like that so we get x equals zero or x equals negative one or x equals negative seven fourths so in a solution set notation we put negative seven 4 comma negative 1 comma 0. so those are your three solutions now again if you want to check this go back to the original equation you're going to plug in a negative 7 4 for x and plug in a negative 1 for x and plug in a 0 for x in each case you're going to verify that the left and the right side are equal hello and welcome to algebra 2 lesson 48 in this video we're going to learn about rational expressions and rational functions so the majority of you worked with rational expressions back in algebra 1. at this point you just need to know that working with a rational expression is in many ways identical to working with fractions so let's just start out by talking about fractions a little bit so we should know at this point that a rational number is the quotient of two integers so some examples of a rational number something like two-thirds so i have an integer two over an integer three something like five sevenths integer five over an integer seven something like negative two elevenths an integer of negative two over an integer of eleven so very easy to define a rational number the one restriction is that what we cannot have the integer zero in the denominator because division by zero is considered undefined so just like we have rational numbers which are the quotient of two integers we have rational expressions so a rational expression is the quotient of two polynomials so just to give you two examples let's say it's something like 6x squared minus 5 and this was over x cubed this is a polynomial it's a binomial in the numerator this is a polynomial it's a monomial in the denominator so this is a rational expression or as another example let's say you had something like 3x to the seventh power minus 5x cubed plus 2x minus 1 over let's say you just had something like 4x to the fifth power minus 7. so you've got a polynomial in the numerator over a polynomial in the denominator so this is considered a rational expression now in your textbook you're probably going to see something that looks a little bit confusing they will start out by saying that a rational expression is what it's p over q where q does not equal 0. and i want to talk about this for a minute because we're going to go into a section where we're talking about this denominator not being allowed to be equal to 0. so p is a polynomial and q is a polynomial but q lies in the denominator and we say q cannot be equal to 0 because again we can never divide by zero because division by zero is undefined so specifically what we're saying is that if we have a rational expression generally speaking we have variables in our denominator sometimes you won't but in most cases you will so we want to say that that variable cannot be equal to any value that results in the denominator becoming 0. so i want to start out by just talking about finding the domain of a rational expression so i want you to recall that the domain we talked about this when we talked about functions is the set of allowable x values so the domain or the set of allowable x values of a rational function includes all values that result in a non-zero denominator so what's allowable here for x is anything that does not result in the denominator being 0. now for the first problem we're going to look at and we're just going to find the domain it's very very simple if i look at f of x is equal to 3x squared minus 2 over x and i say well what's the domain what is the domain all you need to think about if you get this problem is look at the denominator and say what can i not plug in for that variable because again i don't want the result to be 0. well it's very straightforward here i just cannot plug in a 0. if i plugged in a 0 for x i'd be dividing by 0 which is not allowed so the domain would be the set of all x values such that x does not equal zero okay and again i put this in set builder notation so let me just write this out this is the set of all x we have this such that that real number you choose which is x it can't be equal to zero so i can choose any other value that i'd like all right so that's your domain for this one for this one it's going to be a little bit more complicated we have f of x is equal to we have x squared minus 4x over 36 minus 4x squared what i want to think about is a value or it could be values for the variable x that would result in this whole thing becoming 0. so in order to do that what do i need to do i need to set up and solve an equation what i need to do is say well 36 minus 4x squared cannot be equal to zero or i could put equal zero and then if i find a value or values that make this equation true then they have to be excluded because if i plug in for x there and i evaluate i'll end up with 0 and i don't want that for my denominator so how could we solve something like this i know we talked about the quadratic formula in algebra 1 we haven't really got to in algebra 2 yet so i'm going to stick with things that we can solve via factoring so this is the difference of two squares this is 6 squared minus 2x squared so we could factor this into 6 plus 2x that quantity times 6 minus 2x that quantity we set this equal to 0 we can use our zero product property so in other words this factor here six plus two x would be equal to zero and then or we'd have six minus two x is equal to zero let's go ahead and subtract six away from each side of the equation so this would cancel let me scroll down get some room going here and what we're going to have is 2x is equal to negative 6. divide both sides by 2 and we'll get that x is equal to negative 3. over here let's subtract six away from each side of the equation this will cancel we'll have negative two x is equal to negative six what we'll do is we'll divide both sides of the equation by negative two and we'll have that x is equal to 3. let me put my or there so x equals negative 3 or x equals 3. so you have to remember what we're doing here let me let me erase everything so let me just put this off to the side as like a little note we solved 36 minus four x squared equals zero and the solution set contained two elements negative three and three so what that's telling me is that if i plug in a 3 or a negative 3 in for x there so if i plugged in a 3 let's just start with that you would get a 0 as your denominator and that's not allowed so you would have 36 minus 4 times 3 squared is 9. so 4 times 9 is 36. so you get 36 minus 36 which is 0. it would be the same thing if i plugged in a negative 3 because you'd have negative 3 squared which is positive 9 again you get 36 minus 36 which would be 0. so then the domain here the domain is the set of all x values such that x does not equal negative 3 or 3. so negative 3 or 3. so let's take a look at one last one so we have g of x is equal to 4x plus 12 over 7. so in this case is there going to be any restriction on the domain no there's not i don't have a variable in the denominator so i'm basically good to go so for something like this we would say the domain the domain is the set of all real numbers okay pretty simple overall all right so now let's talk about kind of the next topic that comes up when you talk about rational expressions for the first time how can we simplify a rational expression well it goes back to simplifying a fraction so if i have something like 2 over 4 we know that we can cancel the greatest common factor between numerator denominator now a lot of us at this point we know that 2 and 4 share a common factor of 2 so you can say okay 2 divided by 2 is 1. 4 divided by 2 is 2 so you end up with one half like that but if you didn't know that off the top of your head which you're not going to know this in case of a lot of rational expressions you would factor it right so you would say okay well 2 is a prime number so really the best i could do is say 2 times 1 and 4 is 2 times 2 so then when you look at it like this it's obvious that you cancel a common factor of 2 between numerator and denominator and you're left with one half okay so this is basically the process that we're going to use to simplify a rational expression let's go ahead and take a look at the first example so we have 12a minus 16 over 12. so the first thing i want to do is i want to look at that numerator and i want to factor out the greatest common factor so between 12 and 16 it's going to be 4. so if i pull a 4 out i'll have 4 times the quantity inside of parentheses i'll have 3a minus 4. then down here i have 12 which is divisible by 4. so i could write this as 4 times 3. now one thing i want you to realize is that this is multiplication right here this is multiplication so this 4 is a factor and this quantity 3a okay 3a minus 4 is a factor okay 4 is multiplying that entire thing so we are able to cancel common factors this is a factor and this is a factor so between the numerator and denominator they share a common factor of 4 so that can be cancelled so what's left now you have the quantity 3a minus 4. i don't need the parentheses anymore i can just get rid of them so 3a minus 4 and then over just 3. now here's where students make a gigantic mistake we just talked about common factors and i know it's confusing because there's a minus sign in here but it's this entire thing that's being multiplied by 4 there so those two are factors here i just have 3a minus 4 over 3. there's no common factor between the numerator denominator other than 1. right i could do this if i wanted to i could put this in parentheses and say times one i could put this times one and then if you wanted to cancel one with one it wouldn't do anything right it's still just equal to one but that's an example of something you could actually do but because i don't have any common factors i can't just go through and do this i can't say okay well i've got a 3 here and a 3 here that's a very common mistake that you're going to see let me just label this is a common mistake and you see it in algebra 1 you definitely still see it in algebra 2 it's something you have to realize that is illegal your answer here your simplified result is 3a minus 4 over 3 right nothing else you can do all right let's take a look at the next problem so we have x squared minus 9x plus 8 over 6x minus 6. what you want to do here is you want to factor the numerator and denominator and see what you can cancel so in the numerator we have a trinomial so we're going to factor that as the product of two binomials so it's an easy scenario i have x squared so i'm going to put x here and x here and then just find two integers whose sum is negative nine and whose product is positive eight well that would be negative eight and negative one then we'll put this over we have six x minus six i could factor out a 6 from that and i'd be left with x minus 1 inside of parentheses so again this is multiplication here the quantity x minus 8 is multiplying the quantity x minus 1. so this is a factor and so is this then down here this is multiplication 6 is multiplying the quantity x minus 1. so this is a factor and then so is this this quantity x minus 1. so do i have a common factor between the numerator and denominator you have the quantity x minus 1. this can be cancelled with this they're common factors so what i'm left with is x minus eight over six now this is something that you really need to understand you might be given a second task on a problem like this and they might say state the restricted values as well as to simplify is this right here this x squared minus 9x plus 8 over 6x minus 6 the same as x minus 8 over 6. not technically because if i think about this right here x cannot be equal to 1 because if i plug the 1 in there 6 times 1 is 6 and then 6 minus 6 would be 0. so with this original rational expression we have a restriction where x cannot be 1. in this simplified version right here x can be whatever you want because there's no variable in the denominator by canceling a common factor we have a lost some information okay we have lost some information so if you're going to be given a task like this you want to find the restricted values before you do any simplification okay so we would say that in this case x cannot be equal to 1. well let's take a look at the next example so we have 3x cubed minus 14x squared plus 16x and this is over 15x cubed minus 39x squared plus 18x so we want to simplify here so in the numerator what can i do to factor well you would notice that everything has an x so i could pull that out to start and i would have 3x squared minus 14x plus 16. and this is over in the denominator everything has a 3 that's common and everything also has an x so let's pull out a 3x and i'd have 5x squared minus 13x plus 6. now what i'm going to do is factor this and this each of those into the product of two binomials so in the numerator and kind of scroll down get some room going i have 3 x squared minus 14 x plus 16. so i'm going to factor this using factoring by grouping if i think about i want two integers whose product is 3 times 16 which is 48 and whose sum is negative 14. so i know i need a negative and another negative so think about some factors of 48 we've got 1 and 48 2 and 24 3 and 16 4 and 12 and then 6 and 8. if i had negative 6 and negative 8 that would work so i'm going to rewrite this middle term here i'm going to say this is 3x squared minus 6x minus 8x plus 16. we'll put this equals let me scroll down get a little room going from the first group here i could pull out a 3x i'm going to be left with x minus 2. from the second group to match the signs i'm going to pull out a negative 8 that would leave me with x minus 2. so i have that common binomial factor of x minus 2 and i want to bring that out so if i factor that out i have x minus 2 that quantity times the quantity 3x minus 8. and so this is the factorization so let me just copy this okay so let's drag that up there we have this x that's outside so we have x times the quantity x minus 2 times the quantity 3x minus 8. that's your numerator now let's work on the denominator so i have my 3x and then 5x squared minus 13x plus 6. let's factor that so i want two integers whose product is 5 times 6 or 30 and whose sum is negative 13. that one's really easy you do negative 10 and negative 3. so you would have 5x squared minus 10x minus 3x plus 6. again i just used these two integers to rewrite that middle term so then i'm going to factor by grouping so from the first group i could pull out 5x i'd be left with x minus 2. from the second group i could pull out a negative 3 and i'd be left with x minus 2. so i have that common binomial factor of x minus 2. and so what that's going to give me is the quantity x minus 2 times the quantity 5 x minus 3. so i'd have 3x times the quantity x minus 2 times the quantity 5 x minus 3. all right so let's erase all this if i wanted to know the restricted values before i cancel anything i would set this 3x times the quantity x minus 2 times the quantity 5x minus 3 equal to 0 and i'd use my zero product property to get a solution now you can do that if you want you'd find your restricted values because once we start canceling things you're going to lose information okay you're going to lose information so if you're asked to do a simplification and also to find the restricted values here's where you would do that you would find the restricted values and then you would cancel okay afterwards so between the numerator and denominator what are the common factors we see that we have an x here that we can cancel with an x here then we have the quantity x minus 2 that cancels with the quantity x minus 2. so what i'm left with this is gone i have 3x minus 8. so no need to keep it inside of parentheses 3x minus 8 then over now we have 3 times 3 times the quantity 5x minus 3. and we keep this one inside of parentheses because this is multiplication here and again don't make the mistake of going and saying well i can cancel this 3 with this 3. you can't do that okay it doesn't work that way this is multiplication here so these two are factors but up here this 3x and this negative 8 those are not factors that's a subtraction problem so if i had 3x minus 8 multiplied by 3 like that well then i can cancel this 3 with this 3. but that's not what we have here okay so please be very careful very cautious when you're canceling things it's got to be multiplication if i go back up here you can see that this is multiplication and this is multiplication this is multiplication this is multiplication this is multiplication so this cancels with this those are common factors and this cancels with this again those are common factors so in the numerator i'm just left with this which is what i got 3x minus 8. in the denominator i'm still left with a 3 multiplied by the quantity 5x minus 3 which is what you have down here so you end up with 3x minus 8 over 3 times the quantity 5x minus 3. all right let's take a look at one more problem so we have 6x squared plus 9xy minus 2xy minus 3y squared this is over negative 3x plus y so how would i factor the numerator here well i could use grouping it's a four term polynomial in the first group i could pull out a 3x what i'd have left is a 2x plus a 3y in the second group i could pull out a negative and then y so what i have left is a positive 2x and then a positive 3y so plus 3y so you can see that you have a common binomial factor of 2x plus 3y so let's put this equals in the denominator i still have negative 3x plus y and so let's go ahead and factor this so i pull out that common binomial factor in the numerator i'll have the quantity 2x plus 3y then times the quantity you'll have 3x minus y and this is over you have negative 3x plus y this 3x minus y looks similar to negative 3x plus y the signs are just different when the signs are different but you have the same terms they end up canceling and becoming negative 1. it's kind of like having 5 over negative 5 right it's the same thing they're just opposites so these would cancel become negative 1. so it's the same concept here and to prove that to you what i'm going to do is i'm going to enclose this inside some parentheses and i'm going to factor out a negative 1. so keep the numerator the same you have the quantity 2x plus 3y times this is multiplication the quantity 3x minus 1. so each one is a factor so this is the factor and this is a factor now in the denominator if i pull out a negative 1 so i put a negative 1 out in front of parentheses the signs would just change so this would be 3x and then this would be minus y so from negative 3x becomes 3x from positive y becomes negative y now this is multiplication so negative 1 is a factor and so is the quantity 3x minus y inside of parentheses is being multiplied now it is completely obvious when i cancel this with this then i'm left with this okay i'm left with that negative 1. so if you need to do that a few times for it to make sense for you go ahead and do it but the quicker thing to do is to realize that if i have this and i have this if i cancel them i'm going to get negative 1. you have 3x and negative 3x so basically the same thing with the sign is different you have negative y and positive y same thing with the signs are different so if you cancel that you're going to end up with negative 1. so my results here my result here in the numerator i'm just going to have that 2x plus 3y in the denominator i have a negative 1. so i can put over negative 1 but if if we divide by negative 1 really i don't need the fraction bar anymore and get rid of that i can put a negative 1 out in front like this or i can just use the distributive property and say i have negative 2x minus 3y as my answer hello and welcome to algebra 2 lesson 49 in this video we're going to learn about multiplying and dividing rational expressions so again we've come across a topic that we learned back in algebra 1 but we want to review this for the purposes of algebra 2. before we get into anything more complicated we've got to make sure that we understand these topics that we learned in algebra 1. so i want to start by saying that when we multiply or divide rational expressions we follow the same rules as with fractions so if i was to multiply two fractions together let's say it was something like 1 4 times i don't know let's say 4 7. well we would check between the fractions to make sure that they were simplified first and if they are then i would check to see if i could cross cancel i can cancel this 4 with this 4. so that leaves me with just a 1 and a 7. so i'd have 1 over 7 or 1 7 as my answer it's essentially the same process we're going to follow when we start multiplying and dividing rational expressions so as an example let's say i had n squared minus 3n minus 18 over 5n plus 15 and we multiply this by 8n squared minus 32n over n minus 4. so the first thing i want to do is i want to factor everything completely and when i factor with these rational expressions in a lot of cases it's going to work out to where you can cancel between numerator denominator and then i can also cross cancel as well so let's just start out by factoring everything so i'm going to put my equal sign to factor n squared minus 3 n minus 18 i can factor that into the product of two binomials i know this would be n and this would be n and then i need two integers whose sum is negative three and whose product is negative eighteen so we think about that you come across negative 6 and positive 3. negative 6 plus 3 would be negative 3 and then negative 6 times 3 would be negative 18. so we put negative 6 and positive 3. all right then we want to factor this guy right here i would just look to pull out a 5 and that would leave me with what an n plus 3. so you can already see before you go any further that i can cancel this factor of n plus 3 with this factor of n plus 3. remember this is multiplication so these are factors and these are factors and we cancel common factors all right then times over here in the numerator i can start by pulling out an 8n so i'd be left with n minus 4 and then in the denominator i have n minus 4. so you'll notice you can cancel a common factor here of n minus 4. and to make that obvious put this in parentheses and put times 1. and then you'll realize this is a factor and so is this or you could say these are both factors these are factors as well so you cancel those common factors all right so now that i have done that i've canceled between numerator and denominator of each i'm going to look to see can i cross cancel anything before i multiply so let me just rewrite what's left i'd have n minus 6 over 5 multiplied by 8n and you could put over 1 if you want so can i cancel anything between 5 and 8n no i cannot can i cancel anything between n minus 6 and 1 no i cannot so i can leave this in factored form and say this is 8n times the quantity n minus six over five all right let's go ahead and take a look at the next example so we have a plus three over two a plus six this is multiplied by 30 a minus 48 over 64 a squared minus 40 a cubed so again i just want to factor everything with a plus 3 i cannot factor that so i'm just going to leave it as a plus 3. with this one 2a plus 6 you know you could pull a 2 out so you'd have 2 times the quantity a plus three and obviously you can now cancel between the numerator denominator this factor of a plus three i'm going to put times one here just to make that obvious for you that these are factors then times over here i have 30 a minus 48. so 30 and 48 share a common factor of 2 and also 3 so that means they each share a common factor of 6. so let's pull a 6 out that would leave me with 5 a minus 8 inside and then in the denominator i have a 64 and i have a 40. so those are each divisible by 4 they're also each divisible by 8. that's about as best we can do so let's pull out an 8 and then we have a squared and a cube so 8 a squared is what we can pull out so i'd be left with 8 then you'd have your minus 5 a now we know we can do some canceling between here and here but let's hold off on that for a second i have 5a minus 8 i have 8 minus 5a those are opposites if you'll notice i have 5a and i have 5a forget about the sign for a second i have 8 and i have 8. each one has an opposite sign here 8 is negative here 8 is positive here we have negative 5a here we have positive 5a so if you have opposites like that essentially you can just cancel them and say the result is negative 1. if i was to do a little side note here and just show you this i could take this right here let me just start with 5 a minus 8 and put it in parenthesis and i'll just put times 1 and make it completely crystal clear what's going on for this one let's just rearrange the order let's say this is negative 5a plus 8 like this now let's say i factored out a negative 1 from here so if i pulled out a negative 1 what would happen this would turn positive and this would turn negative so now we have five a minus eight and five a minus eight this would cancel with this and leave me with this i now have one over negative one which is just negative one so again if you see that situation you don't need to go through this long lengthy process i've got the same terms but with different signs i know those are opposites so i can cancel this with this and say it's a negative 1. now i can do some more canceling if i want i know that 6 and 2 have a common factor of 2 and i know 6 and 8 have a common factor of 2. so whatever you want to cancel that with it it's up to you it's going to be the same in the end i'll just cancel this six with this two i'll say this is a one and now this is a three so this right here is completely canceled this is a one so i don't even need to worry about one times anything is itself in the numerator here i have three times negative 1 which is negative 3. in the denominator i just have 8 a squared so my answer here is negative 3 over 8 a squared well let's take a look at one more of these and then we're going to jump in and look at some division i think you can see that it's pretty easy to do it's just a lot of a lot of cancelling a lot of trying to get things simplified all right so we have 7x squared plus 73x plus 90 over 14x squared minus 22x minus 60. and this is multiplied by 2x minus 6 over 10x squared minus 40x so the first thing i want to do is try to factor everything and for 7x squared plus 73x plus 90 we can think about two integers whose product would be 7 times 90 which is 630 and whose sum is 73. i know the two such integers would be positive let's think about 630 for a second it ends in a zero so i know it would be 63 times 10. now 63 and 10 add to be 73 so i've found my two integers pretty easily let's erase this i'm going to use those two integers to rewrite this middle term here so i'm going to say that we have 7x squared plus i'm going to say we have 63x plus 10x plus 90. again i took this middle term 73x and i wrote it as 63x plus 10x didn't do anything illegal now i'm going to factor by grouping so from the first group i can pull out a 7x and i'd be left with x plus 9. from the second group i could pull out a 10 and i'd be left with x plus 9. so i could factor out this common binomial factor of x plus 9 and what's that going to leave me with i would have x plus 9 that quantity multiplied by the quantity 7x plus 10. and so let me erase this and i'll write this in factored form i've got x plus 9 that quantity multiplied by the quantity 7x plus 10. all right so let's work on the denominator now in the denominator everything is even so i could start by just pulling out a 2 and i'd be left with 7x squared minus 11x minus 30. now how can we factor this well give me two integers whose product is 7 times negative 30 or negative 210 and whose sum is negative 11. well this is another one if i just think about positive 210 for a second i just think about 210 it ends in a zero so i know it would be 21 times 10. so i could play with the signs with 21 and 10 and get myself a sum of negative 11 and a product of negative 210. if i had negative 21 and positive 10 that would do it for me negative 21 times positive 10 is negative 210 negative 21 plus 10 is going to give me negative 11. so let me go ahead and factor this part down here if i had 7x squared i'm going to rewrite the middle term using these two so minus 21x and then plus 10x and then minus 30. so let me scroll down for a minute i'll come right back in a second so from the first group i could pull out a 7x i'd be left with x minus 3 inside from the second group i could pull out a 10 and i'd be left with x minus 3 inside so i would factor out this common binomial factor of x minus 3 and that would give me the quantity x minus 3 multiplied by the quantity you'd have 7x then plus 10. all right so let's erase this part we'll have 2 times the quantity x minus 3 then times the quantity 7x plus 10. all right so let's erase all this and we can see that we can cancel a common factor of 7x plus 10. that's canceled remember this is multiplication those are all factors so 2 is a factor the quantity x minus 3 is a factor and the quantity 7x plus 10 is a factor these are factors and then the quantity x plus 9 is a factor and then the quantity 7x plus 10 is a factor so we cancelled common factors then times for these guys it's going to be easier to factor pull out a 2 from the first one so i'd have x minus 3 inside in the denominator here i could pull out a 10x and i'd be left with x minus 4 inside so what can i cancel here well i know between 2 and 10 i can cancel a common factor of 2. so this would be 5. i can now cross cancel a factor of x minus 3 between here and here so that's gone and nothing really else that i can do i have x plus 9 here nothing to cancel with that my denominator or the denominator over here this is completely cancelled so really nothing else i can do so let's just put equals we have x plus 9 for our numerator that quantity in the denominator we have 2 times 5 times x which is 10x times the quantity x minus 4. and again i'm just going to leave this in factored form because this is generally preferable all right so the next thing we want to talk about would be dividing rational expressions so to divide rational expressions multiply the first rational expression by the reciprocal remember that means we flip it of the second all right for the first problem we're looking at negative 2a plus 2 over a minus 1 and this is divided by we have 40 minus 5a over 5a minus 40. so if i look over here what can i factor well i can factor out a negative 2 or a positive 2. because the sign here is negative let's go ahead and factor out a negative 2. that's going to leave me with a minus 1 inside down here i just have a minus 1. if you want you can enclose this inside of parentheses and put times 1 so that it's crystal clear that you can cancel this with this i'm left with negative 2 over 1 there then this is divided by so that means i'm going to multiply by the reciprocal of this so that means 5a minus 40 would come up to the numerator 5a minus 40 and 40 minus 5a would go into the denominator now before i go through and factor anything what do you notice here i have 5a and negative 5a i have negative 40 and positive 40. these two are opposites and again if you want to switch the order around and say this is negative 5a plus 40 like that and then factor out a negative one from one of them it doesn't matter which one if i factor out a negative one then i would erase these two signs here put my negative 1 out here so this becomes positive and this becomes negative because negative 1 times 5a is negative 5a negative 1 times negative 40 is positive 40. so that would get me back to what a hat so now when you look at these two it's obvious that this cancels with this i can put it inside of parentheses and put times 1 if you want and what i'm left with is 1 over negative 1 or just negative 1. so what i get for my answer here i have negative 2 over 1 which is just negative two multiplied by one over negative one which is negative one so negative two times negative one is simply two and that's my answer all right let's take a look at another one now so what we're gonna do here i'm gonna factor the numerator of this first guy and it looks like i can pull out a seven so if i pulled out a seven i'd be left with seven n minus four this is over four n and then i'm dividing by so i wanna multiply by the reciprocal so bring 5n up here and this is over from this numerator here it's going into the denominator but i want to just go ahead and factor it i can pull out a 5n so i pull out a 5n and i'll be left with 7n minus 4. so now what can i do well it's obvious that i can cancel this 5n with this 5n i can cancel this quantity 7n minus 4 with this quantity 7 and minus 4. so this has cancelled and become 1. so all i'm really left with is 7 over 4n and so that's going to be my answer 7 over 4n now let's take a look at one more problem so we have 6x squared plus 10x over 15x squared plus 13x minus 20. this is divided by 45x to the fourth power over 5x minus 4. all right so from the numerator of the first fraction here i know i could pull out a 2x so that would leave me with a 3x plus a 5. in the denominator down here i can't start by factoring anything out 13's a prime number nothing is going to be common so i'm going to try to factor this using grouping so 15 times negative 20 is negative 300 so i want a product of negative 300 and a sum of 13. so a sum of 13. i think about factors of 300 let's just think about 300 for a second i know they're going to have to be somewhat close together because it's going to be a sum of 13. so i'm going to exclude things like 1 times 300 or 2 times 150 i want to get kind of closer together so if i just cut this off and say i know it's 30 times 10 i could work from there okay now 30 and 10 won't work but i can decrease this one and increase this one so the next two factors would be 25 times 12. now 25 and 12 if i play with the signs can produce a positive 13 as a sum and a negative 300 as a product i would want positive 25 and negative 12. so let's use that to rewrite that middle term so we're going to say this is 15x squared plus 25x minus 12x minus 20. let me erase this so we have a little bit of room to work and i'll scroll down a little bit for now so from the first group here i could pull out a 5x that would leave me with 3x plus 5. from the second group i could pull out a negative 4. both sides here are negative both sides here are positive so that's why i'm pulling out a negative along with that 4. so that would leave me with 3x plus 5 inside and so i have a common binomial factor of 3x plus 5. so let's go ahead and factor that out and we'll just write our answer above so we'll have 3x plus 5 that quantity multiplied by the quantity we have 5x and then minus 4. all right so let's erase all this and we can already see that we could cancel this factor of 3x plus 5 with this factor of 3x plus 5. nothing else to cancel between numerator and denominator here then we're going to multiply by the reciprocal of this so 45 x to the fourth power would go into the denominator nothing i can really do with that in the numerator i just have five x minus four nothing i can really do with that now what i can do is i can cancel this with this and again if if you're confused by that just put parentheses around it like this and say it's times 1. then you realize it's a common factor the quantity 5x minus 4 is cancelled with the quantity 5x minus 4. so all i'm left with here is 2x times 1 or just 2x over 45x to the fourth power now can i simplify this any further yeah i can cancel one of these x's this would be x to the third now with this x here between 2 and 45 there are no common factors other than 1 so really can't do anything with that so we'll end up writing this as 2 over 45 x cubed hello and welcome to algebra 2 lesson 50. in this video we're going to learn about adding and subtracting rational expressions so again we've come across another lesson that we talked about back in algebra one so we should at this point have a fair understanding of how to add and subtract rational expressions if you don't you can pick it up from this lesson we're just going to go a little bit quicker in algebra one we had two separate lessons we first talked about how to find the lcd for a group of rational expressions and then we moved into a lesson where we talked about how to add and subtract rational expressions here we feel like you have enough understanding of this it's just a review so we're going to compact it into one lesson so i have here when we add or subtract rational expressions we follow the same rules we learned with fractions so essentially when the denominators are the same you just add the numerators and place the result over the common denominator so if i had something like 1 4 plus i know let's say three-fourths i have a common denominator of four so that stays the same and i had one plus three which is four and then the thing you wanna do whenever you're done is simplify we know four over four is 1 right so we would simplify that now the other scenario the more difficult or challenging scenario is when the denominators are different so the first thing is you have to find the lcd and then you have to write your fractions as equivalent fractions where the lcd is the denominator and then you can proceed with your addition or subtraction so if we thought about something like let's say 5 6 plus let's say 2 3 just to make it really easy we know the lcd would be 6 right or hopefully you know that so i would take 5 6 and i would leave it unchanged because the denominator is already the same as the lcd for two-thirds i would perform a little trick i would multiply it top and bottom by the number two two over two is the same thing as one if i multiply by one i'm not changing something's value so if i do 2 times 2 that's 4 it would be over 3 times 2 which is 6 so we can say this is 5 6 plus 4 6 and now i can do my addition my denominator is the same it's a 6 5 plus 4 is 9 so i end up with 9 6 which of course we can reduce 9 and 6 share a common factor of 3. so if i cancel a factor of 3 here this would be 3. if i cancel a factor of 3 here this would be 2. so we end up with an answer of 3 halves so those are your two scenarios when working with fractions and it's the same exact thing you're going to come across with rational expressions it's just much more tedious because you're working with variables the variables have powers there's all kinds of things that are going on all right so let's start out with some real easy ones just to kind of get our brain going so let's suppose we see x minus 4y over 6xy plus x plus 3y over 6xy so your denominator is the same you have 6xy and 6xy just like when we work with fractions again if the denominator is the same keep the denominator the same so for the answer i'm just going to keep 6xy in the denominator and i'm just going to add the numerators so i would have x minus 4y it's addition so i'm just going to put plus and then write this x plus 3y i'll make this a little longer so what would this be equal to well essentially i'm just going to combine like terms x plus x is what that's just 2x and then negative 4y plus 3y would be negative y or just put minus y and this is over that common denominator of 6xy now the last thing you always want to do you want to see can i simplify this further usually what you would do is just try to factor everything now with 2x minus y i can't really pull anything out if i had something i'd pull out like let's say i had a 2 here well i could pull a 2 out i could cancel that with the 6 but that's not the scenario i have i have 2x minus y over 6xy nothing i can do to simplify further so this is our answer right here all right let's take a look at another easy one so let's suppose we have m minus six n over 15 m squared n plus five m minus n over 15 m squared n so again the same denominator so if i have a common denominator already all i need to do it's very very simple i'm just going to add the numerators so i would have m minus 6n plus 5m minus n over that common denominator of 15 m squared n so i'm just going to combine like terms in the numerator so m plus 5m is 6m and then negative 6n minus n is negative 7n and this is over again 15 m squared n so is there anything i can factor from the numerator here that i could cancel with the denominator the answer to that is no you have six and you have seven number wise no variable wise i have an m and then an n so nothing i can really pull out so this is going to be your final answer here six m minus seven n over 15 m squared n all right so let's look at one final easy one and then we'll look at some scenarios where we don't have a common denominator given to us so we have 3x minus 3y over 8xy squared minus x plus 6y over 8xy squared so i'm looking at this and again i notice that i have this common denominator of 8xy squared so that just stays the same i have subtraction here so this is where you got to be real careful i am subtracting this whole thing away so remember if i'm subtracting something away i can add the opposite of it so for the numerator i have 3 x minus 3y then what i'm going to do is i'm going to put minus okay minus and i'm going to put inside of parentheses this x plus 6y that's going to remind me to distribute that negative to each term and i'm changing the sign of each term i'm going to add the opposite of it and this is over we have that common denominator of 8 x y squared all right so let's go ahead and work this out we have 3x minus 3y i'm going to distribute this to each term so i can think about this as having plus a little negative 1 here negative 1 times x is minus x and then negative 1 times 6y is minus 6y and this is over you have 8 x y squared all right scroll down get a little room going all right so the next thing i'm looking at is i have 3x minus x that's going to give me 2x then i have negative 3y minus 6y that's going to give me negative 9y and this is over that common denominator of 8xy squared so nothing i can do to simplify any further i just report my answer as 2x minus 9y over 8xy squared all right so let's look at one where there's not a common denominator we'll start out with something pretty simple we have four over p plus four plus three p over p minus four so before we even do this let's just think about two fractions for a second let's say i had something like one third plus i don't know let's say two-fifths to get the lcd what do i do i take the denominators and i factor them now three and five are each prime number so there's not really anything i can do to factor them they're just 3 times 1 or 5 times 1. so what i want to do when i build my lcd my lcd that's what it's the least common multiple of these two numbers so of 3 and 5 and so to do that when i have my factorization i list each prime factor the only exception is when i have a duplicate prime factor so if i have a prime factor that occurs in each prime factorization i go with the largest number of repeats now in this case i just have a 3 and a 5. so i would put 3 times 5 and that would give me 15. so i would convert each of these into an equivalent fraction where 15 is the denominator and then i could do my addition so hopefully you remember how to do that i know for some of you it's been quite a while since you've had to do that but it's going to be the same process here i look at my denominators so i have p plus 4 and i have p minus 4. so the question is can i factor p plus 4 no i cannot can i factor p minus 4 no i cannot so just like when i had a denominator of 3 and a denominator of 5 and i couldn't factor them i just multiplied the two together i'm going to do the same thing here so the the lcd is going to be what it's going to be p plus 4 multiplied by p minus 4. and generally we want to leave this in factored form i know you could use foil here and kind of write it in a different format but we want to leave it like this all right so if this is the lcd and i have let me kind of just drag this up here so it's out of the way so if i put equals here i'm going to have 4 over p plus 4. now the lcd is what it's p plus 4 that quantity times p minus 4. so in order to have a denominator of p plus 4 times p minus 4 i've got to multiply the denominator by p minus 4 and also the numerator by p minus 4. make sure you're using parentheses here because again this is the quantity p minus four okay don't make that mistake of saying okay i've got 4p minus 4 like that that's wrong okay you don't want to do that then you're going to have plus so next you see you have 3p over you have p minus 4 and what am i missing here well i'm missing the p plus 4. so let's multiply the top and the bottom by the quantity p plus 4. and let's scroll down get a little room going so if i do 4 times the quantity p minus 4 i'd have 4p minus 16. then plus you have 3p times p that's 3p squared then 3p times 4 is plus 12p and this is over you've got p plus 4 that quantity times p minus 4 that quantity so if i combine like terms in the numerator and kind of scroll down get a little room going i'm going to start out with 3p squared then next i look i see 4p and 12p well that would combine to give me 16p and then lastly i'd have minus 16. so then this is over we have p plus 4 times p minus 4. so i'm not done because with these i want to see if i can factor something i have left this in factored form to see if i can cancel so let's try to factor this right here into the product of two binomials and let's see what we get we're trying to see if we can cancel anything with the denominator so give me two integers whose product is 3 times negative 16 or negative 48 so this would be my product i'm looking for and we want a sum of 16. so we know we need one positive and one negative so let's think about the factors of just positive 48 forget about the fact that it's negative so for 48 you've got what you've got 1 and 48 those are obviously too far apart you've got 2 and 24 those would never work it is divisible by 3 it's 3 times 16 but those won't work now you could also do 4 and 12. now it seems like 4 and 12 would work because you've got a positive 16 there so you could say okay well 4 plus 12 is 16 but you can't make the sign work with the product if i made one of those positive and one of those negative you will not get to that sum so then the next thing we try it's not divisible by 5 it is divisible by 6 it's 6 and 8. so 6 and 8 but that won't work so after 6 and 8 we don't have any more factors so i would say this polynomial is prime so let me erase this and i will say that this is prime and because this is prime there's nothing else i can do to simplify it and so this is my answer 3p squared plus 16p minus 16 over the quantity p plus 4 times the quantity p minus 4. all right let's take a look at another kind of easy one so we have 4 over 2n squared minus 12n plus 7n over 2. so i've got a denominator that's just a 2 and i've got a denominator that's a 2n squared minus 12in so i can't really factor 2. it's a prime number it's just 2. if i factor this what could i do well i could pull out a 2n and pull out a 2n and then i would multiply that by the quantity n minus 6. so because this already has a 2 and this has a 2 when i think about my lcd it would just have 1 to it so my lcd would have a 2 and then it would have the rest of this times n times the quantity n minus 6. so this guy right here already has all that all i need to do is change this rational expression here so let me again move this up out of the way we'll say this equals i'll kind of erase this real fast so for the first guy i'm just going to write 4 over i'm going to write this in factored form so 2n times the quantity n minus 6 then plus for this guy i have 7n over 2 and i'm multiplying by what i'm missing so i'm missing an n and i'm missing the quantity n minus six all right so let me put equals and we'll just scroll down get a little room going we'll have four plus we're multiplying three of these guys together so the order isn't going to matter i could do 7 n times n first that would give me 7n squared so let me put 7n squared here and then times the quantity n minus 6. and let me just do this down here the denominator we know will be 2n times the quantity n minus 6. all right so we'll continue we have 4 plus you have 7n squared times n that's 7n cubed and then 7n squared times negative 6 would be negative 7 times 6 is 42 and then your n squared would come along for the ride and this is over we have 2n times the quantity n minus 6. so this isn't going to be something we can factor in the numerator but i can reorder it put that polynomial in standard form we'll write 7 n cubed minus 42 n squared let me kind of make that a little better and then lastly plus 4 then all over 2n times the quantity n minus 6. again i leave this in factored form in the denominator just to show that nothing is going to cancel all right let's look at one that's a little bit more tedious so we have 7 over we have 2n cubed minus 6n squared then we have minus 8 over n minus 3. so remember i have this minus here so i've got to be very very careful what i can do i can put that this is plus and then i can make this a negative 8. if i put plus and then i change something to its opposite meaning i multiply it by negative 1 i'm good to go right that's preferable to what we normally do because students will get confused so once i've rewritten this problem i think about the denominators so this is 2n cubed minus 6n squared i could factor out a 2n squared and i'd be left with n minus 3. and then here i just have an n minus 3. so what's common to both they each have this n minus 3. i'm just missing the 2n squared i'm just missing this 2n squared so again we come to the scenario where this first guy here is unchanged we're just going to have 7 over we have 2n squared times the quantity n minus 3 and that's in factored form then i'm going to put plus remember i change this to negative 8 so we'll have negative 8 times what am i missing here i'm missing 2n squared so times 2n squared over you would have n minus 3 times 2n squared so i'll put 2n squared over here like this times that quantity n minus 3. so now i have a common denominator and so i can just go through now so i have 7 plus negative 8 times 2n squared is negative 16n squared and then this is over we have 2 n squared times the quantity n minus 3. so let me just reorder the terms i'm going to put negative 16 n squared plus 7 over 2 n squared times the quantity n minus 3. so not really anything i'm going to be able to do to simplify this any further so we just leave our answer like this we have negative 16n squared plus 7 over 2n squared times the quantity n minus 3. all right now let's get into one of the more tedious problems and you'll typically see more of these in algebra 2 than you did in algebra 1. so you have 6p minus 8 over 3p squared plus 20p minus 7 plus 13 p minus 11 over p squared plus 6p minus 7. so i want to just start by getting a common denominator so i'm going to factor this guy right here and i'm going to factor this guy right here so i know this one's easier so let me just start with that i know this would be p and this would be p two integers whose product is negative seven and whose sum is positive six i could do positive seven and negative one positive 7 and negative 1. if i did 7 times negative 1 that's negative 7 7 plus negative 1 is positive 6. so that checks out for this guy right here i've got a non-1 as my leading coefficient so it's a little bit more challenging to factor we can use reverse foil just to get a little practice let's go with 3p and p remember this is a prime number so i go with that to start now i need to work out my outer and my inner to where the sum is 20 p and the product is negative 7. now it's easy because negative 7 or just 7 in general is a prime number so really the only possibilities are positive 1 and negative 7 or negative 1 and positive 7. so i can try different combinations i could say okay if i had positive 1 and negative 7 like this what would that give me well the outer just think about the numbers 3 times negative 7 is negative 21. the inner 1 times 1 would be 1. so if i sum those i would end up with a negative 20. well i want a positive 20. so all i have to do is just change the signs around everything is in the right position so if i make this negative and this positive i'm going to be good to go right because my outer would be 21p 3p times 7 would be 21p my inner would be negative p those would sum to be 20p all right so i've got my factorizations going now what do they share well they each share p plus seven so that's already there all i need to do to think about the lcd again everything gets listed but what they share or what's common in both just gets listed once so i would put three p minus one then times p plus seven and i've got a p plus seven here and a p plus seven here so it only goes in once you go with the largest number of repeats between any factorization then i've got a p minus 1. so that's my lcd so in other words for this first one here i am missing the p minus one so i'm going to multiply six p minus eight times the quantity p minus one then plus over here what am i missing well i have p plus seven i have p minus one i'm missing three p minus one so i'm going to multiply 13 p minus 11 times again what i'm missing which is this three p minus one then for the denominator we already know that's going to be the lcd so that's 3p minus 1 times p plus 7 times p minus 1. and again these can get quite tedious very very quickly so let's go ahead and use foil here six p times p is six p squared the outer six p times negative one would be negative six p the inner negative eight times p will be negative eight p we can go ahead and combine like terms here we know that's negative 14p and then negative 8 times negative 1 is positive 8. and we have plus over here 13p times 3p is 39p squared the outer 13p times negative 1 is going to be negative 13p the inner negative 11 times 3p is minus 33p and then the last negative 11 times negative 1 is positive 11. all right so what can we do to combine like terms here let me write this over that common denominator of 3p minus 1 times p plus 7 times p minus 1. so i have 6p squared and 39p squared so that would give me 45 p squared then next i have negative 14p i have negative 13p that's negative 27p and then minus another 33p which would be negative 60p and then lastly i have 8 plus 11 which is 19. so this is all over we have 3p minus 1 times p plus 7 times p minus 1. so is this guy in the numerator factorable we think about 45 times 19 that's 855 it's a pretty big number so i think about 855 as the product i would be looking for for the two integers but it would have to sum to negative 60. well this would be 5 x 171 5 is prime 171 is 19 which is prime times nine nine is three times three so not a whole lot of factors for 855 if i played with the side let me just let me just think about this for a second if i played with the signs could i make it work the answer to that is no so let's just quickly go through these factors real quick so you can see the first thing would be 1 855. we really don't need to list that but we will anyway we know those are too far apart so this will never work so let me just line that out the next thing you would come across is look at your factor tree you'd have a 3. so 3 times what well what's left 3 times 19 times 5 which is 285 again those are way too far apart so even playing with the signs i never get anywhere so the next thing we would come across we see we have a 5 so i could put 5 and then what would be 19 times 3 times 3 which is 171 but again i can't play with the signs and make that work so then i would go to 3 times 3 or 9 times what you'd have 19 times 5 or 95 but again that's not going to work i can't play with the signs and and make that get to negative 60. so next i think about 5 times 3 or 15 15 and what we'd have 19 times 3 or 57 again those aren't going to work so you can line that out then the last thing you could try is 19 times 45 so 19 and 45 and those obviously wouldn't work so i'm out of possibilities here so i can erase this and just declare that this guy over here is a prime polynomial so this is prime and so i'm not going to be able to factor this and cancel with anything down here so this is how i report my answer 45 p squared minus 60 p plus 19 over the quantity 3 p minus 1 times the quantity p plus 7 times the quantity p minus 1. all right let's try one that's going to be a little bit tedious overall and i gave you one that you could simplify in the end you know once you have some practice doing that so i have 5x minus 5 over x cubed plus 3x squared minus x minus 3 minus 2x plus 7 over x squared plus 4x plus 3. so if you see minus just put plus and then negative 1. so this negative 1 is going to multiply by the numerator remember if it's a fraction i don't put negative 1 in the numerator and the denominator because those would cancel so you just want a negative 1 either in the numerator or the denominator preferably you're going to do in the numerator you don't want to mess up your denominator because you need to get a common denominator you don't want to mess with that so just put plus and then negative 1 times your numerator and then you can forget about the messy subtraction all right so for my denominator in this one i can use factoring by grouping right i have a four term polynomial from the first group i could pull out an x squared and that would leave me with x plus three in the second group i could pull out a negative one and that would leave me with x plus three if i factor out the common binomial factor of x plus 3 this guy would factor into what you'd have the quantity x plus 3 times the quantity x squared minus 1. now x squared minus one you might not notice this but that's what that's the difference of two squares so this can be further factored so really this is x plus three times the quantity x plus one times the quantity x minus 1. always look for further factoring this is something i see on tests all the time where you get x squared minus 1 just left like this and the student just kind of moves on so let's just erase all of this and we'll bring this up here so this is our factored denominator now to factor this denominator i'm going to set this up as the product of two binomials i'll put x here and x here and then i want two integers whose sum is four and whose product is three so we know that would be plus three and plus one now i have over here x plus three and x plus one so those are common to each so really all i'm missing is this x minus 1 here over here so this guy right here can stay as it is i can keep 5x minus 5 as it is and all i really need to do is take this guy right here and multiply the numerator denominator by what i'm missing which is x minus 1. so let's put equals we've got this i'm going to put plus first i'm going to use my distributive property negative 1 times 2x is negative 2x and then negative 1 times 7 is minus 7. so it would be this multiplied by this what i'm missing this quantity x minus 1. okay so now that i have that set up this is over my common denominator which is x plus 3 that quantity times x plus 1 that quantity times x minus 1. so again all i did was i multiplied the numerator here by x minus 1 which i'm doing right there that's the multiplication and i multiply the denominator although i didn't show you that i'm multiplying the denominator by x minus 1 also if i did that down here i would have that common denominator that consists of x plus 3 that quantity times x plus 1 that quantity times the quantity x minus 1. so that's where i got that all right so now that we fully understand let's kind of scroll down and let's complete this process so we've got 5 x minus 5 and then plus we want to use foil here so negative 2x times x is negative 2x squared then negative 2x times negative 1 is positive 2x then negative 7 times x is minus 7x the negative 7 times negative 1 is plus 7. so let me write this over that common denominator we've got the quantity x plus 3 times the quantity x plus 1 times the quantity x minus 1. all right in the numerator i've got my leading term of negative 2x squared nothing to combine with that i've got 5x 2x and negative 7x 5x plus 2x is 7x 7x minus 7x is 0. then i'm just left with negative 5 plus 7 which is positive 2. so this is over we've got the quantity x plus 3 times the quantity x plus 1 times the quantity x minus 1. and i promised you that we could finally cancel something so in the numerator if you look if i were to pull out a negative 2 what would that leave me i would have negative 2 times in parentheses i'd be left with x squared the sign here would change so it would be minus and if i pulled out a 2 it would be 1. now remember x squared minus one that's the difference of two squares so i can further factor that into the quantity x plus one times the quantity x minus one and this is over so we'll have the quantity x plus three times the quantity x plus one times the quantity x minus one so look at what we can cancel we can cancel this x plus one with this x plus one and this x minus one with this x minus one and in the end i'm left with a negative two over the quantity x plus 3. you don't need to put parentheses around it you can just leave it as it is so this is going to be my simplified answer hello and welcome to algebra 2 lesson 51. in this video we're going to learn about complex fractions so again we talked about this back in algebra 1 and for the majority of stuff with regard to rational expressions we've already seen it before so how would we simplify something like this well first and foremost we know it as a complex fraction that's a fraction where there's another fraction in either the numerator the denominator or both the numerator and the denominator well we really have two methods to kind of work with this the first method would be to attack kind of the numerator and denominator separately then to do kind of the main division so this would be your main division or your main fraction bar let me kind of make that arrow a little better so what i could do here is i could say okay well i need a common denominator so i can multiply this by 3 over 3 and so this would give me what i would have 25 plus 75 over 9 and then this is over 10 25 plus 75 is a hundred so you would basically have a hundred over nine you think about this as divided by ten and you could do 10 over 1 if it makes you more comfortable with the process because what do we do we take the first fraction which is 100 over 9 and we multiply by the reciprocal of the second fraction which is 1 over 10. so i can see that i can cancel this with this and i get a 10 and so i end up with what 10 9. so what's the other way to kind of do this kind of the faster way at least in most scenarios well you can find the lcd for all the denominators involved okay so for all the denominators involved so i have a 9 i have a 3 and i have a 10. so what's the lcd well 9 is what it's three times three three is prime and ten is two times five okay so what i would do is i would include each different factor up to the largest number of repeats it appears in any factorization so here i have three times three here i have three so it appears twice in the factorization of nine only once here so my lcd would include two factors of three or nine then times i have 2 doesn't appear any well so it just goes in there then times 5 again doesn't appear anywhere else so it just goes in there 9 times 2 is 18 18 times 5 is going to give us 90. so what i would do is i would multiply the numerator and denominator of the complex fraction by 90. now this is where it gets a little complicated because you're like well what's the numerator and denominator of the complex fraction again if this is the main division i can multiply this by 90 and this by 90. and so what's going to happen let me just erase this real quick again i'm just multiplying by one and let me make these into one multiplication symbol and 90 over 90 is 1 i'm allowed to do that so what would this give me well if i distributed this 90 to this 25 over 9 the 90 would cancel with the 9 give me a 10. so i would basically have 10 times 25 which is 250 then plus i'd do the same thing again 90 times 25 over 3 90 would cancel with 3 and give me 30 30 times 25 would give me 750 then this would be over 90 times 10 which would be 900 so what does this leave me with it looks like it's not going to work out but it will 250 plus 750 is a thousand this is over 900 well if i just cancel two zeros i end up with what 10 9 just like i saw with the other method so really it doesn't matter what you use it's just a matter of what you're more comfortable with now this is a very easy scenario to deal with because we're working with just regular fractions when we turn up the heat here and we look at these complex fractions that have rational expressions in them it's going to be a little bit more tedious so let's look at the first scenario here where we have x squared over 5 plus 25 over xy plus 2x over 25 over x squared so this is your main division this is your main division so if i want to i can go through and i can get common denominators here i can do the addition and then i can do the division with this but i'm going to tell you right away the lcd method is generally this isn't every time but it's generally going to be faster when you're doing these with rational expressions it just is so let's just focus on that what is the lcd for all denominators i've got a 5. so throw that in there i've got an xy plus 2x this could be factored into what x times the quantity y plus two so we'll multiply this by x times the quantity y plus two and then this one is just x squared i've already got an x here do i need x cubed no i do not because if i broke this down this would be x times x and this would be x so i'll go with the largest number of repeats so all i'm going to do is make this x squared okay please don't make the mistake of saying okay i've got one x here and i've got an x and another x so this is x cubed it doesn't work that way again the largest number of repeats in this case that's going to be 2 of them so my lcd is what it's just 5x squared i'll put these together 5x squared times the quantity y plus 2. so if i just multiply the numerator which is this guy right here i'll put it in parentheses 5x squared times the quantity y plus 2. if i multiply my numerator by this and my denominator by this this is my denominator i'll be able to just simplify this all right so let me put my multiplication symbol and now we're ready to go so i will have what i will have x squared over five multiplied by five x squared times y plus two right that quantity you can see that these fives would cancel and so what i'm left with is x squared times x squared which would be x to the fourth power so let's go and write this as x to the fourth power times the quantity y plus two now then i have plus i'm going to multiply this times this so we'd have 25 over in factored form that's x times the quantity y plus 2 times we have 5 x squared times the quantity y plus 2. so what's going to happen is this is going to cancel with this this will cancel with one of these i'll have 25 times 5 which is 125 times x so plus 125x okay then this is over now for this one if i multiply this by this we know just the x squared would cancel so i'd be left with 5 times 25 which is 125 times the quantity y plus 2. now is there anything i could factor out from the numerator well i could pull out an x right if i wanted to so let's just go ahead and do that you want to try to leave these things in factored form if you can so if i pulled out an x i would have x cubed times the quantity y plus 2 and then plus 125. then this is over so we'd have 125 times the quantity y plus 2. nothing else i can really do to factor nothing i can really cancel i just leave it like this so that's obvious i can't cancel anything else let's take a look at another one so we have y plus 3 over x plus 6 minus y plus 3 over x minus 5 and this is over we have 3 over x minus 5 minus y plus 3 over x plus 6. so for the lcd it's going to be 1. the lcd is 1. nothing can really be factored we would just list things so i have an x plus 6 here and here so only one of those goes in when i build my lcd then i have an x minus 5 here and here so again only one of those goes in when i build my lcd so what i'd want to do is multiply the numerator of the complex fraction which is going to be this up here and the denominator of the complex fraction which is this down here by this quantity x plus 6 times the quantity x minus 5. so this is x plus 6 times x minus 5. all right so if i use my distributive property here i could distribute this to this and so what would happen here we need to write everything out this is going to cancel with this so what i'm going to have is x minus 5 that quantity times the quantity y plus 3. let me erase this and this if i did the same thing here now x minus 5 that would cancel with this and i have minus you'd have y plus 3 that quantity times the quantity x plus 6. now that minus is important i'm subtracting all of this away you can put this in brackets if you want to make it crystal clear or if you're good enough to remember that you can leave it alone but i think it'd be important here to just put it in brackets that way we don't make a simple sign mistake all right then this is going to be over what so i'm doing the same process here so this would get multiplied by this the x minus five that would cancel here and here so you would have three times the quantity x plus six then minus if i erase this and this this would get multiplied by this so this would cancel with this and i would have y plus 3 that quantity multiplied by x minus 5 that quantity again we have that minus sign so let's put it inside of rockets okay so let's scroll down a little bit now what should we do here let's go through and do our multiplication and see what we can cancel what we can get rid of x times y is xy x times 3 is plus 3x negative 5 times y is minus 5y and then negative 5 times 3 is minus 15. now i'm going to put minus and i'm going to put my brackets there and i'm going to do my multiplication y times x is xy y times 6 is plus 6y 3 times x is plus 3x and then 3 times 6 is plus 18. so then this is over we have 3 times x which is 3x then 3 times 6 which is plus 18. and then right here we put minus and then inside of brackets y times x is xy and then y times negative 5 is minus 5y and then 3 times x is plus 3x and then 3 times negative 5 is minus 15. all right let's scroll down and get a little room going so what can i do here so let me just start out by just doing this i have xy i have plus 3x i have minus 5y i have minus 15. now this minus is going to get distributed to every term inside the parentheses we don't want to make a sign mistake so this would be minus xy it would be minus 6y it would be minus 3x and it would be minus 18. okay we'll come back and simplify that in a second and make that a little bit better down here i have 3x plus 18 and i'm doing the same thing i'm going to distribute this to every term every term so i'll have minus xy i'll have plus 5y i'll have minus 3x and i'll have plus 15. okay so we have xy minus xy those would cancel just get rid of that you have 3x and you have negative 3x those cancel get rid of that you have minus 5y minus 6y those would give you minus 11y i have negative 15 and minus 18. so that's going to give us negative 33. so then this is over down here what can we cancel again we have 3x and negative 3x that cancels so we'd have negative xy plus 5y and then 18 and 15 would be 33. so can we do anything to factor the numerator well i could pull out an 11 or a negative 11. let's just pull out a negative 11. that would give me y plus 3. in the denominator could i pull anything out could i factor anything the answer to that is no so i'm just stuck with this negative xy plus 5y plus 33 and this is the best that we can do right we could do different variations of that we could leave it like this i could have pulled out a positive 11 from the numerator you know all kinds of things that we could have done but really that's the answer so we have negative 11 times the quantity y plus 3 over negative xy plus 5y plus 33. all right let's take a look at one more of these again i think this is something that most of you already know how to do we have 1 3 plus 5 over y plus 7 and this is over we have x minus 6 over 9 plus x minus 6 over 3y plus 21. so 3 i can't factor y plus 7 i can't really factor 9 i could factor into 3 times 3 and then 3y plus 21 i could factor in a 3 times the quantity y plus 7. so what do i have here i have a three here two threes here and a three here so if i'm building my lcd i would put two threes in because that's the largest number of repeats so i'd put nine there then i have a y plus 7 and a y plus 7. it only occurs once in each of those so i just got to put 1 in so once i have my lcd it's very very simple just again multiply the numerator of the complex fraction by nine times the quantity y plus seven and do the same thing to the denominator so we're multiplying by nine times the quantity y plus seven all right so let's erase this and we can get start so multiply this by this we know that 9 would cancel with 3 and give me 3. the only thing we have up here is a 1. so it basically be 3 times the quantity y plus 7 then plus if i use my distributive property here the y plus sevens would cancel so this would cancel with this and i'd be left with nine times five which is 45. in the denominator we use our distributive property the nines would cancel you'd have x minus six that quantity multiplied by the quantity y plus seven then plus the next thing we'd see we would multiply this by this now again this factors into three times the quantity y plus 7. so what would happen is the 9 would cancel with the 3 and give me a 3 and the y plus 7s would cancel you would essentially have 3 times the quantity x minus 6. let me put my equals here and put equals here and then let's scroll down a little bit so in the numerator 3 times y is 3y plus 3 times 7 is 21 and then plus 45. in the denominator x times y is x y then x times seven is plus seven x then negative six times y is minus six y then negative six times seven is minus forty two then plus three times x is 3x and then 3 times negative 6 is minus 18. all right so let's keep simplifying here scroll down a little bit more so all i can really do in the numerator 21 plus 45 5 plus 1 is 6 2 plus 4 is 6 so this would be 66 so this would be 3y plus 66 over in the denominator what can i do i have xy i have 7x plus 3x which is 10x so plus 10x i have negative 6y and then i have negative 42 minus 18 which is minus 60. so the question is can i factor anything and cancel well i know in the numerator i could factor out a 3. so i could factor out a 3 i would have y plus 22 okay in the denominator could i factor that using grouping in the first group i could pull out an x that would leave me with y plus 10. in the second group i could pull out a negative 6 that would leave me with what a y plus 10. so if i factor out that common binomial factor i would have the quantity y plus 10 multiplied by the quantity x minus 6. so it does factor but it just doesn't do me any good i can't do anything with it i have a 3 nothing to cancel with i have the quantity y plus 22 nothing to cancel with so i just leave it in factored form you could report your answer this way but you're kind of leaving it open to yourself hey could i have factored stuff so you want to just leave it in factored form to say hey there's no way that i could have canceled anything see here's the proof so 3 times the quantity y plus 22 over y plus 10 that quantity times the quantity x minus 6. hello and welcome to algebra 2 lesson 52 in this video we're going to learn about solving equations with rational expressions so again we've come to a subject that is something we learned back in algebra one but we want to review in algebra two to make sure we completely understand it and that's so that when we get to tougher material we just have a better foundation right we'd rather see something twice and completely understand it versus once and just kind of halfway understand it so i want to start by just jogging your memory a little bit and i want to talk to you a little bit about how we solved linear equations with fractions so when solving a linear equation with fractions remember we are able to clear the fractions by multiplying both sides by the lcd again the least common denominator of all denominators involved so again we saw this at the beginning of algebra two when we reviewed our topic of solving linear equations in one variable we also saw it in depth in algebra one right we spent a lot of time on it but i just want to go through and look at a quick sample problem and i want you to look at this and try it on your own so you have negative 1 12 minus one half x is equal to three halves x plus two thirds plus one fourth so obviously if you get something like this you are able to work with the fractions you can go through and do the same process just working with fractions but it's usually going to be easier to multiply both sides by the lcd and just clear the fractions so if i look at all the denominators involved i have a 12 a 2 a 2 a 3 and a 4. now 2 and 3 are prime numbers 12 is what it's 4 which is 2 times 2 and then times 3. 4 i'm going to write as 2 times 2. so when i build my lcd what's going to go in there well again it's the largest number of repeats if i have something that's duplicated so i have 1 2 here 1 2 here 2 2's and 2 2's the largest number of repeats is two so when i build my lcd it's going to be two times two and then the only other prime factor i have i have a three here one occurrence and i have a three here one occurrence so i'm just going to throw one of those in and my lcd is 1 it's 2 times 2 which is 4 times 3 which is 12. so i would multiply both sides of this equation by 12. very easy to do and so let's set that up we'd have 12 times the quantity you have negative 1 12 minus 1 half x and then this is equal to put parentheses around the other side you have 3 halves x plus 2 two-thirds plus one-fourth again i'm multiplying this side by 12 as well so what are we going to have here 12 times negative 1 12 we know the 12 would cancel with the 12 i get negative 1. then 12 times negative 1 half x 12 would cancel with 2 and give me 6. so i'd have minus 6x so this equals so i've got to do the same thing over here 12 is going to multiply by 3 halves x 12 would cancel with 2 and give me 6. 6 times 3 is 18 so i'd have 18x there and then here i'd have 12 times 2 thirds so 12 would cancel with 3 and give me 4. 4 times 2 is 8 so plus 8. then you'd have 12 times 1 4 12 will cancel with 4 give me 3 so you just have plus 3. so now i would simplify this 8 plus 3 is 11 so let's just write 11 there i can't really do anything else on the left or the right so i'm going to move all my variable terms to the left all my numbers to the right so let me add 1 to both sides and let me subtract 18x from both sides so this will cancel and this will cancel negative 6x minus 18x is negative 24x and then 11 plus 1 is 12. so this equals 12. let's go ahead and divide both sides of the equation by negative 24. so what's going to happen is this will cancel with this and just give me x and this will be equal to you've got 12 over negative 24. so i know it's negative and then 12 over 24 is going to be what that's one half so this is negative one half now you could check this if you want you can pause the video and go back i can assure you i've checked and it does work but essentially i'm just jogging your memory on how you go about doing these things you want to multiply both sides by the lcd so you can clear the fractions all right so now that we've kind of thought about this a little bit i want you to think about applying the same technique if you had an equation with rational expressions the main thing you have to be careful about we must think closely about restricted values so let me highlight that remember we talked about restricted values at the beginning of the chapter where we started talking about rational expressions in general we have to check solutions to ensure we do not end up with a denominator of 0. remember we can never divide by zero so if you get x equals some value and when you plug it in it gives you a denominator of zero you have to reject that solution that solution is not valid let's just take a look at the first problem we're going to start out with a very easy one here just to kind of get our feet wet so we have 1 over m plus 5 plus 1 over m squared plus 5 m this equals 4 over m squared plus 5m so the first thing you always want to do is you want to find your lcd sometimes the lcd is very easy to find other times it takes quite a bit of work because of the amount of factoring you have to do but for this one it's very very easy m plus 5 doesn't factor so this is just m plus 5. this guy and this guy are the same we know that that would factor into what m times the quantity m plus 5. so a lot of you can already tell that the lcd is just m times the quantity m plus 5. when you build the lcd again you look at your factors every factor would go in there but if there's a duplicate right or something that's common to more than one of these then what you're going to do is you're going to go with the largest number of repeats well i've got one m plus five here one here and one here so i would just throw one in when i build my lcd then you look at m m occurs here and here only occurs once in each so i only put one in when i'm building my lcd so the lcd or you can also say the lcm right the least common multiple of these guys is going to be m times the quantity m plus 5. so that's your least common denominator or your least common multiple of these denominators okay so now that we know that the next step is very very easy we're just going to multiply both sides of the equation by this right here just like we did with that example where we cleared the equation of the fractions m times the quantity m plus five is going to be multiplied by one over m plus 5 and then plus this is going to be multiplied by that so 1 over m times the quantity m plus 5 and that's just this in factored form all i did was factor out an m so times we're going to have m times the quantity m plus 5. and this equals you've got a 4 over m times the quantity m plus 5. again i just factored that that's all i did times you have m times the quantity m plus 5. so super easy example what you're going to see is that one when you multiply here this would cancel with this you'd have m times 1 or just m then plus the next thing is this cancels completely with this so i would just have a 1. then this equals over here this cancels completely with this so i would just have a 4. so m plus 1 equals 4 is about as simple as it gets in terms of solving equations we just subtract 1 away from each side and you get m is equal to 3. so let's write that using a solution set also so we get some practice with that and you want to go back and you want to check for two things the first thing would be make sure the left and right side is equal because you could get an answer that doesn't make your denominator zero but it might not still work the other thing is you've got to make sure that your answer or your solution does not result in a denominator that's 0. if you get that you've got to reject that solution it's not valid so let's go back up let me just erase everything here and i'm going to plug in a 3 everywhere there's an m and let's see what we get so we'd have 1 over 3 plus 5. 3 plus 5 is 8. so this would be 1 8 plus you'd have 1 over 3 squared is 9 5 times 3 is 15 15 plus 9 would be 24 this should be equal to we have 4 over 3 squared is again 9 plus 5 times 3 is 15 so again this would be 24. now you don't need to reduce this fraction all you want to do is get a common denominator over here so multiply this by 3 over 3 and you would get what you would get 3 fourths plus one twenty fourth is equal to four twenty fourths well yeah that checks out three plus one is four four over twenty four is four twenty fourths so you would get four twenty fourths is equal to 4 24 and then if you want to after you've done that you can go ahead and simplify there's just no point in simplifying at that point so you could say okay well 24 divided by 4 is 6. 4 divided by 4 is 1 so this would be 1 6 is equal to 1 6. all right so let's turn up the heat just a little bit we'll look at one that's a little bit more challenging so we have x squared plus x minus 2 over x squared minus 16 plus we have x over x plus 4 and it equals 1 over x minus 4. so again a lot of you who have done this before can already tell what the lcd is a lot of these problems kind of set up the same way so this factors into what this is x plus 4 times x minus 4 right this is the difference of two squares again you see a two term polynomial with a minus you've got to be thinking the difference of two squares then you have x plus four and then you have x minus 4. so the lcd is what it's x plus 4 that quantity times x minus 4. so what i'd do is i would multiply both sides of the equation by that so what i'm going to do because it's hard to fit this on the screen let's just go through and piece by piece multiply so if i multiply this times this we know this would cancel because that is in factored form the quantity x plus 4 times the quantity x minus 4. so what would happen is i would just be left with the numerator which is x squared plus x minus 2. now if i multiply this by this the x plus 4 would cancel with the x plus 4 and i'd have x minus 4 times x so you'd put plus and x out in front make sure this is in parentheses x minus 4. then equals if i multiply this times this the x minus 4 would cancel you'd have 1 times the quantity x plus 4. so you can put that in parentheses or not put it in parentheses there's really nothing else over there so it wouldn't matter all right so now that we have that done let's just go through and simplify so we'll have x squared plus x minus 2 then we have x times x which is x squared so plus x squared we have x times negative 4 which is minus 4x this equals again we have x plus 4. so what i can do again if i see that i have x and x on two different sides of an equation i can get rid of that right because if i subtracted x away from each side this would go away so then what else can i do on the left side x squared plus x squared is 2x squared then we have minus 4x so minus 4x and then we have minus 2 and this equals 4. so we're going to be dealing with a quadratic equation and i know we talked about the quadratic formula back in algebra one we have yet to review it in algebra two so i just gave you something you could factor right so we could solve this using the zero product property so let's subtract four away from each side of the equation so this one this side would now be zero and over here negative two minus four would be negative six so let's factor this guy so the first thing you'd notice here is you can pull a two out you'd be left with x squared minus two x minus three this equals zero and of course the two would still be out in front you set this up as a product of two binomials so you'd have x here and x here and just give me two integers whose sum is negative two and whose product is negative three well you could go with negative three and positive one so minus three and positive one negative three times one is negative three and negative three plus 1 is negative 2. so that would work so to solve this now what we're going to do we're going to take each factor with a variable so this factor here and this factor here we're going to set them equal to zero we're going to solve so you would have x minus three is equal to zero or x plus one could be equal to zero you add three to both sides of the equation here you get x equals three over on this side you subtract 1 away from each side of the equation you get x is equal to negative 1. so my solution or in this case would be solutions would be what x could be negative 1 or it could be 3. now i put that as the solution set now but i can't actually say that it's the solution until i check to make sure that it doesn't make the denominator or any denominator equal to 0 in that rational expression now i've set up my solution set here but it might not be true because again remember i've got to check to make sure that no rational expression has a denominator of 0 when we substitute these values in for our variable x so let's check that out real quick all right so again the proposed solutions were we had negative 1 and 3. so let's check negative 1 there so i'd plug in a negative 1 there and there and there and there and there and there so negative one squared is one so you would have one plus negative one which is zero then minus two so the numerator is negative 2. in the denominator negative 1 squared and this would be in parentheses because i'm plugging in for the x so negative 1 squared is 1 1 minus 16 is negative 15. so this would basically be negative over negative which is positive so positive two-fifteenths then we have a negative one a negative one over you have negative one plus four which is positive three so you can put negative one-third there you have a plus like that or you could just simply say it's minus 1 3 whatever you want to do then this is going to be equal to we have 1 over you have negative 1 minus 4. negative 1 minus 4 is negative 5. so you can say this is negative 1 5. all right so let's get a common denominator going here if i multiply this by 5 over 5 what i'd have is 2 minus 5 over 15. now 2 minus 5 is what that's negative 3 so you'd have negative 3 over 15 negative 3 over 15 which of course would simplify to negative one-fifth so you get negative one-fifth equals negative one-fifth so this one checks out doesn't make any denominator zero left and right side are equal good to go on that the next one we wanna check is three okay we want to check 3. all right so you'd have 3 squared which is 9 plus 3 minus 2 over 3 squared again which is 9 minus 16 plus you'd have 3 over 3 plus 4 which is 7 and this equals 1 over 3 minus 4 which is negative 1. so 1 over negative 1 is just negative 1. let's write that like that 9 plus 3 is 12 12 minus 2 is 10. 9 minus 16 is going to give me negative 7. now what you can do here you have a positive over a negative i want a common denominator with this so all i need to do is bring this negative into the numerator and erase it from the denominator negative 10 divided by positive 7 is the same thing as 10 divided by negative 7. so now i have negative 10 plus 3 which would give me negative 7 so this would be negative 7 over 7 which is also negative 1. so the left and the right side are equal none of my denominators became 0 so this works as a solution as well so our solution set here contains two elements negative one and three each one of those can be substituted in for x again no denominator becomes zero and the left and the right side would be equal for either solution let's take a look at one final problem this is going to be one that's going to be kind of tedious going to take a little bit of time but it's typical when you get to algebra two to get more tedious problems so we have one over b cubed plus five b squared plus six b plus we have one over b plus 2 and this is equal to we have 1 over 2b squared plus 10b plus 12. so obviously we want to get the lcd so for this one i would start by just pulling out a b that would give me b squared plus five b plus six then i would say okay well give me two integers whose sum is five and whose product is six so we'd have b times and inside of parentheses i have b here and b here and to answer that question the numbers would be two and three right so if i had positive two times positive three i'd get six positive two plus positive three would give me five so b plus two and b plus now this guy right here is just b plus 2. i can't factor that at all so that's just what it is over here for the 2b squared plus 10b plus 12 we could factor out a 2 and we'd be left with b squared plus 5b plus 6. now what do we notice from that this is the same thing as this so what we can do is say this is two times and just copy this b plus two times b plus three so what's our lcd going to be well the lcd would what it would contain one copy of b plus two i've got b plus two here here and here so just one of those when i build the lcd then i've got a b so let's put that out in front i don't have that anywhere else but we're going to put that in then we have b plus 3. so just one of those so b plus 3 is going in because it's here and here and only occurs once in each and then we also have a 2. so let's put that out in front so the lcd is 2b times the quantity b plus 2 times the quantity b plus 3. all right so if i was to multiply the lcd by this first guy right here what would cancel remember in factored form this is b times the quantity b plus 2 times the quantity b plus 3. here i have 2 times b times the quantity b plus 2 times quantity b plus 3. so everything in the denominator would cancel and i would just be left with a 2. times this 1 which would give me 2. then you'd have plus if i multiply this times this the b plus 2 would cancel so this would be gone what would happen is i'd have 2b times the quantity b plus 3 times 1. so 2b times the quantity b plus 3. then this equals if i multiply 2b times the quantity b plus 2 times minus b plus 3 times this guy over here what's going to happen is this will cancel what will be left is just the b so b times 1 which is just b all right so let's scroll down get a little room going and we can solve this guy so we can have 2 plus 2b times b is 2b squared and 2b times 3 is 6b this equals b so let's subtract b away from each side of the equation we'll have 2b squared plus 5b then plus 2 is equal to 0. now i can't factor a 2 out because i have this 5 here so what i'm going to do is i'm going to factor this using reverse foil and i'm going to set each factor equal to 0 use my zero product property and we'll get our solutions so i know if it's 2b squared one of these would be 2b the other would be b and i just got to work out the outer and inner products now so i know that i want the last two to multiply together to be two so really i know that's only one and two i can't put a 2 here because y well the reason for that is there's no common factor here so there shouldn't be one in the binomial this would produce a common factor of two so i've got to rearrange that and say plus one and plus two now does that work well the outer would be what 4b the inner would be what 1b 4b plus 1b is 5b then 1 times 2 is 2. so we've correctly factored this guy so now all we need to do is say 2b plus 1 is equal to 0 or b plus 2 is equal to 0. let's see where that leads us if i subtract 1 away from each side of the equation i'll have 2b is equal to negative 1 divide both sides by two you'll have b is equal to negative one half over here we have or if i subtract two away from each side of the equation we'll have b is equal to negative two this isn't final we've got to check these solutions to make sure they're valid for right now let's just list it as negative 2 comma negative 1 half and then we'll check and make sure that those are valid all right so to check negative 2 i'm going to plug a negative 2 in everywhere there's a b so here here here here here and here but before we even go through all that work i want you to specifically notice this right here if i plug in a negative 2 there we're gonna get negative two plus two negative two plus two is zero so this right here would be one over zero you cannot have that so if you notice something like that just stop you have to reject that solution so that is not valid you have to reject it so what we would try now is negative one half so we would have one over you'd have negative one half cubed so that would be negative one-eighth then plus you'd have negative one-half squared which should be positive one-fourth times five so it'd be five-fourths then you'd have six times negative one-half which would be negative three so then you'd have plus you have one over you have negative one-half plus two so negative one-half plus two we'll simplify all this stuff in a minute let me just drag this down here so we have some room and then this equals we have one over if you squared negative one half you'd have a fourth a fourth times two would be a half then plus 10 times negative one half is negative five so you put minus five if you want and then plus twelve all right so let's simplify this see what we get so we'd have one over you have negative 1 8 plus 5 4 minus 3. so to get a common denominator there let's just go ahead and multiply this by 2 over 2 and multiply this by 8 over 8. so you would have negative 1 plus 10 minus 24 over a common denominator of 8. so negative 1 plus 10 is 9 9 minus 24 is negative 15. so if this is negative 15 here you've got to think about this as a division problem it's basically 1 divided by negative 15 over 8. that's the same thing as if i said i had 1 times flip this guy 8 over negative 15. so this would be 8 over negative 15 or negative 8 15 then you have plus here in the denominator i would get a common denominator by multiplying by 2 over 2. so you basically have negative one plus four which is three over two so this would be one over three halves which again is like one multiplied by the reciprocal of this so this is two thirds here now to get a common denominator here let's multiply this by 5 over 5. 2 times 5 is 10. so this is 10 15. now 10 plus negative 8 is 2 so the left side is just 2 15. this should be equal to this over here so let's see what we get so you get one half minus five plus 12. so let's multiply this by two over two let's multiply this by two over two so we're gonna have one over one minus ten plus 24 all over 2. 1 minus 10 is negative 9 negative 9 plus 24 is positive 15. this would be positive 15 here and again i have 1 divided by 15 halves so it's like 1 times the reciprocal of this 1 times 2 15 which is 2 15. so you get 2 15 equals 2 15. so you know this solution is valid let's put a check mark there so really you can only say that b is equal to negative one half only your solution set would just contain one element negative one half again this negative two here that we found through solving the equation is not a valid solution hello and welcome to algebra 2 lesson 53 in this video we're going to learn about applications of rational expressions all right so at this point we should be very comfortable with solving word problems all we're going to do in this lesson is look at some examples where we deal with word problems where we have to set up equations that have rational expressions involved so not any more difficult than what we've been doing in the past just a slight variation and i just want to note that we did cover this in algebra 1. it's just something we recover in algebra 2 so we make sure we fully understand something like this before we get into something more challenging so i want to start out with this problem here so we say that a riverboat travels 75 miles against the current and the same amount of time as it goes 125 miles with the current if the speed of the current is 5 miles per hour what is the speed of the boat in still water so what are we trying to find we're trying to find out this right here what is the speed of the boat in still water so meaning if there was no current at all well to do this we need to think about our distance formula let's scroll down and think about the distance formula for a second distance is equal to the rate of speed multiplied by the amount of time traveled we all know if we're in a car and i use this example all the time and we're driving let's say 30 miles per hour that's our rate of speed and we do this for let's say four hours what's the distance going to be we could work that in your head just do 30 times 4 that's 120 this would be 120 and then for the units it would be miles because we're dealing with miles per hour and hours so we would go 120 miles in that scenario now let's think about this formula with a nice little table if we re-read through the problem we have two scenarios we have against the current and we have with the current so we have against the current which i'll just label as against and we have with the current which i'll just label as with so let's put distance is equal to rate times time like that and we'll make ourselves a nice little table i went over just a little bit so let's kind of clean that up okay let's make a little line here and now we're ready to go all right so the distance when we're going against and the distance when we're going with let's see if we can fill that out so it says that the riverboat travels 75 miles against the current in the same amount of time as it goes 125 miles with the current so when i'm filling this out the distance for against is going to be 75 right short for 75 miles and then with width it's going to be 125. again it's short for 125 miles and if you wanted to you could go up here and just say kind of slide this over a little bit this is in miles okay now let's think about the rate of speed and the time so for the rate of speed we're going to need to introduce a variable all it tells us here is that the riverboat travels 75 miles against the current in the same amount of time as it goes 125 miles with the current now it says that the speed of the current is 5 miles per hour but that's not a rate of speed for the boat and how fast it's traveling that's just a rate of speed for the current the main question here is what is the speed of the boat in still water so let's go ahead and say that the unknown is how fast the boat is going to travel if there's no current involved and then we can take this information here this 5 miles per hour the speed of the current and we can add or subtract based on the scenario we're in so if we say that we're going to let a variable like x be equal to the speed of the boat in still water [Music] then what happens well if we're fighting the current which again the current let me just kind of make this on a side note the current is what it's five miles per hour so if the rate of speed in still water is x and i'm fighting the current then i've got to subtract five away from whatever x is let's just say and i'm not saying this is the answer let's just say the boat goes 30 miles per hour in still water well if the current is 5 miles per hour and it's fighting us well then we're going to take 30 minus 5 and say we're going 25 miles per hour if we're going against the current so i would subtract 5 away here now when we go to with the current now the current's pushing on the boat it's making us go faster so i would add 5 so i would have x plus 5 here so now we have our rate done so now let's think about tau so all it tells us as far as information is that the two times are equal it says the same amount of time let me highlight that same amount of time but it doesn't give us a time for either one but fortunately with the distance formula if you have two of them you can get the third one right they're doing a little bit of manipulation so let's erase this for a second and let's just scroll down a little bit if i wanted to solve for time which is what i want to know and i have distance equals rate times time i could divide both sides by r and i would have time on one side of the equation by itself so time is equal to distance divided by rate i have a distance here of 75 i have a rate of x minus 5. so my time for this one going against the current is 75 divided by x minus 5. for with the current it's 125 which is the distance divided by the rate of x plus 5. so now we're beginning to see what we can do here we've got our table filled out so we have all the information we're going to need to gain a solution the main thing here is that it says the two times are equal to each other so the time here which is represented with 75 over x minus five is equal to or the same as this time here which is 125 over x plus 5. again the way i'm getting this in the problem it says in the same amount of time so again the two times are equal so that's how we get our equation so now all we need to do is solve this equation for x and remember x is representing the speed of the boat in still water all right so we can use the lcd method or we can realize that we basically can just cross multiply here whatever you want to do cross multiplying would be faster but let's use our lcd method so we get some practice with that the lcd would be what it would be x minus 5 times x plus five so i'd multiply this by x plus five times x minus five and over here by x plus five times x minus 5. and what is this going to do for us well in this case this would cancel with this over here this would cancel with this and so you're left with what you're left with 75 times the quantity x plus 5 is equal to 125 times the quantity x minus 5. and again you could have got the same thing by just cross multiplying if i erase this what could i have done if i multiply this by this that's what i've got here and then it would be equal to this times this that's what i've got here right so the easier thing to do in that scenario would have been to just cross multiply all right so 75 times x is 75x and then 75 times 5 is 375 so plus 375. this equals 125 times x is 125x and then 125 times negative 5 is negative 625. all right so what we want to do here let's subtract 125x away from each side of the equation so that'll cancel let's subtract 375 away from each side of the equation so that'll cancel so 75x minus 125x is negative 50x this will be equal to negative 625 minus 375 is negative one thousand so to finish this off let's just divide both sides of the equation by negative 50. so this will cancel you'll have x is equal to negative over negative is positive and you could cancel a 0 here you'd have 100 divided by 5 which is 20. so x equals positive 20. so again when we have a word problem we can't just stop and say x equals 20 and say oh we're done it doesn't work that way we have to think about what this actually means remember x was the speed of the boat in still water that's what we're trying to find to answer the question what is the speed of the boat in still water well the speed of the boat or we could say the boat travels at 20 miles per hour in stillwater now does that make sense in terms of the problem well a riverboat travels 75 miles against the current so 75 miles against the car if the boat travels 20 miles per hour in still water if it's fighting the current and the current's 5 miles per hour that means it's going 15 miles per hour so to go 75 miles at 15 miles per hour that would take five hours so it would take five hours now it says same amount of time as it goes miles with the current so in five hours with the current if i go with the current here i'd take 20 miles per hour and i would add the speed of the car to it so now it's 25 miles per hour what would that do in 5 hours well 25 times 5 is 125 so that's accurate right it would go 125 miles in the same amount of time as it goes 75 miles so the boat would in fact travel 125 miles with the current in the same amount of time the five hours as it takes to go 75 miles traveling against the cart so we have the correct answer here again the boat travels at 20 miles per hour in still water let's take a look at another one that involves this distance formula or again a motion ward problem so jeff traveled to his parents house by car on the trip there he averaged 80 miles per hour on the trip back due to road construction he only averaged 50 miles per hour if he spent 9 more hours driving on the way back what was the total distance to his parents house so you might look at this problem and just be really confused on how to get started it doesn't seem as simple or as straightforward as the last problem but again if you go back to that distance formula if you just say distance is equal to rate times time and you make yourself a little table okay this is a great way to just organize your information so we have a trip to his parents and a trip back from his parents so there's the trip there and there's the trip back so those are your two different scenarios okay so let's write distance here is equal to rate times time all right so let's go back up and see what we can figure out so the first information we're given is that on the trip there he averaged 80 miles per hour so he averaged 80 miles per hour so that is something we're going to write under rate now it says on the trip back he averaged only 50 miles per hour so let's put those in under rate so for the trip there it was 80 miles per hour on the trip back it was only 50. now when we start talking about distance remember the main question here is what was the total distance to his parents house what was the total distance to his parents house so we don't know that so if it's an unknown let's assign a variable to represent it so let's say that we're going to let x be equal to the distance and this is in miles to his parents house so we put x for the trip there and x for the trip back because the distance is the same now what about time again we saw this in the last section where we said that we could find time by algebraically manipulating this formula if i divide both sides of this equation by r i would have that distance over rate is equal to time so i have a distance of x over a rate of 80 so that's my time for the trip there for the trip back i have a distance of x over a rate of 50 so x over 50 is my time for the trip back now how do we set up an equation the times are not equal like they were in the last problem we can't say this is equal to this it's not that straightforward here we have to read back through the problem and think about this so jeff traveled to his parents house by car on the trip there he averaged 80 miles per hour on the trip back due to road construction he only averaged 50 miles per hour now this is going to be the key to setting up an equation you always look for information that they give you that allows you to do this if he spent nine more hours driving on the way back just stop for a second we know that the time driving on the way back was nine hours more than whatever it was driving there so we know that the trip back is going to take us nine hours more so that means x over 50 which is the time for the trip back would be equal to or the same as the time for the trip there which is x over 80 plus an additional nine hours right because again the trip back is equal to or the same as the time for the trip there which is x over 80 plus it takes another 9 hours so this would be the equation that you want to run with now to solve this i just want to find my lcd so what's the lcd if i look at the denominators here i have 50 and i have 80. if i factored 50 i would get one i would get 5 times 5 times 2 right basically 25 times 2. if i factored 80 80 is what it's 16 times 5 so 16 is 2 to the 4th power times and then again you'd have 5. so for my lcd i would have 2 factors of 5 or 25 multiplied by four factors of two or sixteen so this is four hundred so we basically take 400 and multiply it by what you have x over 50 and then this equals you'd have x over 80 plus 9 and you're multiplying this side by 400 as well so this is going to clear all the denominators for us if i multiply 400 times x over 50 the 400 would cancel with the 50 and give us an 8. so this would be 8 times x or just 8x and this is equal to if i do 400 times x over 80 the 400 would cancel with the 80 and give us a 5. so you'd have 5 times x or 5x and then plus if you did 400 times 9 the quick way to do that 4 times 9 is 36 put 2 0's at the end so you'd have 3600 there all right so to solve the equation let's subtract 5x away from each side so this will cancel 8x minus 5x is 3x this is equal to 3600 we'll now divide both sides of the equation by three and we get that x is equal to twelve hundred all right so let's go back up to the top we said that the distance to his parents house or the total distance to his parents house was represented with x we just found out that x was twelve hundred so we'll say the distance to his parents house was 1200 miles and it's pretty easy to check that again if you can just move the distance formula around based on what you need it's easy to check things like this so we think about averaging 80 miles per hour and you're trying to go 1200 miles that's that's something you can calculate all the time right if you're doing a road trip so i know that again distance equals rate times time so if the rate is 80 and the distance is 1200 again to calculate the time i would divide the distance by the rate so what is 1200 divided by 80 that's going to give me what well it's going to give us 15. so the amount of time for the trip there was 15 hours so for the trip there let's just notate that for the trip there it was 15 hours now what about the trip back so we averaged 50 miles per hour so we want to make sure that this is nine more hours than the trip there that's how we know we got the right answer so we would take 1200 and divide it by 50 and this gives us what well this is going to give us 24 that's exactly what we want right so the trip back is 24 hours and again if you look back through the problem that's what it tells us to expect it tells us that on the trip back due to road construction he only averaged 50 miles per hour he spent nine more hours driving on the way back so on the way back he spent 24 hours driving on the way there he spent 15 hours driving 24 is 9 more than 15 so we have the correct answer here the distance to his parents house was 1200 miles all right let's shift gears for one moment and we're going to look at a problem that deals with rate of work or work rate most of you who have done these type of problems before you know they're super easy they're a little complicated to kind of wrap your head around when you first start doing them but there's a very easy formula that you follow for them and if you follow that it's going to be very very simple for you to get the answer quickly so it takes cody nine hours to tar a roof trevor can tar the same roof in five hours if they work together how long would it take them so when you work a work rate problem you think about everybody's contribution in terms of one unit of time so if we're working with hours our unit of time is hours so in one unit of time we're thinking of one hour so for one hour we think about cody as being able to tar one-ninth of the roof then for trevor it takes him five hours so in one hour he's done one-fifth of the roof so this is trevor so once i have these amounts i think about let's add these two fractions together to see in one hour how much of the job would they do if they worked together so what is 1 9 plus one-fifth nine is three times three and five is a prime number so i would multiply this by five over five i'll multiply this by nine over nine we'd end up with what five plus nine over 45 which is going to give us 14 over 45. so if i scroll down just a little bit let's get a little room going we could say that in one hour 14 over 45 this amount of the job is completed if they work together so in one hour fourteen forty-fifths of the job [Music] is complete now if we want to see how long it would take to actually complete the job we could let a variable like x be equal to the number of hours to complete the job and if you wanted to be more specific on this i just wanted to kind of shorten it you could say the number of hours to complete the job if they're working together so what would i do once i have this set up i have in one hour the amount that working together they have completed so that's 14 over 45. if i multiply this by the number of hours it takes for them to complete the job what happens is this times this will give me one completed job so once you've set this up and i know it can be a little confusing when you first start seeing these the whole key is starting out by finding out how much they can do together in one unit of time so in this case it's hours so we figure out in one hour trevor and cody can do fourteen forty-fifths of the job then we say a variable like x so you could use y or z or whatever you want represents the number of hours for them to work together and complete the job so the amount they complete in one hour so this is amount completed in one hour multiplied by the number of hours would be equal to or would give us a complete job so all we have to do is solve this for x and that's super easy to do you're basically going to just multiply both sides by the reciprocal of that fraction and when we do that if i multiply this by 45 over 14 and multiply this by 45 over 14 you're going to end up with x is equal to 45 over 14. so this is how long this many hours right here is how long it takes for them to complete the job if they're working together so we could say working together it would take 45 over 14 that number of hours to tar the roof now if you do the division on 45 divided by 14 you get kind of a nasty decimal and to not have to round things i just leave it as a fraction but it's about 3.21 if you just wanted to estimate about 3.21 hours so a lot of teachers would accept that as an estimate just to make the answer cleaner but just to be precise let's say 45 or 14 that number of hours now if you wanted to check this you could add trevor's contribution and cody's contribution and you should end up with one completed job so cody's contribution is what he does one-ninth of the job in an hour so if he works for 45 over 14 hours how much is he going to do this would be his contribution or cody's contribution if we add that to trevor's contribution in an hour he does one-fifth of the job again he's going to also work for 45 over 14 that number of hours so would this equal one completed job well you can look and see that it would this would cancel with this and give us one it would give us a 5 so you'd have 5 14 plus this cancels with this and gives us a 9. so 9 14 does this equal 1 5 plus 9 is 14 so this would end up being what this would be 14 over 14 which is 1 1 equals 1 so that equation checks itself out hello and welcome to algebra 2 lesson 54. in this video we're going to learn about radical expressions so again we've come to a topic that we learned thoroughly in algebra 1 and we just need to review here before we get to more challenging material in algebra 2. so we're going to start out with the easiest scenario which is where we deal with square roots so i want you to recall that the square root of a number and let's just say that number is q right just as a placeholder for now so this is any number that when multiplied by itself gives us the number q back now by using a variable like q a lot of you will stop and say well that doesn't really make a whole lot of sense so what you can do you can just cross q out and you can just pick a number that you know is a perfect square so as an example we know that 4 times 4 is 16 so 16 is a perfect square so we could say 16 and 16. now if we re-read it we would say recall that the square root of a number 16 is any number that when multiplied by itself gives us the number 16 back now we think about 16 as what most of us would say it's 2 to the fourth power it's 2 times 2 which is 4 4 times 2 is 8 and 8 times 2 is 16. we already know that 16 is 4 times 4 but we might forget that negative 4 times negative 4 is also 16. so that's where this definition really comes in where it says it's any number okay so it could be more than one so the square root of 16 would be four but also it would be negative four now when we talk about square roots we notate the positive square root or what we call the principal square root differently from the negative square root so i would say the square root of 16 like this to ask for four i would say the square root of 16 like this to ask for negative 4. so this one right here is the positive or we'd say the principal square root of 16 whereas this guy right here because we put that negative out in front is asking for the negative square root of 16. and the difference between the two is just very very slight if i look at this one i say okay well what positive number when multiplied by itself gives me 16. well we know that off the top of our head that 4 or more specifically positive 4 times positive 4 would give me positive 16. when i look at this one i look at this negative out front i say well what negative number when multiplied by itself would give me 16 well negative 4 times negative 4 would be 16. so those are the two different notations that you'll run across if you see something like what is the square root of 25 this is just asking for the principal square root so they just want 5 back if you see what is the negative square root of 25 like that they're asking for negative 5. as we learned back in algebra 1 when we started looking at the quadratic formula there's a shortcut notation for this i don't need to separate these two like this i can just put plus or minus like this in a compact form and then put square root of 25. this is asking for the positive or principal square root of 25 it's also asking for the negative square root of 25. so this i could put is equal to plus or minus 5. so i've accounted for both possibilities the positive square root of 25 is 5 the negative square root of 25 is negative 5. so this is again just a more compact way to write things so let's just look at a quick example here so we have the principal square root of four so that's asking for what positive number when multiplied by itself would give me four all of us know that would be 2 right 2 times 2 is 4. now with this one we have the negative square root of 4 so we're saying hey what negative number when multiplied by itself would give me 4 and that's negative 2. now again if you want to practice that more compact notation you could say plus or minus the square root of 4 and that's equal to plus or minus 2. all right so let's look at something with higher roots now so we should know at this point that in general when we see a square root it's really what if i have the square root of 4 like this it's missing a number there that we normally display for higher roots because square roots are so common square roots are so common that we just leave that off but that number that's missing is a 2. that 2 is known as the index or the order so in this generic example here we have the nth root of a so right here this is the index or you could say it's the order okay this a right here is known as the radicand that's the value that's under the radical symbol so we'll label this we'll say this is the radical symbol and the whole thing if i looked at everything involved here that's just known as a radical so if i say hey you've got this radical i'm talking about everything so this whole thing right here is a radical so again when we see square roots like the square root of 4 we don't display an index or order because it's understood to be 2. so if i wanted the square root of let's say 100 i would write it like that now it wouldn't be incorrect if i wrote the square root of 100 like this and put a 2 here it's just not common to see that now as we move into higher level roots we see something like this i'm going to start out with something known as a cube root so the index is a 3. this is my index and my radicand is an 8. so this is my radicand and before we think about this let's think about the square for a second if i want the square root of 4 i'm thinking about what number or numbers when multiplied by itself is going to give me 4. now the reason i'm doing that is because if i think about the square root of 4 again this is a 2 so i'm saying hey what number multiplied by itself twice is going to give me this value here when i look at a cube root now i'm thinking about a number that when multiplied by itself 3 times because the index is a 3 is going to give me that radicand back so with the cube root of 8 with the cube root of 8 again i'm looking for a number that would multiply by itself 3 times would give me 8. now for most of you would know this is 2 right 2 times 2 is 4 4 times 2 is 8. right so that would be your answer there but if you didn't know something like that you can always go back to your factoring you could say i know 8 is what it's 4 times 2 and i know 4 is 2 times 2 these are all prime factors so i know what's 2 times 2 times 2 or 2 cubed so again the cube root of 8 is 2. now you'll notice that i didn't give you two sets of notation for this one here and that's not because it's a higher root it's because this index is odd so i want to just cover some rules that you need to know something you want to definitely jot down so if you see something like the nth root of a and i know we don't like working with generic examples but just something you can substitute in to your given example you say that you have the nth root of a n is even so n is let's say two or it's four or it's six or it's eight or it's ten something that's divisible by two and then also in this scenario a is greater than or equal to 0. so basically a is not a negative value if this occurs we're going to have two sets of notation just like i showed you so you're going to have the nth root of a like this this is your principal root then you have the negative of this so negative nth root of a this is your negative that's the negative root so in that scenario like if we had let's say the square root of 4 for example square root of 4 is what that's 2. that's your principal root then you have the negative square root of 4 which is negative 2. but again it's not just with a square root it could be a fourth root or an eighth root something easy to do let's say we had the fourth root of 16. so the fourth root of 16 many of you know that 2 to the 4th power is 16. 2 times 2 is 4 4 times 2 is 8 8 times 2 is 16. so we know this would be 2 but then you'd also have to account for the negative 4th root of 16 and this would just be negative 2. if i multiply negative 2 by itself 4 times i would get positive 16 as a result all right the next scenario i want to talk about if we have the nth root of a and n is even and a is less than zero so now a is a negative value so you'll recall this from algebra when we said we have the square root of i don't know let's say negative 16. what happens here well it's not a real number we're going to learn how to deal with this in algebra 2. we didn't talk about an algebra one but we will have a way to evaluate this using something known as imaginary numbers but we haven't gotten to that yet so for right now we just say that this is not real this is not a real number so again if n is even so 2 4 6 8 10 12 whatever it is and a is less than zero a is a negative value then you're going to write that it's not a real number so let's say you have the fourth root of let's say negative 625. again this is even this is negative so it's not a real number all right the final scenario if you have n that's odd okay so your index is odd a can be whatever it can be positive it can be negative it can be zero can be whatever you want it to be you're only going to have one root and the reason for this deals with the rules for positives and negatives if i multiply a negative by itself an odd number of times i get a negative so if i had something like let's say the cube root of negative eight so what number when multiplied by itself three times gives me negative eight well that's negative two negative two times negative two is positive four then positive four times negative two is negative eight just keep in mind that you are allowed to have a negative radicand if this guy is odd okay it's just the case where you can't have a negative radicand when that guy is even so a lot to kind of throw at you especially if you didn't learn this back in algebra one it's just one of those things where you just have to practice enough and it's something you're going to have down pretty much right away all right so let's just take a look at some really easy examples so we've seen this already if i had the principal square root of 25 we know that would be what i'm just asking for a positive number that when multiplied by itself gives me 25 we know that would be 5. then what about the square root of negative 25 now again if i have a negative radicand and i have an index that's even in this case the index isn't shown but it's an invisible index of a 2 what happens here this is not a real number this is not a real number now don't get this confused with what we're going to see now here we have the negative square root of 25. underneath the radical symbol i have a positive value so i'm okay this is positive it's the negative outside that's telling me hey i want the negative square root of 25. that's a big source of confusion so make sure you understand the difference between the two the negative square root of 25 is just asking for what negative number when multiplied by itself will give me 25 and that's negative 5. right negative 5 times negative 5 is 25. all right for the next example we have the principal fourth root of 81. so we're asking for what positive number when multiply by itself four times is going to give us 81. well some of you will know right away that's three if you didn't again you can use a factor tree so take 81 and break it down most of you will know that 81 is what it's 9 times 9 9 is 3 times 3 so just a quick factor tree will show you 3 times 3 times 3 times 3 or 3 to the fourth power is 81 so that means if i'm looking for a positive value now when multiplied by itself 4 times is going to give me 81 we know that's going to be 3. all right what about the cube root of negative 64. because this index is odd i know i'm going to have one answer and i know that the negative is allowed this is okay again because this is an odd index a negative times another negative times another negative or three negatives or an odd number of negatives in general will produce a negative so all you got to ask yourself is take 64. forget about the negative 4 second 64 factors in a what it's 4 times 16 16 is 4 times 4. so without going into 4 is 2 times 2 i know i have 4 times 4 times 4 or 4 cubed that gives me 64. all i got to do is just drag this negative over here and say this is negative 4 right because the negative times the negative times the negative will give me a negative back so negative 4 times itself three times will give me negative 64. all right for the next one we have the negative fourth root of ten thousand so what are we looking for here we're looking for a negative number but when multiplied by itself four times is going to give us ten thousand now so i'm just gonna put a negative out in front because i could just forget about it now now i just think about ten thousand ten thousand is a power of ten so it's really easy right it's a one followed by one two three four zeros so that means it's 10 to the fourth power so the way i get that is if i'm evaluating something like this 10 to the 4th power you write a 1 and you follow it by this many zeros so this is 1 2 3 4. so kind of a cool trick if you don't know that so i automatically know that if 10 to the fourth power is ten thousand the fourth root of ten thousand forget about the fact that it's the negative fourth root we've already accounted for that the fourth root of ten thousand would just be ten so we just put negative ten here as our answer right negative ten times negative ten is a hundred a hundred times negative ten is negative one thousand negative one thousand times negative ten is ten thousand all right let's take a look at another one let's say you saw the cube root of negative one hundred twenty-five so again if i'm thinking about something like this this is negative 3 negatives makes a negative so i just start out with putting a negative there then you can just forget about it you can say okay well 125 125 would factor into what 25 times 5 25 is 5 times 5 so 5 cubed would be 125. so what i'm looking for here is negative 5. negative 5 times negative 5 is 25 25 times negative 5 is negative 125. all right last one we have the principal fourth root of negative 16. so again i have a negative radicand and i have an even index so when that happens we don't have a real number this is not a real number and again the reason for that is i can't take a negative and multiply it by itself four times and end up with a negative right it's going to give me a positive an even number of negatives will always give me a positive and so that's why this value doesn't exist using real numbers now let's take a look at another rule that's pretty time saving for us if we have the nth root of a to the nth power so something like if you had the fourth root of let's say two to the fourth power what would you guess would happen there well you know from your studies of algebra one or hopefully you remember that this ends up canceling with this and you're left with this right so if n is even so in other words if this is even and this is even they're the same value and n is greater than zero so even and greater than zero then what we're going to find is we have the nth root of a to the n we'll say this is equal to the absolute value of a so as an example let's just say we took something like five so we take the sixth root of five to the sixth power now you don't even need to calculate what five to the sixth power is you can just say okay well if i took five and i raised it to the 6th power whatever number that is once i take the 6th root of it i'm going to be right back to 5. so this will cancel with this and i'm left with this so this would be equal to 5. as another example let's say i was working with a negative and to make it real simple let's just make it negative 2. so if i had the square root of negative 2 squared what would that be equal to well some of you might say oh that's not a real number you have a negative down there well think about squaring this first if i did negative 2 squared first that would give me 4 and if i took the square root of 4 i would get 2. so that's where this definition comes into play notice how we say it's the absolute value of a so if we're saying a here is negative 2 well the absolute value of negative 2 is 2. and that comes from that again that squaring operation or in this case it could be anything that's even so raises them to the fourth power the sixth power the eighth power the tenth power whatever it is an even number of negatives will give us a positive value then when we perform that root operation we're going to go back to the absolute value of whatever that was right because we took a positive took the root of it and we went back to whatever the absolute value of that was all right the next one we want to talk about is the other scenario where n is odd so if n is odd and n is greater than 0 then what's going to happen in this case so this is odd and it's greater than 0. in this particular scenario these two will just cancel each other out and you're just left with a so the only difference is when you have an even involved you got to say in case you're a in case this radicand was negative you've got to account for the possibility that you do an even number of factors of that guy right because you're raising it to an even power it's going to end up being a positive then you take the root and knocks it back down to the absolute value of that so that's all we're saying here again n is even and n is greater than zero all we're saying is if this guy was negative an even number of negatives would have made it positive so you get back to the absolute value of that in this scenario because n is odd and odd number of negative factors would give you a negative it's okay so as an example let's say i have the cubed root of i don't know negative 2 cubed well this is just going to be what this would cancel with this and i just have negative 2. all right so let's say we saw something like the sixth root of negative seven to the sixth power so again without even going through and saying okay well negative seven times negative seven without giving you six factors of negative seven and then taking the sixth root of that you automatically know that this would cancel with this but i'm not left with negative seven i'm left with the absolute value of negative seven which is seven right so it's a little bit more complicated than just canceling things i don't want you to make the mistake of just cancelling things saying okay this is negative 7 because that would be wrong taking negative 7 to the power of 6 would have made it positive and then taking the 6 root of that would have brought you back to positive seven so it's important to realize that this is equal to the absolute value of negative seven which is seven so here's an example where we have an odd index so we have the cube root of negative nine cubed so in this case you could just kind of cancel these and say i'm just left with negative 9. all right negative 9 to the third power would still give you a negative result right so when i take the cube root of that it would still be a negative so when the index is odd i can just basically cancel these when the index is even i want to cancel them and then i want to use the absolute value here so that's the only difference you're watching out for so now we've kind of reviewed radicals let's talk about fractional exponents or otherwise called rational exponents so once you recall if you have something like a raised to the power of 1 over n this is equal to or the same as the nth root of a so as an example let's say i had 4 to the power of one-half just kind of following this what would this be well this part right here becomes your index so this part right here would be the index and index of 2 is a square root so we normally don't display that but for the purposes here i'm just going to put a 2 here and then 4 just comes over here so 4 raised to the one-half power is like taking the square root of 4 and that should make sense because of the rules for exponents if i have four to the power of one-half and i multiply by four to the power of one-half what would that give me base stays the same you add your exponents one-half plus one-half is one so this is equal to four if i did this using this and i said okay well this is the square root of 4 times the square root of 4 we would expect that to be what to be 4 right the square root of 4 is 2 the square root of 4 again is 2 2 times 2 is 4. so it has to make mathematical sense and it does if i wanted something let's say as a cube root okay a cube root so let's say i had eight to the power of one-third well this would be the cube root of 8. again whatever's here in the denominator becomes my index or order and then just becomes my radicand so the cube root of 8 is the same as writing 8 to the 1 3 power or you take something way more advanced let's say we had i don't know 620 and let's just raise this to the power of 1 8th as an example well this is what this is the eighth root of 620 because this part right here goes here and this part right here goes here so the other thing we need to know with this let's say we have a raised to the power of m over n this just gets split up into the nth root of a and this is raised to the power of m so a good example of this would be something like 9 to the power of let's say 3 halves so just following this what would you see again the denominator becomes the index this becomes the radicand and then this right here we're going to raise everything to that power so if i took the square root of 9 i would get what i would get 3. and then if i cubed 3 i get what 3 times 3 is 9 times 3 is 27. now there's other ways to do that you might see your book tell you to take this guy and raise it to the third power first so i could also say that 9 to the 3 halves power is what it's 9 cubed okay so 9 is raised to that power and then we take the square root of that but this produces a bigger number to take a square root of and so it's not usually what you want to do 9 cubed is 729 so it would be the square root of 729 which is going to give you 27. all right so let's just look at some examples we have 4 to the power of one half we know that's just what you erase something to the power of one half you're just asking for the square root so it's the square root of four now again you take this right here that's in the denominator that's your index so the index would be a two on a square root we just don't show that this right here this 4 that number that's the base when we talk about an exponent becomes the radicand so you basically get the square root or the principal square root of 4 which is 2. if you look at 216 to the power of 1 3 we're asking for the cube root of 216. again this denominator here becomes the index 216 becomes the radicand so what is the cube root of 216 so if you don't know that off the top of your head just look at the last two digits it's a 16 so you know this is divisible by 4. so this would be 54 times 4 4 is of course 2 times 2. 54 is 9 times 6 6 is 3 times 2 9 is 3 times 3. now what do i have here i have 1 2 3 factors of 3 and 1 2 3 factors of 2. so we've got 3 cubed times 2 cubed basically 3 times 2 is 6 so it's 6 cubed so this is equal to what it's equal to 6 because 6 cubed would give me 216. so let's take a look at a harder example so we have 81 to the power of 3 halves 81 is going to be my radicand my index is going to be what it's going to be a 2. it's always the denominator and then i'm raising this whole thing to the power of 30. so the square root of 81 is 9 9 cubed as we just found out was 729 all right let's take a look at 625 to the power of 3 4. so again 625 is my radicand and the index here is what it's a 4 and then we want this all raised to the third power so what is the fourth root of 625 most of you know that is 5 minus 5 is 25 25 times 5 is 125 125 times 5 is so if this becomes 5 when i cube that i go back to what i go back to 125. what about if we mix some things up and we ask for 16 to the power of negative 1 4. so what am i asking for here well again i want the fourth root i want the fourth root of 16 and then i would raise everything here to the power of negative 1. so i can go ahead and take this first well this is 2 so this would be 2 to the power of negative 1 which is what it's just take the reciprocal of the base so 1 over 2 raise this to a positive 1 which is basically just one half now the other way to do this if you wanted to you could say okay i have 16 to the power of negative 1 4. well you could have started out by saying i have 16 raised to the power of negative 1 and then i'm going to take the fourth root of that same answer either way this would be what this would be the fourth root of take the reciprocal of this this is 1 over 16. and it's a little bit more complex here because it's a fraction but just think about the fourth root of 1 over sixteen well one half times one-half times one-half times one-half is going to give you one-sixteenth you know one to the fourth power is one so that's easy and then what to the fourth power sixteen well that's two so you end up with one half either way all right what about something like negative 32 to the three-fifths power well again we're looking at we have negative 32 and then i'm going to take the fifth root of that and i'm going to raise the whole thing to the third power so what's the fifth root of negative 32 that's going to give us negative 2 and we want to cube that negative 2 cubed is negative 8. but the next one we're going to look at 5 raised to the power of 5 3 over 5 raised to the power of 2 3. now just as we saw when we worked with exponents in the past if you have the same base which we have a base of five a base of five and you're dividing you subtract the exponent in the numerator minus the exponent in the denominator so five just stays the same and we would do five thirds minus two thirds same rules apply no matter what you're working on no matter how complex it is you always go back to those basic rules that you learned so five minus two is what that's three so you would have 5 raised to the power of 3 over 3 which is 5 to the power of 1 right it's just 5. let's say we take a look at m squared over m to the power of 1 3 times m to the power of 2 3 times m to the power of 1 3. so again same rules apply so m squared over i'm just going to use my product rule for exponents m would stay the same and we would add all these exponents so we know that one-third plus two-thirds is three-thirds three-thirds plus one-third is four-thirds so what would happen here is if i'm dividing m stays the same and we would do two which is the exponent in the numerator minus four thirds which is the exponent in the denominator to get a common denominator i would write two as what six thirds and we'd end up with m to the power of 6 minus 4 is 2 and then over 3. so you can leave it in this format or you could write that you have the cube root of m that's squared or you could say you have m squared and take the cube root of that all of these are the same what about m cubed squared times m squared over m to the power of one half well using my power to power rule here i know this would be m to the sixth power and times m squared all i would do is add exponents here so m would stay the same six plus two is eight so this would basically be m to the eighth power over m to the power of one half so m stays the same and i would take 8 and i would subtract away a half so let's write this as 16 over 2 minus 1 over 2 this would equal m to the power of 15 over 2. again you could leave it like that or you could write that you have the square root of m raised to the 15th power or you could say you have m to the 15th power and then you could take the square root of that right all of these this one this one and this one mean the exact same thing all right for the last one we have x squared raised to the two thirds power times x to the power of one half over x to the power of negative three halves so if i look at this i use my power to power rule i would multiply 2 times 2 and get 4 so this would be x to the power of 4 3 then we multiply by x to the power of one half and of course this is over x to the power of negative 3 halves so let's put this is equal to let's deal with the numerator first so we have what we have x staying the same and we're going to add the exponents so if i add four thirds plus one half multiply this by three over three multiply this by two over two g times four is eight so you would have what you'd have eight over six plus one times three is three over six a plus three is eleven so this would be x to the power of eleven sixths and this is over x to the power of negative three halves now i don't need to do anything fancy here other than leave x the same and then just subtract so i would do 11 6 minus a negative notice that that's a negative and i'm subtracting it away so that's plus a positive three halves so to get a common denominator let's multiply this by three over three and this would be nine sixths okay nine six so this would be x to the power of eleven plus nine is 20 over 6. now if you get a scenario like this you can reduce this we know that 20 and 6 are each divisible by 2 so we can really say this is 10. over 3. and then what we want to do is say okay we have the cube root of x and this is raised to what the 10th power or again you could also say you have x to the 10th power and you're taking the cube root of that hello and welcome to algebra 2 lesson 55 in this video we're going to learn about simplifying radicals so again we have another lesson where we're just kind of covering or refreshing our memory on things that we learned back in algebra 1. so i want to start by talking about the product rule for radicals and the quotient rule for radicals before we get into actually how to simplify a radical so the first thing again the product rule for radicals if we see something like the nth root of a times the nth root of b we could say this is equal to the nth root of a times b now when we look at again very generic things like this things we'd see in our textbook a lot of you get confused you say what does that really mean we're saying if the index is the same so in other words if i had something like the square root of 2 times the square root of 3. so the index here is a 2 the index here is also a 2. this is going to be equal to the square root of in this case you have a in this case you have b so we combine them and say a times b in this case we have 2 in this case we have 3 so we combine them and say 2 times 3 so this is the square root of 6. that's all it is okay in exponent form remember if we raise something to the power of 1 over n it's the same thing as taking the nth root of it so the nth root of a can be written as a to the power of 1 over n and the nth root of b can be written as b to the power of 1 over n so you multiply the two together and we're just showing this with exponents it's the same thing between here and here we have a times b inside of parentheses and it's raised to the power of 1 over n so as an example here i could say this is two to the power of one-half times three to the power of one-half this would be equal to two times three to the power of one-half or six to the power of one-half which again we know is the square root of 6. let's take a look at some examples let's say we have the square root of 5 multiplied by the square root of 6. we can combine these two and just say it's the square root of 5 times 6 which is the square root of 30. again notice how your index here is a 2 your index here is a 2 so same index so we're allowed to combine them in that way for the next one we have the cube root of 4 multiplied by the cube root of 7. so same index you got a 3 here and a 3 here so that would be the cube root of 4 times 7 which is 28. now for this one i gave you an example where you can't use this method you have the fourth root of two multiplied by the cube root of seven now there is a way to combine this and we'll get to that at the end of the lesson but you can't do it using the product rule for radicals to use the method that we've been doing you've got to have the same index and again this is a 4 and this is a 3. so you're not going to be able to combine those using this method all right here's an example with some variables involved so suppose we had the 7th root of 20x cubed multiplied by the seventh root of 10x so it would be the seventh root of what 20 times 10 is what that's 200 and then x cubed times x is x to the power of 3 plus 1 which is four so you'd have the seventh root of 200x to the fourth power so you get the idea again we talked about this back in algebra one even if you didn't see an algebra one it's very very simple if the indexes are the same and you're multiplying just set up that root with that same index and just multiply the two radicands together okay that's all we're doing all right now let's talk about the quotient rule for radicals which is pretty much very similar if you have the nth root of a over b you can say this is equal to the nth root of a divided by or over the nth root of b so it just allows us to split things so in other words if i had the square root of let's say 1 4 i could split this up into the square root of 1 over the square root of 4 and then it's a lot easier to figure this out instead of looking at a 4th and saying well what number times itself would give me a fourth i could say what number times itself will give me one that's one what number times itself will give me four that's two so this ends up giving me one half so we can also show this in exponent form we have a over b raised to the power of 1 over n we just say this is equal to a raised to the power of 1 over n over b raised to the power of 1 over n it's no different than if i had something like let's say 10 over 2 and this is squared this is the same as 10 squared over 2 squared right either way if i do 10 over 2 that's 5 5 squared is 25. if i do it this way 10 squared is a hundred 2 squared is 4 100 divided by 4 is also 25. so same result either way all right so let's look at a few examples of this we have the square root of seven twenty-fifths so all we're gonna do is break this up we have the square root of seven over we'd have the square root of twenty-five now 7 is not a perfect square so i'm going to leave that as just the square root of 7 and then the square root of 25 is 5. so we'd end up with the square root of 7 over 5. now we have the cube root of 8 over 216 so i'm going to break this up into the cube root of 8 over the cube root of 216 the cube root of 8 is 2 the cube root of 216 is 6. and we can reduce this fraction don't stop and say oh well i have to finish there it's the same rules that apply throughout all of mathematics if you have 2 over 6 that's the same as 1 3 right cancel a common factor of 2. right then what about the fifth root of negative 1 over 32 so this would be the fifth root of negative 1 over the fifth root of 32 so the fifth root of negative 1 is just negative 1 because negative 1 times itself 5 times would give me negative 1 back and this is over the 5th root of 32 is 2. so this is negative 1 half as the answer all right so now let's talk about what we came here for which is how to simplify a radical and again it's something you need to do to get full credit on your answers just kind of like when you reported fractions but you didn't simplify them and your teacher would take points off you need to follow these rules because this is what's expected of you so for a simplified radical the first and the most important thing is that the radicand contains no factor that is raised to a power which is greater than or equal to the index now that might seem kind of challenging if you just read it but really it's quite simple if i had something like the square root of 20 for example you think about factors of 20 there's a perfect square inside of there so all we're saying is that it would be simpler if i rewrote this and said hey 20 is really 4 times 5 or 2 times 2 times 5. so 4 is a perfect square so i could write this as the square root of 4 times the square root of 5 and i can say well hey this square root of 4 that's really just 2. so now i can further simplify this and say this is 2 times the square root of 5. so all we're saying is that this is preferred to this because we're taking out the perfect square which was 4 that was a factor of 20 right we wrote the square root of 4 as 2. all right so the next one is a little confusing we say the radicand has no fractions so this ties into your quotient rule for radicals so let's say you saw something like the square root of 5 over 36 so this radicand contains a fraction it also has 36 involved 36 is a perfect square so it's kind of two things you want to take care of so for this you would split it up into the square root of 5 over the square root of 36 so this would end up being the square root of 5 and 5 is not a perfect square so we can't do anything with that and the square root of 36 is just 6. so now we started with something that had a fraction in the radicand we just have the square root of 5 over 6 and my radicand here of 5 does not contain a fraction all right so the next one we have no denominator contains a radical so if you see a denominator that contains a radical most of you remember this from algebra 1 you have to use a process known as rationalizing the denominator we're not going to get to that in this lesson so we'll cover that kind of later on but it's a very very simple process if i end up with something like let's say 1 over the square root of 2 i just basically multiply the numerator and denominator by square root of 2 and what happens square root of 2 times square root of 2 is what that becomes 2. then the numerator is 1 times square root of 2 that's square root of 2. so all they're saying is that versus having it in this format 1 over square root of 2 we'd rather see square root of 2 over 2. we don't want a radical in the denominator here there's a radical in the denominator here it's radical free in the denominator so this is preferred to this although they are mathematically equal to each other all right then the last one we'll talk about there can be no common factor between the index of the radical and the exponent in the radicand so in other words if you saw something like this let's say you saw i don't know the fourth root of let's say 4 squared so for this one you could just convert it into exponent form and you can say you have what you have 4 squared that's raised to the power of 1 4. now power to power rule tells you you have what you have 4 raised to the power of 2 times the 4th which is 2 4 and you can see what they're talking about you don't want to have a fraction there that has a common factor right because you can simplify that two-fourths is the same as one-half so this is four to the power of one-half or in other words it's the square root of four which is two all right let's just run through some examples here real quick again this is something we talked about in algebra one for most of you will you'll remember this it just comes back to you very very quickly but others who've never seen it it's something you just pick up okay it's not a very challenging process so we have the fourth root of x squared again for something like this you just write everything in exponent form you'd have x squared and then it's raised to the power of 1 4. so in this form again if i have x that stays the same and then we multiply two times a fourth we know that's two fourths and we just did this in the previous example where two-fourths was reduced to one-half so this is x to the power of one-half or if you wanted to you could write the square root of x what about something like the 20th root of y to the fifth power again let's write y to the fifth power raised to the power of 1 over 20. again this index here becomes your denominator okay so that's how i got that so now again power to power rule y stays the same you'd multiply 5 times 1 over 20 is 5 over 20. of course 5 and 20 are each divisible by 5 so i could reduce that 5 divided by 5 is 1. 20 divided by 5 is 4. so this is y raised to the 1 4 power or again this is the fourth root again that 4 comes from that denominator that's where i'm getting it so it's the fourth root of y all right so now let's take a look at the cube root of 320 and we just want to simplify this so what we're going to do is we're going to factor 320 and we're going to look for perfect cubes right because this is a cube root if i had a fourth root i'd look for a perfect fourth if i had a square root i'd look for a perfect square right so on and so forth so factoring this what can you tell right away it ends in a zero so when it was divisible by 10. so this would be 32 times 10. now without going any further a lot of you remember that 32 is what it's 2 to the fifth power 10 is 2 times 5. so really 2 to the fifth power times 2 is two to the sixth power so this is two to the sixth power times five now let's just stop for a minute let's write out two to the sixth power so it's two times two times two times two times two times two so i can put this into three groups of two times two right because this would be a group and this is four this would be a group and this is four and this would be a group and this is four that's exactly what i'm looking for i'm looking for a perfect cube and the perfect cube would be 4 times 4 times 4 which is 64. so i can write this as what the cube root of 64. multiplied by the cube root of 5. now the cube root of 5 is not going to be able to be simplified any further but what we're saying is that we could replace this with a 4. cube root of 64 is 4 so when i just write a 4 there then we put times the cube root of 5. so that's your simplified answer now another way to do this let's kind of erase this real quick and in my opinion it'd be a little bit quicker you could do this using exponents so let's say that you want to just start out by saying okay this is 320 raised to the power of one-third now if you realize that this is 64 times 5 and 64 is 2 to the sixth power you can say okay well i could write this as 2 to the sixth power multiplied by 5. this is all raised to the power of 1 3. again all i'm doing is just using rules that we learned back when we talked about exponents so i can further break this down into 2 to the sixth power raised to the power of 1 3 multiplied by 5 raised to the power of 1 3 i can use my power to power rule and say 2 to the 6th power is raised to the power 1 3 that would be 2 raised to the power of 6 3 then multiplied by 5 raised to the power of 1 3. now 6 over three is two so this would give me two squared times five to the power of one third we know two squared is four so this would be four times five raised to the power of one third and five raised to the power of one third is the cube root of 5. so this is nothing more than 4 multiplied by the cube root of 5 which is exactly what we just saw a minute ago so kind of either way you want to do this a lot of times i personally like to use exponents because they're a little bit quicker for me but if you want to do it the other way that's fine too again you're just trying to get the right answer alright as another example let's say we looked at the square root of p to the 8th power this is a good one to do with exponents so we would say we have p to the eighth power and this is raised to the power of one half right remember if you have the square root of something it's like having the radicand raised to the power of one half so using my power to power rule i'd have p raised to the power of eight and then that's raised to the power of one half so p stays the same we multiply eight times a half that's eight over two so this is p to the fourth power right p to the fourth power all right let's take a look at another one so we have 4 multiplied by the cube root of 16 x to the 8th power so let's go ahead and take 16 and factor we know this is what 4 times 4 we know 4 is 2 times 2. so we know for a perfect cube we would take 3 of these and we would say it's 8 times 2. so let's put that we have 4 multiplied by the cube root of eight multiplied by the cube root of two so i've separated those two then multiplied by then the next thing i come across is this x to the eighth power so to figure out the perfect cube there look at the exponent there the exponent is an eight the index here is a three so you've got to think about what's the closest thing to eight going down that's going to be divisible by three well that's going to be 6. so in other words i would split this up into the cube root of x to the sixth power times the cube root of x squared x to the sixth power times x squared is x to the eighth power so once this is all broken down we can further simplify this and this but then when we look at these two this one and this one we're not going to be able to do anything else so in other words you will have four multiplied by the cube root of eight is two and then the cube root of x to the sixth power all you really have to do is divide here you say six divided by three would be two another way to do that again if you're uncomfortable doing that just go ahead and write that off to the side using exponents you have x to the sixth power raised to the power of one-third you know this would end up being one x stays the same multiple exponents 6 times 1 3 is 6 over 3 and then 6 over 3 is 2. so this is x squared and again that's the reason why i asked the question what's the closest number to 8 again it's got to be going down it can't be going up we can't go into 9 or 10 or something like that we don't we don't have that so we have 8 of these what's the closest number going down that's divisible by 3. it's not 8 it's not 7 6 is divisible by 3. so that's kind of what you're doing in your head then the next thing we have we have this cube root of 2 which can't be simplified and this cube root of x squared which can't be simplified because you think about x squared there there's not enough x's to have a perfect cube right if i had x cubed that'd be a perfect q to be x times x times x but i just have x times x it's not enough to have a perfect cube so that can't be simplified so now all i do is multiply things together in the front 4 times 2 is 8 then times x squared so that was what we could take out then what's left we'd have the cube root of 2 times x squared or just 2x squared and just notice how inside here your radicand you can't simplify that any further 2 is not a perfect cube and 2 doesn't factor right it's just 2 times 1. x squared again as we just talked about is not a perfect cube so you can't do anything further with this so you just report your answer as 8x squared times the cube root of 2x squared all right for the next one we're going to look at one that's kind of tedious we have 4 times the 7th root of negative 384 x to the 10th power y to the fourth power z to the fifth power so what can we simplify right away so we have four multiplied by we have the seventh root of what let's deal with this negative i could put the seventh root of negative one multiplied by the seventh root of 384 and we can deal with that number part right now if we do a factor tree you would notice that the final two digits 84 are divisible by four so therefore the number is 384 is 4 times 96. now 4 is 2 times 2 96 is what it's also divisible by 4 it's going to be 24 times 4. 4 again is 2 times 2 24 is 4 times 6 4 is 2 times 2 6 is 2 times 3. so you've got 1 2 3 4 5 6 7 factors of two times one factor of three so this is two to the seventh power times three so let's write this as the seventh root of 128 which we know is two times the seventh root of three now we're not gonna be able to simplify this any further so we would just end up leaving that as it is then we deal with the variables here now if i have an exponent that is greater than or equal to this index here i want to pull enough out to where i can simplify so if i have x to the tenth and i have a seventh root what i want to do is i want to split this up into the seventh root of x to the seventh right because that would simplify then multiplied by the seventh root of x cubed x to the seventh times x cubed is x to the tenth that's all i did there then multiplied by you'd have the seventh root of y to the fourth you can't simplify that because four is smaller than seven right there's just not enough then multiplied by the seventh root of z to the fifth same thing the five is smaller than the seven so there's not enough the reason we could simplify the 10 is because it was greater than the 7. so i pulled 7 out that's what this is here and then 10 minus 7 is 3 so that's how i got this 3 here so now what can we simplify we can simplify this and this and this everything else we can't so if i think about the 7th root of negative 1 that's just negative 1. so you'd have negative 1 multiplied by you have a 4 there the 7th root of 128 is 2 and then let's do this one the seventh root of x to the seventh power is just x right because x times itself seven times is x to the seventh power so those are pulled out to the front and then we have everything else that we'll just combine so times we'll have the seventh root of three then times x cubed then times y to the fourth then times z to the fifth and the easy way to tell if you've simplified something when you have variables involved just look at the exponents again if the exponents are smaller than the index you're good to go right this is a three this is a four this is a five they're all smaller than seven as far as the number part goes you can look at 3 and say okay 3 is a prime number there's nothing i can do there so what i'm left with negative 1 times 4 times 2 is negative 8 then times x then times the 7th root of again three x cubed times y to the fourth times z to the fifth all right so the last thing i'm going to cover today we're going to talk about how we simplify products and quotients of radicals with different indexes so a lot of students don't even know this is possible but this is definitely something you can do so if i have something like the square root of 5 multiplied by the cube root of 7 what could i do here i can't just say oh you know square root of 5 times 7. sometimes you'll see students go okay well that's this is a 2 this is a 3 2 times 3 is 6. so it's the 6th root of 5 times 7. those are wrong okay what you need to do is convert this into exponent form so you basically have 5 raised to the power of one-half multiplied by 7 raised to the power of 1 3. then you're just going to get a common denominator goal that's all you need to do so if i have something like one-half and i have one-third the common denominator would be a six right i'd multiply this by three over three multiply this by two over two and i'd end up with what 3 6 and i'd end up with 2 6. so all i would do is write this as 5 and now it'll be raised to the power of 3 6 and then times you'll have 7 raised to the power of 2 6. so the index would be the same if i convert this back into kind of a radical i will now have the sixth root of 5 cubed multiplied by the sixth root of seven squared now i have the same index here now i can use my little trick i can say this is the sixth root of 5 cubed which is 125 multiplied by the sixth root of 7 squared 7 squared is 49. now this is a really big number in here but at least we can combine it this way so if you do 125 times 49 it's 6125 so we'd have the sixth root of 6125 all right let's take a look at one more example like this so suppose you had the fourth root of 2 multiplied by the 6th root of 3. again i would just write this in exponent form i would say this is 2 raised to the power of 1 4 multiplied by 3 raised to the power of 1 and then we just think about what between 1 4 and 1 6 the common denominator the least common denominator be more specific would be 4 is 2 times 2 6 is 2 times 3 so you basically have 2 times 2 times 3 or 12. so i would multiply 1 4 times 3 over 3 i would multiply 1 6 times 2 over 2. so 1 times 3 is 3 over 4 times 3 which is 12. over here 1 times 2 is 2 over 6 times 2 which is 12. so 3 12 and 2 12. so i would write this as we'll have 2 raised to the power of 3 12 multiplied by 3 raised to the power of 2 12. let's erase this let's erase this so now looking at it i could rewrite it i could go back into radical form this is going to be the 12th root of 2 cubed multiplied by the 12th root of 3 squared so this equals what we'll have the 12th root of 2 cubed is 8 multiplied by 3 squared which is 9 this end up being the 12th root of 72. hello and welcome to algebra 2 lesson 56 in this lesson we're going to learn about finding the distance between two points so again we have another topic that we covered back in our algebra 1 course so for most of you this lesson will just be a review if you haven't seen this topic before don't worry this lesson is going to be very very easy this concept is something you can pick up pretty much right away so again the purpose of the lesson today is just to be able to figure out the distance between any two points on a coordinate plane and basically the way we're going to do this is we're going to use something known as the distance formula now i know some of you will hear the term distance formula and you're immediately thinking about the distance formula we used with motion word problems right you're thinking about distance traveled equals the rate of speed times the time traveled so this distance formula has the same name but it performs a very different application okay so essentially you're going to pick one of the points and you're going to choose it to be x sub 1 y sub 1 you're going to pick another point you're going to choose to be x sub 2 y sub 2 and you're going to plug into this formula and it's going to spit out a distance between the two points very easy to use but before we kind of get to that i want you to understand where it comes from so basically it's a direct application of the pythagorean formula we can say the pythagorean theorem so that's the very famous a squared plus b squared equals c squared that you see on your screen so the pythagorean formula basically relates the lengths of the sides of a right triangle so you might not have taken a geometry class before in case you haven't basically a right triangle is a triangle with a 90 degree angle so this symbol right here tells you that you have a 90 degree angle or a right angle which is why we call this guy a right triangle so this is a right triangle okay so when you see this symbol you know that this formula is going to apply so with a right triangle we have two legs which are normally labeled as a and b now these are going to be the two shorter sides of the triangle so you see we have leg a here and leg b here then we have the longest side which is always opposite of this 90 degree angle this is known as the hypotenuse and this is always going to be labeled as c so essentially when we square each of the two legs and then sum those amounts it's always going to be equal to the hypotenuse squared so in other words the measure of leg a so from here to here if i square that guy that's where i get this a squared from and then i add that to the measure of leg b so from here to here that guy squared so that's where b squared comes from this is equal to or the same as the hypotenuse or c squared so from here to here okay so that's where the c squared comes from so a squared plus b squared equals c squared so how can we use this to basically find out the distance between two points on a coordinate plane well what's going to end up happening is if you know two of the sides you can use this formula to find the third side or basically what's unknown so let's say i had a point that was here and a point that was here so what's going to happen is i'm going to be able to find the measure of leg b and the measure of leg a the measure of leg b will just be a horizontal distance so from here to here okay so we're going to be looking at the x values for that we'll talk about that in a minute when we get to the kind of example and then the measure for leg a that's going to be this vertical distance so from here to here okay so that's going to be us looking at kind of the y values those are easy to find we can easily find the value for leg b we can easily find the value for leg a so we can plug that into the formula and solve for the unknown which is going to be c the hypotenuse or again in this case the distance between the two points all right so let's take a look at the example so we have the point 5 comma 0 and the point negative 3 comma 6. so let's start by just plotting those so we have 5 comma 0 and we have negative 3 comma 6. so if we want to plot these let's go to 5 comma 0 first so that guy we're just going to go 5 units to the right we're not going to move it all vertically so this is 5 comma 0. so let me label that this is 5 comma 0 and negative 3 comma 6. so i'm going to go 3 units to the left and 6 units up that's going to be right there so i'm going to call this negative 3 comma 6. so let me draw a line connecting the points so this line right here that represents the distance between the two points will be c or the hypotenuse now what i'm going to do to make a right triangle i'm going to take an x value from one of the points and i'm going to take a y value from the other so in other words i could take 5 from this guy and 6 from this guy so i could make a point 5 comma 6 that would complete my right triangle so if i go 2 5 and then up to 6. so that right there would give me a right triangle now i'm not going to do it that way because i can do it another way it's going to be easier for us to see so i'm going to take an x value of negative 3 and a y value of 0. so i'm going to go to negative 3 comma 0 on the number line it's going to be right here so that's another way you can kind of complete that guy so take an x value from one point and a y value from the other that's how you can get your kind of third vertex for the right triangle so let's kind of fill this in now okay so not perfect but you get the idea let me draw this symbol here real quick that tells us we have a 90 degree angle okay so i'm just going to go ahead and label this kind of horizontal measure from here to here i'll label that as b and i'm going to label this vertical measure from here to here i'm going to label that as a okay so how can we find the measure for b and the measure for a we we know that we're going to end up solving for c but b and a should be easy to find so for b what do i want this is a horizontal line so think about the x axis or the x values what is the x value here and what is the x value here well let me kind of write this point in here this is negative 3 comma zero so the x value here is negative three and the x value here is five so just think about this before we kind of do any procedure just think about this if i go from negative three to zero i've traveled three units and then if i go from zero to five i've traveled another 5 units so 3 plus 5 is 8. so that tells me the distance or the measure for b is going to be 8. we're not going to pull out a coordinate plane each time and count things we need a better procedure so basically if i'm looking at this distance here i want the difference in x values but because we want to guard against getting a negative answer right because we don't want a negative distance we can do this inside of absolute value bars so i could say that i have the absolute value of negative 3 which is the x coordinate here minus 5 which is the x coordinate there this is equal to the absolute value of negative 8 which is equal to 8 which is what i got or you could also do it the other way okay the order isn't going to matter so you can go ahead and say that you have 5 minus a negative 3 which is 5 plus 3. so the absolute value of that would be the absolute value of 8 which is also 8. okay now similarly when we look at kind of the measure of a now i'm looking at a vertical line so with a vertical line i'm thinking about y values okay because you're thinking about this as being parallel to the y-axis the y-axis is going up and down okay so now we're thinking about y values so here i have a 6 for a y-value and here i have a zero okay so for those two points again from here and here so i would do 6 minus 0 inside of absolute value bars and that would give me the absolute value of 6 which is 6 or i could flip that around and say the absolute value of 0 minus 6 which is the absolute value of negative 6 which is also 6. again the absolute value operation is just there to make sure that we don't get a negative distance okay so once we have the value for b and a we know that we can solve for c so let's go to a different page so again i'm just going to write that a squared plus b squared equals c squared so for a we know that this guy was 6. for b we know this guy was 8. so if i plug in there what's going to happen is i'm going to have 6 squared plus 8 squared is equal to c squared let me kind of move this up a little bit and i'll erase this we don't need this anymore kind of write that in so 6 squared is 36 plus 8 squared is 64. this equals c squared 36 plus 64 is 100 so you get 100 is equal to c squared now we would take the square root of each side to get c by itself so you take square root of this side you're used to doing this you'll say plus or minus the square root of this side so what you're going to see in a minute is one of these solutions you can throw up right so you'll have plus or minus 10 is equal to c okay so basically c is going to be equal to plus or minus 10. now does it make sense for c to be equal to negative 10. if i go back does it make sense for the distance from here to here to be negative 10 no because the distance can't be negative so when we do this guy we just need the principal square root right we don't need to think about any negatives there because that would basically be nonsense right so see there the distance between these two points is going to be 10. so that's going to be my distance so if i go back this guy right here if i wanted to label it i could say this is 10. we already know that b is 8 and we know that a is 6. so now that we've seen an example of this on the coordinate plane let's make it very very simple so all you really need is this formula here and i'll kind of explain where this comes from so if you have points x sub 1 y sub 1 and x sub 2 y sub 2 you just plug into this formula and you'll find the distance between the two so let's say we go with our original example so we had five comma zero and we had negative three comma six right those were our two points so let's label this first one let me kind of drag this over here so it's not in the white let's label this first one as x sub 1 y sub 1. let's label this second one as x sub 2 y sub 2. and again it doesn't matter which one your label is which okay you can plug in whichever one you want for either so my distance is going to be equal to the square root of so you have your x sub 2 which in this case is negative 3 minus your x sub 1 which is 5. this quantity is squared then plus you're going to have your y sub 2 which is going to be 6 minus your y sub 1 which is going to be 0 this quantity squared now before we do anything i want you to remember where we got this from this negative 3 minus 5 was what this was the horizontal distance if we go back from right here right it was the difference in x values so that's all we need to do we take the difference in x values now in this case we did it in absolute value bars but when we come here we don't see absolute value bars that's because we're squaring the result okay so if this was negative when we squared it becomes positive if it was positive when we squared it stays positive okay so that's why you don't need absolute value bars there so negative 3 minus 5 is negative 8 negative 8 squared is 64. so this is 64 here then here we have the difference in y values so 6 minus 0 so if we go back up again that's the same thing we did here we had a y coordinate of 6 we had a y coordinate of 0. we take the difference between the two again when you do it on this you take the absolute value right but when you get to the formula you don't need the absolute value operation because again you're squaring it okay so 6 minus 0 is 6 6 squared is 36. so at this point we have what we have our a squared plus our b squared and essentially if i erase this i could say that d squared which just took the place of c squared we just re-labeled it as d for distance okay so that's clear we'll just kind of change this back so if d squared is equal to this was our b squared plus our a squared well what happens is to get d by itself i just take the principal square root of each side right i don't need to go plus or minus because the negative solution will be thrown out you don't want a negative distance so at this point we pretty much just solve so 64 plus 36 is 100 and then the square root of 100 is 10. so d equals 10. so much much quicker to just use the formula but i want you to understand where it comes from so that in a pinch if you can't remember it you can go back through and say okay well if i had you know a squared plus b squared equals c squared okay well i know that one of these is going to be the difference in x values one of these is going to be the difference in y values you know so on and so forth you can derive the formula all right let's take a look at another example so we have negative 6 comma 3 and negative 8 comma 2. so let's just label this as x sub 1 y sub 1. let's label this as x sub 2 y sub 2 and what i'm going to do is just plug into the distance form so d is equal to the square root of so you have your x sub 2 minus your x sub 1 so negative 8 minus a negative 6 so that's plus 6 to make that better and this is wrapped inside of parentheses and squared and plus your y sub 2 which is 2 minus your y sub 1 which is 3. again wrap this out of parenthesis and squared so negative 8 plus 6 is negative 2 negative two squared is four so this is four two minus three is negative one negative one squared is one so this is four plus one or five so the distance between the two is going to be the square root of five okay so the distance is the square root of five let me make that a little a little bit better and this basically is an irrational number so to get an exact value you'd leave it in this form if you wanted to approximate it you could punch it up on a calculator and you get about 2.24 okay if you want to approximate it but again if you want an exact value go with just square root of 5. all right let's take a look at another one so we have negative 2 comma 4 and negative 4 comma 4. i'm just going to change the order up and i'm going to say this first point is x sub 2 y sub 2 and i'll say the second point is x sub 1 y sub 1 okay so i'm going to plug into my distance formula so the distance is equal to the square root of so for the formula you have x sub 2 which is negative 2 minus x sub 1 which is negative 4. so minus a negative is plus a positive so negative 2 plus 4 this is wrapped in parentheses n squared then plus we're going to have our y sub 2 which is 4 minus our y sub 1 which is negative 4 so minus a negative is plus a positive again this is wrapped in parentheses and squared so negative 2 plus 4 is 2 2 squared is 4. so this is 4 4 plus 4 is 8 8 squared is 64. so if we take the sum of these two values 4 plus 64 is going to be 68 so this is going to end up being the square root of 68. now you can simplify this because 68 factors in a 4 times 17. so what i could say is that this is the square root of 4 times 17 square root of 4 is 2 so i can say this is 2 times the square root of 17. all right let's take a look at one more so we have the distance between these two points we have 3 comma negative 5 and negative 3 comma 4. so let's label this first point as x sub 1 y sub 1. let's label this second point as x sub 2 y sub 2. so the distance between the two again the distance is equal to the square root of you've got x sub 2 which is negative 3 minus x sub 1 which is 3 okay that's wrapped in parentheses and squared then plus you've got your y sub 2 which is 4 minus your y sub 1 so minus a negative 5 would be plus 5. again this is wrapped in parentheses and squared so negative 3 minus 3 is negative 6 negative 6 squared is 36 so this is 36 and then 4 plus 5 is 9 9 squared is 81. so 36 plus 81 is 117. so this would be the square root of 117. this can be simplified one plus one is two two plus seven is nine so we know this guy's divisible by nine it's going to end up being 13 times nine so 13 times nine square root of nine is three so i can go ahead and write this as the square root of 13 as my answer hello and welcome to algebra 2 lesson 57 in this video we're going to learn about adding and subtracting radical expressions so again most of this chapter where we're talking about radicals is nothing more than a review from algebra one but again it's something we want to have a refresher on before we move into anything more challenging so we're going to start by talking about this topic of adding and subtracting radical expressions so we have here that we add and subtract radical expressions using the distributive property so this is something similar to when we combine like terms so for example if we had something like let's say 2x minus 5x what would we do well we would work with the coefficients we would say 2 minus 5 is negative 3 and then times that common variable which is x now we had some conditions with this we couldn't say have 2x squared minus 5x and combine those because those are not like terms right so we can't combine them but if we had something like 2x squared minus 5x squared then we could combine them it's got to be the same variable which in this case it's x and x it's got to be raised to the same power this is x squared this is x squared so when that occurs the variable just stays the same so you have x squared and then you just work with the coefficients 2 minus 5 is negative 3. so i did the same thing for this and then for this now i'm going to apply that same logic up here if i have 2 times square root of 7 minus 5 times square root of 7 the square root of 7 part is just going to stay the same it's the same index right a square root has an index of 2 and it has the same radicand the radicand is 7 in each case so that just stays the same then we just work with what's multiplying the radicals so 2 is multiplying the square root of 7. 5 is multiplying the square root of 7. so we just do 2 minus 5 which gives us negative 3. so you'd end up with negative 3 times the square root of 7. another way to kind of look at this if you wanted a more kind of technical approach to it you could factor out the square root of 7. so i could say let me factor out the square root of 7 from each so i'd have the square root of 7 out in front and what would be left inside we'd have this 2 here then minus you'd have this 5 here and then 2 minus 5 is negative 3 and it's multiplying this square root of 7. so this is kind of the more technical approach probably what you'll see in a textbook but really all you need to do as long as the index is the same and the radicand is the same all you really need to do is work with the numbers that are multiplying that common radical again in this case it's just 2 minus 5 which gives us negative 3 and we multiply by that common radical which is square root of 7. all right let's take a look at another one so let's say we run across negative 9 times square root of 15 plus 11 times square root of 15. again i've got the same index it's a square root in each case and i've got the same radicand i've got a 15 and i've got a 15. so this part is going to stay unchanged we don't need to do anything with this we just work with the numbers out in front so it would be negative 9 plus 11 that would give us 2 and then times that common radical the square root of 15. now as i alluded to for this to work we've got to have like radicals like radicals have the same index and the same radicand now again we talked about something similar to this when we're working with polynomials remember we had like terms same variable raised to the same power well for like radicals it's the same index with the same radicand so let's say we see something like this 4 times square root of 5 plus 8 times square root of 7. can i do the same technique to simplify here no i cannot i have a square root in each case so the index is the same but my radicand here is a 5. my radicand here is a 7. so these are not like radicals right so not like radicals here [Music] and it's the same thing as when we saw something like you know 4x squared plus 5x cubed those are not like terms so we can't do anything to simplify there all right here's another example so we have negative 3 times the cube root of 7 minus 2 times the fourth root of 7. so in this case the radicand is the same so you've got a 7 here and a 7 here but the indexes are different this is a 3 right is a cube root this is a 4th root so these are not like radicals so these are not like radicals and again this is the same thing as when we came across non-like terms when we're working with polynomials so we're not going to simplify this any further than that all right let's look at some problems where we can do some simplifying and we'll come across some things where we have to simplify before we can do addition or subtraction so we have 7 times the square root of 7 plus 8 times the square root of 7. so again all i need to do is look at the numbers that are multiplying the radicals so we have 7 plus 8 that's 15 times that common radical that's the square root of 7. all right what about negative 8 times the square root of 10 plus 5 times square root of 10. again all i'm looking at is negative 8 and 5. so negative 8 plus 5 is negative 3 and then times the square root of 10. all right now we have 2 times square root of 24 minus 8 times square root of 24. so 2 minus 8 would give me negative 6 and then times the square root of 24. now you don't want to report your answer in this format because y the reason is if you look at the square root of 24 this is not simplified we know that if we factored 24 we could get a perfect square from 4 right 24 you could make is 4 times 6 4 is a perfect square so we want to simplify this so let's make this negative 6 times square root of 4 times square root of 6 this would give us negative 6 times the square root of 4 is 2. so i could replace that with a 2 negative 6 times 2 would give me negative 12 and then times the square root of 6. now this is simplified because the square root of 6 i can't do anything else with that 6 factors into 3 times 2. no perfect squares in there so we're basically good to go all right for the next one we have 2 times square root of 20 minus 3 times square root of 5 plus 2 times the square root of 5. so you might look at this and say well you've got the same index in each case but i have a 5 is a radicand here and a 5 is a radicand here but a 20 as the radicand here so i can't combine this with any of these before you do that make sure that you factor things because if i look at 20 we know it's what it's 5 times 4 4 is a perfect square so that can come out so really what i can do is i can say this is 2 times the square root of 5 times the square root of 4 minus 3 times the square root of 5 plus 2 times the square root of 5 this is equal to we know that the square root of 4 is 2. so if this is 2 you'd have 2 times 2 which is 4 times the square root of 5 minus 3 times the square root of 5 plus 2 times the square root of 5. now that we've simplified these are all like radicals so i can just work with the numbers out in front so in other words i would say 4 minus 3 is 1 1 plus 2 is 3. so this would be 3 times that common radical which is the square root of five all right what about something like this we have two times the square root of six minus three times the square root of five minus three times the square root of forty-five now square root of six i can't simplify any further square root of five i can't simplify any further but square root of 45 we can do something with that for the square root of 45 45 is 9 times 5. 9 is a perfect square so what i can do we can write 2 times the square root of 6 minus 3 times the square root of 5 minus 3 times the square root of 9 times the square root of 5. the square root of 9 is 3. so you basically have 3 times 3 which is 9. so let's rewrite this as 2 times the square root of 6 minus 3 times the square root of 5 and then minus we said 3 times 3 would give us 9 times the square root of 5. although i don't have like radicals everywhere i can combine these right these are like radicals so what we can say is we'd have 2 times the square root of 6 negative 3 times square root of 5 minus 9 times the square root of 5 you just take negative 3 and subtract away 9 that would give us negative 12 and then it's times that common radical which is the square root of 5. so we end up with 2 times square root of 6 minus 12 times the square root of 5. all right let's take a look at another one so we have 2 times the square root of 20 minus the square root of 45 plus 3 times the square root of 54. so i can simplify this this is what 5 times 4. we can simplify this this is 5 times 9. and we simplify this this is 6 times 9 and so there's a perfect square as a factor of each of these so let's go ahead and simplify so we would have 2 times the square root of 5 times the square root of 4 minus the square root of 5 times the square root of 9 plus 3 times the square root of 6 times the square root of 9. so we know that this is 2 this is 2. we know that this is 3 right this is 3. and we know again that this right here is 3. so what do we get you'd have 2 times 2 which is 4 times the square root of 5 minus you'd have 3 times the square root of 5 plus you'd have 3 times 3 which is 9 times the square root of 6. so although we can't combine everything we can perform this operation here so we're going to do 4 minus 3 which would give us 1. now if one is in front i don't need to write it i could put 1 times the square root of 5 or i could just put square root of 5 right 1 times anything is just itself then plus we're going to have 9 times the square root of 6. now i can't simplify this any further these are not like radicals so this is just my answer all right so let's take a look now at 3 times the cube root of 24 minus 3 times the cube root of 54 minus 2 times the cube root of 16. so now we're working with cube roots so you think about 24 you're looking for a factor that would be a perfect cube right a number that when multiplied by itself three times gives you that number back so when i think about 24 are there any factors that would be a perfect cube well yeah eight would be right eight is two cubed so eight times three fifty-four let's think about that 54 has 27 right so 54 is 27 times 2 and 27 is what it's three cubed three times three is nine nine temperature is 27 and 16 is what it's eight times two 8 is a perfect cube again it's 2 cubed so let's go ahead and simplify we'll have 3 times the cube root of 8 times the cube root of 3 minus 3 times the cube root of 27 times the cube root of 2 minus 2 times the cube root of 8 times the cube root of 2. so we look at what we can do here we know that this would simplify it would be 2. we know that this would simplify right here it would be 3. and again we know that this would simplify it would be 2. so let's go through and crank this out 3 times 2 would be 6 so you would have 6 times this cube root of 3. then minus 3 times 3 is 9 times this cube root of 2 then minus you'd have 2 times 2 which is 4 times the cube root of 2. now we don't have like radicals everywhere but what we can do is we can combine these two because you have cube root of 2 and cube root of 2. so you start out by just writing 6 times the cube root of 3. then you have negative 9 minus 4 which is negative 13 or minus 13 times that common radical which is the cube root of 2. all right for the next one we have negative 2 times the cube root of 162 minus 4 times the cube root of 48 minus 4 times the cube root of 5 minus the cube root of 6. so let's take a look at 162. we would first want to factor that so what would this factor into well it's divisible by 2 so let's just start with that if i divided this by 2 i would get 81. so 2 is prime 81 is what it's 9 times 9. we know 9 is what it's 3 times 3 so we do have a perfect cube here we have 3 times 3 times 3 which is 27 times what's left which is 3 times 2 or 6. so let's factor this as 27 times 6 so we'll say that this equals negative 2 times the cube root of 27 times the cube root of 6. then minus you have your 4 out in front now for 48 we could factor that as 8 times 6. we already know that 8 is a perfect cube so let's do the cube root of 8 times the cube root of 6 then minus you have 4 times the cube root of 5 then minus the cube root of 6. so we know that we can simplify this guy right here this is 3. we know we could simplify this guy right here this is 2. so what would we get if we do that negative 2 times 3 would be negative 6 then times the cube root of 6. then you have minus we know this is 2 so you'd have 2 times 4 that's 8 then times the cube root of 6 then minus 4 times the cube root of 5 then minus the cube root of 6. now what can we combine when we think about like radicals we have the cube root of 6 the cube root of 6 and the cube root of 6. so i'm just looking at what's multiplying in front so negative 6 negative 8 and then this is understood to be negative 1 right so it's just like if i had a negative x this is understood to be a negative 1 times x same thing here if i have a negative out in front it's like having a negative 1 multiplying that so i would have negative six minus eight that's negative 14 then minus another one that's negative 15 times this cube root of six then we would have minus 4 times the cube root of 5. and then these are not like radicals so we can't simplify this any further all right so now we have negative 2 times the cube root of 32 minus the cube root of 108 plus 2 times the cube root of 256 plus 4 times the cube root of 4. all right so we think about 32 we know that there's a perfect cube there 32 is 8 times 4. right again 8 is a perfect cube so we put negative 2 times the cube root of 8 times the cube root of 4 then minus for the cube root of 108 we don't think about 108 that often so let's factor that guy so 1 plus 0 plus 8 is 9. so we know this number is divisible by 9. it's also divisible by 3. so if we went ahead and divided it by 9 we could do 9 times 12. then we could see that 9 is 3 times 3 and 12 is what it's 4 times 3 and you could stop there because you know you basically have 27 times 4. 27 is a perfect cube so we could say we have minus the cube root of 27 times the cube root of 4 then plus we'll have 2 times the cube root of for 256 we don't really need to factor it at this point you should know it's 2 to the 8th power so if you're thinking about a perfect cube you want an exponent that would be divisible by 3 okay so 8 wouldn't be so you start going down 7 wouldn't be 6 would be so you could break this up into 2 to the sixth power times 2 squared right 2 to the sixth power is 64. so you can put the cube root of 64. times the cube root of 2 squared which is 4. let's erase this then plus you'd have 4 times the cube root of 4. nothing i could really do with that all right so now we look at what we can simplify so this guy right here is going to be 2. so negative 2 times 2 is negative 4 and then times what's left which is the cube root of four then we can simplify this guy right here that's going to be three so you'd have negative three so minus three times the cube root of four then plus this guy right here the cube root of 64 is 4 so you'd have 2 times 4 which is 8 times the cube root of 4 then lastly you have plus 4 times the cube root of 4. so now we have like radicals everywhere so i'm just going to go through and see what we can combine we have negative 4 here and positive 4 here so essentially this would cancel with this right those would become 0. all i'm really left with is negative 3 plus 8 which is 5 times this common radical which is the cube root of 4. hello and welcome to algebra 2 lesson 58 in this video we're going to learn about rationalizing the denominator so for the majority of you you took algebra 1 before jumping into this algebra 2 course and i know most of you have seen rationalizing the denominator before but again when you get into algebra 2 or even college algebra there's just a lot of topics that you go over again so it's not something that you should be bored with take the time to again cover this again and make sure you have a complete understanding of it and if it's something you're seeing for the first time this is not something that's super difficult some of the examples will get a little bit challenging though all right so let's start off with something kind of simple so we have 5 times the square root of 3 over the square root of 2. so the first question we would ask is is this simplified so i would relate this to when we work with fractions if we had something like 3 over 12 we didn't report that right we wanted to simplify this and say okay well 3 divided by 3 is 1 12 divided by 3 is 4. so the simplified answer here would be 1 4. right this is simplified but mathematically these two have the same value okay they're both equal to 0.25 in decimal form it's going to be kind of a similar experience over here our rule is we don't want a radical in the denominator so here we have the square root of 2 which is a radical so in order to simplify this what we do is a little trick we're going to multiply the numerator and denominator by the square root of 2. now what's that going to do for us the square root of 2 times the square root of 2 is the square root of 4 and the square root of 4 is 2. so what we did is we took an irrational number like the square root of two let me write that this is irrational and we made it into a rational number which is two henceforth the term rationalizing the denominator we go from an irrational number that's a denominator to a rational number that's our denominator now i can't just do something to the denominator without doing the same thing to the numerator remember just like we work with fractions same non-zero number over itself is one so this has a value of one and if we multiply by one something is unchanged so you'd essentially have five times square root of three times square root of two so we'd have 5 times square root of 3 times square root of 2 which simplifies to 5 times the square root of 6. so some of you will say well how come there's a radical in the numerator why is that allowed the reason for this is when this rule was invented on how to simplify a radical they didn't work with calculators now i can take this guy right here and this guy right here and i can punch it up on a calculator in an equal amount of time if i didn't have a calculator this one right here on the left this guy would take longer to calculate than this guy now the rule has still stuck around and you're expected to know how to do this in almost every math class so it's something you just need to learn how to do but mathematically this guy right here and this guy right here they have the same value it's just like when we looked at 3 12 and 1 4 right they are the same mathematically one is simplified one is not all right let's take a look at another so suppose you see something like this 4 times square root of 9 over 2 times square root of 6. well the first thing is you know you can cancel this 4 with this 2 right 2 divided by 2 is 1 4 divided by 2 is 2. so this would turn the problem into 2 times the square root of 9 over the square root of 6. now the next thing is the square root of 9 is 3. so we can go ahead and simplify that and say we have 2 times 3 which is 6 over the square root of 6. okay now in order to report a simplified answer we don't want a radical in the denominator so again we're going to use our little trick we're going to multiply this by the square root of 6 over the square root of 6 and what's going to happen is you'll have 6 times the square root of 6 over square root of 6 times square root of 6 is what it's 6 right it would be the square root of 36 square root of 36 is 6. now between the numerator and denominator i can cancel a common factor of 6. this is multiplication here so i'm canceling common factors and what we're left with is simply the square root of 6. all right let's take a look at another one so we have 4 minus the square root of 5b squared over 5 times the square root of 15 b cubed so a lot of things that simplify first you always want to think about what can i simplify so we have 4 minus now this right here you think about b squared the square root of b squared is nothing more than b again a quick way to do that is with exponents if i have b squared and i'm taking the square root of that so this is like raising something to the one-half power you do power to power rule here you have b raised to the power of 2 over 2 which is just b to the first power or b what's going to happen is i'm going to pull a b outside and it'll be times the square root of 5 and this is over down here i have 5 which is multiplied by you've got 15 b cubed now 15 is 5 times 3 so i can't really do anything with that so that would stay underneath but with b cubed again if this guy right here is equal to or greater than the index you know you can simplify that right i can separate that up into b squared multiplied by b so if i wanted to go kind of a long way about this i could say this is the square root of b squared times the square root of b this guy right here is b so that could come out and i could erase this and if i'm multiplying by the square root of b i could just write this underneath so we've simplified that part as much as we can but again in the denominator there is a radical so i'm not going to multiply by the entire denominator i'm just going to multiply by what i need to to get rid of that radical so let's start that down here we have 4 minus b times the square root of 5 over 5b times the square root of 15b now what i'm going to do i'm going to multiply the numerator denominator by this guy right here so the square root of 15b over the square root of 15 b now there's a minus sign here if i'm going to do this properly i'm going to use my distributive property this is going to be multiplied by this and then it's going to be multiplied by this so you'd have the square root of 15 b multiplied by 4. so 4 can just go out in front then minus i've got b times the square root of 5 to the square root of 15 b so b is out in front times the square root of 5 times the square root of 15 b and this can be simplified further but we'll do that in a minute so this is over now down here i've got this 5 b multiplied by you have the square root of 15 b times the square root of 15 b which is just 15 b okay so let's further simplify now so in the numerator we have 4 times the square root of 15 b nothing i can really do with that 15 is 5 times 3 b is just b can't really do anything then minus then for this guy right here i have b times the square root of 5 then i'm going to break this up into the square root of 5 times the square root of 3 times the square root of b now square root of 5 times square root of 5 is 5. so i might as well just erase this and put 5 and then there's really nothing else i can do with the rest of that so if i want to clean this up i can put minus 5b minus 5b times the square root of 3b times the square root of 3b then this is over you've got 5b times 15b really all i can do is say that's 75 75 b squared the last thing i want to do is see if there was anything that would cancel between numerator and denominator remember you've got subtraction here this is not multiplication so i would have to factor something from both terms in the numerator i'd have to factor something out to see if it would cancel here so when i look at this there's not really anything i could do that would make an impact some of you will say well in the numerator you could factor out the square root of 3b you could but there's nothing to cancel that with down here so this would be as simple as we can make this all right let's take a look at another one so we have 2 plus 3 times the square root of 5v over 2 times the square root of 20 v to the 4th power so in the numerator nothing really to simplify so just 2 plus 3 times the square root of 5v now in the denominator we have 2 times for the square root of 20 v to the fourth power we'll break that up into the square root of 4 times the square root of 5 times the square root of v to the fourth power so we can do some simplifying now 2 plus 3 times square root of 5 v over the square root of 4 is 2. so 2 times 2 would be 4. so let's put a 4 out in front then the square root of v to the 4th power is v squared so let's put a v squared next to that and then we're left with this square root of 5. so this is pretty easy to clean up i have a radical in the denominator that i want to kind of get rid of so all i'm going to do is multiply the numerator and denominator by that square root of 5. all right so what's going to happen again i've got to use my distributive property because i have this plus here so square root of 5 times 2 is just 2 times square root of 5 then plus the square root of 5 multiplied by 3 times the square root of 5v you can really say this is 3 times square root of 5 times square root of 5 is 5. so 3 times 5 would be 15. so let's just write 15. and then the thing that would be left would be the square root of v okay so times the square root of v then over down here square root of 5 times square root of 5 is 5. 5 times 4 v squared would give us 20 v squared you're looking if you want to cancel here because there's this plus sign you'd have to factor something out from both these terms here to be able to cancel it with something in the denominator we don't really have anything you have 2 and you have 15. you have square root of 5 and your square root of v so nothing really i can pull out there that i could cancel with that 20 v squared so this is simplified 2 times square root of 5 plus 15 times square root of v over 20 v squared all right so now let's talk about something that's a little bit more complex which is dealing with this when we have higher level roots so when we get to higher level roots we need to pay attention to what is needed we think about something like this we have 3 times the cube root of 4r squared over 5 times the cube root of 25r this is one thing that you really got to pay attention to so you're looking at this you're saying okay i want to simplify this a lot of students will just start out and go okay you know what i'm going to multiply this by cube root of 25 r over cube root of 25 r now 25 times 25 is 625 that is not a perfect cube it's a perfect fourth but not a perfect cube so again when you see something like this you want to factor things and kind of think about stuff so in the numerator there's nothing really i can do to simplify so three times the cube root of 4 r squared we know 4 is a perfect square but not a perfect cube r squared is a perfect square but not a perfect cube so we move on to the denominator we have our 5 and then we're going to break this up so times the cube root of 5 times the cube root of 5 5 times 5 is 25 and then times the cube root of r so you're thinking about how you could get a perfect cube there you've got 5 times 5 there you just need another 5 to get a perfect cube 5 times 5 is 25 5 times another 5 be 125 that's a perfect cube so when you do this you want times the cube root of 5 but you also have to think about this r here you have the cube root of r you need to make that a perfect cube also so if you just have r to the first power you're going to need to multiply that by r squared so that you can get r cubed and have a perfect cube so we would add under here r squared and you do the same thing in the numerator and the denominator so this is how you go about setting these things up especially when you get into these higher level routes you want to take the time to break things up and see what's needed if you don't you might end up costing yourself a lot of extra time or you might just outright get the wrong answer all right so let's go through and multiply so in the numerator we'll have 3 which is multiplied by the cube root of 4r squared times the cube root of 5r squared so this would be the cube root of 4 times 5 which is 20 times r squared times r squared which is r to the 4th power of course we can simplify this and we'll deal with that in a second over we'll have 5 multiplied by we'd have 5 times 5 which is 25 times another 5 which is 125 so it would be the cube root of 125 which is 5 so 5 times 5 and then you'd end up with the cube root of r cubed which is r so my denominator is radical free so in the denominator we would basically just have 25 r in the numerator you'd have 3 multiplied by 20 is what it's 5 times 2 times 2. there's no perfect cubes in there so you can just leave that 20. so we'll put the cube root of 20. multiplied by when we think about the cube root of r to the fourth power we know we can simplify this again if the exponent here is the same or greater than the index you can simplify you could split this up into the cube root of r cubed multiplied by the cube root of r this right here is just r so i could put an r there and i can get rid of this and then this r can go underneath here now before i conclude my problem this is multiplication here so i don't need to factor anything i can just cancel right i can cancel common factors this r with this r ended up with 3 times the cube root of 20 r over 25. all right for the next one let's take a look at 5 times the 4th root of 3x to the 4th power over three times the fourth root of 75 x cubed so when we look at this in the numerator if we have x to the fourth power we're taking the fourth root of that we know that's just x so we can simplify that to starting just say i have 5 x right i just pulled this out times the fourth root of 3. okay then this is over we have 3 multiplied by we have the fourth root of 75 x cubed now x cubed is not a perfect fourth so we can't do anything with that and 75 is not a perfect fourth but again we're going to think about what we need to do to get a perfect fourth so 75 is what it's 5 times 5 which is 25 times 3 and then x cubed is x times x times x so i know to get a perfect fourth here i just need one factor of x right that's it one more factor of x we'll be good to go there for this guy i need two factors of five and i need three factors of three so that's five times five which is 25 times 3 which is 75 times 3 again which is 225 times 3 one more time which is 675 so we'll need to multiply this by the fourth root of 675 and then x okay we'll do that to the numerator and denominator all right let's erase this and let's crank this out so we'll have 5x multiplied by the fourth root of 3 times the fourth root of 675x over you'll have 3 multiplied by the fourth root of 75 x cubed times the fourth root of 675x we know the fourth root of x to the fourth power would just be x so let's just do that first 75 times 675 ends up being 50 625 but we don't need to think about that because we built this by looking at the prime factors we remember it was 5 times 5 times 3 to get to 75. so i only used two prime numbers in this i used a five and a three so basically this number that we end up with is five to the fourth power times three to the fourth power or in other words it's 15 to the fourth power so the fourth root of 15 to the fourth power is just 15. so times 15 3 times 15 is 45 so let's write this as 45x now before we do anything else we notice that 5 would cancel with 45 this would be a 9 and x would cancel with x now what are we left with now if we look in the numerator we've got the fourth root of 3 multiplied by the fourth root of 675x so we know we're not going to be able to do anything with this but we think about 675 that came from two factors of five or 25 and it came from three factors of three or 27. so we look through here and we look for a perfect fourth let me put times the fourth root of x just so we don't forget that so we would have what one two three four factors of 3 or 81 let me just complete this i'm going to put over 9. so you would have the fourth root of 81 times the fourth root of 5 times 5 or 25 times the fourth root of x over 9. this right here has a value of 3. so you basically have a 3 here so i could cancel this with this and leave a 3. and as my answer is going to go we would end up with the fourth root of 25 x over 3. all right so for the next one we have 5 times the cube root of 2 n to the 4th power plus 3 times the cube root of 3n over 5 times the cube root of negative 6n so what we want to think about here we have 5 times let's break this up a little bit we know if we see an index of 3 and an exponent of 4 we can do something there so let's say we have the cube root of 2 times the cube root of n cubed times the cube root of just n plus 3 times the cube root of 3n over we have 5 times the cube root of i'm going to split this up into negative 1 times the cube root of 6n this can be simplified and this can be simplified so what we want to do we'll have 5 times the cube root of 2. this would end up being what this would end up being just n so i can just write this n out in front i'm just going to slide this down a little bit and then times you'll have the cube root of n and then plus you have 3 times the cube root of 3n and then over this again can be simplified the cube root of negative 1 is just negative 1. so i'd put negative 5 out here times the cube root of 6n well down here again we don't want this radical in the denominator so what we want to do we think about 6 6 is 2 times 3. so really i'm not going to get it any more simple n is just n so if i have 6n all i'd want to do is multiply this by 6 n that amount squared or 36 n squared that would give me a perfect key it would give me 216 n cubed which is a perfect cube so we'd multiply the numerator by the cube root of 36 n squared and the denominator by the cube root of 36n squared and what are we going to get again if i kind of combine this to make it a little easier put this n underneath then this would get multiplied by this so you'd have 5n out in front multiplied by the cube root of 2n times 36n squared 36 times 2 is 72 and times n squared is n cubed then plus this is going to multiply by this now so you'd have 3 out in front the cube root of 3n times the cube root of 36n squared you would have the cube root 3 times 36 is 108 and times n squared is n cubed all right then this is over down here you're going to have negative 5 out in front and again the cube root of 6n times the cube root of 36n squared is going to give you what it would be the cube root of 216n cubed which would basically be 6n so negative 5 times 6n would be negative 30n and now we're just looking to see can we simplify anything further so before we try to factor anything in the numerator let's think about 72 for a second 72 is what it's 8 times nine we know eight is a perfect cube it's two times two times two nine is not it's three times three so what we could do is we could write this as five n times the cube root of eight times the cube root of nine times the cube root and we think about this right here this n cubed the cube root of n cubed so these two can be simplified the cube root of n cubed is n so we can just make this n squared and the cube root of 8 is 2. so we can just make this 10. so we get 10n squared times the cube root of not then you're going to have plus over here we think about 108. now what is a hundred eight it's divisible by two it's also divisible by three so we know it's divisible by six so let's factor it and say if i divide 108 by 6 i would get 18. 6 is 3 times 2 18 is 6 times 3 6 is 3 times 2. you've got 1 2 3 factors of 3 and 2 factors of two so it's essentially 27 which is a perfect cube times four so we can say this is three times the cube root of let's just go ahead and do 27 n cubed times you'd have the cube root of 4. now again this part right here can be simplified the cube root of 27n cubed is going to be 3n so 3n multiplied by this 3 that's already here would give us 9n so this would be 9n multiplied by the cube root of 4. then over we'll have negative n now here's the question again because this is addition can i factor anything out and then cancel well number wise the answer to that is going to be no 10 is 5 times 2 9 is 3 times 3 nothing i can really do there but variable wise yes i have n squared and i have n so i could factor an n out and i'd have 10 n times the cube root of 9 plus 9 times the cube root of 4 over negative 30 n then this would cancel with this and what i'm left with we'd have 10 n times the cube root of 9 plus 9 times the cube root of 4 and then this is over we'd have negative 30. right so this would be our answer hello and welcome to algebra 2 lesson 59 in this video we're going to learn about rationalizing a binomial denominator so in the last lesson we reviewed how to rationalize the denominator and we looked at the simple scenario something like 6 over the square root of 2. so we should all know at this point if we see this this is not considered simplified because the rules tell us that we don't want a radical in the denominator now depending on what you get you might get a square root in the denominator cube root fourth root fifth root whatever it is you need to imply a little trick so with this guy right here since it's a square root i'm just looking to multiply by the square root of two and i'm going to do that to the numerator and the denominator that's what makes it legal right square root of 2 over square root of 2 is 1 multiply by 1 you leave something unchanged when we do this we will be radical free in the denominator square root of 2 times square root of 2 is 2. 6 times square root of 2 is just 6 times the square root of 2 and of course we can simplify this further 6 divided by 2 is 3 so this would end up giving us 3 times the square root of 2. but the main idea here is just to understand that this right here is considered simplified and it is preferred to this right here they are the same value they are equal to each other but it's just like when we saw something like one-half and two-fourths mathematically in decimal form these are both point five but we prefer one-half to two-fourths right one-half is considered simplified and it's the same thing with this scenario here now in this lesson we're going to look at some more complex scenarios the first thing we're going to do before we kind of jump into that we need a little back story here so we're going to kind of backtrack a little bit we're going to learn a definition and then we're going to kind of move ourselves forward and i'm going to show you how to do this when you have two terms in the denominator so when we have two binomials such as a plus b and a minus b if the first terms are identical and the last terms are identical and only the signs differ you have conjugates okay so here you have a and you have a here you have b and you have b and then your signs are different you have a positive and you have a negative you'll recall that we already worked with these before we just didn't refer to them as conjugates at the time when you multiply the two together using foil the outer and the inner cancel each other out and so you're just left with the first term squared minus the last term squared right we call the difference of two squares so if i had something like a plus b multiplied by a minus b you could do this the long way a times a is a squared the outer a times negative b is minus a b the inner b times a would be plus b a or plus a b however you want to write that and then the last b times negative b is minus b squared so again this becomes the difference of two squares you see that these two would cancel each other the outer and the inner terms cancel each other and you're left with the first term squared so you've got a squared minus the last term squared so you've got minus b squared there so we put equals a squared minus b squared and we know this from working with special products so let's look at some quick examples so without doing foil we already know that if we have x minus y times x plus y it's what again i've got x here and here i've got y here and here i've got different signs i've got a minus and i've got a plus so i would just say this is x squared right the first guy squared minus the last chi squared which is y squared okay the difference of two squares for this guy we have z plus four times z minus 4 so again if i multiply this out i get z squared the first guy squared minus 4 squared the last chi squared now you can write 4 squared or you could just write 16 because we know 4 squared is 16. all right so we look at x squared minus 9 times x squared plus 9. again this would be x squared squared right this is what we have here in the first position of each so it's x squared squared i just think about if i multiply x squared times x squared that would be what x to the fourth power or x squared squared then minus this guy squared 9 squared is 81. all right so let's apply this knowledge and let's look at an example here so suppose you see 4 over negative 5 minus 3 times the square root of 3. you can't just do this times square root of 3 over square root of 3 like we've been doing okay that's not going to leave your radical free in the denominator what you need to do is multiply the numerator and denominator by the conjugate of the denominator so what is the conjugate going to be again the first terms would be the same so i have a negative 5 here so for the conjugate i'd have a negative 5. the sines will be different if that's a minus this would be plus and then this guy would be the same 3 times square root of 3. so these two would be conjugates okay so what i'm going to do i'm going to multiply by negative 5 plus 3 times square root of 3 over negative five plus three times square root of three and i know some of you don't believe this but it's going to end up leaving you with a radical free denominator all right so let's take a look you just think about this putting parentheses around this four would get multiplied by negative five which would give us negative twenty and then plus four would get multiplied by three times the square root of three four times three is twelve so you'd have twelve times the square root of three there then down here you basically think about this as using foil but again because this is something we know already the outer and the inner they're going to cancel right i would do negative 5 times negative 5 that would give us 25 the outer negative 5 times 3 times the square root of 3 would be negative 15 times the square root of 3. the inner negative 3 times square root of 3 times negative 5 would be positive 15 times square root of 3 and then negative 3 times square root of 3 times positive 3 times square root of 3 would be negative 9 and then square root of 3 times square root of 3 is 3 so times 3 so this is basically negative 27. now again the outer and the inner are going to cancel these are opposites those are gone so you're left with 25 minus 27 and what does that give you it's going to give you negative 2. in the numerator we have negative 20 plus 12 times the square root of 3. so i'm radical free in my denominator now i'm not done simplifying but i am radical free now what else can we do well one thing is a lot of people don't like having a negative in the denominator so you could factor out not only a 2 but a negative 2 from the numerator so let's go ahead and do that if we factor out a negative 2 we'd have a negative 2 out in front and then inside of parentheses would have a positive 10 minus 6 times the square root of 3 and this is over negative 2. so what's going to happen is we're going to cancel this negative 2 with this negative 2 and we're left with our answer here so this is going to simplify into 10 minus 6 times the square root of 3. now nothing else i can do to make that any simpler let's take a look at another one again once you kind of get used to these they're very very easy not any more difficult it is a little bit more work because of the simplification but again it's conceptually not any more difficult so we have 2 over 4 times the square root of 2 minus 2 times the square root of 5. so again all i want to do is i want to multiply the numerator denominator by the conjugate of the denominator so this will be the same and this will be the same we just have to change the sign so we would have 4 times the square root of 2. instead of minus we're going to have a plus and then 2 times square root of 5. okay do the same thing for the denominator and we're going to multiply so 2 times 4 times square root of 2 would be what this would give us 8 times the square root of 2. then plus if we have 2 times 2 times the square root of 5 2 times 2 is 4 then times square root of 5. and then this is over down here again let's use our shortcut let's not go through and use foil we know it'll be the first guy squared so if i square 4 i get 16 if i square the square root of 2 i get 2. then minus it would be this guy squared if we square 2 we get 4 if we square the square root of 5 we get 5. so what we end up with is what we have 8 times the square root of 2 plus 4 times the square root of 5 over 16 times 2 is 32 minus 4 times 5 is 20. so we say this is equal to we'll have 8 times the square root of 2 plus 4 times the square root of 5 over 32 minus 20 is 12. i could simplify a little further because i could factor a 4 out from the numerator cancel with a common factor of 4 in the denominator so if we did that we'd have 4 times the quantity inside of parenthesis you'd have 2 times the square root of 2 plus you'd have the square root of 5. 5 over you could write 12 as 4 times 3 or you could just keep it as 12 doesn't really matter you can cancel this with a factor of 4 here so this would be 3 now and so our final answer would be 2 times the square root of 2 plus the square root of 5 and this is over 3. let's take a look at another one so we have 4 over negative 1 minus the square root of 3x to the fourth power now if you notice something that you can simplify like you have x to the fourth power under a square root symbol go ahead and simplify that first so let's say this is 4 over you have negative 1 minus let's write this as the square root of 3 times the square root of x to the fourth power so you should all know at this point that again if the index is 2 and the exponent in here is a 4 you just divide 4 divided by 2 is 2. in other words the square root of x to the fourth power is x squared because x squared squared would give you x to the fourth power so we can rewrite this and say we have 4 over negative 1 minus this would end up being again x squared so we'll put x squared out in front times the square root of 3. so again i have two terms for my denominator and all i want to do is multiply the numerator denominator by the conjugate of this denominator so negative 1 will stay the same and x squared times the square root of 3 will stay the same we just need to change the sign if this is a minus it will become a plus so negative 1 plus x squared times square root of 3. all right so let's put this inside of parentheses and let's see what happens 4 times negative 1 is negative 4. then four times x squared times square root of three would just be plus four x squared times square root of three then this is over four in the denominator here again we could skip through using foil and just use our shortcut we know we just square the first one so if i squared negative 1 we would get 1. then we'd have minus we would square this guy right here x squared squared is x to the fourth power and then the square root of 3 squared is 3 so times 3. so what is this going to be equal to we'll have negative 4 plus 4 x squared times the square root of 3 over we would have 1 minus and you could just write this as 3x to the fourth power now is there anything else i can do to simplify here well i could factor out a 4 or a negative 4 from the numerator but nothing i can really cancel that with in the denominator there's not really anything else we can do to simplify so we just report our answer as negative 4 plus 4x squared times the square root of 3 over 1 minus 3x to the fourth power all right let's take a look at another one so suppose we see 6a over you have the square root of 3a squared minus the square root of 2a squared again if you see stuff you can simplify go ahead and do so so we would have 6 a over if i just pulled this out remember the square root of a squared is just a so we would have a times the square root of 3 minus same thing here square root of a squared is a so a times the square root of 2. now again all i want to do is multiply the numerator denominator by the conjugate of this denominator so let's just rewrite this real quick we'll have 6 a over we have a times the square root of 3 minus a times the square root of 2. so we're going to multiply this by again the conjugate of the denominator so we're going to have a times the square root of 3 plus right instead of minus a times square root of 2. so what are we going to do we'll have 6a times a times square root of 3. so that would give us six a squared times the square root of three then plus six a times a times square root of two would be six a squared times square root of two and then this is over again for these we could use our special products formula so i'll take the first guy and square it so i'd have a squared times the square root of 3 squared is 3. then you'll have minus we take this guide squared we'll have a squared multiplied by the square root of 2 squared which is 2. now there is something i can cancel in every term of the numerator and in every term of the denominator you have an a squared so you can go ahead and factor that out so if we factored out an a squared we'd be left with 6 times square root of 3 plus 6 times square root of 2 over down here if we factor out an a squared we'll be left with 3 minus 2. so what does that give us this is going to cancel with this and so what we're going to end up with is 6 times the square root of 3 plus 6 times the square root of 2. now this is over 3 minus 2 which is 1. now i can put it's over 1 or i could just leave it like it is because anything over 1 is just itself so our answer is just 6 times the square root of 3 plus 6 times the square root of 2. all right again let's take a look at another one so we have 2 plus 8 times the square root of x to the 4th power over 3 times the square root of 3x cubed minus 6 times the square root of 5x so again if there's something you can simplify go ahead and do that before you start so we could simplify this we could simplify this and that looks like that's it so we'd have 2 plus 8 times what the square root of x to the 4th power is x squared so 8 x squared and if you wanted to you could reorder that and say you have 8x squared plus 2 doesn't really matter then this is over for this guy right here if we have the square root of x cubed we could say that we had the square root of x squared times the square root of x right we can break that up the square root of x squared is x so we can write this as 3x times you'll have this guy and you'll have this guy so the square root of 3 times the square root of x we'll just write that as the square root of 3x and then minus 6 times the square root of 5x now again once we've got to this step all we want to do is multiply the numerator and denominator by the conjugate of this denominator so we have 3x times square root of 3x minus 6 times the square root of 5x so we're going to multiply this by we'll have 3x times the square root of 3x instead of this being minus we'll put a plus and then 6 m square root of 5x this is over again 3x and the square root of 3x plus 6 times the square root of 5x we're going to use foil in our numerator so 8x squared times 3x times the square root of 3x you would do 8 times 3 that's 24. you would do x squared times x that's x cubed and then times the square root of 3x for the outer you would have 8x squared times 6 times the square root of 5x 8 times 6 is 48 so plus 48 and then you'd have x squared and then times the square root of 5x for the inner we have 2 times 3x times the square root of 3x so plus 6x times the square root of 3x and then for the last we have 2 times 6 times the square root of 5x that's plus 12 times the square root of 5x all right so then this is over in the denominator again i'm going to use my special products formula 3x times the square root of 3x squared so 3x squared would be 9x squared the square root of 3x squared would just be 3x so times 3x and if we do this calculation now we'd have 27x cubed so let's just go ahead and write that then minus from the second guy squared 6 squared is 36 and then the square root of 5x squared would just be 5x so 36 times 5 is 180. so we'll put minus 180 and then times x now is there anything i can do to further simplify here well if we look not everything has an x i'm missing that from right here right when we talk about in terms of being able to factor out an x and cancel so that's not going to work but everything is divisible by 3. so we could pull out a 3. from the numerator and from the denominator so that would leave me with 8x cubed times the square root of 3x plus 16x squared times the square root of 5x plus 2x times the square root of 3x plus 4 times the square root of 5x then this is all over if i pulled out a 3 i'd have 3 times you'd have 9x cubed minus 60x so let's go ahead and cancel this 3 with this 3 and what are we going to be left with we're going to have 8x cubed 8x cubed times the square root of 3x plus 16 x squared times the square root of 5x plus 2x times the square root of 3x plus 4 times the square root of 5x and this is all over we'll have 9x cubed minus 60x all right for the last problem we're going to look at one that's a little bit complex we get a lot of emails on this type of problem and it's something that when we taught this back in algebra 1 we saw a lot of questions on this because it's a very popular bonus question for a homework or for even a test so let's say you saw something like 4 over the square root of 3 plus the square root of 7 minus the square root of 6. you don't have two terms in this denominator you have three so how do we rationalize here well we're going to use the same concept but what we're going to do is we're going to group two of the numbers together so let's say we grouped the first two numbers together and let's just say we call this a then we have minus and then let's just call this one b so we have a minus b if i had two terms in the denominator and one was a and one was b and i had this minus sign here the conjugate of that would be a plus b so again with this being a this square root of 3 plus the square root of 7 i would just change the sign here this is minus so now it's going to be plus and then my b is the square root of 6. then this is over again you do the same thing so square root of 3 plus square root of 7 plus square root of 6. okay so let's do our multiplication we're going gonna see where we're at so four times the square root of three would be four times square root of three then plus four times the square root of seven is just four times square root of seven then plus 4 times the square root of 6 is 4 times square root of 6. then this is all over now you can go through this the long way and you could take let's say the square root of 3 multiplied by every term here then you could take square root of 7 and multiply by every term here you can take negative square root of 6 and multiply by every term here and then you can cancel all those things or again you can realize the shortcut i can take this and multiply it by this so in other words i can take this first part or will be labeled as a and i could square it so i could have the square root of 3 plus the square root of 7 squared then i could subtract away this second guy which is the square root of 6 squared right i could treat it again just like this was a and this was b so if i had a minus b times a plus b again that's what i would do right that's my special products formula no different here all right so let's simplify here so we have 4 times the square root of 3 plus 4 times square root of 7 plus 4 times the square root of 6. this is over we can use a special products formula for this as well i would have the first guy squared so the square root of 3 squared is 3 plus 2 times the first guy times the second guy so 2 times the square root of 3 times the square root of 7 would be 2 times the square root of 21 and then you would have plus the last guy squared so the square root of 7 squared is 7. now 3 plus 7 is 10. so you might as well just put this as 10 plus 2 times the square root of 21 then minus lastly i have the square root of 6 squared so if i square the square root of 6 i get 6. so 10 minus 6 is 4. so this is 4 plus this guy now you might say hey that didn't work we still have a radical in the denominator yeah we do but what is this this is a binomial denominator we already know how to do that so we took an extra step and now we're at a point where we can just multiply again by the conjugate 4 minus 2 times square root of 21 over 4 minus 2 times square root of 21. and again i know this is a very tedious problem but it's something that again if you get this as a bonus on a test it's really worthwhile to know how to do something like this so what we're going to do is we're going to take 4 times the square root of 3 and multiply it by 4. so that's going to give us 16 times the square root of 3 then we'll take 4 times the square root of 3 and multiply it by a negative 2 times the square root of 21 4 times 2 is 8 square root of 3 times square root of 21 is going to be the square root of 63. then next you'll have plus you have 4 times the square root of 7 times 4 that's 16 times square root of 7. then next we have 4 times square root of 7 times negative 2 times the square root of 21 that'll be negative 8 times the square root of 147 then we have plus 4 times square root of 6 times 4 that would be 16 times square root of 6 and lastly let me kind of rewrite this so i can fit it all on the screen so we have 16 times square root of 3 minus 8 times the square root of 63 plus 16 times square root of 7 minus 8 times the square root of 147 plus 16 times the square root of 6. just erase this and we'll kind of drag this up okay so the last one was 4 times square root of 6 times negative 2 times square root of 21 so that's negative 8 times the square root of 126. okay now this is all over again if i'm multiplying here and i have conjugates i just need to do the first guy squared so 4 squared is 16. minus the last guy squared 2 squared is 4 square root of 21 squared is 21. so what's 4 times 21 that's going to give us 84. so 16 minus 84. and if we do 16 minus 84 we'll get negative 68 so let's write negative 68 here you notice that the numbers that are multiplying these radicals are all even numbers you've got 16 you've got a negative 8 16 a negative 8 16 and a negative 8. now this is negative 68. so if we think about between numerator and denominator there would be a common factor of 4 or you could also do negative 4 just depends on what you want to do i prefer to cancel a common factor of negative 4 i just like to report my denominator as positive so in the numerator instead of 16 i'll have negative 4. so we'll put negative 4 out in front times negative 4 and square root of 3 and then this would become positive 2. so plus 2 times square root of 63 this would become negative 4 so minus 4 times the square root of 7. this would become positive 2 so plus 2 times the square root of 147. this would become negative four so minus four times the square root of six then lastly this will become positive two so plus two and the square root of 126. then this is all over negative 68. now if we cancel a common factor of negative 4 between numerator and denominator this will end up being positive 17 down here now we're still not done because there's more things that we can do so i have negative 4 times the square root of plus 2 times the square root of 63 is the square root of 9 to the square root of 7. we know the square root of 9 is 3. so you'd have 2 times 3 which is 6 times the square root of 7. then you have minus four times the square root of seven then you have plus two times the square root of 147. now let's stop for a second 147 we don't work with that very often but this is 49 times 3. now we know that 49 is what it's 7 times 7. so i could say we have the square root of 49 which is 7 times the square root of 3. so 7 times 2 would be 14 so this would be 14 times the square root of three then you have minus four times the square root of six then lastly plus two times the square root of 126. 126 is 2 times 7 times 3 times 3 or times 9. so 9 is a perfect square so again if we pull that out the square root of 9 is 3 so i could make this 2 times 3 or 6 times the square root of 14. okay now this is all over 17. all right so what do we have here that we can combine you've got a negative 4 times the square root of 3 and 14 times square root of 3. so those are like radicals so we could say that's ten times the square root of three then we've got six times the square root of seven minus four times square root of seven those are like radicals we say that's plus two times the square root of 7 then we have minus 4 times the square root of 6 nothing we can do with that and then lastly plus 6 times square root of 14 nothing we can do with that and then this is all over again 17. so that's going to be your final simplified answer nothing else we can do here and again i know it was a long problem i know it was very tedious but it's something you should get used to doing in case you get that as a bonus question on the test something you can do and make up for any problems you encountered that you couldn't solve so we've got 10 times the square root of 3 plus 2 times square root of 7 minus 4 times square root of 6 plus 6 times square root of 14 all over 17. hello and welcome to algebra 2 lesson 60. in this video we're going to learn about solving equations with radicals so when we start talking about solving an equation with a radical involved we need to think about a new rule so if both sides of an equation are raised to the same power so just as an example let's say i have squared both sides of an equation all solutions of the original equation meaning before i squared both sides are solutions to the new equation but the catch to this is going to be that it doesn't go in reverse each solution to the new equation in other words once i've squared everything if i solve that new equation it's not necessarily going to work in the original okay so we have to check for something known as extraneous solutions and these solutions do not work in our original equation so for you to understand this i wrote a little simple example here and this is very common in any textbook you're looking at you'll see something like x equals three and right now we know that if we replace x with anything other than three we would get a false state right i can only say that 3 can be replaced for x so i get 3 equals 3. all right so now that we understand that if we squared both sides of the equation let's just say this became x squared and if i square 3 i would get not what happens here well 3 is still a solution if i plug in 3 there 3 squared equals 9 that's true 9 equals 9. so it seems like there's no problems everything is good to go but remember if i square a negative it becomes a positive so this is where kind of information is lost okay we've squared both sides and we've lost the information on whether or not it was a positive or a negative so in other words what i could also have as a solution here is negative 3. so if i plugged in a negative 3 there and i squared it right the negative and the 3 would be squared this would also be equal to positive 9 so you'd get a true statement 9 equals 9. so with the equation x squared equals 9 there's two solutions your solution set would contain negative 3 and positive 3. but again with the original there's only one solution x here can just be three so if you've squared both sides to get to this and you end up with these two solutions you've got to plug them back in to the original equation and you've got to figure out if they work or they don't work you plug in a 3 you see that it works you know that's a valid solution you plug in a negative 3 you'd see that negative 3 does not equal 3. so negative 3 is an extraneous solution it's something you have to reject all right so let's look at the procedure to solve an equation with radicals basically all you want to do is isolate one of the radicals first for your easier problems the one you get introduced to you basically just have one radical and normally it's just a square root so with those you just push the radical on one side you push your number or whatever else on the other side and then you're going to raise both sides of the equation to a power equal to the index of the radical so in other words if it's a square root the index is a 2 so you square both sides if it's a cube root the index is a 3 so you cube both sides so on and so forth now with your harder problems the ones you're going to see at the end of the section or let's say you're doing a college algebra course or even in this algebra 2 course you're going to get problems where there's more than one radical in there so you might have to go back and isolate the radical again so we put here repeat the previous two steps if necessary sometimes you got to do it on a tougher problem sometimes you don't if you get something simple then you're just going to solve the equation and check all solutions in the original equation again you've got to check all solutions in the original equation okay it's very very important that you do that because if you don't again you might end up reporting some answer that works in your transformed equation okay your new equation but it doesn't work in the original right your extraneous solutions all right so let's start out with something very very easy we have that the square root of p plus 4 is equal to 5. now we're going to do this kind of as it sits now and then i'm going to show you kind of with exponents so that you understand both ways so i can again square both sides of this equation and what happens here the squaring undoes the square root so you can basically cancel those out and then what's left under the radical symbol your radicand is what's going to be there so you'd have p plus 4 is equal to if i square 5 we know we get 25. very very easy equation to solve if we subtract 4 away from each side of the equation we'd find that p is equal to 21. now let me just erase this real quick let's think about the original equation we were presented with if we plugged in a 21 there would we have a true statement 21 plus 4 is 25 the square root of 25 or at least the principal square root of 25 is in fact equal to 5 so you would get 5 is equal to 5. so this checks out right you can say your solution set contains just the element 21. now let me show you this using exponents and a lot of students will see this in like oh yeah i get it now okay this is a good way to kind of think about these if we know already that taking the square root of something is like raising it to the power of one-half what we can do is we can say that we have p plus 4 and inside of parentheses this whole thing is raised to the power of one half and this equals 5. now using what we know about exponents again if we square both sides here let me just put this inside of brackets and we're going to square this and we're going to square this what happens here well we're using our power to power rule if you're raising a power to another power you multiply the exponents so what is one half times two well that's one right these are going to cancel and it's going to be basically raised to the first power so in other words this would cancel with this and i'm just left with p plus four so i have p plus 4 is equal to 25 and again we solve that and we get that p is equal to 21. so just another way to kind of think about it if you see this on a test and you're drawing a blank on what to do you can convert your square root into raising it to the power of one-half and then it's obvious okay what do i need i need to square it because squaring something that's raised to the power of one-half would get me to an exponent of one all right for the next one we're going to look at the square root of 3x minus 16 is equal to the square root of 19 minus 2x so in this case we have a square root on one side and a square root on the other you just square both sides so i'm going to square this side and i'm going to square this side and what's going to happen well again this would cancel with this this would cancel with this and we're left with the radicand in each case so you basically just have 3x minus 16 is equal to 19 minus 2x now we know how to solve something like this very very simple we can add 2x to each side of the equation and we can add 16 to both sides of the equation so this is going to cancel over here and this is going to cancel over here 3x plus 2x is 5x and this is equal to 19 plus 16 is going to give us 35. we divide both sides of the equation by 5 we get that x is equal to 7. again we want to check to make sure that this is a valid solution so let's erase all this and we're going to plug in a 7 for x here and here and let's see if we get the left side equal to the right so we would have the square root of 3 times 7 is 21 21 minus 16 is 5 and this is equal to 19 minus 2 times x 2 times 7 is 14 19 minus 14 is 5 so you get the square root of 5 over here as well now the square root of 5 is an irrational number but you have the square root of 5 equals the square root of 5. same thing on the left and the right so we have a true statement here so our solution set for this equation just contains the element 7. all right let's take a look at a harder type problem so now we have 8 plus the square root of 34 minus 3b is equal to b what you want to do is you want to isolate your radical on one side so i want to move 8 over here so let me just subtract 8 away from each side of the equation and what's going to happen is you'd have the square root of 34 minus 3b this is equal to b minus 8. now i've isolated my radical so all i need to do now is just square both sides that's going to get rid of that radical so we'll square this side as well this will cancel with this and i just have my radicand so i just have 34 minus 3b and this is equal to now don't make this mistake we talked about this when we talked about polynomials i see this all the time you learn in polynomials and then you forget it you go back and you say well this is b squared minus 8 squared it is not you have to expand this right using foil or using your special products formula this is b minus 8 times b minus 8 okay it is not b squared minus 8 squared or b squared minus 64. that is wrong okay so we know the formula for this this is b squared minus 2 times b times 8 which is 16 b then plus this last guy 8 squared which is 64. and again this is from our special products formulas those are the things that again we talked about this because you learn that and then you get a problem like this later on and you're just like boom i got it you don't have to sit there and do foil every time so now that we have that let's say this is equal to 34 minus 3b let us add 3b to both sides of the equation so that's gone and let's also subtract 34 away from each side of the equation so that's gone so i'm going to rewrite this this would be zero over here because everything's canceled but i'm going to move all this to the left so we would have b squared negative 16 b plus 3b is going to give us negative 13 b and then 64 minus 34 is 30. so plus 30 and of course this equals 0. again all i did was i just move this to the left and i move that to the right now we have a quadratic equation we have in algebra 2 so far only talked about how to solve this equation if it's factorable so we only give you examples here where it's factorable but again in a few lessons we're going to start talking about the square root property the quadratic formula completing the square things we learned back in algebra 1 at the end so most of you can just do this with the quadratic formula if you want to do that that's fine at this point we're still doing it with factoring so if i factor this guy i'd have a b here and a b here two integers whose sum is negative 13 and whose product is 30. well that's going to be negative 10 and negative 3. so we solve this using again our zero product property or you could say your zero factor property and all you're doing is you're taking each factor this b minus 10. that's a factor and this b minus 3 that's a factor and you're setting them equal to 0 and you're solving so it's b minus 10 is equal to 0 or it could be true that b minus 3 is equal to 0. either one would work as a solution so let's add 10 to each side of the equation we'll get b is equal to 10. then or let's add 3 to each side of the equation we'll get b is equal to 3. now these are the solutions for again our transformed equation they're not necessarily going to work in the original so the solution set here is 3 comma 10. but let's go back up to the top and let's see if these work out all right so let's go ahead and check these so if i had a 3 plugged in for b you would get 8 plus the square root of 34 minus 3 times 3 is 9 is equal to 3. 34 minus 9 is going to give you 25. so the principal square root of 25 is 5 so you'd have 8 plus 5 which is 13 is equal to 3. well that's not true 13 doesn't equal 3. so we've got to reject that we're going to reject it it's an extraneous solution it works in the transformed equation but it does not solve or satisfy the original now let's try 10. so if we plugged in 10 here and here what will we get we'd have 8 plus the square root of 34 minus 3 times 10 is 30 is equal to 10. 34 minus 30 is 4 so you'd have 8 plus the square root of 4 equals 10. you can all see at this point that this is going to work square root of 4 is 2 so you'd have 8 plus 2 equals 10 and of course this leads to 10 equals 10. so this one does work as a solution so the solution set for this equation is going to contain one element just the element 10. okay so you could say b is equal to 10. let's take a look at another one i'm going to try one that's a little bit harder or not harder but just more tedious so i've got one radical that's isolated on one side on the other side i've got a radical plus a number so what we want to do is again just square both sides and you're not going to be radical free the first time you do this here so if i square this side this will cancel with this and we'll have the radicand which is 9k minus 2. then this is equal to over here i'm going to go ahead and use my special products formula you have to be just very careful about this this right here you're going to treat it just like you would let's say this is x and this is y okay you'll see what i mean so for that formula the first guy gets squared so it's 4 squared plus 2 times the first guy times the second guy 2 times 4 is 8 times the second guy which is the square root of 4 minus k then plus you've got the last guy squared so if i squared the square root of 4 minus k i'd be left with 4 minus k all right so now that we have that we want to do some simplifying the easiest thing is 4 squared we know that's 16. let's go ahead and just write that and then 16 plus 4 is 20. so we could say that we have 9k minus 2 is equal to again 16 plus 4 is 20 plus 8 times the square root of 4 minus k and then minus k now we know that we can add k to both sides of the equation and that will cancel we know that we can subtract 20 away from each side of the equation so that will cancel and so what we'll have is what we'll have 10k minus 22 is equal to 8 times the square root of 4 minus k now i have just a radical times a number on one side is everything divisible by 8 here because if i divided by 8 i can get rid of that out in front but 10 is not divisible by 8 and 22 is not divisible by 8 so i wouldn't do that here okay you can do it but i think it would cause more work because you can end up with some kind of nasty fractions to work with so let's go ahead and just square both sides of the equation and what are we going to get over here we're going to use our special products formula we know we would square 10k so that's 100 k squared then minus 2 times s times this 2 times 10 is 20 20 times 22 is 440 and then of course times k then the last guy is plus this guy squared 22 squared is 484 and this is equal to remember this is multiplication here so it's kind of like you're saying you have 8 squared multiplied by the square root of 4 minus k squared you just got to make sure that you raise both of these to the power of 2. so in other words you can kind of put this inside of parenthesis and say 8 again will be squared and then this guy inside of parenthesis the square root of 4 minus k that's raised to the power of 2 as well all right so let's continue simplifying we've got 100 k squared minus 440 k plus 484 is equal to 8 squared of course is 64. and then the square root of 4 minus k squared again this cancels with this so you can have this is multiplied by 4 minus k and this has to be in parentheses because again this guy right here this 8 squared is multiplying this whole thing okay so don't make the mistake of just saying oh we've got 64 times 4 that's not going to work okay you've got to multiply it by the 4 and by the negative k all right so let's continue here so we've got 64 times 4 which is 256 and then minus 64 times k which is just 64k so over here we still have our 100k squared minus 440k plus 484. all right so let's add 64k to each side of the equation that'll cancel let's subtract 256 from each side of the equation that'll cancel so we'll end up with 100 k squared and then negative 440 k plus 64k is negative 376 k and then 484 minus 256 would be plus 228 and this equals zero now again so far in algebra we've only talked about kind of doing these problems these quadratic equations with factoring the first thing you would notice is that everything is divisible by 2. 100 is divisible by 2 so is negative 376 so is 228. now the other thing is that everything is divisible by 4. so a cool little trick to work with smaller numbers we can divide both sides of this equation by 4. if i divide 0 by 4 i just get 0. so what does that matter right so this will still be 0. if i take a hundred and divide by four i get 25 so this would be 25 k squared then minus 376 divided by 4 is 94 so this would be minus 94k and then plus 228 divided by 4 is 57 so now we have 25 k squared minus 94 k plus 57 equals zero all right so with a quadratic equation again you can use the quadratic form if you want we haven't gotten to that yet so we're going to solve this by factoring here so i want two integers whose sum is negative 94 and whose product is 25 times 57 which is 1425. so if you factor this again we know this came from 25 times 57 this is 5 times 5 this is 19 times 3. so right away you know that you want two negatives right two negatives sum to a negative two negatives multiply to be a positive so thinking about that 5 times 5 times 3 is 75 it's 25 times 3 as 75. 75 plus 19 is 94. it's got to work out the signs right you just want all that to be negative so what you would want is two integers negative 75 and negative 19. so we're going to use those two to rewrite that middle term okay so we'll have 25 k squared minus 75k minus 19k plus 57 equals zero from the first group here we can pull out a 25k so we'll pull that up so inside we'll be left with k minus three from the second group here because the signs are in a different order right this is plus this is minus this is minus this is plus we're going to pull out a negative 19 and that'll leave us with k minus 3 inside and so we have this common binomial factor this k minus 3. that can be factored out so let's go ahead and take that out and what we're going to have is we'll have k minus 3 again we're factoring that out times what's left is going to be 25k minus 19 and this equals zero again zero factor property so you've got k minus three equals zero add three to each side you get k equals three or 25 k minus 19 equals 0 add 19 to each side of the equation you'll get 25 k is equal to 19 divide both sides by 25 and you'll get that k is equal to 19 20 fifths we need to check both of these in the original equation so k equals 3 where k equals 19 20 fifths all right so let's check again you want to check k equals 3 or k equals 19 20 fifths all right let's do 3 first that one's easier to do so if you plug in a 3 here and here you would have the square root of 9 times 3 which is 27 let's go ahead and write that 27 minus 2 which is 25 so the square root of 25 is equal to 4 plus the square root of 4 minus 3 is 1 so it's the square root of 1. square root of 25 is 5 equals 4 plus square root of 1 is 1 so 4 plus 1 is 5. 5 equals 5. so yeah this one's going to check out now the other guy over there that's going to be much more tedious but again on these you have to check them so it's not something you could just kind of skip through so we'd have the square root of nine multiplied by nineteen twenty-fifths minus two this should be equal to four plus the square root of 4 minus 19 20 fifths all right so over here 9 times 19 is 171 so this would be the square root of 171 over 25 then minus 2. now i want a common denominator going so i'm just going to multiply this by 25 over 25. so you might as well just say this is 50 over 25. so if i did this operation 171 minus 50 would be what that would be 121. so this would be the square root of 121 over 25 and then we know the square root of 121 is 11 we know the square root of 25 is 5. so you might as well just write this is 11 fifths so this is equal to you've got 4 plus the square root of for 4 minus 19 over 25 again if i get a common denominator going multiply this by 25 over 25 it's essentially a 100 minus 19 which is 81 so this would be the square root of 81 over 25. we know the square root of 81 is 9. we know the square root of 25 is 5. so this should be equal to 4 plus 9 5. so get a common denominator going here let's multiply this by 5 over 5 you'll get 11 fifths is equal to 4 times 5 is 20. so you end up with 20 plus 9 which is 29 over 5. so 11 fifths does not equal 29 fifths that's false so we have to reject that solution so let's go back up and let's reject this guy because it doesn't work in the original equation so your only solution here is going to be that k is equal to 3 or again using a solution set notation you just put 3 as your only element and so again that's why it's so important to check these and i know you get some of these that are going to be really really tedious but they throw those in there just so you get some good practice all right let's take a look at one that's a cube root now so we have the cube root of 31 minus x is equal to the cube root of x squared plus 11. so with this one it's very easy again if i have a cube root i want to raise everything to the third power or i want to cube it so what happens is this cancels with this right you can think about this as i have the one third power raised to the third power one third times three is one right so they just cancel each other out i'm left with just my radicand which is 31 minus x over here same thing this cancels with this i'm left with my radicand which is x squared plus 11. and so all we really need to do here let's add x to each side and let's subtract 31 away from each side that'll cancel becomes zero you will have that zero is equal to x squared plus x 11 minus 31 is going to give us negative 20. and we could rewrite this as x squared plus x minus 20 is equal to zero now again we're going to solve this using our zero factor property so let's factor this guy and see what we can get i know this is x and this is x two integers whose sum is one and whose product is negative twenty would be positive five and negative four now we'll set each factor equal to zero so you'll have x plus five equals zero or x minus 4 is equal to 0 if we then subtract 5 away from each side of the equation we'll get x is equal to negative 5. over here if we add 4 to each side of the equation we'll get that x is equal to 4. so these are the two proposed solutions all right so the two solutions we found were x equals negative 5 and then x equals positive 4. just going to plug them in like we have been so if i plug in a 4 there and there what will we get we'll have the cube root of 31 minus 4 which is 27 is equal to the cube root of 4 squared which is 16 plus 11 which is 27. so obviously the cube root of 27 does equal the cube root of 27 this is basically 3 equals 3. so this guy's going to check out the next one we want to try is negative 5 okay negative 5. so let's plug that in there and then there so 31 minus a negative 5 is like 31 plus 5. 31 plus 5 is 36 so you'd have the cube root of 36 then over here you have to be careful because you have negative 5 squared negative 5 squared is 25 25 plus 11 is 36 as well so we'd have the cube root of 36 equals the cube root of 36 same value on each side so this solution works as well so your solution set for this equation contains two elements we have negative five and four hello and welcome to algebra 2 lesson 61. in this video we're going to learn about complex numbers so the purpose of this lesson is to deal with the situation where you have to take the square root of some negative value and this comes up immediately in the next lesson when we start working with quadratic equations again a lot of you will recall back in algebra 1 we ended our course by learning how to solve quadratic equations using the quadratic formula but it doesn't matter what method you're using if you're required to take the square root of a negative value until you have the lesson where you learn about complex numbers you basically have to just stop the problem and say hey there's no real solution so let's talk a little bit more about why there's no real solution previously we learned that when we square a number the result is non-negative so we all know if we take something like negative 1 and we square it we're going to end up with 1 right positive 1. if we took negative 4 and squared we're going to end up with positive 16. so on and so forth the only thing you can have as a result of squaring something in the real number system is either 0 or some positive value but you'll never get a negative value from squaring again in the real number system now here's an example of something where we run into this problem if you see x squared plus one equals zero let's just start by subtracting one away from each side of the equation so that would cancel and we would have x squared is equal to negative one now what value can you think of that you could plug in for x that when it's squared would produce a negative one you can't right again if i have a real number and i square it the result is either going to be zero or it's going to be positive so using real numbers i cannot get a solution so up to this point we just stop and say there's no real solution but in mathematics we always find ways around things okay so we're going to introduce today the imaginary unit known as i and the definition i'm going to give you here is that i squared is equal to negative 1. now if we were to manipulate this let's say we have this i squared and this equals negative 1 if we take the square root of this side and the square root of this side what's going to happen is this would cancel with this and we'll just have i and this is equal to the square root of negative 1. so we can now use this property this definition to proceed when we want to take the square root of a negative value so let's look at an example of this let's say you see the square root of negative nine and again a lot of students get confused because they're like oh i already know how to do that that's equal to negative three no it's not okay don't get confused between the negative square root of nine and the square root of negative nine these are different this is just the negative of the square root of nine this is just negative three with this guy right here i'm taking the square root of a negative number this has no real solution but again if we define i to be equal to the square root of negative 1 then we can use a little trick we could say this is equal to the square root of negative 1 times the square root of not let me erase this real quick we can replace this right here with i because i is defined to be the square root of negative 1. so i'm going to write that out in front i'm going to put i out in front times the square root of 9 and the square root of 9 is just 3. so this is 3 times i or 3i that's all we're doing we've just found a workaround for this problem of taking the square root of a negative number now some people will write this as the square root of 9 times i like that the reason that i don't do that the reason that most textbooks don't do that is you don't want to make a mistake maybe this is extended a little too far and you think the i is under the square root symbol you don't want to run into that you don't want to make that mistake so to make it crystal clear that i is multiplying the square root of 9 we usually stick it out in front all right let's just look at a few practice problems just get our feet wet here so let's suppose you see the square root of negative 10. so what we're going to do is we're going to break this up into the square root of negative 1 times the square root of 10 and again we know the square root of negative 1 by definition is i so just replace it with i and then times the square root of 10. square root of 10 you can't do anything else with that 10 is 2 times 5 no perfect squares there so we just leave this as i times the square root of 10. then we have the square root of negative 20. if you see the negative you can just go ahead and replace it with an i out in front times what would be left would be the square root of 20 okay you don't need to go through this every time but just for the sake of completeness let's just keep doing it so we'll put the square root of negative 1 times the square root of 20. so again this is by definition i times the square root of 20 is what is 4 times 5. 4 is a perfect square and times the square root of 5. we know the square root of 4 is 2. so if we completely simplify this we'll have 2i because again the square root of 4 is 2 2 times i is 2i then times the square root of 5. but the next one we have the square root of negative 15. again the shortcut is just to take this say hey this is i and then times the square root of 15 and you'd be done but the long way again if you just want to realize what's going on here this is the square root of negative 1 times the square root of 15. the square root of negative 1 is by definition i then times the square root of 15. and what about the square root of negative 3. again this is the square root of negative 1 times the square root of 3. this is i by definition so this is i times the square root of 3. all right now we have the negative of the square root of negative 50. so be very very careful here you have a negative out in front so let's leave that out in front then i'm going to break this up into the square root of negative 1 times the square root of 50 we know is 25 times 2. 25 is a perfect square so let's break that up we'll put 25 under our square root symbol and times the square root of 2. and we know this has a value of 5. so let's just go ahead and replace it we know this is basically i so you would have negative i times 5 or negative 5 i and then times the square root of 2. all right so now let's talk about some kind of immediate applications of this we want to make sure that you're crystal clear on the product rule for radicals so when working with multiplication and division we must change the form first so here's what i mean by that once people know that they're allowed to take the square root of a negative number the first thing you start seeing on your tests when you give this out you'll see somebody go okay square root of negative 2 times square root of negative 18 that's going to be the square root of negative 2 times negative 18. this is the square root of 36 which equals 6. no that is wrong okay this is wrong okay the correct way to do this problem is to break it up section by section this right here again becomes i this right here becomes i so if we break it up and say we have i times the square root of 2 multiplied by i times the square root of 18 what do we get well i times i by definition is i squared which is negative 1. so let's just put i squared for now then times you'll have the square root of y let's break this down you'll have 2 times 18 is 2 times 3 times 3. so essentially we know that 1 2 would come out and 1 3 would come out 2 times 3 is 6. so what you'd end up with is 6 times i squared and i squared again by definition is negative one so this would be six times negative one which equals negative six now again if you did this the wrong way your sign would be incorrect if you did this like this you'd end up with positive six that's again not correct this is the correct way to do it and then this is incorrect okay this is this is wrong so you've got to make sure that you change the form first don't just go through and apply the product rule for radicals as you know it because that's not going to work when you're taking the square root of a negative the same thing is going to apply when we get to a division problem as well all right let's take a look at some examples so we have the square root of negative 8 times the square root of negative 4 times the square root of 2. so i can't just say hey i have the square root of negative 8 times negative 4 times 2. it wouldn't work out that way what i've got to do is change the form first before i do my multiplication so we all know if we see a negative we can just replace that with i mean we could just replace that with i so what we do is we say we have i times the square root of 8 would be what it would be the square root of 4 times the square root of 2 then you'd have times you've got i times the square root of 4 then times you've got the square root of 2. so what is this equal to well we know that this right here has a value of 2. we know this right here has a value of 2 and we know if we took the square root of 2 and multiplied by the square root of 2 those two together would have a value of 2. so what we're going to have is i times i or i squared then times 2 times 2 times 2 or 8. so this ends up being 8 times i squared i squared by definition is negative 1. so 8 times negative 1 which is negative 8. and again this is the correct way to do it if i had gone through and just multiplied and said hey this is going to be the square root of negative 8 times negative 4 times 2 i would have the square root of 64 which is positive 8 right i would not have ended up with negative eight so i would have gotten the wrong answer all right what about the square root of negative five times the square root of eight so again if i see this i'll replace it with i so this is i times the square root of five multiplied by for the square root of 8 we'll put the square root of 4 times the square root of 2. so we'll end up with i out in front times the square root of 4 is 2. so let's put 2 i out in front actually times the square root of 5 times square root of 2 square to 5 square root of 2 would be the square root of 10 so this would be 2i times the square root of 10. all right now let's take a look at a division problem so we have the square root of negative 100 over the square root of negative 4. so again i don't want to just go through and say hey this is the square root of negative 100 over negative four what we want to do we want to go ahead and set this up as i times the square root of a hundred again just taking that out and replacing with i over i times the square root of four just taking that out replacing with i so at this point you do have i over i same thing over itself it's still going to cancel okay still going to cancel so then you have the square root of 100 which is 10 over the square root of 4 which is 2 then 10 over 2 is going to give you 5. now what about the square root of 28 over the square root of negative 7. so for this guy right here what are we going to see the square root of 28 we'll write it as the square root of 4 times the square root of 7. and for this guy right here we have the square root of negative seven so let's take this out we'll have i times the square root of seven so we know that this would cancel with this and we know the square root of four is just two so this would be two over i now a lot of people will just stop and say hey that's my answer but remember i has a definition of what i is equal to the square root of negative 1. we do not allow radicals in the denominator so don't report an answer with i in the denominator because you're reporting an answer with the square root of negative 1 in the denominator so what you want to do you want to rationalize we want to rationalize the denominator so 2 over i we could just say we'll multiply by i over i what happens is i times i is i squared then you have 2 times i in the numerator i squared has a definition of negative 1 so you get 2i over negative 1 or negative 2i as your answer all right so let's talk a little bit about something known as a complex number so we're used to just the real number system now we're going beyond that we're talking about the complex number system so a complex number is formed using the imaginary unit i along with our real numbers so you'll generally see this in your textbook written just this way they say let a and b be any two real numbers and we'll see our complex number so this is known as a complex number and it's written as a which is a real number plus b another real number times i the imaginary unit so a is referred to as the real part and then bi is referred to as the imaginary part in some textbooks you will see that only b is described as the imaginary part your textbook might say that the textbook i'm using actually says that bi is the imaginary part it's basically irrelevant it's not something you want to argue over you just need to understand the structure of this so that when we start working with complex numbers we know hey this is how we can add or subtract them and we're going to this part right now so in order to add or subtract complex numbers we use our commutative associative and distributive properties so we kind of think about this the same way that we think about adding like terms so if i have something like 8 plus 2x and then i add 4 plus 7x we all know what to do we take 8 and 4 and add those together that's 12 and we take 2x and 7x we'd add those together that's 9x very very clear what to do it's the same thing here okay so i have eight plus four so let's group that together eight plus four and then we have two i plus seven i i can't combine 8 and 2i together okay that's not going to work it's just like trying to add 8 and 2x what we do here just to make this completely clear we had 8 and 4 that's 12. and then for this guy 2i plus 7i would be 9i to make that crystal clear you could factor out the i right that's common to each one there and say you have i times the quantity 2 plus 7 and so this would be 12 plus 2 plus 7 is 9 times that i there so you end up with 12 plus 9 i so it's just like adding like terms it's the same principle here now when we see it like this where we have a the real part plus bi the imaginary part this is known as standard form for a complex number so when you write something you want to write it that way you don't want to put 9i plus 12 you want to put 12 plus 9i let's take a look at another one we have negative 2 plus we have negative 2 plus i minus the quantity 6 minus 12 i so we have negative 2 and then plus negative 2 plus i this will again get distributed to each term so you'll have minus 6 and then minus a negative 12i would be plus 12i so negative 2 plus negative 2 is negative 4 and then negative 4 plus negative 6 is negative 10 and then we have i plus 12 i which is 13 i and again we wrote this in standard form this is a plus bi your real part plus your imaginary part let's take a look at one more of these we have 7 plus 4i plus negative 8 plus 8i plus 2 minus 2i so again if i just go through and say okay the real parts 7 plus negative 8 is negative 1 negative 1 plus 2 is positive 1. then your imaginary parts you've got 4i plus 8i minus 2i so 4 plus 8 is 12 12 minus 2 is 10 so this is plus 10 i and again this matches the a plus bi this is your standard form all right now let's talk about multiplying two complex numbers together we basically use the same methods we would use for polynomials so if we had a single term polynomial or a monomial multiplied by a two-term polynomial or a binomial what would we do we just use our distributive property you'd have negative 2i times negative 5 that'd be positive 10i and then you'd have negative 2i times negative 2i that would be positive 2 times 2 is 4 i times i is i squared now i squared has a definition of negative 1. so this is 10 i plus 4 times negative 1 which is basically 10 i minus 4. now this is not in standard form we want the real number first so we're going to switch the order and say this is negative 4 plus 10 i all right now we have something that looks like the product of two binomials so for something like this we can use foil we have negative 2 plus 6i that quantity multiplied by the quantity 6 minus 3i so if i look at the first terms negative 2 times 6 would be negative 12. if we look at the outer terms negative 2 times negative 3i would be positive 6i if we look at the inner terms 6i times 6 would be positive 36i and then the last terms 6i times negative 3i would be minus 18i squared so we have 6i and 36i that can be combined so we'll have negative 12. again 6 plus 36 would be 42 so plus 42 i and then minus 18 i squared so we're not done we have negative 12 plus 42 i here's the big thing i squared is negative 1 so this would be minus 18 times negative 1. so we know that negative 18 times negative 1 is positive 18. so essentially you'd have negative 12 plus 18 plus 42i negative 12 plus 18 is 6 so we end up with 6 plus 42 i again this is in standard form you've got your real part plus your imaginary part all right so let's take a look at one more of these and just like if i was multiplying three binomials together i would multiply two of them first find that product then multiply the result by the third so we have 3 plus 5i that quantity multiplied by the quantity negative 7 plus 5i multiplied by the quantity 3 plus 7i so let's start by multiplying these two together and what's that going to give us so for foil 3 times negative 7 is negative 21 the outer 3 times 5i would be positive 15i the inner 5i times negative 7 will be minus 35i and then the last 5i times 5i is 25i squared now we know i squared is negative one so let's just go ahead and write plus 25 times negative one and again this is times this guy right here we'll do that in a second let's just simplify first and then we'll multiply by this so this would be equal to what you'd have negative 21 25 times negative 1 is negative 25. so let's put minus 25 and then 15 minus 35 is negative 20 so minus 20 i so negative 21 minus 25 is negative 46. so let's erase this and we're going to write negative 46 then minus 20 i and this will be multiplied by 3 plus 7 i so we can use foil again here so negative 46 times 3 is going to give us negative 138 the outer negative 46 times 7i is negative 322 i the inner negative 20i times 3 is going to give us negative 60i and then the last negative 20i times 7i is going to be negative 140 i squared all right now remember i squared is negative one so let's just replace this with negative one and so what we'd have is negative 140 times negative one which is positive 140. so positive 140 plus negative 138 would give us 2. then negative 322 i minus 60i would be negative 382 i so as our answer in standard form we have 2 minus 382 i all right now let's talk about division now when you work with division you're going to need to think back to when you rationalized a binomial denominator okay we did that in a previous lesson so recall we talked about this thing known as conjugates so we have a plus b and a minus b those are conjugates right if you multiply them together you use foil you end up with a squared minus b squared right the middle two terms will drop out it's not going to be any different when you're working with complex numbers so here you'll have a complex conjugate so a plus bi times a minus bi those two are complex conjugates and the result is a real number now normally if you have a plus b multiplied by a minus b you get a squared minus b squared with this a plus bi multiplied by a minus bi you're going to end up with a squared plus b squared this is a big source of confusion here the sign is negative here it's positive why do you think that is it's because of this i times i which is i squared which is a negative 1 which changes the sign so if we went through and did this a times a is a squared the outer a times negative bi is minus abi the inner bi times a is positive abi and the last bi times negative bi is minus b squared i squared now we know already that this would cancel so what do we have we have a squared minus b squared i squared is negative 1 so times negative 1. all that's going to do is change the sign if you have a negative times a negative you end up with a positive so this gives me a squared plus b squared okay so that's where the difference in the sign comes in all right so the quotient of two complex numbers is a complex number if we start out with something simple like 7 over 3i we already talked about earlier the fact that i is what it's the square root of negative 1. so we don't want to leave it like that we want to multiply by i over i basically the same thing we've been doing to not have a radical in the denominator this is going to give us 7i over 3i squared again i squared is negative 1. so this is 7i over negative 3. and of course you could also write this as negative 7 i over 3 but these two answers are basically the same all right so what about something like 4 over 6 i well the first thing you would do is just reduce right they're each divisible by 2 so this would be 2 over 3i and then again because i represents the square root of negative 1 we just multiply by i over i we'll end up with 2 i over 3 i squared i squared is negative 1 so we'll end up with what 2i over negative 3 or negative 2i over 3. all right now let's take a look at one where we're going to need to multiply by the complex conjugate so we have 2 plus 6i over 4 minus 3i again i is the square root of negative 1. so really we could write this as 2 plus 6 times the square root of negative 1 over 4 minus 3 times the square root of negative 1. we don't want radicals in the denominator so we want to rationalize all right so if we want to do that what would we do again we worked with binomial denominators before so we multiply both the numerator and denominator by the conjugate of the denominator in this case is going to be no different so the terms stay the same 4 would be the same and 3i would be the same it's just that the sign is going to be different here it's a subtraction so this is going to be addition now in the denominator it's a breeze because we know the formula it's a squared plus b squared and again the reason it's plus is because of that i squared being equal to negative 1 it's going to change the sign so we'll go through this the quick way we'll go through it the long way let's do it the long way first so if we do foil that first two times four is eight then the outer two times three 3i is plus 6i the inner 6i times 4 is plus 24i then the last 6i times 3i would be plus 18i squared now we know this has a value of negative 1 so we can replace it but let's just continue for now in the denominator again the quick way is to realize that this follows a formula that's a squared so 4 squared is 16 plus b squared b in this case is 3. forget about the odd it's just the 3. so 3 squared is 9. so you'd have 16 plus 9 which is 25. now i'm going to do this the long way and you're going to see that this denominator down here is going to end up being 25 in the end so 4 times 4 is 16. the outer 4 times 3i is plus 12i the inner negative 3i times 4 is minus 12i and the last negative 3i times 3i would be minus 9i squared again this i squared here has a value of negative 1. so keep that in mind so we scroll down get a little room going if this i squared is negative 1 really i have 18 times negative 1 which is negative 18 negative 18 plus 8 would be negative 10. so this is negative 10 then plus 6i plus 24i is 30i and this is over you'd have 16 plus 12i minus 12i this is going to cancel and you basically have negative nine times negative one because again this is negative one negative nine times negative one is positive nine so you've got 16 plus nine which again is 25. so that's the long way to do it the short way again a squared plus b squared if you just remember that you can calculate that denominator very quickly now we're not done because we can simplify this further everything there is divisible by 5. so let's factor out a 5 we'd have negative 2 plus six i inside the parentheses over we'll write this as five times five we'll cancel a common factor of five between numerator and denominator and we'll report our answer as negative two plus six i over five or if you wanted to you could say this is negative two-fifths plus six-fifths times i either answer is acceptable all right for the next one let's use our little shortcut for the denominator we have negative 10 plus 7i over 1 plus 6i again we'll multiply the numerator and denominator by the complex conjugate of the denominator so this would be 1 instead of plus it'll be minus and then 6i over 1 minus 6i to multiply the two numerators let's use foil so the first terms negative 10 times one would be negative 10. the outer negative 10 times negative 6i would be positive 60i the inner 7i times 1 would be positive 7i and then the last 7i times negative 6i would be minus 42i squared now again this i squared has a value of negative 1. so keep that in mind for the denominator let's go ahead and just use our shortcut we know that if we have a plus bi multiplied by a minus bi this is equal to what it's a squared okay the first value squared plus b okay not the i forget about the i b squared so following this i would take 1 which represents a here and square it that's just 1. then i would take 6 which represents b here forget about the i that's already calculated for you with the sign so 6 and then i would square it that's 36. all right and again if you don't believe me on that foil that out and you'll see that you get the same thing all right so we have negative 42 times negative 1 which is 42 42 minus 10 which is 32. so you'll have 32 then plus 60i plus 7i is 67i then over 1 plus 36 is 37. so you'll end up with 32 plus 67i over 37. now nothing we can do to simplify that further another thing again if you wanted to you could say this is 32 over 37 plus 67 over 37 and then times i all right so the last thing we want to talk about and i know this lesson's kind of dragged on a little bit we want to learn how to simplify powers of i so this is something for sure you're going to see in this section and basically just cycles through these four values so we start off with i to the first power or just i we know this is nothing more than the square root of negative one but we just write it as i for now then we see i squared we know by definition this is negative one then we have i cubed now what is i cubed using the rules for exponents it's i squared times i to the first power i squared is negative 1 and i to the first power is just i i mean you could rewrite it as a square root of negative 1. but for the purposes of what we're doing here this is just going to be equal to negative with a negative 1 times i let me erase all this and we'll make it more compact this is just negative i and then lastly we have i to the fourth power so i to the fourth power is going to be i squared times i squared which is negative 1 times negative 1 which is 1. now when you work with something that has an exponent okay that's larger than 4 let's say i saw something like i to the 18th power and you get this on a test and you're like freaking out what do i do how do i simplify that well think about 18 18 is what it's not divisible by four i mean it's divisible by three it's divisible by two it's a lot of different things you can do here but the quickest thing to do is think about well 18 is 2 away from a number that is divisible by 4. 16 is divisible by 4. so using my rules for exponents i could say this is i to the 16th power times i squared if you have i raised to a power that is divisible by 4 you can just say that hey this is equal to 1. it's just that quick and the reason you can do that is to kind of go through this i could write this as i to the fourth power raised to the fourth power right 4 times 4 is 16. i've done nothing illegal r to the 4th power by definition is 1 so this is nothing more than 1 to the 4th power which is 1. so if you have i raised to a power that's divisible by 4 you could just replace it with 1. so this is basically 1 times i squared now i've simplified this to the point where i could just look at my little chart here i know i squared is negative one so this is one times negative one which is just negative one it's really that simple if i looked at something really really big let's say i saw something like i to the power of i don't know let's say 603 well i would know that if i came across 600 that's divisible by 4. so i could write this as i to the power of 600 times i to the power of 3. so again if you see i to a power that's divisible by 4 okay 600 being divisible by 4 we can just mark it out and say this is 1. so this is 1 times i cubed and i cubed if i look at my little chart is negative i okay very very simple as another example let's say that we looked at something like a negative exponent let's say negative i don't know let's say 9. so what do we do we know with negative exponents we write 1 over i to the ninth power so same rules apply here we think about simplifying this so 9 is what if you think about 9 1 less would be eight eighths divisible by four so let's go ahead and do that so we have one over we have i to the eighth power times i i to the eighth power again if eight is divisible by four you can mark this out and just put one because i can always just say this is i to the 4th power squared i the 4th power is 1 1 squared is just 1. so any time you get i raised to a power and that power is divisible by 4 cross it on just put 1. okay that's all you got to do so what i'm left with is a 1 over i again this represents the square root of negative 1. so we rationalize something like this so times i over i you would get i over i times i is i squared which is negative 1 which equals negative i and again this is a more complex problem that you might see a lot of students will get this far okay they'll get one over i but they won't take the extra step of rationalizing the denominator to get this negative i as a result so this right here although it's not technically wrong it's not how we report our answer and so you will get marked off on the test for that okay so you've got to make sure you report this and not just that all right let's take a look at another one let's say we had i don't know let's say i to the power of 57 well we know that 57 is divisible by 3. so would we want to do that or do we want to do this let me ask you what's better should i go i to the 56th power 56 is divisible by 4 times i to the first power we know this would be 1 so this is 1 times i or just i that was very quick let's say i said okay well i know i cubed is negative i let's try it this way let's say this is i cubed raised to the power of 19. is that going to get me anywhere really all this is telling me is i have negative i raised to the 19th power does that get me where i want to go no it really doesn't okay i don't want to do that i want to i want to stick to things that are going to be very very easy because although i can figure this out i can say this is negative 1 to the 19th power times i to the 19th power where does that really get me negative 1 to the 19th power is still negative 1 i to the 19th power is 1. well i'm going to have to split this up i could say it's i to the 16th power or we can even do 18th power times i now what i'm doing here i to the 18th power i could say this is negative 1 times i squared to the ninth power times we'll have i so what do we end up with i squared is negative one negative one to the ninth power is still negative one so you'd have negative one times negative one times i negative one times negative one is 1 1 times i is i look at all that work i had to do to figure out what i figured out very quickly by just let me erase this real quick and i'll go back to what i just did in case you forgot by using the divisibility rules for 4 if i just said okay i to the 56 times i to the first this is 1 times just i so 1 times i gives me i look how quick that was versus what we just did okay so don't make that mistake that a lot of students make try not to use these unless you need to if it's a number for your exponent that's larger than 4 look for the next number down that's divisible by 4. so if it's 13 i'm looking to write this as i to the 12th power times i to the first power this is 1 and you get 1 times i if it's i to the 49th power we know 48's divisible by 4. so i to the 48th power times i to the first power this is 1 1 times i is just i hello and welcome to algebra 2 lesson 62. in this video we're going to learn about the square root property and completing the square so i want you to recall that a quadratic equation has a squared term okay a squared term and then no terms of higher degree so as a generic example and one that you'll probably see in your textbook we have ax squared let me highlight that exponent of 2 plus bx plus c equals 0. so this is a quadratic equation in standard form so you'll see that the exponents go in descending order you have the exponent that's 2 then you have the exponent that's 1. and one thing you need to understand here is that for a b and c a is the coefficient of x squared b is the coefficient of x and c is the constant they're allowed to be any real number that you want with one exception a the coefficient of x squared is not allowed to be equal to zero because if we plugged in a zero there you'd have zero times x squared so that term would basically go away you'd be left with bx plus c equals 0. so without a squared term there it would no longer be a quadratic equation and so that's why that's not allowed we previously learned how to solve a quadratic equation by factoring and for those of you that took an algebra 1 course you've already learned how to solve a quadratic equation by using this process known as completing the square and then more easily using a process known as the quadratic formula so basically we're going to go through that again hopefully you did see it in algebra 1 and this is just a refresher if you didn't it's not too hard to pick this up here so we're going to start out by just going through what to do when it is factorable so we have 2x squared minus 3x minus 9 equals 0. so what we want to do is we want to factor the left side so we know how to factor a trinomial into the product of two binomials now at this point in algebra 2 you should have a preferred method for doing this either it's reverse foil or it's factoring by grouping whatever you want to do for myself i see that this is pretty easy to do with reverse foil you have 2 which is a prime number times x squared so i know that's going to factor into 2x and then x so i know those first two positions i just need to work out the outer and the inner to sum to this term negative 3x so the way you do that is you look at negative 9 and you say well what are the possible factors of negative 9 well it's negative 1 and positive 9 or it's positive 1 and negative 9. the only thing is those would never work out for you and the reason for that is they're too far apart if i put a positive 9 here and a negative 1 you'd end up with an outer of 18x and an inner of negative x those are too far apart to ever give you a negative 3x in terms of a sum if i flip the signs around and i did positive 1 and i did negative 9 it's going to be the same thing it's just too far away so we can go ahead and eliminate those and say that those are not going to work okay so what would work well in terms of negative 9 you have another possibility you have positive 3 okay you have positive 3 and you have negative 3. now where would i want to put negative 3 where would i want to put positive 3 if the result here is a negative 3x i want the term that's going to be larger in terms of absolute value to be negative so i would think about the 2x multiplying by the negative 3 and i would think about the 1x multiplying by the positive 3 so that my result again when i do the sum is going to be negative and we can check this we do our foil 2x times x is 2x squared my outer 2x times negative 3 would be minus 6x my inner 3 times x would be plus 3x and then my last 3 times negative 3 is negative 9. so if you sum these two now you would have 2x squared minus 3x minus 9 which is exactly what we started with right here the main thing here let me just erase this is just to understand how we go about solving this by factoring i want you to realize that this right here inside of parentheses is a factor and so is this it's one of the hardest things for students to wrap their head around this is multiplication here the quantity 2x plus 3 is multiplied by the quantity x minus 3 and this equals 0. so we use something known as the zero product property what does that tell us well if you have something like a times b and this equals zero well what are the possibilities we know when we multiply by zero we get zero as a result so either a is zero b is zero or they're both zero okay it's got to be one of those things so what i can do is i can extend that here and say well either 2x plus 3 is equal to 0 or this other factor here x minus 3 is equal to 0. so when we solve each equation here separately so i would subtract 3 away from each side of this equation we would get 2x is equal to negative 3. then we would divide both sides by 2 and we would get x is equal to negative three halves then over here we add three to each side of the equation and we would have x is equal to three so you have two solutions here x equals negative three halves or x equals three let me just kind of notate that up here we'll say x is equal to negative three halves or x is equal to three using solution set notation we could put negative 3 halves comma 3. now if you wanted to you could pause the video at this point and you could stop and you could check these you could plug a negative 3 halves in for each x and verify that you get the same value on both sides of the equation in this case it would be 0 equals 0. you could also plug in a 3 and verify that you get 0 equals 0. i'll leave that up to you to do on your own what i want to do now is talk about what can we do when we can't solve a quadratic equation by factoring well again we have a few different methods available and we're going to start out by just talking about something known as the square root property so here's the square root property and again if you took algebra 1 you'll remember this it tells us that if x squared is equal to k then x is equal to the square root of k or x is equal to the negative square root of k remember we have those two different notations this is the principal square root and this is the negative square root you might say well why do you need that well just think about if you had something like x squared is equal to 4. we all know at this point that i could plug in a 2 there and say 2 squared is equal to 4 so that would be a solution or i could also plug in a negative 2 there and say negative 2 squared is equal to 4. the real thing is here when you square something you lose that information about the sign was it positive was it negative because both of them would work and you really don't know so you have to account for both possibilities and you say x is equal to the principal square root of k or x is equal to the negative square root of k so here's another good example let's say we saw something like x squared is equal to not so x squared is equal to 9. so what we would do is we would take the square root of this side and then we would say plus or minus that's a shorthand for saying positive square root or principal square root or negative square root you put plus and then you put a minus sign underneath and then we'll say the square root of 9. so we all know at this point that this would cancel with this and we would just have x by itself and this equals plus or minus the square root of nine which is three now don't get fooled here this is two different solutions this tells me right here that i have x is equal to positive three or x is equal to negative 3. again if i plugged in a 3 here and i squared it i'd get 9. if i plugged in a negative 3 and squared it i'd also get not so we're counting for both of those possibilities with this plus or minus here all right so let's take a look at an example that we can use this on so we have 4x squared plus 2 equals 102. so you might say whoa that doesn't look like something i use the square root property on if you have a perfect square equal to some number you can use the square root property immediately now we don't have that here but we could easily do that if we just subtract 2 away from each side and then we would have what we would have 4x squared is equal to 100. now you have the option 4 is a perfect square but typically it's a little bit easier if you just have a coefficient of 1 here so let's go ahead and just divide both sides of the equation by 4 and we'll find that x squared is equal to 25 and then if we use our square root property we could say take the square root of this side and then plus or minus the square root of this side so this would be x is equal to plus or minus 5 right square root of 25 is 5. so again this is two solutions this is x is equal to 5 or x is equal to negative 5. so we can write negative 5 comma 5 like that and for this one let's just stop and check for a second so if i plugged in a 5 there what would we get you'd have 4 times 5 squared is 25 then plus 2 equals 102. yeah 4 times 25 is 100 100 plus 2 is 102. you'd end up with 102 equals 102. and again if i plugged in a negative 5 it'd be no different negative 5 squared is also 25 so it leads to this exact same scenario now let's try without dividing both sides by 4. if we just subtracted 2 away from each side of the equation we'd have 4x squared is equal to 100 what happens if i take the square root of this side and i go plus or minus the square root of this side well the square root of 4 is 2. the square root of x squared is x so i'd have 2x is equal to plus or minus the square root of a hundred which might as well just say plus or minus ten now this leads to two different equations it's two x is equal to ten or two x is equal to negative ten if you solve these for x we divide both sides by 2 you're going to get x is equal to 5 or divide both sides by 2 here you'll get x is equal to negative 5. again exact same thing i got there it's just a little quicker because we divided both sides by 4 okay and we got x squared by itself so you pretty much want to do that to save yourself the extra work that we had to do here all right let's take a look at another one so we have 2x squared plus 3 equals negative 19. so again if i have something squared equal to a number i can use that square root property so let me subtract 3 away from each side of the equation so that'll go away we'll have 2x squared is equal to negative 19 minus 3 is negative 22. i want x squared by itself i want 1x squared so let's divide both sides of the equation by 2 and this will cancel with this we'll have x squared is equal to negative 11 and then now all i need to do is take the square root of each side so i'll take the square root of this side and i'll go plus or minus the square root of this side so we know this ends up being x and this is equal to plus or minus the square root of negative 11. now we didn't work with negative square roots back in algebra one but here we've already done a lesson when we talked about this i can say that this is x is equal to the principal square root of negative 11 or x is equal to the negative square root of negative 11. now when we think about this we know that if you have a negative inside you can pull it out and just say this is i right so i can erase that negative and i can say this is i times the square root of 11. i can erase this negative and i can say this is negative i times the square root of 11. now what if i checked this to prove it to you and i want to do that real fast so let's just go with this first one you'd have 2 times x is squared this whole thing is what x is equal to so you'd have i times the square root of 11 and this is all squared then plus 3 equals negative 19. so you'd have 2 times i squared has a definition of negative 1. let's just write i squared for right now i'm just going to put an arrow and say this is negative 1 times the square root of 11 squared is 11. okay then plus 3 equals negative 19. again i know this is negative 1. so 2 times negative 1 is negative 2 negative 2 times 11 is negative 22. so essentially what you have here is negative 22 plus 3 equals negative so you end up with negative 19 equals negative 19 so yeah that works out let's erase this real quick and what would happen if i had a negative i there if i had a negative there well it wouldn't matter because if i squared that negative i'd get a positive it would lead to the exact same scenario where i end up with negative 19 on this side and negative 19 on this side again two times you can think of this as negative one squared which would be one times i squared which is negative 1 times the square root of 11 squared which is 11 plus 3 equals negative 19. so 2 times 1 is 2 2 times negative 1 is negative 2 negative 2 times 11 is negative 22 negative 22 plus 3 is negative 19. so again you get negative 19 equals negative 19. so each of these work as a solution so the solution said you could put negative i times the square root of 11 comma i times the square root of 11 or if you wanted to write this in a more compact format you could put plus or minus i times the square root of 11. all right let's take a look at something that's a little bit more challenging so we have the quantity x plus 12 squared minus 15 equals 181 so can we use our square root property here yes we can if this guy right here is squared that's a perfect square i took the square root of it i'd get x plus 12. so all i need to do for this type of problem is add 15 to each side of the equation so that this guy is isolated so this would cancel itself out and i would have x plus 12 that quantity squared is equal to 181 plus 15 is going to be 196. now all i need to do now is just take the square root of each side so i'm going to take the square root of this side and then i'm going to put plus or minus the square root of this side so this would cancel with this and i'm going to be left with this x plus 12. so i'm left with just x plus 12 and this equals plus or minus the square root of 196 is 14. so 14. so i've got two different scenarios here i've got x plus 12 is equal to 14 or x plus 12 is equal to negative 14. and you've got to go through both of those because those two different solutions are needed to get a complete answer so we'll subtract 12 away from each side here we'll get x is equal to 2 then or we'll subtract 12 away from each side here we'll get x is equal to negative 26. so those are your two solutions x can be 2 or x could be negative 26. all right so when we run across something like x squared plus 16x plus 14 equals zero we can't immediately use our square root property right some of you will see this and you'll say okay we'll just subtract 14 away from each side of the equation take square root of both sides and we're done it's not going to work for that you need to use a more elaborate process for this and that process is known as completing the square and of course you can use the quadratic formula if you want to but we're not going to get to that until the next lesson so for completing the square we'll just go through that real quick and this is one of those things you want to write it down and just practice the steps to where they just become second nature for you it's kind of a tedious process and a lot of students don't like it but you want to make sure that you memorize it because once you get to using the quadratic formula you're not really going to go back to this because of how tedious it is and then if you need to use it later on you will have forgotten this process so the first thing to do is to make sure the squared term has a coefficient of 1. so if we go back up and i'll kind of jump back and forth this already has a coefficient of 1 so we're good to go there the next thing you want to do is you want to move all the variable terms to one side and the constant to the other so for this one i could just subtract 14 away from each side of the equation this would cancel we would have x squared plus 16x and then i'll move this kind of down because we're going to put something in that place right there this will be equal to negative 14. so all the variable terms are on one side these are the variable terms and the constant is on another side by itself this is your constant all right the next step is the one that's the most confusing you want to multiply the coefficient of the x term just the coefficient by one half then square it so here's the coefficient for the x to the first power term it's a 16. if i multiply 16 by one half and then i square it what am i going to get 16 times a half is 8 8 squared is 64. now you're going to use this in one second i just want to relay a simple trick to remembering this step when i was in high school i struggled with remembering this i didn't practice the steps like i should have and one day our teacher said something to it she said cut it in half and square it cut it in half meaning multiply by half and then square the result you might not remember multiply by a half and square it but cut it in half is easy to remember so cut it in half and then square it okay so that's what we've done here we've cut the number 16 in half by multiplying by a half that gave us 8 and then we squared it to get 64. now what do we do with 64 well what we do is we add the result which again in this case is 64 to both sides of the equation now when we do that let me just go back up real quick what's going to happen is the left side of the equation is now a perfect square trinomial if some of you know what that is some of you are like what is this guy talking about what is a perfect square trinomial so as a little side note let's just show you what that is real fast we talked about special products and special factoring a lot if you have something like x squared plus 2xy plus y squared this is x plus y that quantity squared all a perfect square trinomial is is something that you could take that looks like this and you can factor it into a binomial squared that's all it is completing the square just gives you one side of your equation as a perfect square trinomial which allows you to factor it into a binomial squared which then allows you to use the square root property because i can just take the square root of this guy and i'll just have what's in parentheses for this guy right here this x squared plus 16x plus 64 again if i think about special factoring i know that this guy goes here right so just x goes there and then this guy the square root of that would go here so all i need to think about is what is the square root of x squared that's x and then plus what is the square root of 64 that's 8 and that's squared so i've factored my left side into the quantity x plus 8 squared now let me erase this real quick i'm going to just drag it up and this is equal to over here i just need to do a sum negative 14 plus 64 is going to give me positive 50. now let's go back down and again factor the left side and solve the equation using the square root property okay solve the equation using the square root property so we've already factored the left side so all we need to do is take the square root of this side and then put plus or minus the square root of this side so this will cancel with this and we just have x plus 8 and this is equal to plus or minus the square root of 50. now this is where it gets even more tedious because the square root of 50 50 is not a perfect square so we've got to think about how to simplify that let's just start out by saying that this is x plus 8 is equal to the principal square root of 50 or x plus 8 is equal to the negative square root of 50. let's scroll down a little bit so the square root of 50 is what it's the square root of 25 times the square root of 2. square root of 25 is 5 so it's basically 5 times the square root of 2. this is 5 times the square root of 2. this guy right here i might as well say it's negative 5 times square root of 2. and so what do we have here you would have x plus 8 is equal to 5 times the square root of two we could subtract eight away from each side of the equation that would go away and you'd simply be left with x is equal to five times square root of two minus eight then or over here i would have x plus 8 is equal to negative 5 times square root of 2. we would subtract 8 away from each side of this equation and we'd end up with x is equal to negative 5 times the square root of 2 minus 8. now if you wanted to write this more compactly what you could do you could say that x is equal to negative 8 plus or minus 5 times the square root of 2. so this accounts for 5 times the square root of 2 and accounts for negative 5 times the square root of 2. all right so in solution set notation we'll say negative 8 plus or minus 5 times the square root of 2. now you're free to check each solution right you've got two of them i've already checked them i know they work it'd be good practice for you to do it in the interest of time and making this video a little bit shorter i want to just stop right here and move on to the next problem all right so let's take a look at another one so we have 10x squared minus 8x minus 21 equals 5x squared so the first thing we're going to do we're going to have this in the format of ax squared plus bx plus c equals 0. so let's just subtract 5x squared away from each side of the equation that's going to cancel and become 0 and this is equal to 10x squared minus 5x squared is 5x squared and this is minus 8x minus 21 again it equals 0. now in this particular case the coefficient for x squared is not 1. you have a 5 hanging out there you don't want that okay so what you're going to do is you're going to divide both sides of the equation by 5. so i'm going to divide this by 5 this by 5 this by 5 and this by 5. so we know that this would give me x squared minus 8 fifths x minus 21 fifths is equal to 0 divided by 5 is just 0. now the next thing that i want to do i want to move all my variable terms to one side all my numbers to the other so i'm simply going to add 21 fifths to both sides of the equation so you'll have x squared minus eight fifths x is equal to 21 fifths now the next thing i want to do is i want to take this coefficient for the x to the first power variable i want to cut it in half meaning multiply by half and then square it i'm going to take that result and add it to both sides of the equation so over here if i did that if i had negative eight fifths and i multiply by a half and then i square this what am i gonna get well this would cancel with this and give me a four so it would be negative four fifths squared the negative square would be positive so don't worry about that 4 squared is 16 5 squared is 25. so what are we going to have let's kind of scroll down we're going to have x squared minus eight fifths x and then plus sixteen twenty-fifths this equals twenty-one over five plus sixteen twenty-fifths this left side is a perfect square trinomial the right side is just going to be a constant i just need to multiply this by 5 over 5 so i have a common denominator and i can figure out what this side is going to be so 21 times 5 is 105 if we add 105 and 16 we get 121 so this is 121 over 25. now on the left side again this is a perfect square trinomial that we've created so following the pattern from special factoring i know it's something in this case you have a minus so minus something else that quantity is squared it's always the square root of this guy so the square root of x squared is x and then the square root of this guy the square root of 16 over 25 well we know if we take the square root of 16 over 25 we get four fifths so four fifths goes there so we've got x minus four fifths that quantity squared is equal to 121 over 25 and then now i can take the square root of each side so i'll take the square root of this side and then plus or minus the square root of this side so we'll have x minus four fifths is equal to you've got the principal square root of 121 over 25 which is 11 fifths and you'll say or you've got x minus four fifths is equal to the negative square root of 121 over 25 so that's negative 11 fifths all right so now let's solve these two equations if we add four-fifths to each side over here what are we going to have you'll have x is equal to 11 plus 4 over that common denominator of 5. so you'll say x is equal to 15 over 5 which is 3. then over here you'll have four-fifths added to each side of the equation so we'll have x is equal to negative 11 plus 4 over the common denominator 5 negative 11 plus 4 is negative 7 negative 7 over 5 is negative 7 5. so x equals 3 or x is equal to negative seven fifths so in solution set notation we'll go ahead and say this is negative seven fifths comma three and again if you wanna check these go back and plug negative seven fifths in for x in the original equation verify the left and the right side are equal then plug three into the original equation for x verify the left and the right side are equal hello and welcome to algebra 2 lesson 63 in this video we're going to learn about the quadratic formula so for the majority of you that did take an algebra 1 course you remember the quadratic formula it's generally going to be the final thing that you learn in algebra 1 before you kind of take your break and you come back the following year and you start algebra 2 or in some cases you might start geometry just depending on where you live but for the quadratic formula a lot of you will find that it's kind of a relief right because when you work with quadratic equations there's kind of two different scenarios there's kind of the less tedious ones that you can solve using factoring and then there are the more tedious ones to where you can't solve it using factoring so you've got to use this kind of messy and tedious completing the square method so a lot of us don't like using it i personally don't like use i don't know anybody likes using it but it's something that is necessary to learn because you will be tested on that at some point in your life but moving kind of forward you don't need to rely on either of those you don't have to factor things anymore you don't have to use completing the square because there's a catch-all method known as the quadratic formula so with a quadratic formula we just take this generic ax squared plus bx plus c equals zero again where a b and c are real numbers and a is not allowed to be zero and we just solve it using completing the square so it's done for us to where we can just plug in for a b and c and then we just simplify we have our solution so i want to take you through that real quick you might never have to derive the quadratic formula in your class but you also might get in a situation where you do have to derive it so let's start out with ax squared plus bx plus c is equal to zero so how would i complete the square if i had that just pretend like you had numbers for a b and c what would you do well you want to start by making sure that this guy right here is a one right now it's an a so i don't know what it is all i know is it's not allowed to be 0. but to make sure that it's a 1 i can divide both sides of the equation by a so a divided by a is 1. so now i just have x squared then plus b over a times x plus c over a and this would be equal to 0 over a which is 0. now what's the next step we want to make sure all the variable terms are on one side and then all the numbers on the other so we want these guys right here on one side and then we want this guy on the other side so let's move this over here so let's subtract c over a away from each side of the equation so that's going to give us x squared plus b over a times x this will be equal to negative c over a the next step is to complete the square meaning we want the left side to turn out to be a perfect square trinomial so to do this again we look at the coefficient for our variable raised to the first power so in this case that coefficient is b over a we want to cut it in half or again multiply it by half so b over a multiplied by a half and then we want to square the result okay now i always say cut it in half and then square it so what will we get here you get b over 2a and that would be squared remember everything here has to be squared so b is going to be squared over 2 which is squared which is 4 and then a is squared that's a squared so what i'm going to do is i'm going to add this to both sides of the equation so that the left side is a perfect square trinomial so we will have x squared plus b over ax plus b squared over 4 a squared this is equal to b squared over 4 a squared minus c over a all right let's scroll down get a little room going now what do i want to do next well i know that this is a perfect square trinomial this is a perfect square trinomial what does that mean in case you don't know it means that i can factor this into the product of a binomial squared so something plus something squared would get me that now if you remember from your special factoring in your special products lessons this would be what i would take this is x squared so this would be x and this is b squared over 4a squared we just figured out what this was because we squared and let me just go back up we squared this to get to this so it's just b over 2a so then right here i'm going to have b over let me kind of erase these 2a so this is equal to we have b squared over 4a squared minus if i wanted to get a common denominator going i would multiply this by 4a over 4a so 4a times c would be 4ac and then 4a times a would be 4a squared so let's copy this i'll bring it down and just continuing so this guy on the left is x plus b over 2a again this is squared and this equals we have a common denominator now so we can write b squared minus 4ac over the common denominator of 4a squared so what i want to do now is i want to use my square root property i'm going to take the square root of this side and i'm going to say this is plus or minus the square root of this side so this will cancel with this we'll have x plus b over 2a and this is equal to let me scroll down get a little room going we'll have plus or minus you have and i'm just going to break this up the square root of b squared minus 4ac over the square root of 4a squared let's just write the square root of 4a squared now 4a squared is a perfect square squared of 4 is 2 square root of a squared is a so let's keep simplifying this will be x plus b over 2a which is equal to plus or minus the square root of b squared minus 4ac over again the square root of 4a squared is 2a now the next thing i want to do let me subtract b over 2a away from each side of the equation and let's scroll down get a little room going so what we're going to have is x is equal to plus or minus the square root of b squared minus 4ac then minus b this whole thing is over 2a now the quadratic formula is here it's just not usually written like this there's one more step to do so we have x is equal to i'm going to take this negative b and put it all the way at the far left so i'm going to put minus b or negative b then plus or minus the square root of b squared minus 4ac and this whole thing is over that denominator of 2a so this is the quadratic formula this is the quadratic formula now all i need to do to use this is have my quadratic equation written in standard form so ax squared plus bx plus c equals 0 and then i just plug in for a b and c and that will give me the solutions now this part right here this b squared minus 4ac this is called the discriminant now the discriminant is going to tell you whether you have a real solution or whether you have a solution that's going to involve the imaginary unit right if you have a negative square root it's also going to tell you if you have two solutions or just one if this guy right here ends up being zero you're just gonna have one solution right because this plus or minus plus or minus zero is is nothing right you just if you add zero or you subtract away zero you're not changing anything if this guy ends up being positive you're going to end up with two solutions two real solutions and if it's negative you're going to end up with two solutions that involve that imaginary unit i alright so let's take a look at the first example so we have 2x squared minus 5x plus 3 equals 0. so this is already written in standard form force this is in standard form so if it's not in standard form the first thing you got to do is write it like this again ax squared plus bx plus c equals 0. so what you got to get used to is just kind of matching things up so a here is 2. so a is equal to 2. b here is negative 5. so b here is negative 5. and the sine is very important remember i could write this as plus negative 5 like that so b is negative 5 and then c is equal to 3 right c is 3. so i'm going to take these inputs and just plug it into the quadratic formula and i'm going to have my answer really really quickly again for the quadratic formula x is equal to negative b plus or minus the square root of b squared minus 4ac all over 2a all right so again i'm just going to plug in so we'll have x is equal to negative for b i have negative 5. so the negative of negative 5 again don't confuse the signs that's a negative out in front this is a negative 5. then plus or minus the square root of you have b squared negative 5 squared is 25 minus 4 times for a you have two for c you have three and then this is all over two times a so two times for a we have two all right let's scroll down a little bit so x is equal to what x is equal to the negative of negative 5 is 5 and then plus or minus the square root of we'll have 25 minus 4 times 2 is 8 8 times 3 is 24. this is 24 and this is all over 2 times 2 which is 4. so then x is equal to you have 5 plus or minus the square root of 25 minus 24 is 1 so you have the square root of 1 there over 4 and then let's scroll down a little bit more so then we'll have x is equal to 5 plus or minus we know the square root of 1 is 1 and then this whole thing is over 4. so this is going to give us two different solutions you have x is equal to five plus one over four and then you also have x is equal to five minus one over four so your two solutions here five plus one is six six over four you have x is equal to six over 4 which is going to reduce to what they're each divisible by 2 6 divided by 2 is 3. 4 divided by 2 is 2. so x equals 3 halves or 5 minus 1 is 4 4 over 4 is 1 so x is equal to 1. so in solution set notation we'll write 1 comma 3 halves so these are your two solutions now you can pause the video you can go back and you can check and you can verify that one works as a solution and that 3 halves works as a solution the other thing you might notice is that this 2x squared minus 5x plus 3 equals 0 is factorable so you might want to solve it using factoring and verify that you get these as the same answer so if i was to factor this guy what would i get well this is a 2x squared so i put 2x here and x here now what goes here and here well my final term is a 3. now don't just go with 3 and 1 because again the middle term is negative so that means i'm going to need a negative 3 and a negative 1 because negative times negative would give me that positive there and negative plus negative would give me a negative there so i've got to make sure i'll line this up right i want to put a negative 3 here and a negative 1 here so that would be the correct factorization because the outer will be negative 2x and the inner would be negative 3x negative 2x plus negative 3x is negative 5x you can eyeball this and see that this guy right here if we solve it by factoring you have x minus 1 equals 0 or you have 2x minus 3 equals 0. those are your two factors and if we solve each of these you add 1 to each side of the equation here you get x equals 1. that's exactly what we got right there if you solve this guy you add 3 to both sides of the equation you get 2x is equal to 3 divide both sides by 2 you get x equals 3 halves and again that's exactly what we got right there so you can see that whether you use factoring or the quadratic formula you get the same result and you could complete the square if you wanted to it's all going to give you the same answer it's just a matter of what your personal preference is and i know for most students once they learn the quadratic formula it's just something they're going to use forever right it's just the quickest and easiest method to use in most situations all right let's take a look at another one so suppose you saw 8x squared plus 6 equals 0. so you might say well i'm missing my my bx term right my term where i have x to the first power well that's okay you could rewrite this and say you have 8x squared plus 0x plus 6 is equal to 0. remember this guy this guy and this guy are allowed to be any real number that you choose with the one exception that this guy cannot be zero we never said that this guy couldn't be zero only that this guy couldn't be so my a is going to be equal to 8 right that's the coefficient for x squared my b is going to be equal to 0 that's my coefficient for x to the first power and my c is going to be equal to 6 right that's my constant and again all i'm going to do is i'm just going to plug in so x is equal to the negative of b plus or minus the square root of b squared minus 4ac this whole thing is over 2a so let's plug in we want to plug in a 0 here we want to plug in a 0 here we want to plug in an 8 here we want to plug in a 6 here and we want to plug in an 8 here so we'll have x is equal to negative 0 is just zero then plus or minus the square root of zero squared is zero so minus four times eight times six and this is all over two times eight which is sixteen let's scroll down get a little bit more room so we'll have x is equal to plus or minus the square root of now before we go through and multiply we know this is not going to be a perfect square so let's just separate this up into negative 1 times 4 4 is a perfect square and then let's take 8 and break it up into 4 times 2 4 times 2 6 is 3 times 2 so times 2 times 3. this i can go ahead and combine and say this is 4 and so what am i going to be left with this is over 16 well i have a 4 a 4 and a 4. i know that each 4 can be pulled out square root of 4 is 2. so this is essentially 2 times 2 times 2 which is 8. so x is going to be equal to plus or minus 8 times the square root of now negative 1 that can come out square root of negative 1 is i so plus or minus 8 i times the square root of what's going to be left is a 3 and then this is over 16. now i can cancel between numerator and denominator here i have this 8 that can cancel with the 16 so this would be a 2. so in the end i'd end up with x is equal to plus or minus i times the square root of 3 all over 2. so two separate solutions right you have x is equal to i times square root of 3 over 2 or x is equal to negative i times square root of 3 over 2. but again this is kind of the shorthand way to write that and then for solution set notation why don't we go ahead and use this compact one right here so inside my separation let's say we have plus or minus i times the square root of 3 over 2. and again feel free to check this you can just pause the video plug it in and verify that you get the correct answer you have two different solutions to check but it's going to be something that's going to be a little bit tedious for you to do all right let's take a look at another example so we have 14x squared minus x plus 9 this is equal to negative 4x plus 9x squared so the biggest mistake that students make is they would see something like this and say okay i'm going to plug in a 14 for a i'm going to plug in a negative 1 for b and a 9 for c no you have to put this in standard form so you want the left side to look like this something times x squared plus something times x plus some constant and this is equal to zero so the right side is zero the left side looks like this that's not what we have here so we need to kind of manipulate this a little bit we would add 4x to each side of the equation and we would subtract away 9x squared from both sides of the equation so what that's going to give us this would cancel and become 0. on the left side 14x squared minus 9x squared would be 5x squared then negative x plus 4x would be positive 3x then plus 9 and this equals 0. so this is now in standard form and something we could just plug into the quadratic formula so we could take 5 and plug it in for a 3 and plug it in for b and 9 and plug it in for c so a equals 5 b equals 3 and c equals 9. so what's the quadratic formula at this point you should start to memorize it it's x is equal to you have negative b plus or minus the square root of b squared minus 4ac all of this is over 2a so again i'm plugging in a 5 for each a i'm plugging in a 3 for each b and i'm plugging in a 9 for the c so what we'll have we have x is equal to negative 3 plus or minus the square root of 3 squared is 9 minus 4 times you have 5 times 9. okay this is all over 2 times 5 which is 10. all right so we're going to have x is equal to you have negative 3 plus or minus the square root of 4 times 5 is 20 20 times 9 is 180. so if you had 9 minus 180 that's going to give you negative 171. then again this is all over 10. so how do we simplify this because we're not done we can make this look a little better let me copy this we'll bring it to the next page so realize that if you see a negative inside of the square root symbol you can pull out an i let's start by just doing that let's say this is x is equal to negative 3 plus or minus i times the square root of 171. now before i write that what is 171 what would that factor into well one plus seven is eight a plus one is nine so i know it's divisible by not it would be nineteen times not nine times seven is ninety plus nine times nine which is eighty-one eighty-one plus ninety would be 171. so i could write the square root of 9 times 19 like that this is all over 10. we know this guy is a perfect square so the square root of 9 is 3. i could pull that out so we could say that we have x is equal to negative 3 plus or minus 3i times the square root of 19 and this is all over 10. now again you could leave it like this but again you want to realize this is two different solutions this turns out to be what turns out to be x is equal to negative 3 plus 3i times the square root of 19 over 10 and then also you have x is equal to negative 3 minus 3i times the square root of 19 over 10. okay those two different solutions in solution set notation we'll just write this in the more compact form so negative 3 plus or minus 3i times the square root of 19 all over 10 and then we're done all right let's take a look at another so we see x squared plus 8x minus 25 equals negative 6. again we want this in standard form so we would just add 6 to each side of the equation so that would give us x squared plus 8x then negative 25 plus 6 would be negative 19 so minus 19 and this equals 0. so now this matches our ax squared plus bx plus c is equal to 0. where a is going to be 1 b is going to be 8 and c is going to be negative 19. so a is 1 b is 8 and c is negative 19. and again once you get good at this you just kind of eyeball things and plug in and you're basically done with the problem in a few minutes so we'll have x is equal to negative b b is eight so negative eight plus or minus the square root of b squared b is eight so if i squared eight i get 64 minus four times a a is one times c c is negative 19. and this is all over 2 times a again a is 1 so this is just 2. all right so what do we have here we'll have x is equal to negative 8 plus or minus the square root of you have 64 then negative 4 times 1 times negative 19 is going to be 76 if you had 64 and 76 what are you going to get it's going to give you 140. so the result of this right here is going to end up being 140. so this is over 2 so we'll have x is equal to negative 8 plus or minus the square root of 140 all over 2. now we're not done we can keep simplifying 140 is what well it ends with 40. so we know it's divisible by 4. it would be 4 times 35 and 35 is 5 times 7 so we can take this and say this is x is equal to negative 8 plus or minus if i did the square root of 4 that would be 2 then times the square root of 35 can't really make that any simpler than this is over 2. now one thing we can do we can factor out a 2 from here and here and cancel with this 2. so we can say we have x is equal to you could say you have 2 times the quantity negative 4 plus or minus the square root of 35 over 2 and i'll cancel this 2 with this 2 and i'll have my final answer which will be simplified we'll have x is equal to negative 4 plus or minus the square root of 35. now whatever you want to do here you can leave it like this or again you could say you have x is equal to negative 4 plus the square root of 35 or x is equal to negative 4 minus the square root of 35 right this one and these two it's telling you the same thing it's just this one is more compact now in solution set notation i'm just going to write that we have negative 4 plus or minus the square root of 35 that'll be the only thing i'll put in there again this notates two different solutions all right for the next one we have negative nine x squared minus four x plus eight equals four again very easy at this point we just subtract four away from each side of the equation we put this in standard form it's negative nine x squared minus four x plus 4 equals 0. and again just by eyeballing this now i should know that this is a this is b and this is c alright so a is negative 9 b is negative 4 c is positive 4. so i'm going to plug into my formula i have x is equal to negative b the negative of negative 4 is positive 4 then plus or minus the square root of b squared negative 4 squared is 16 then minus 4 times a a is negative 9 times c c is positive 4. now this is all over 2a again is negative 9. 2 times negative 9 is negative 18. all right so the next thing we want to do we have x is equal to 4 plus or minus the square root of negative 4 times negative 9 is positive 36 positive 36 times 4 is 144. so you would have 16 plus the result of this which again is 144 and this is all over negative 18. so 144 plus 60 is 160. so this would be x is equal to 4 plus or minus the square root of let's write 160 as 16 times 10 we know 16 is a perfect square again this is all over negative 18. square root of 16 is 4 so let's write x is equal to 4 plus or minus 4 times the square root of 10 this is all over negative 18. now we can simplify this let me drag this up here we go x is equal to i could factor out a 2. so 2 times the quantity 2 plus or minus 2 times the square root of 10 over negative 18. i could cancel a common factor of 2 between here and here so this would be negative 9 now and what we'll end up with is x is equal to 2 plus or minus 2 times the square root of 10 over negative 9. so again this is two different solutions you'll have x is equal to 2 plus 2 times the square root of 10 over negative 9 or x is equal to 2 minus 2 times the square root of 10 over negative not and again for solution set notation let's just use the compact form we'll put 2 plus or minus 2 times the square root of 10 all over negative 9. hello and welcome to algebra 2 lesson 64. in this video we're going to learn about solving equations that are quadratic in form so typically you're going to see this topic come up right after you learn how to solve quadratic equations using the quadratic formula it's a very easy topic you might struggle with it at first because you do have to make a little substitution but once you kind of get that down it's very very easy so when a non-quadratic equation can be rewritten as a quadratic equation we say it is quadratic in form so quadratic in form that's what we're looking at so as an example of this let's say you saw something like 9x to the fourth power plus 3x squared minus 20 equals zero well this equation although it's not a quadratic equation it's quadratic in form meaning we could make a substitution and we could create a quadratic equation and therefore we could solve it the same way we solve a quadratic equation now when you look at a generic quadratic equation let's say we look at ax squared plus bx plus c equals 0. what do we see typically it's going to have three terms now that's not always the case but just typically you'd see three terms one of them is going to be a constant and then you're going to have generally two variable terms involved again not always but generally speaking now what you want to notice here is that the variable x that's raised to the smaller power it's raised to the power of 1. the variable x that's raised to the higher powers raised to the power of 2 or squared this higher power 2 is double that of the lower power 1. so up here you can see the same pattern you have three terms you have the higher power 4 that is double that of the lower power 2. so what we can do when we see this again if it's quadratic in form we can make a little substitution and we can create a quadratic equation okay so let me just erase this real quick and i want to walk you through the first one completely so what you want to do is you want to pick a variable doesn't matter what it is just don't choose the same one in the problem so i don't want to choose x because that's in my problem let's just choose z something easy so z is going to be equal to and it's going to be equal to the variable raised to the smaller power so in this case that's x squared so z equals x squared so what i want to do is i want to go and i want to plug in a z everywhere i see x squared but the problem with that is and what you kind of need to be walk through here is that yeah i only see x squared here but by using my rules for exponents i could create x squared squared there this is x squared squared so really i would want to rewrite this problem first and say that i have 9 x squared squared again by rules of exponents x would stay the same 2 times 2 would give me 4. so that's x to the 4th power i just rewrote it then plus 3 x squared then minus 20 and this equals zero now if i substitute z in everywhere i see x squared what am i going to get so this would be z and this would be z so i would have 9z squared plus 3z minus 20 equals 0. so let's forget about everything above what i just wrote let me scroll down and let's just think about the new equation that we have if we saw that we know how to solve it right it's a quadratic equation so we can use our quadratic formula we can complete the square you know etc etc so if we think about what is a that's 9 that's the coefficient for the squared variable this is a what's b that's 3 that's the coefficient for the variable raised to the first power and what's c c is going to be negative 20 that's your constant so the quadratic formula in case you forgot it's negative b plus or minus the square root of b squared minus 4ac all over 2a so we're just going to plug in right so we know that b is going to be 3 so i'd have negative 3 and then plus or minus the square root of b squared again b is 3 so 3 squared is 9 minus 4 times what's a a in this case is 9 then times c c in this case is negative 20. this is all over 2 times a again a is 9 so 2 times 9 is 18. so and i forgot to put z equals so let me put that over here z equals this let's scroll down get a little room going so we'd have z is equal to negative 3 plus or minus the square root of 9 and then i'm going to treat this as negative 4. so negative 4 times 9 is negative 36 and then negative 36 times negative 20 is 720 so that would be positive 720 so i'd be adding 9 plus 720 and really we could just do this right now and say this is 729 okay and this is over 18. now the square root of 729 is 27. so we'll say z is equal to negative 3 plus or minus again this whole thing is going to simplify to 27 over 18. so that's going to give us two different solutions so we'll say that z is equal to negative 3 plus 27 over 18. and then or z is equal to negative 3 minus 27 over 18. let's scroll down a little bit more z is equal to negative 3 plus 27 is going to be 24 so you'd have 24 over 18. each is divisible by 6. 24 divided by 6 is going to be 4. 18 divided by 6 is going to be 3. that's going to give us 4 thirds then or over here negative 3 minus 27 is going to be negative 30. so z is equal to negative 30 over 18. each of these is going to be divisible by 6. so negative 3 divided by 6 is negative 5 and then 18 divided by 6 is 3. so z could be 4 3 or z could be negative 5 thirds now let's stop for a minute a lot of people make it to this point and they stop and they hand in their test or they finish the problem and they say i'm done you're not done you're not solving for z you're solving for x remember if we go all the way back up here our original problem was 9x to the fourth power plus 3x squared minus 20 equals zero so i still don't know what x is i just know what z is and z equals x squared so i've got to substitute again and a lot of people just think oh my it's a lot of work it's it's kind of a tedious process but it is simple so we found that z was equal to four thirds or z was equal to negative five thirds so let me erase all this real quick i don't need it anymore and since z equals x squared i can simply say x squared is equal to well z equals four thirds and x squared equals z so x squared is equal to four thirds or again z is equal to negative five thirds x squared equals z so i can say x squared is equal to negative five thirds so i can solve this using the square root property i can take the square root of this side i can go plus or minus the square root of this side so this would cancel here and i'd have x and this is equal to plus or minus the square root of 4 over the square root of 3. let's just write it like that so we know that the square root of 4 is 2. so let's just go ahead and say this is plus or minus 2 over the square root of 3. and of course we don't want a radical in the denominator so we'll say this is x is equal to 2 over the square root of 3 or x is equal to negative 2 over the square root of 3 and if we rationalize the denominator i multiply this by square root of 3 over square root of three i end up with x is equal to two times square root of three over three or x is equal to in this case times square root of three over square root of three negative 2 times square root of 3 over 3. so that's one pair of solutions there you've got another pair coming so let's copy this we're going to put this on the next sheet okay so we'll just kind of leave that there for now we'll go back up so now we'll deal with this guy x squared equals negative five thirds square root of this side plus or minus the square root of this side so this guy will cancel with this guy i'll have x is equal to plus or minus the square root of negative 5 over the square root of 3. now the square root of negative 5 of course i want to pull the negative out of that right so i pull an i outside and i get rid of the negative and really the square root of 3 and the square root of 5 i can't do anything with either one of those so i just want to rationalize and i don't need to split this into two different solutions for now i can do that when i get down there i'll say this is times the square root of 3 over the square root of 3. to rationalize and so i'll say we have x is equal to plus or minus i times the square root of 5 times 3 which is 15 over square root of 3 times square root of 3 is 3. so x equals plus or minus i times the square root of 15 over 3. so let's go down let me copy this real quick and we'll turn this into two different solutions we'll say x is equal to i times square root of 15 over 3 or x is equal to negative i times the square root of 15 over 3. if you want to you can do a solution set here and let me just kind of drag this up and i'll do one so you can see how you write that let's use a more compact form let's make it easier we'll put plus or minus 2 times the square root of 3 over 3 comma we'll put plus or minus i times the square root of 15 over three so those would be our two solutions really it's four solutions it's two times square root of three over three negative two times square root of three over three i times square root of 15 over three or finally negative i times square root of 15 over 3. all right let's take a look at another problem and we see that we have one that fits the same pattern we saw last time 8x to the fourth power plus 10x squared minus 25 equals zero so again you've got three terms involved just like you'd see with ax squared plus bx plus c then the next thing you're looking at is the larger exponent is a four the smaller one is a two so it's double right the four is doubled out of the 2. so all i need to do to make this into a quadratic equation is make a substitution where let's say q is equal to the variable raised to the smaller powers in that case this is x squared and again you're going to have to rewrite this using your rules for exponents so that you can make that substitution so i would say this is 8 x squared squared plus 10 x squared minus 25 equals zero i want to plug in a q everywhere i see an x squared and then it's very easy from there i'm going to say this is 8 q squared plus 10 q minus 25 equals 0. all right so we can again solve this using the quadratic formula i know what my a b and c are here this is of course a this is my b and negative 25 is my c so we want again for the quadratic formula negative b plus or minus the square root of b squared minus 4ac all over 2a so q will be equal to negative b b is 10. so negative 10 plus or minus the square root of b squared again b is 10. so 10 squared is 100 minus 4 times for a it's 8 for c it's negative 25 so times negative 25 and then this guy is all over 2a a is 8. so 2 times 8 is 16. so q is equal to negative 10 plus or minus the square root of you've got 100 sitting out in front you basically have negative 4 times 8 that's negative 32 and then negative 32 times negative 25 is 800 so you'd have a 100 plus 800 which would be 900 so then this is all over 16. so let's scroll down and get some room going here so to finish this up we have q is equal to you have negative 10 plus or minus the square root of 900 is 30. this is over 16. so two different solutions we're going to get here let's go ahead and draw those out real quick you have q is equal to negative 10 plus 30 so negative 10 plus 30 over 16 or q is equal to negative 10 minus 30 over 16. okay so for the first one here q is equal to negative 10 plus 30 is 20 so you'd have 20 over 16 each is divisible by 4. if i divide 20 by 4 i'd get 5. 16 divided by 4 is 4. then or you'd have q is equal to negative 10 minus 30 is negative 40. so negative 40 over 16 each here is going to be divisible by 8 negative 40 divided by 8 is going to be negative 5. 16 divided by 8 is going to be 2. so you end up with negative 5 halves there so you've got q equals 5 4 or q equals negative 5 halves again we're not done right we made a substitution to solve this using the quadratic formula so we've got to go back and substitute again and then we can get our answer all right so we come back to our original problem which was 8x to the fourth power plus 10x squared minus 25 equals zero so again we substituted for x squared we said q equals x squared we find out that q was 5 4 where q was negative 5 halves so i can say that x squared equals 5 4 or x squared is equal to negative 5 halves so let's scroll down a little bit so i'm going to solve this by taking the square root of each side so if i take the square root of this side and i say plus or minus the square root of this side so i would have x is equal to plus or minus the square root of five over two then coming over here to deal with this one i have x squared equals negative five halves so i would take the square root of this side and then plus or minus the square root of this side so we know that this would be or x is equal to i'd have plus or minus the square root of negative five over the square root of two this one's a little harder to clean up because i have that square root of 2 down there so i'm going to have to rationalize the denominator and then also i have a negative radicand so for this one i can just say that x is equal to plus or minus i know that would come out as i write that negative there so i times the square root of 5 over the square root of 2 and then to rationalize i can multiply by the square root of 2 over the square root of 2. and let me just write this up here what i'd end up with is that x is equal to plus or minus i times the square root of 5 times square root of 2 is the square root of 10 over you have square root of 2 times square root of 2 which is 2. so let me erase this let me kind of write that a little better so again we had plus or minus i times the square root of 10 this is all over 2. you can always write this in a solution set so let's say in solution set notation we have plus or minus the square root of 5 over 2 comma we have plus or minus i times the square root of 10 over 2. and again this is four different solutions you have x equals square root of 5 over 2 you have x equals negative square root of 5 over 2 you have x equals i times square root of 10 over 2 and then you have x equals negative i times square root of 10 over 2. let's take a look at another one we have 3x to the fourth power minus 24x squared minus 27 equals 0. so again you see that you have three terms and you have your pattern where you have a constant you have x raised to a smaller power you have x raised to a larger power that the larger power is double that of the smaller power right 4 is double that of 2. and you're going to see examples where you have x to the sixth power and x cubed or you might see x to the 18th power and x to the ninth power right so you're just looking for this guy right here to be twice this guy right here okay the larger to be double that of the smaller so when i want to make my substitution here let's just use u let's say let u be equal to x squared so i'm going to plug in a u everywhere i see an x squared but again i've got to kind of transform this a little bit so it makes it easy for me to visualize what's going on i'm going to write this as 3x squared squared minus x squared minus 27 equals zero plug in a u here and here so that gives me three u squared minus 24 u minus 27 equals 0. all right so now we know we have a quadratic equation and very very easy to solve at this point if you look you can see that right away you could factor a 3 out and you could solve this with factoring right so i could pull a 3 out and i'd have u squared minus 8u minus 9 and this is equal to 0. now because this is an equation i can divide both sides by the same non-zero number and i'm not changing the solution so i can divide both sides here by 3 and 0 divided by 3 is still 0. so really i'm just kind of reducing some of the tediousness of the problem because i'm working with smaller numbers right so it's just something you want to do so this is u squared minus 8u minus 9 equals 0 and i can solve this using factoring pretty easily give me two integers whose sum is negative eight and whose product is negative nine well that's negative nine and positive one so u minus nine and then u plus one so the solutions i know that i would set this up to be u minus nine equals zero or u plus one equals zero and of course i add 9 on both sides of the equation here i get u equals 9. then or i subtract 1 away each side here i get u is equal to negative 1. so let's copy this all right so we know that u equals 9 or u equals negative 1. but again we got to find out what x is so if x squared equals u and u is 9 then x squared really equals 9 or u also equals negative 1 so x squared also equals negative 1. so to find x let's take the square root of each side over here and so i would have x is equal to plus or minus the square root of 9 square root of 9 is 3. so x equals positive 3 or negative 3 then or you have take the square root of each side over here and you would have x is equal to plus or minus the square root of negative 1 is known as i right by definition so plus or minus i so pretty simple for this one in solution set notation we just say plus or minus 3 comma plus or minus i again you've got four solutions there you've got x can be positive 3 x can be negative 3 x can be positive i x can be negative i all right let's wrap up the lesson with one that's a little bit different so you have 2 times the quantity 3x plus 2 squared minus 12 times the quantity 3x plus 2 minus 32 equals 0. so most of the problems you encounter when you're working on this topic are going to be pretty straightforward like the ones we looked at you might get some variations where you have x to the sixth power and x cubed or you might get again x to the 18th power x to the ninth power again you're looking for that higher power to be double that of the small okay that's what you're looking for here it's not quite so obvious but if you just think about this right here i have three x plus two and three x plus two i have an exponent that's two and an exponent here that's really a 1. so the 2 is double that of the 1. okay if i made a substitution if i said let's let g okay let's let g be equal to this quantity 3x plus 2 well i'd plug in a g there and a g there i'd have a quadratic equation i would have 2 g squared minus 12 g minus 32 is equal to zero okay pretty straightforward now with this guy right here i have a common factor of two so again to make this easy on myself i can divide both sides of the equation by two i can factor out a two and divide both sides of the equation by two however it's more comfortable for you or you can work with it as it is but again if you can do something to make the number smaller it always makes your calculations faster especially if you don't have a calculator so let me just go through and divide everything by two i would have g squared minus six g minus 16 equals 0. all right so let's erase this and i'm going to scooch this up here so we have some room and can i factor this guy looks like i can so give me two integers whose sum is negative six and whose product is negative sixteen well i could do negative eight and positive two so g minus eight and g plus two okay so we all know at this point that g could be eight or g could be negative two but again for completeness we would say g minus eight equals zero or g plus two equals zero add eight to both sides of the equation here g equals 8. then or subtract 2 away from each side of the equation here g equals negative 2. let's erase all this real quick let's erase this i'm just gonna keep my solutions now we originally said that g was equal to or g is the same as three x plus two so i can say that since g equals eight i can plug eight in here and i can just say that eight is equal to or the same as the quantity three x plus two so we know how to solve this very very simple subtract 2 away from each side you would have 6 is equal to 3x divide both sides by 3 and you would get that 2 is equal to x or x equals 2 however you want to think about that then you can do the same thing plug in a negative 2 for g there and you would say that this is or negative 2 is equal to 3 x plus 2 subtract 2 away from each side of the equation you would have that negative 4 is equal to 3x divide both sides of the equation by 3 and you would say or x is equal to negative 4 thirds so x here is 2 or x is negative 4 thirds let me put x on the left side because that's how i like it and of course if you wanted to use a solution set you could say negative four thirds comma two hello and welcome to algebra 2 lesson 65 in this lesson we're going to learn about graphing parabolas so up to this point we've only graphed lines as we move forward in math we must be prepared to deal with more complex shapes so the focus of today's lesson is to learn about something known as parabolas now first and foremost the graph of any quadratic function or you could say the graph of any quadratic equation is known as a parabola so what we're going to start out today with is this f of x equals x squared if this is generally what you're going to see in your textbook or in your class this is going to be the first parabola you deal with and it's what we're going to use as a comparison when we're talking about shifts okay and we're going to get to that in a second so what you'll notice about a parabola is that the shape kind of looks like a u but depending on the function that you're given this guy could be skinnier it could be wider it can open up as it does here or it might also open down we're going to see examples of these not only in this lesson but in the next lesson where we continue talking about parabolas now this function is already graphed for us and you might say well why is that the case well we aren't really going to talk too much about how to sketch the graph of a parabola at this point the process of graphing is pretty straightforward you gather some points you plot them and then you sketch the one thing i want you to know is that the process of graphing something that's not linear is a little bit more difficult you're going to be less accurate and generally if you want something that's going to be accurate you either need to use a graphing calculator or some type of graphing software so to eliminate the inaccuracies i just have the curve already sketched for us and we can go through and get some points going just so it kind of feels like we're completing the full process so if we had f of x equals x squared and we wanted to sketch the graph we'd get some points going so let's just start with x equaling negative 2. so if i plugged in a negative 2 for x i would square it i would get positive 4. then let's do x equals i don't know negative 1. plug in a negative 1 for x squared i get 1. then let's do x equals 0 square 0 you get 0. then let's do x equals 1 square 1 you get 1. and then lastly let's get x equals 2 square 2 and you get 4. so we could plot these points so negative 2 comma 4. so negative 2 4 that's right there then negative 1 1. so negative 1 1 then 0 0 so 0 0 or the origin then 1 comma 1 so 1 comma 1 and then 2 comma 4. so 2 comma 4. you can see where we plotted our points and again if you were sketching this freehand you'd have to get a lot of points kind of going this way to really get this thing accurate so as you move forward you really want to start relying on graphing calculators and graphing software i realize that in some classes they're going to make you sketch some graphs it's okay your teacher is going to understand that it's not going to be 100 accurate okay so now that we have the general idea of what a parabola is let's talk about a few important things so the first vocabulary word that you're going to encounter is that of a vertex so the vertex and let me highlight that is the lowest or highest point on our graph so if we have a parabola that opens up like in the example we saw the vertex is the lowest point on the graph so in this case it's here if we have a parabola that opens down which we'll see at the end of the lesson it's going to be the highest point on the graph so one other quick point i want to make about parabolas a parabola is symmetric about its axis this just means we could fold our graph and the two curves would coincide so you can imagine taking this y-axis here and just drawing a line down there i know that's not perfectly straight if you folded the two curves you could see that they would coincide so that's what they're talking about there so now let's talk about shifts this is the first thing you're going to come to when you start talking about parabolas so we've seen the graph of f of x equals x squared so the question is going to be how can we use this to graph something like f of x equals x squared plus 3. well without looking at the graph let's just do some points for each one so we saw the points for this one we know it's negative 2 and then 4 negative 1 and then 1. zero and then zero one and then one and then two and then four so let's use the same x values and see what we get for f of x or y values if i had a negative two plugged in for x what would be the result for y or f of x well it's going to be the same thing as here it's just going to be 3 units higher right because i'm taking this and then i'm just adding 3. so i can just take this right here 4 and say 4 plus 3 would give me 7 or i could manually do it negative 2 squared is 4 4 plus 3 again is 7. so i can just take all these values and i can just add 3 to the y or f of x values so negative 1 would correspond to a y value of 4 0 will correspond to a y value of 3 1 would correspond to a y value of 4 and then 2 would correspond to a y value of 7. again all i'm doing is i'm taking the result from here and i'm just adding that 3 to it now what do you think that does in terms of the graph well for the same x value okay for the same x value my y value in every case has been increased by three so that's going to create a vertical shift up by three units and you can see this graphically so this is the original this guy right here is f of x is equal to x squared we see the vertex occurs at zero comma zero we saw this we talked about it already this guy right here is the graph of f of x is equal to x squared plus three the vertex or the lowest point occurs here at zero comma three everything has just been shifted up by three units so the vertex went up three units if you took a point like 1 comma 1 which is right here if that's shifted up 3 units then that now occurs at 1 comma 4. let's take a look at another example so again you have this f of x equals x squared now we're going to look at f of x equals x squared minus 1. some of you can already guess that this will be a vertical shift down of 1 unit right because you have this minus 1 here so the y values will just be decreased by one unit so in other words if i go through here and say negative two negative one zero one and two we remember this is four this is one zero one and four i can copy this so negative 2 negative 1 0 1 and 2. all i'm going to do is take these y values and i'm going to decrease them by 1. because i'm plugging it into the same thing the only difference is that minus one at the end so i plug in a negative two i squared i get four i take away one i get three so this is one less so minus one this right here wouldn't be one it would be zero one less this zero would be negative one this one would be zero and this four would be three so for the same x value the y value is one less so this is a vertical shift of one unit down and you can see that here so again this is the original this is f of x is equal to x squared and then this guy right here is your f of x is equal to x squared minus one it's a vertical shift down here's your vertex at the origin zero comma zero on the next equation where we have f of x equals x squared minus one the vertex moves down one unit so it's at 0 comma negative 1. again if i pick a point like 1 comma 1 which occurs on the original f of x equals x squared if i have an x value of 1 here in this equation it goes down by 1 unit for the y so that's going to be right here so just shift it down by one unit okay so pretty easy to understand that one so generically when working with parabolas if you have f of x equals x squared plus k okay so plus k we know what f of x equals x squared looks like we're adding this k to it so this will be shifted up k units if k is greater than zero so we saw that when we had f of x equals x squared plus three it was shifted up three units then down the absolute value of k units if k is less than zero so when we had x squared minus one we think about that as plus negative one we're really saying the absolute value of negative 1 is 1 so we're shifting it down one unit right we just saw that so this is what you want to write down and kind of commit to memory right if you're working with f of x equals x squared plus k your vertical shift will be up if k is greater than zero or down if k is less than zero right so if i had f of x equals x squared plus 10 i know that's a vertical shift of 10 units up if i had f of x equals x squared minus 30. i know that's a vertical shift of 30 units down so very very easy once you understand what's going on that's a typical question you'll get on a test all right so just some more kind of information the vertex for a function of the form f of x equals x squared plus k is going to occur at 0 comma k so you're thinking about the lowest point here well if i plugged in a 0 for x 0 squared would give me 0. i would be left with f of x or you could think about that as y is equal to k so 0 is the x location k would be your y location or your f of x location all right so now that we talked about the easier scenario which are vertical shifts let's move on to horizontal shifts now these might give you a little trouble mentally but again it's like anything else once you kind of get used to it it's really not that bad so we have f of x equals x squared and what we're going to compare this to we have f of x equals x minus 1 squared so let's just think about this and again this is something that really gives students a lot of trouble let's say i started with my typical negative 2 negative 2 squared we know that's 4 and then negative 1 negative 1 squared is 1 0 0 squared is 0 1 1 squared is 1 and then 2 2 squared is 4. when we think about points for this guy right here if i had a negative 2 plugged in for x what would happen to y well i plug in a negative 2 negative 2 minus 1 would be negative 3 negative 3 squared would be not okay so you say well i don't really see what happened okay so let's keep going so you plug in a negative 1. negative 1 minus 1 would be negative 2. negative 2 squared would be 4. so negative 1 and then you'd have four if i plugged in a zero zero minus one be negative one negative one squared would be one if i plugged in a one one minus one would be zero zero squared is zero and then if i plug in a 2 2 minus 1 would be 1 1 squared is 1. so what's going on here well what's going on is that the x value must be 1 unit larger to obtain the same y value this is going to create a horizontal shift to the right by one unit because again let's just think about this the x value needs to move one unit to the right or be one unit larger to obtain the same y value so just if we take a look at the points an x value of negative two corresponds to a y value of four an x value of negative 2 corresponds to a y value of 4. here i've got to increase the x value by 1 or go from negative 2 to negative 1 to get a y value of 4. so i've got the same y value here and here but my x value increased when i went to this guy by one all right i went from negative two to negative one so again that's what's going to create this horizontal shift to the right okay by one unit it's because to get the same y value or to be in the same vertical location i've got to move to the right by one unit so here you can see the two graphs on the same coordinate plane this blue guy right here that's our f of x equals x squared and this green guy right here this is f of x is equal to the quantity x minus 1 squared and again all this is a horizontal shift to the right by one unit for this guy the vertex occurs at zero comma zero so for this guy i'm just going to shift to the right by one unit so the vertex occurs at one comma zero okay so everything is to the right by one unit so again if i look at a point like one comma one this has moved one unit to the right so now i'd have an x value of 2 with a y value of 1. if i have an x value of 2 that corresponds to a y value of 4 for f of x equals x squared for this guy i now need an x value of 3 to correspond with a y value of 4. so again i'm just shifting everything to the right by 1 unit let's take a look at another example of this horizontal shift so we have f of x equals x squared again you're always going to see that and we're comparing this to f of x equals the quantity x plus 2 squared so now we're adding 2 to x so think about this this is going to go in the opposite direction now the way you want to think about it is that the x value must now be two units smaller to obtain the same y value okay two units smaller and we can just do some points to see that so again we go with our negative 2 negative 1 0 1 and 2. you probably memorize this by now this is going to be 4 this is going to be 1 0 1 and 4. all right so let's use the same x values over here so we'd have negative two negative one zero one and two so negative two plus two would be zero zero squared is zero then negative one plus two would be one one squared is 1 0 plus 2 is 2 2 squared is 4 1 plus 2 is 3 3 squared is 9 and then 2 plus 2 is 4 4 squared is 16. now again when i look at this here let's just take 0 comma 4 as an example i have an x value here of 0 that corresponds to a y value of 4. here when i have a y value of 4 i'm thinking about an x value of 2 so 2 comma 4. so in other words this guy right here this 0 is 2 units less than this 2 here 2 units less in terms of x we're moving 2 units to the left on the x axis to get to that same y value of 4 and you can see it when you look at the graph so here's your vertex here for again f of x equals x squared and then here's your vertex for f of x is equal to the quantity x plus 2 squared you can see everything has just shifted 2 units to the left here it says 0 0 here it's at negative 2 comma 0. so my x location just went two units to the left or decreased by two you can think about this point one comma one that we keep going to well if i decrease my x value by two so i go one two units to the left i know that i'm gonna have my y value of 1. you can see that there if i take a point like 2 comma 4 that's here if i decrease my x value by 2 and i go from 2 to 0 i'm going to see 0 comma 4 right so the same y location occurs when the x value is two units left or two units less so this is what's creating the horizontal shift all right so here's the kind of cheat sheet for this when working with parabolas again if you see f of x is equal to the quantity x minus h and this is squared this will be shifted h units to the right if h is greater than zero okay h units to the right so if it's x minus 1 then it's going to shift one unit to the right because again the x value needs to be one unit larger to obtain the same y value so that's what creates that shift then the other scenario we say the absolute value of h units left if h is less than 0. so this gets a little bit tricky you're thinking about things like let's say you saw the quantity x plus 3 in there well really i could write this as f of x is equal to the quantity x minus a negative 3 right because we're always keeping that minus symbol in there so this is squared but if you see plus 3 you know that you're just going to move left by 3 units okay you don't need to go through all this really what this is technically saying is that you take the negative three you take the absolute value of that you get three you're going to move three units left but again this is ultra complicated when you see a minus sign and a positive number behind it you know you're moving right by that number of units when you see a plus sign and a positive number so like x plus 3 for example you know you're going to move left by 3 units so the vertex for a function of this form if we have f of x equals the quantity x minus h squared this is going to occur at h comma 0. so in other words if i plug in an h here for x that's going to be h minus h which is 0 0 squared is 0. so that's giving me the lowest point for y or for f of x so the lowest point for the function all right so if you're on a test and you're really really confused you don't know what's going on horizontal shifts give students the most trouble you can kind of go through it and with the vertical shifts they really get it right away but a lot of students with the horizontal shifts just like i don't know what's going on i can't figure this out just find the value that causes x minus h to be zero this will give you the shift so if it's let's say i had f of x is equal to the quantity x minus 30 squared what would make this equal to zero well if i just plugged in a 30 for x 30 minus 30 would give me zero zero squared would be zero so if i'm plugging in a positive 30 you think about positive as moving right right so i'm moving right 30 units so it's a 30 unit to the right shift horizontal shift to the right 30 units however you want to think about that if i replace this with 25 we know the shift would be 25 units to the right because i'd plug in a 25 in for x and i would get a zero there if i change this to a positive let's say a positive 7. how can i make this 0 well plug in a negative 7 for x negative 7 plus 7 is 0. so you think about negative as moving to the left a negative direction so i'm shifting to the left seven units so this is a little trick it's not something to confuse you if you're lost when you're talking about horizontal shifts just find the value that causes the quantity x minus h to be zero okay so forget about the square forget about the f of x forget about the notation what would cause this to be zero give that answer and you found your shift if it's negative you're moving to the left if it's positive you're moving to the right so here i'd need a negative 7 for x so i'm moving to the left 7 units if this was plus 89 i would need to put a negative 89 in there so i'm shifting 89 units to the left right so on and so forth so in tougher examples we combine a vertical and horizontal shift so you'll see these examples where you see f of x equals the quantity x minus 2 squared then plus 3. so some of you get it right away you just combine the two shifts so if i think about the horizontal shift i just look at this and i say okay i have a negative two there a minus two i know that corresponds to a shift to the right so i shift to the right by two units then this plus three here is a shift up by three units so a shift up by three units so this guy would shift so it shifts right two units and up three units okay pretty easy overall so you can see this graphically again this is the original f of x equals x squared and then this guy right here is f of x is equal to we have the quantity x minus two squared and then plus 3. so here's your vertex here's your vertex so that's what i look at and you just think about the fact that okay it's a shift of two units to the right so i'd go one two units to the right and then three units up so one two three units up so you go from there to there so two units to the right three units up this is all you really need to do when you're testing especially if it's presented to you in a graph you could just count the shift to the right or the shift to the left and then count the shift up and down right so very very easy to do if you get it on a graph so when we see the format f of x equals a times the quantity x minus h squared plus k where again a does not equal zero we can say the vertex occurs at h comma k and why is that the case let's go back up we're thinking about how to make this in this case because this is positive as low as possible so if i plugged in an h for x h minus h would give me 0 0 squared is 0 0 times a is 0. so this is essentially gone so i'm just left with k right so that's the lowest it can be plug in an h for x that eliminates that it wipes it out i'm just left with k so that's why our vertex occurs at h comma k and we have a vertical line of x equals h as its axis again that's going to split it in half if you fold on that the two curves are going to coincide and then the graph opens up if a is greater than zero right if a the coefficient of the squared variable is positive and it's going to open down if a is less than zero or again the coefficient of the squared variable is negative all right so overall not too difficult just things you need to get used to a lot of is just memorizing what you need to do and then as we move forward we're going to start seeing some downward facing parabolas so all we've seen so far are the parabolas that open up so the f of x equals x squared that variety now we're going to look at f of x equals negative x squared so the trick here is that if you have a negative coefficient on the squared variable so you're going to have a parabola that opens down so a downward facing u so now if you think about the vertex it's going to be the highest point right so this 0 comma 0 here is the lowest point for the upward facing parabola but the highest point for the downward facing parabola okay and as you move higher in math you're going to start talking about these as minimums and maximums but again we'll get to that later on hello and welcome to algebra 2 lesson 66. in this lesson we're going to continue to learn about parabolas so in the last lesson we covered some basic topics that come up when working with parabolas in this lesson we're just going to go a little bit further and expand on what we know so the first topic we're going to talk about today would be how to find the vertex when our function is in standard form so we see we have f of x is equal to ax squared plus bx plus c and of course here a does not equal 0. so what's the vertex here and in case you didn't watch the last lesson what is a vertex in general so with a parabola with a standard parabola a u shape that opens up we know that the vertex is the lowest point on that graph so in terms of the y location or the y value on that coordinate plane it is as small as it's going to get now if you have the other scenario if you have a parabola that opens down so an upside down u is what that looks like in that case the vertex is the highest point so in terms of the y value it's as high as it's going to get but either way we use the same formula to obtain the vertex now in the last lesson i talked to you about the vertex form so vertex form looks like this it's f of x is equal to you have a times the quantity x minus h this is squared and then plus k so the reason we call this vertex form is that i can look at this function in this form and i can immediately tell you what the vertex is so the vertex here the vertex will occur at the point h comma k meaning whatever is here for h will be plugged in for the x coordinate and whatever's here for k will be plugged in for the y coordinate now before we go any further let's think about why the vertex occurs at the point h comma k although you would use the same thought process either way i want you to think about the fact that if you had an upward facing parabola so a standard u-shape your vertex is going to occur at the lowest possible point so the y value is as small as it's going to be so just for a second replace f of x with y so we can get past that mental roadblock and i just want you to think about what you could plug in for x to make y as small as it could be well because we have this squared right here no matter what i plug in for x i subtract away something for h when i square that it's going to either be 0 or positive right if it turns out to be negative when i square it it's positive if it turns out to be positive y squared it's still positive if it's 0 and i square it i get 0. so the best i can do at making this small is to plug in an h right because h minus h would give me zero so this guy right here once i square zero would be zero i multiplied by a i still have zero so this is knocked out and i'm just left with k so that's why your vertex occurs at the point h comma k so now once we know that let's think about how we could take a function in standard form like f of x equals ax squared plus bx plus c and transform it into vertex form well what we're going to do is we're going to use that method known as completing the square and most of you remember this from when we solved quadratic equations we saw that we had to start out by learning the square root property then we learned how to complete the square and then we found a shortcut known as the quadratic formula it's going to be the same thing here we're going to complete the square then we're going to see a shortcut known as the vertex formula okay so let's jump in and look at an example so let's say we see f of x equals x squared minus nine x plus seven and your question here is to find out what is the vertex so in order to do this we want to put it in vertex form so let's let's just write that form first so this is f of x is equal to a times the quantity x minus h squared plus k all right so if you remember from completing the square we start out by moving all the variable terms to one side and all the numbers on the other side in this particular case we're not going to do that we want to just keep f of x on one side of the equation by itself so we're not really going to mess with the left side at all we're just going to do everything to the right side but i'm still going to group the variable terms together so i'm going to say x squared minus 9x i'm just going to put some parentheses around it to say this is grouped together and then we have this plus 7. now to complete the square you look at your coefficient for the variable raised to the first power in this case that's going to be negative 9. so what do you do you cut it in half meaning you multiply by a half and then you square it so i would have negative 9 multiplied by a half and then the result of this is going to be squared so this would be negative nine halves squared which would be eighty-one fourths now what does this do this eighty-one fourths let me just jog your memory here so let me erase this real quick and i'm just going to put this value off to the side for a minute so 81 4. the 81 4 is added right here that is the constant that makes that trinomial into a perfect square trinomial so remember a perfect square trinomial is able to be factored into a binomial squared that's what we got to get to to get this guy right here we want a binomial squared so let's just go ahead and do that so we would have f of x is equal to we'd have the quantity x squared minus 9x and then this guy right here is what i'm adding so plus 81 4. so now i've got my perfect square trinomial but there's a problem i can't just go around adding things to equations right it's got to be balanced normally if we add something to one side we've got to add it to the other but in this case i told you we're not going to mess with the left side of this equation so what we're going to do is we're just going to subtract it away outside the parentheses so minus 81 4 and then i have my plus 7. so this right here is all together that's my perfect square trinomial that's what i'm going to factor into a binomial squared and then this part right here we can basically say this is together so let's scroll down and get some room going and we'll see what we get so we have f of x is equal to most of you remember from special products how to factor a perfect square trinomial you know that this would be x you have a minus here so you get a minus here and you can always take this last term right here and take the square root of it and so the square root of 81 is 9 the square root of 4 is 2. so you get x minus 9 halves that quantity is squared and again if you don't believe that that factors into that pause the video do foil you'll get this back then we'll have minus 81 4 and then plus 7. now i want to combine these two so i want to get a common denominator going so i'll multiply this by four over four seven times four is 28 so you would have 28 fourths there and then now it's just a matter of adding negative 81 and 28 so if we do that we get negative 53. so we'll say that we have f of x is equal to the quantity x minus nine halves squared minus 53 fourths okay so let's scroll down and i want to write the vertex form one more time so we have f of x is equal to a times the quantity x minus h squared plus k the vertex occurs the vertex occurs at the point h comma k here's where you have to be ultra careful you do have a minus sign here and here so whatever occurs after that minus sign in this case that's nine halves is going to be your h that's going to be your x coordinate for that ordered pair so the vertex the vertex occurs at the point you'll have nine halves within you have a plus sign here but a minus sign here so you've got to be very careful go ahead and translate this into plus a negative 53 4 so you don't make a mistake and so negative 53 4 represents k so that will be your y value so negative 53 fourths so the vertex for this function is going to occur at the point nine halves comma negative 53 fourths all right let's try another one now so we have f of x is equal to x squared plus 3x minus 10 and again i want to put this in vertex form so for vertex form it's f of x is equal to a times the quantity x minus h this is squared then plus k now the vertex again the vertex will occur at the point h comma k all right so let's go ahead and take this guy we have f of x is equal to x squared plus 3x minus 10. again i'm going to group these variable terms together normally i would move negative 10 over to the other side by adding 10 to both sides of the equation but again if you're doing this you want f of x by itself so i'm just going to leave it alone i want to add one half of this coefficient right here which is three squared so one half times three squared that would be three halves squared which would be nine fourths so i'm going to add 9 4 inside those parentheses but again i can't just do that i can't just go around just adding things to equations i've got to make it balanced so i'm going to subtract 9 4 away from the same side of the equation because 9 4 minus 9 4 would be 0. okay so now that that's done this can be factored into a binomial squared so we would have f of x is equal to this would factor into x plus 3 halves that quantity squared and then you'd have your minus 10 minus 9 4. of course i'd want to get a common denominator going so let's put this as plus negative 10. we'll multiply this by four over four so you'd have negative 40 fourths so negative 40 fourths minus nine fourths so you'd end up with minus 49 fourths so minus 49 4. so we have this in vertex form but again you have to be very careful this is a minus here this is a plus here so if you just go through and say okay my vertex my vertex will occur at three halves comma 49 fourths you will be wrong the signs just won't be correct so i want to erase this i don't want to match this up with this i can make a plus into minus a negative and i can make a minus into plus a negative so now it matches exactly and we'll see that h is negative three halves so i just put a negative out in front here let me kind of make that a little better so negative three halves and k okay k is going to be negative 49 fourths so negative 49 fourths so the vertex here occurs at negative three comma negative 49 fourths all right let's take a look at one more of these and then i will show you how to derive the vertex formula and from that you will find a very very easy method to find the vertex when your function is in standard form so we have f of x equals 3x squared plus 5x plus 7. so the one problem here is that we have this 3 that's the coefficient for x squared you will recall when we complete the square we say that we want the coefficient of the squared variable to be one right you'll recall that step from when we completed the square when we're solving quadratic equations so what we would do is we would divide the whole equation by in this case it would be 3. but we're not going to do that here what we're going to do here is we're just going to factor it out from this grouping here so in other words if this is a group i could factor a 3 out from that group and if i factor the 3 out from 3x squared i would have x squared if i factored a 3 out from 5x what would i have 5x divided by 3 is what i'm looking for there because 3 times 5x over 3 the 3s would cancel and i'd be right back to 5x then i'd put plus 7 and i can kind of move that out here and i can go through the process now just completing the square so i want to take one half of my coefficient for the variable raised to the first power and square it so 5 thirds times a half squared so this would be 5 6 squared 5 squared is 25 6 squared is 36. so now i want to add 25 over 36 to this side right here here's where you have to be careful though so i'm going to show you a common mistake so plus 25 over 36 close the parentheses minus 25 over 36 so students will do that and then they'll continue to solve the problem this is going to give you the wrong answer why do you think that's the case in case you haven't seen it already well you're multiplying 3 by this guy right here so i'm not really adding 25 over 36 i'm multiplying 3 by 25 over 36 so i've got to do the same thing to this guy otherwise i'm not balancing the equation so what i usually do is i'll expand my parentheses out here and then i'll do another step so i'll do f of x is equal to 3 times the quantity so i'll type my parentheses down i'll say x squared plus five thirds x plus 25 over 36. i'll close my parentheses there but this allows me to remember that 3 is going to be multiplied by this so i'll put plus negative 25 over 36 multiplied by three now i have added and subtracted the same value to that right side of the equation so i'm good to go so then plus seven okay so let's scroll down get a little room going all right so we're gonna have f of x is equal to three times if i factor this guy i'm gonna have x plus five six in this quantity squared then over here i have negative 25 times three that's negative so you'd have negative 75 and this would be plus 7. so i would get a common denominator going and to do that i'd multiply this by 36 over 36 and so 7 times 36 is 252 so you would have 252 minus 75 which is 177. so you could say this is plus 177 over 36. now i can reduce this fraction 177 is divisible by 3 1 plus 7 is 8 plus 7 is 15 and so is 36 right 3 plus 6 is 9. if i divide 177 by 3 i get 59. so this is 59 and then 36 divided by 3 is obviously 12. okay so i have this in vertex form but again i've got to match everything up to where i don't make a sign mistake so the vertex form again it's f of x is equal to a times the quantity x minus h and this is squared and then plus k this is a minus right now this is a plus so i've got to change this to minus or negative so that this value right here can be plugged in for h then i have plus k so in this case i have plus 59 12 that's okay right this will just match up completely that'll be my value for k so the vertex the vertex will occur at negative 5 6 that's my h here and so that's going to be my x value comma k is 59 12. so 59 12 will be my y value so the vertex occurs at the point at the point negative 5 6 comma 59 12. all right so now what we're going to do is we're going to take a generic quadratic function in standard form and we are going to convert it into vertex form and from this we're going to derive something known as the vertex formula this is basically the same thought process as we use when we derive the quadratic formula right we have these parameters a b and c that we can just easily grab and plug into the formula that we have and we can solve things very very quickly so if i was to have this guy right here and again i want this in the format of f of x is equal to a times the quantity x minus h squared plus k the first thing i would do is i would factor out the a from these first two terms right here so f of x is equal to bring the a outside i would have x squared plus to factor an a out from here i would basically divide b times x by a so in other words this would be b x over a and again if a multiplied by b x over a the a's would cancel be back to bx so then we can close this and then we'll have plus c okay so now that we've done that we want to complete the square we want to take this guy right here and just make it into a perfect square trinomial so that we can factor it into a binomial squared okay that's the goal so let's scroll down get a little room going all right so i want to take my coefficient for the x to the first power variable in this case that's going to be b over a so b over a and i want to cut it in half so multiply by half and then i want to square the result so this would be b over 2a so this would be b over 2a and this would be squared giving me b squared over 4 right 2 squared is 4 times a squared okay so now let me erase this and let me kind of scooch this up again we have to be careful if i just shove this in right here remember a is going to be multiplying by that so i can't just balance it outside of the parentheses okay so what i want to do to remind myself i'll have f of x is equal to a times the quantity x squared plus b over a x plus b squared over 4 a squared then i'm going to put minus b squared over 4a squared this is all inside the parentheses then i'll close the parentheses okay right now the way it's set up a would multiply all of these so i am adding and subtracting the same value to the same side of the equation and so i'm basically adding zero right i'm not changing it it's balanced okay that's what you have to do if you make the mistake and you do this and you forget to multiply by a you're going to get the wrong answer so make sure you put your parentheses here as a reminder and then we'll put plus c here now in the next step we'll tighten this down but the way we're going to do it we're going to say f of x is equal to a times the quantity x squared plus b over a x plus b squared over 4 a squared i'll close the parentheses because this is a perfect square trinomial but then i'm going to remember that a is going to need to be multiplied by this negative b squared over 4a squared so plus we'll have a multiplied by negative b squared over 4 a squared scroll down get some room going and then of course we have plus c okay so now we have a perfect square trinomial here so i'm going to factor that guy into a binomial squared so f of x is equal to a times the quantity this would be x plus square root of b squared is b square root of 4a squared is 2a and then this quantity is squared then plus over here i just need to do some simplification so this would cancel with one of these i'd have negative b squared over 4a now i can flip this around and say i have c minus b squared over 4a and then i can get a common denominator by multiplying this by 4a over 4a and what's going to end up happening is i have plus 4ac minus b squared so 4ac minus b squared over the common denominator of 4a now there's one other thing we want to do here remember when i show you vertex form it's f of x is equal to a times the quantity x minus h squared plus k so here k is this guy right here so that's going to be the y coordinate when you're talking about your vertex and then h there's a problem getting it straight from this because you have a plus sign here and here you have a minus so we want to just make one minor change here and put this as minus a negative b over 2a and now this guy right here will be my h that's what's going to get plugged in for my x location when i'm building the vertex but it's going to get even simpler than that so let's scroll down just a little bit and let's write our vertex so this is the vertex the vertex formula and you definitely want to write this down because it's going to save you a lot of time so the vertex formula you take a quadratic function in standard form and you know what a b and c are and you can plug into this formula and find out the vertex so the vertex the vertex will occur at the point negative b over 2a and that came from right here again normally it's h h here is this negative b over 2a then comma you could plug in 4ac minus b squared over 4a but there's an easier solution i want you to think about the fact that if i plugged in a negative b over 2a for x what would happen there well you would get 0 and again this would cross itself out you'd just be left with this this is what would be left so instead of writing all this all i want to say is that i want to plug this in for x how do we do that with function notation well we say f of whatever we want to plug in for x so in this case that's negative b over 2a and that's going to be my value for y or you could say your value for k however you want to think about this so hopefully that makes sense for you again the vertex formula the vertex occurs at the point negative b over 2a this is what's going to be the x location on the coordinate plane and then f of negative b over 2a this is your y location the coordinate plane and again just so you're not confused the way we got that is if i plugged in a negative b over 2a in for x this part right here would go away it would basically be 0 and i would be left with my value for k which in this case is 4ac minus b squared all over 4a so that's why we put that there it's just easier than writing 4ac minus b squared over 4a if you want to put that there you can it's the same answer either way okay so let's try a few out this is very very simple so we have f of x is equal to 8x squared plus 16x minus 7. so we have a quadratic function in standard form if you get one that's not in standard form you need to put it in standard form okay so if you see that trick question on the test make sure it's in standard form so what's my a that's the coefficient for the squared variable what's my b that's the coefficient for the variable to the first power and for the formula i gave you you don't need c but just in case you want it c is going to be negative seven that's the constant so a is eight b is 16 c is negative seven okay so i want to find the vertex so the vertex occurs at negative b over 2a comma f of negative b over 2a okay so what's negative b b here is 16. so negative 16 over 2a a is 8 so 2 times 8 is 16. negative 16 over 16 is negative 1. so the vertex the vertex occurs at negative 1 comma then what i want to do i want f of negative b over 2a we know negative b over 2a is negative 1 so we want f of negative 1. and this is equal to eight times you'd have negative one squared plus sixteen times negative one and then minus seven so if we had negative one squared that would be one so this is basically eight times one or eight and then if you had 16 times negative one that's negative 16 so this would be minus 16 and then minus 7. so 8 minus 16 is negative 8 then negative 8 minus 7 is negative 15. so this would be negative 15 here so f of negative 1 is negative 15. so just that quickly we found our vertex no messy completing the square no long lengthy process pretty straightforward so all you have to do is kind of write this down and understand how to find a b and c which is very very simple and you will have your vertex right away so once again the vertex here is at the point negative 1 comma negative 15. all right let's try another one so we have f of x equals 3x squared minus 4x plus 6. so again the vertex formula the vertex will occur at negative b over 2a comma f of negative b over 2a okay so for negative b over 2a this is a this is b and this is c so negative b b is negative 4. so the negative of negative four is positive four over two a a is three two times three is six so we'd have four sixths we could reduce that to two thirds so the vertex will occur you'd have two-thirds for the x location comma then you would have f of two-thirds f of two-thirds right it's f of negative b over two a we found that negative b over two a was two-thirds so it's f of two-thirds so this would be equal to three times two-thirds squared minus four times two-thirds plus six if i square two-thirds i would get four-ninths so this would be four-ninths and then 3 times 4 9 this would cancel with this and leave me with a 3 down here so this would be 4 thirds so i'd have 4 thirds here then negative 4 times 2 thirds you think about that as negative 8 thirds so we'll say minus eight thirds then you have plus six six if i multiply by three over three to get a common denominator that would be plus eighteen thirds and then i could say four minus eight is negative 4 negative 4 plus 18 is 14. so this would end up being 14 thirds so f of 2 thirds equals 14 thirds and that's going to be my location for y for that ordered pair so the vertex occurs at the point two thirds comma fourteen thirds all right so now let's talk about the discriminant so you'll recall when we solved quadratic equations we saw ax squared plus bx plus c equals 0. and we derived the quadratic formula so if we want to solve something in this form we can say that x is equal to negative b plus or minus the square root of b squared minus 4ac all over 2a now what we're going to focus on here is this part under the square root symbol this is known as the discriminant now the discriminant gives us information about the number of x-intercepts we're going to get and we'll cover that in a second but for right now we just want to say if b squared minus 4ac is greater than zero we're going to have two real solutions and therefore you'll see in a second two x intercepts if b squared minus four a c is equal to zero you're going to have one solution and therefore one x intercept you might say well why do you only have one solution here if this guy was zero the square root of zero is zero so you have plus or minus zero that's basically nothing right you can get rid of that and you just have negative b over two a that's your solution the last scenario is that b squared minus 4ac is less than 0. so when we see this we don't have a real solution remember we covered what to do in this case you use complex numbers right we bring in our imaginary unit i but when this occurs we will not have an x intercept all right so here we have f of x equals ax squared plus bx plus c and if we allow f of x or y okay think about this as y for a minute to be zero what are we saying remember when we talked about linear equations when y was zero and we solved for x we were finding the x intercept right because it's the point where y is zero so we are going to cross the x axis so if there are two solutions as in b squared minus 4ac is greater than zero we're going to have two x intercepts or two spots where that graph is going to cross the x axis when y is equal to 0. then if b squared minus 4ac equals 0 1x intercept and then if b squared minus 4ac is less than 0 then no x intercepts right so if you don't have a real solution you will not find an x intercept so if we looked at this example f of x equals 5x squared minus 3x plus 10 and we said how many x intercepts will be here well again you just look at your discriminant so that's b squared minus 4ac a is 5 b is negative 3 c is 10. so what is b squared negative 3 squared is 9 minus 4 times a is 5 times c is 10. so just looking at that i know that this is going to be less than 0. 4 times 5 is 20 29 is 200 you would have 9 minus 200 which turns out to be negative 191 and negative 191 is less than zero so again if the part that's underneath the square root symbol is negative you know you're not going to have a real solution so when this is less than 0 there's no real solution and so there's no x intercepts so 0 x intercepts all right if we looked at another one we have f of x equals negative seven x squared minus two x plus one so again we're looking to see how many x intercepts we have so this is my a negative seven this is my b negative two and this is my c one so the discriminant is b squared so negative two squared that's four minus four times a a is negative seven times c c is one so you might as well say this is plus negative 4 to make it easy negative 4 times negative 7 is positive 28 28 times 1 is still 28. so this is 28 and without even doing the addition we know this is positive and so we know that there are two real solutions or two x-intercepts right four plus 28 will give me 32. so two x intercepts all right so another thing to consider if we're graphing a quadratic function if a is greater than 0 the parabola is going to open up if a is less than 0 the parabola is going to open down so a again is the coefficient of the squared variable okay the coefficient of the squared variable so if that's positive you're going to have a u shape if it's negative you're going to have an upside down u shape so two examples here we have f of x equals negative seven x squared minus three x plus two in this case the negative seven here tells me that this guy is going to open down right it's gonna open down for this one f of x equals three x squared minus five x minus two the positive three tells me this guy's gonna open up all right so if you wanted to graph a quadratic function if your teacher made you do it just as an exercise to see if you could the first thing you would do is determine if the graph opens up or down again look at the value for a is it positive it opens up is it negative it opens down then you want to locate the vertex we have the vertex formula for that that's pretty easy so you want to plot the x-intercepts if you have them again f of x or y will be equal to zero so you're solving ax squared plus bx plus c equals 0. what we've been solving forever right we can use the quadratic formula for that and then for the y-intercept you want to find f of 0 okay remember if i'm plugging in for x in function notation i just put what i want to plug in right there so f of 0 means i'm going to plug in a 0 for x and again we know this from graphing linear equations if i plug a 0 in for x i'm finding the point where x is zero and i'm crossing the y axis okay so that's going to give me my y intercept all right then i want to plot additional points as needed to graph a parabola you only need three points but again you're not graphing a line so the shape isn't always a hundred percent clear again that's why we use these graphing calculators and software to kind of do this for so that it's accurate then lastly you will sketch the graph okay so let's give one of these a shot i'm just going to do one and then if you want to do some additional practice on that that's fine but again i'm just going to tell you as you move out of algebra 2 you get into trigonometry and calculus and college algebra most of the time you're not going to hand draw things you're going to be using your graphing calculator or some type of graphing software all right so the first thing is this guy right here f of x equals negative x squared minus 6 x minus 8. this is negative here so i know it's going to open down so opens down so i know that it's going to have this shape right here all right so the next piece of information what is the vertex so the vertex the vertex will occur where negative b over 2a comma f of negative b over 2a and what is a and b so this is a that's negative one and b is negative six so the negative of negative six is six over two times negative one which is negative two so six over negative two is negative three so the vertex will occur at negative 3 comma now i want to plug a negative 3 in for x so i'd have f of negative 3. this would equal negative and then negative 3 squared minus 6 times negative 3. then minus 8. let's erase this so negative 3 squared is 9. so i'd have the negative of 9 just negative 9 and then negative 6 times negative 3 is positive 18 so this would be plus 18 and then it would be minus 8 so minus 8. so negative 9 plus 18 is going to give me 9. 9 minus 8 is 1. so the vertex will occur at negative 3 comma 1. okay so now we want to figure out what our x intercepts are so we would set this equal to 0 so we would say this is 0 and i would just solve negative x squared minus 6x minus 8 equals 0. again i can use my quadratic formula for that x is equal to negative b b is negative 6 so the negative of negative 6 is 6 plus or minus the square root of b squared negative 6 squared is 36 minus 4 times a a is negative 1 times c c is negative 8. this is all over 2a a is negative 1. so 2 times negative 1 is negative 2. so i think about this as plus negative four to make it easy negative four times negative one is four four times negative eight is negative 32 36 minus 32 is four so you'd be taking the square root of four which is two so your two scenarios here is six plus two is eight eight over negative two is negative four and then your other scenario is six minus two is four four over negative two is negative two so those are x values but remember the y value is going to be 0 in each case so we would say that our x intercepts will occur at negative four comma zero and negative two comma zero okay now where's our y intercept going to occur to find that we want f of 0 right i want to plug in a 0 for x so if this was 0 and this was 0 they'd basically be gone right i just have negative 8 there so f of 0 is negative 8 so we would have an x value of 0 and a y value of negative 8 this is your y intercept now we might need additional points let's just go down to the graph and see what we're looking like let's start with the vertex it's at negative 3 comma one so negative three so three is to the left comma one one unit up and then the next point we want our x intercepts we have negative four comma zero and negative two comma zero so negative four comma zero and negative two comma zero so at this point you have three points you could sketch a parabola so you could start there but the problem is you really don't get a good representation of like where this thing is going to go the angle and whatnot so that's why you try to get some additional points so you get more information i know my y-intercept occurs at zero comma negative eight so zero for the x location negative eight for the y location and then we can get one additional point going let's figure out where x would be if y was negative eight so we know there's one place here at zero comma negative eight but what's the other place so let's go and let's plug in a negative 8 for y or for f of x however you want to think about that so negative 8 equals negative x squared minus 6x minus 8. so i would add 8 to each side of the equation this is gone this is gone so you have negative x squared minus 6x is equal to zero i could factor a negative x out there so negative x and then times the quantity you would have an x plus 6 inside and this equals 0. so one solution is that x equals 0. so we already have that with a y intercept x is 0 y is negative 8. we've got that the other solution would come from this factor right here if x plus 6 was equal to 0 if x plus 6 is equal to 0 subtract 6 away from each side of the equation you say that x is equal to negative 6. so an additional point would occur when x is negative 6 and y is negative 8 okay this is an additional point okay so negative 6 comma 8. so that's going to be right there all right so let's go ahead and attempt to graph this guy again i'm not a good drawer but i'll do my best so okay okay so not the best drawing in the world i realize that but if you wanted to see an accurate representation of what this looks like we can look at a computer drawn graph for f of x equals negative x squared minus 6x minus 8. now the scale is a little bit different my y-axis didn't go that far down but again you can see how perfectly drawn this is and that's what you're going to get when you graph these things with some software a graphing calculator your computer maybe some app on your phone whatever you have that's much preferable to hand drawing but again if you need to hand draw it at least you know the steps hello and welcome to algebra 2 lesson 67 in this lesson we're going to learn about quadratic and rational inequalities so we're going to start out by talking about quadratic inequalities and a quadratic inequality is of the form you'll see ax squared plus bx plus c is greater than zero and of course we have the restriction that a is not equal to zero but other than that a b and c can be any real number that you choose of course when we see something generically we have to choose something to go there so here we have a greater than but of course the greater than can be a less than a less than or equal to or a greater than or equal to so before we jump into how to solve a quadratic inequality i just want you to think back for a moment to when we solved a linear inequality in two variables the very first thing we did was to replace the inequality symbol with an equal symbol that gave us a linear equation in two variables and we were able to graph the resulting line now that line had a special name it was called the boundary log and what's important about the boundary line is it separates the coordinate plane into two regions one region would be the solution region so everything in that region satisfied the inequality the other side of the boundary line would be the non-solution region so everything on that side would not satisfy the inequality so from understanding that we come up with this idea of a test point i can figure out where the solution region is by grabbing a point on either side of that boundary line plugging into the original inequality and seeing if i get a true statement if i get a true statement that means my test point is in the solution region if i get a false statement that means my test point is not in the solution region and the other region on the other side of that boundary line is going to be the solution region now another important thing to kind of note here before we move into this process the boundary line remember was part of your solution if you had a non-strict inequality and it was not part of your solution if you had a strict inequality you're going to see that thinking about that process just for a minute is going to bring back some things that are going to help you get through this here all right so when we're solving a quadratic inequality the very first thing you're going to do is you're going to write the inequality as an equation and you're going to solve that equation so this is going to give you a quadratic equation and you can just use the quadratic formula or in some easier examples you'll be able to factor this guy and just use your zero product property so once you've obtained the solutions from your quadratic equation these solutions will be used on a number line so this is going to be an old-fashioned number line one you learned back in pre-algebra and essentially this is going to tell us what the boundary points are or the end points so these will be the points that separate the solution region from the non-solution region so essentially these solutions will set up intervals on the number line and you'll see this in a minute when we get to an example essentially in each interval you can grab a test point okay so we substitute a test point from each interval into the original inequality so obviously if the test point is successful meaning it satisfies the inequality that interval is part of the solution set if it fails if it produces a false result then that interval is not part of the solution set all right now lastly the end points or the boundary points however you want to call it they are included in the solution set for a non-strict a non-strict so you have a greater than or equal to or less than or equal to inequality and excluded for a strict inequality so if you have strictly less than or strictly greater than they're going to be excluded from the solution okay so now let's jump in and just look at a problem once we kind of get going on these you'll see that they're very very easy to solve so we have 3x squared minus 10x plus 3 is greater than 0. so the very first thing i want to do is i want to make this into an equation so 3x squared minus 10x plus 3 is equal to 0. okay so i know that this is a quadratic equation i can use my quadratic formula to solve this pretty easily i just want the values for a b and c so a is 3 b is negative 10 and c is 3. so we could say that x is equal to negative b here b is negative 10. so the negative of negative 10 is 10 plus or minus the square root of b squared negative 10 squared is 100 and then you have minus 4 times a times c a is 3 c is 3 so you say 3 times 3 is 9. so 4 times 9. all right so 4 times 9 is 36 if i had 100 minus 36 that would be 64. so this would be 64 here and this is all over 2a a here is 3 so this is all over 6. square root of 64 is eight so this sets up two scenarios for us we'll have x is equal to ten plus eight which is eighteen over six so that's going to be x equals three or ten minus eight which is 2 over 6 2 over 6 would reduce to 1 3. so x equals 1 3. all right so now we want to take these guys and bring them down to the number line all right so we're just going to paste the answers here so we have x equals 3 or x equals 1 3. so the whole idea here is that those answers are going to set up intervals for us so when x equals 3 that's going to be an endpoint so i can kind of circle three and you can draw just a little vertical line going down and then at x equals one-third you do the same so i don't have a notch for one-third i'd say to be about right there it doesn't need to be precise you're just doing this for visual representation so i can just kind of draw a vertical line there and essentially you now have three intervals here so you have anything left of one third so let's just call that interval a just for just for talking okay then you have anything greater than one-third but less than three let's call that interval b then you have anything that's greater than three okay anything greater than three so that's interval c so now that i have these intervals set up i can just grab test points from each interval see if they work if they do work that interval is in the solution set if they don't work the interval is not in the solution set so from interval a the easiest number to grab would be 0. so i'm going to go back up to my original inequality and i'm going to plug in a 0 for x so a 0 there and there what would happen this is going to cancel this is going to cancel i'll have 3 is greater than 0. so we know that's going to work so anything to the left of 1 3 works as a solution so i'll just write that this is true okay this is true any value you plug in there would work so what about this region where we have the capital letter b so i can just choose something like one okay one is in that interval and if i plugged in a one for x what would i get so one squared is one three times one is three minus ten times one is ten plus three is greater than zero three minus ten is negative seven negative seven plus three is negative four is negative 4 greater than 0 no it is not so that produces a false result and so the values in this interval we'll call it interval b again do not satisfy the inequality and so it's not going to be part of the solution set so i'll write false here false then what about interval c well this is anything greater than 3 so i could just choose 4 and we'll go back here and just plug in a 4 for every x so 4 squared is 16 3 times 16 is 48 minus 10 times 4 is 40 plus 3 is greater than zero so 48 minus 40 is 8 8 plus 3 is eleven eleven is greater than zero so this results in a true statement so we can go ahead and say that this is true over here okay so let's erase this we don't need this information anymore so all i'm going to do is write my solution and i can think about this graphically i can think about an interval notation i can also think about it using set builder notation there's just lots and lots of ways to write the solution graphically i know that 1 3 and 3 are not included if i go back up i have a strict inequality so that means my endpoints will not be included so i would put a parenthesis at one third facing to the left and i would shade everything to the left to indicate that hey one-third doesn't work but anything to the left of one-third would work then at three i'm going to put a parenthesis facing to the right and then i'm going to shade everything to the right so 3 doesn't work but anything to the right of 3 would now that's the graphical solution we saw that before back when we started solving a linear inequality in one variable right we've been doing this forever and another thing we can do we could use set builder notation so we could say the set of all x such that we just list the conditions so one condition is that x is less than one-third then or the other condition would be x is greater than three so any value we choose for x that's less than one-third or greater than 3 will work as a solution in this inequality then another thing we can do we can do interval notation so from negative infinity up to but not including one third it's this set and it's going to be the union with the other set which is not including three but anything larger so before we kind of move on to the next problem i want to demonstrate how you could go about solving this graphically the first thing you would think about is you would think about just f of x or y is equal to 3x squared minus 10x plus 3. say okay this produces this parabola right here i would replace f of x with 0 and i would solve we know the solutions are x equals 3 or x equals 1 3. so that gives me an x intercept that is 3 comma 0 which is right here and it gives me another x-intercept which is one-third comma zero which is right here so again these x-intercepts will represent boundary points or places where you go from a solution region to a non-solution region right it's separating the two now let me just scooch this down a little bit let me scooch this down let me just say this is now greater than zero so what am i asking for in this problem graphically i'm just asking for what x values make y or f of x greater than 0 well i can easily see that starting at 3 but not 3 right because an x value of 3 gives me a y value of 0 and 0 is not included because it's a strict inequality but anything to the right of 3 going this way as we move this way we can see that the y values are larger than 0. it's the same thing over here 1 3 is not included because it makes y 0 and again we have a strict inequality but anything going to the left right as x gets smaller than 1 3 that's also going to give me a y value that's greater than 0. so this is a method you could theoretically use to solve this guy i think it gives you a good insight to look at it a few times but it's just not very practical right if you had to take out graph paper and graph this guy and like set up a parabola it's just a lot of extra work right maybe if you're using a graphing calculator maybe if you're using software it could be easier for you but i think the method that we just used is much preferable to graphing all right let's take a look at one more of these so we have negative 2x squared minus 5x plus 96 is greater than negative 9x of course we want to add 9x to both sides of the inequality that'll cancel you'll have negative 2x squared negative 5x plus 9x is positive 4x and then plus 96 and this is greater than 0. now if you want to to make the numbers you're working with a little bit smaller we can divide both sides of the inequality by 2 or negative 2. but if we're dividing by negative 2 remember we have to flip the direction of the inequality symbol so this would now be a less than so this would be x squared minus 2x so minus 2x and then 96 over negative 2 is 48 so minus 48 and this is now less than 0 over negative 2 which is 0. all right so this is going to be the inequality we're working with here and i'm just going to change this into an equation so i'll change this into x squared minus 2x minus 48 is equal to 0. i can easily solve this using factoring or you can use the quadratic formula it doesn't really matter let's just go ahead and use the quadratic formula so that we're consistent so x equals negative b in this case b is negative 2. so the negative of negative 2 is positive 2 plus or minus the square root of b squared negative 2 squared is 4 minus 4 times a a is 1 here so times 1 times c c is negative 48. okay this is all over 2a a is 1 so just over 2. so the easiest way to do this is just think about this as plus negative 4 here so negative 4 times negative 48 is 192. so you would have 4 plus 192 which is 196 and the square root of 196 is 14. so this would be 14 here so my two solutions would be x is equal to 2 plus 14 is 16 16 divided by 2 is 8 or x is equal to 2 minus 14 is negative 12 negative 12 divided by 2 is negative 6. so x either equals 8 or x equals negative 6. okay so let's take this down to the number line okay so we have x equals 8 so 8 is right here and then x equals negative 6 so that's right here so again just draw yourself a little vertical line at each endpoint and you're dividing the number line into region so left of negative 6 we'll call that region a or interval a between negative 6 and 8 we'll call that interval b and then to the right of 8 we'll call that interval c so in interval a let's just pick negative 7. so does negative 7 work as a solution so plug in a negative 7 here and here negative 7 squared is 49 then you'd have negative 2 times negative 7 which is plus 14 and then minus 48. so is this less than 0 we know without even doing the problem it's not 49 minus 48 would be 1 1 plus 14 is 15. so we know that 15 is not less than zero so that's false so we can say this is false then what can we say about this interval or region b we could just choose zero and if we plugged in a zero for x we know that this would go away you'd have negative 48 is less than zero that's true so this is true and then for region c we could just choose 9. so if i plugged in a 9 for x i plugged in a 9 for x 9 squared is 81 then negative 2 times 9 would be minus 18 and then minus 48 and this is less than 0. 81 minus 18 is 63. 63 minus 48 is 15 so 15 again is not less than zero so anything in region c would fail so this is false now you're working with a strict inequality here so your endpoints or your boundary points are not going to be included so graphically i would put a parenthesis at negative 6 facing to the right and a parenthesis at 8 facing to the left and i would just shade everything in between so anything larger than negative 6 up to but not including 8 would work as a solution for that inequality so in set builder notation we could say the set of all x such that x is greater than negative 6 and less than 8. and then to kind of wrap this up i'm just going to use interval notation i'm going to have a parenthesis negative 6 and then up 2 but not including 8 so another parenthesis so this is my interval again anything larger than negative 6 up 2 but not including 8. all right so now let's talk a little bit about how to solve a rational inequality so with a rational inequality you're really just thinking about the exception here where the denominator is made into zero remember we can never divide by zero so that's the one thing that you really got to pay attention to here everything else is pretty similar so the first thing is to write the inequality such that 0 is on one side and a single fraction is on the other then once you've done that you're going to find all values that make the numerator or denominator zero so i'll take the numerator i'll set it equal to zero and i'll solve it i'll take the denominator i'll set it equal to zero and i'll solve it now any value that creates a zero denominator is excluded from the solution set excluded from the solution set in the case where you have a value that makes the numerator equal to zero you have again you've got an end point there so you've got to pay attention to whether it's a strict inequality or a non-strict inequality all right so again we're using the solutions that we got are the values to create intervals on the number line and these are going to be our endpoints or our boundary points and we're just going to test the number from each interval to determine the solution set so very similar to what we just did again we just have to remember that we can't divide by zero so as a last point we have here that endpoints are included when they do not make the denominator zero and we have a non-strict inequality so if you have a strict inequality they're never included if you have a non-strict inequality they're included only in the circumstance that they do not make the denominator zero if you put something in there that makes the denominator zero you're going to get the wrong answer okay it's a common mistake because you go from quadratic inequalities to rational inequalities in the same section you get used to oh i have a non-strict inequality so i throw it in my answer and then it's wrong all right let's look at the first example so we have x plus 1 over x plus 3 and this is greater than 5. so the very first thing i'm going to do i'm going to subtract 5 away from each side of the inequality so i will have x plus 1 over x plus 3 and this is minus 5 and this is greater than 0 now so how do i write this as a single fraction well i just need to get a common denominator going so let me kind of move this down a little bit let me kind of scooch down get some room going to get a common denominator i would just multiply this by the quantity x plus 3 over the quantity x plus 3. very easy so i would have x plus 1 and then negative 5 times x would be minus 5x and then negative 5 times 3 would be minus 15. this is over the common denominator of x plus 3. of course this is greater than 0. so in the numerator if we combine like terms x minus 5x is negative 4x and then 1 minus 15 is negative 14. so you would have negative 4x minus 14 and this is over x plus 3 and again this is greater than 0. now what i want to do i want to take the numerator and i want to set it equal to 0. so negative 4x minus 14 is equal to 0. okay so i would just add 14 to each side of the equation here that's gone i'll have negative 4x is equal to 14. divide both sides by negative 4. that'll cancel with that i'll have x is equal to we know this would be negative 14 is divisible by 2 and so is 4 so 14 divided by 2 is 7 4 divided by 2 is 2. so this would be negative 7 halves so x equals negative 7 halves that's going to be one endpoint and that's going to be from the numerator so the other endpoint is going to come from setting x plus 3 equal to 0. in this case i could just subtract 3 away from each side of the equation i get x equals negative 3. so x equals negative 3 this comes from the denominator now in this case i have a strict inequality so neither endpoint will be a solution but in the case where it's a non-strict inequality i can take negative 7 halves and it would be a solution right because you think about the fact that if this was a greater than or equal to if i plugged in a negative 7 halves for x the top part would become zero zero divided by anything other than zero would give me zero so i would get zero is greater than or equal to zero which would be true right zero equals zero but because i pull this away and make this a strict inequality this will not satisfy the inequality so it's not part of the solution set the x equals negative 3 will never be part of the solution set because if i plug in a negative 3 down here i get negative 3 plus 3 which is 0 and i cannot divide by zero so it's very important that you pay attention to what's going on i always tell my students to label where everything came from so you don't forget so let's copy this and so these are going to be my end points so negative 3 is right here let's make a little vertical line there and negative 7 halves where's that going to be well it's negative 3.5 really if we thought about it so it's about halfway between negative 3 and negative 4. so let's say it's about right there let's say this is negative 7 halves let's make another vertical line there so it's going to set up a kind of small region let's say this is region a this is b and this is c so in region a let's pick something easy let's just take negative 5. so i'm going to plug back in for x in the original inequality so i'm going to plug in a negative 5 there and there so negative 5 plus 1 is negative 4 so you'd have negative 4 over negative 5 plus 3 is negative 2. so this is supposed to be greater than 5. negative 4 divided by negative 2 is positive 2 so is 2 greater than 5 no so this would be false this is false we can say this one was false okay so it's a pretty narrow region where we have between negative 3 and basically negative 3.5 but we can use negative 3.25 because that would lie between those two so let's go back up and let's say we have negative 3.25 so negative 3.25 so negative 3.25 plus 1 would be negative 2.25 then this is over negative 3.25 plus 3 would be negative 0.25 if i divided negative 2.25 by negative 0.25 i would get 9. so this would be 9 is greater than 5 which is true so for region b i could say this is true and for region c let's just pick 0 if i plugged in a 0 for x what would i get so 0 there and a 0 there essentially these are gone you'd have one third right you have 1 over 3 which is not greater than 5 so that's false so this is going to be false and so we're ready to give our solution we know that the endpoints negative 7 halves and negative 3 are not included so essentially all we would have here is we would graph a parenthesis here at negative 3.5 or negative 7 halves facing to the right and a parenthesis here at negative 3 phases to the left and then we just graph everything in between so that's your graphical solution i can erase the false and true and the a and the b and the c i don't need any of that anymore and we can say that in solution set notation we can say the set of all x such that x is greater than negative seven halves and x is less than negative three and then lastly if we were to use interval notation put a parenthesis negative seven halves comma you would have negative three and then a parenthesis all right pretty easy overall let's just look at one more to kind of wrap things up we have x minus two over x plus two and this is less than or equal to 2. so again i want one single fraction on one side and then i want 0 on the other so i'm going to add negative 2 to both sides or subtract 2 away however you want to think about that so i would have x minus 2 over x plus 2 then minus 2 or i could put plus negative 2 again however you want to think about that this is less than or equal to 0. so to get a common denominator i'll multiply this by the quantity x plus 2 over x plus 2. and what that's going to give me i'll have x minus 2 and then negative 2 times x so minus 2x and then negative 2 times 2 so minus 4 over the common denominator of x plus 2 and this is less than or equal to 0. so x minus 2x is negative x and then negative 2 minus 4 is negative 6. so you have negative x minus 6 negative x minus 6. this is over x plus 2 and this is less than or equal to 0. okay so pretty easy to get to that point and then again all i want to do i want to take my numerator which in this case is negative x minus 6 set it equal to 0. i could solve this by adding 6 to each side of the equation that would go away i'd have negative x is equal to 6 multiply both sides by negative 1 you get x is equal to negative 6. so for the numerator or the numerator x is equal to negative six okay for the denominator we have x plus two so x plus two is equal to zero subtract two away from each side of the equation you get x equals negative two so for the denominator x equals negative two so in this case we have a non-strict inequality you have less than or equal to if i use x equals negative 6 plug in a negative 6 here so you get negative 6 minus 2 that's going to give you negative 8 so you'd have negative 8 over you'd have negative six plus two which would give me negative four this is less than or equal to two negative eight over negative four is two so what you end up with here is a true statement two does equal two so this part right here x equals negative six this is going to work out for us what's not going to work out for us is the denominator part where it says x equals negative 2. this will not be a solution if i plug in a negative 2 for x i'm going to get a 0 denominator so you've got to pay attention to those end points you get one's going to work here one's going to not so let's get this down there so again x equals negative 6 that's going to be right here and i'll draw a little vertical line for myself and then x equals negative 2 is right here i'll draw a little vertical line for myself there so let's call this region a let's call this region b let's call this region c so in region a we can just choose negative 7. so let's erase this and if i plugged in a negative seven here and here what would happen negative seven minus two is negative nine negative seven plus two is negative five negative nine over negative five is nine fifths so nine fifths is this less than or equal to two well yes it is i would need a numerator of ten to produce a result of 2 right 9 divided by 5 is going to be less than 2. so this satisfies the inequality we could say this is true so anything in region a is true what about region b we could choose something like negative four so for negative four if i had negative four minus two i get negative six if i had negative four plus two i get negative two negative 6 divided by negative 2 is positive 3. so you'd have 3 is less than or equal to 2. that's false 3 is not equal to 2 and 3 is not less than 2. so that does not work so anything in region b is going to be false then for region c i can choose 0. so if i plugged in a 0 for x i would basically have negative 2 over positive 2 which would be negative 1 and we would say negative 1 is less than or equal to 2 that's true so for region c we would say this is true now again we think about our solution the x equals negative 6 the one that came from our numerator is going to be part of our solution so let me erase all this and i'll say x equals negative 6 is part of the solution so i'll put a bracket there and then anything to the left of that is going to work anything to the left now at x equals negative 2 negative 2 is not going to be part of the solution right that made the denominator 0. so i'm going to put a parenthesis there and then i'm going to shade everything to the right right so anything that is negative 6 or less and then also anything larger than negative 2 would work as a solution here let's erase this in solution set notation we'll say the set of all x such that x is less than or equal to negative 6 or x is greater than negative two and then an interval notation coming from negative infinity up to and including negative six so you get a bracket there and then the union width will have a parenthesis here because negative 2 is not included out to positive infinity hello and welcome to algebra 2 lesson 68 in this lesson we're going to learn about inverse functions so at some point in your algebra 2 course you're going to run into this topic known as inverse functions and really this is going to lay the groundwork for us and we're going to start talking about exponential functions and logarithmic functions as soon as we conclude with inverse functions so when we think about inverse functions basically we're thinking about functions that reverse each other but before we kind of jump in and start talking about inverse functions i want to just make sure that you understand the basics of a function so we want to just do a quick review at some point in either algebra 1 or at the beginning of algebra 2 you talked about functions and we learned that a function was nothing more than a relation or a set of ordered pairs where each x value corresponds to exactly one y value so there's a clear association with a function i give you an x value and you can tell me what the y value is that's associated with it so as an example let's say we look at this set of ordered pairs here this relation and we say is this a function so for each x value is there a unique y value that's associated with it in other words is there a clear association so to see this let's kind of draw a typical picture that you'd see in this section so let's set this up and say that this is the domain this is the domain or again the set of allowable x values so what exists in our domain here well we just have a set of ordered pairs so remember the ordered pairs are x comma y so the first component or the first entry in each ordered pair represents the x value so we have an x value of 3 we have an x value of 2 we have an x value of eight and then we have an x value of six so our domain consists of those four numbers when we think about the range okay the range that's the set of allowable y values so in this case our range consists of 7 5 we have a 7 again so i'm not going to double list that and then 3. now again with a function for every x there's going to be a clear association with a y value so if i ask you what is y when x is 3 well when x is 3 y is 7. so there's a clear association there when x is 2 y is 5. so when x is 2 y is 5 when x is 8 y is 7. so when x is 8 y is 7 and then when x is 6 y is 3. so when x is 6 y is 3. so this is a function because for every x value there is a clear association with one and only one y value now the common thing that comes up in this section is that hey well i have a 7 here for y and then i have a 7 here for y as well and you'll see that i listed this up here we said that a function is allowed to have y values that correspond to different x values so we have that here and that's allowed because again there's still a clear association if i say hey x is 3 what's y i know the answer is 7. if i say hey x is 8 what's y i know the answer is 7. so it's ok that that occurs because i have a clear association what i cannot have let me kind of scroll back down with a function if i erase this and this and let's say i change this up a little bit and i say now i have the ordered pair three comma five so let's erase two from the domain and let's say that three is now mapped to or linked with seven but three is also linked up with 5. so the reason this violates the definition of a function is that x does not have a clear association with a y value if i say hey if x is 3 what's y you do not know whether to answer 7 or 5 right y could now be 7 or it could be 5. so that's why we always say for each x value there is one and only one y value so there can be a y value that's associated with more than one x value but there can't be an x value that's associated with or linked up to more than one y value now we're going to introduce the concept of a one to one function and you have to understand this to be able to get into this topic of inverse functions so a one two one function let me just highlight that has a stricter definition with a one-to-one function we see that each x value corresponds to one y value that's the same as before okay we've always had that rule but now we're going to say each y value corresponds to one x value all right so when we have a one-to-one function and we'll learn how to determine if we have one or not later on in the lesson but assuming that you do have one we can create a function known as the inverse okay the inverse and this is by interchanging so in other words swapping x and y so your x values become y values your y values become x values so we talk about the set of x values as the domain we talk about the set of y values as the range so the domain of the original function will become the range of the inverse and the range of the original function will become the domain of the inverse so if i have this function f and it consists of the following three ordered pairs two comma three a comma one and nine comma four and let's say we want to set this up to where g is the inverse of f so these two we're just going to interchange x and y so if this is x and this is y then i just want to interchange the 2 and say that this first ordered pair would be 3 comma 2. the second will be one comma eight and the third one would be four comma nine so the domain for f the domain for f is what it consists of two 8 and 9. so those values are the x values in the original function f now we interchanged them when we created the inverse which is g so now the x values became y values so when we think about the range the range for g is what it's going to be these values so it's going to be 2 8 and 9. and you can see that you have 2 8 and 9 in the y location for each ordered pair right so that's now your range when we think about the range for f so the range for f it's what it's three one and four so three one and four so then when we think about the domain four g it's going to be the range from f right so it's going to be this right here so 3 1 and 4. so you see you have 3 1 and 4 in your x positions there so just a little simple example here for you to see what we mean by inverse functions and how the x and the y values are interchanged so first we start out with f again this is a one to one function you have x values of 2 8 and 9. so nothing is duplicated there then you have y values of 3 1 and 4 nothing duplicated there so when that occurs for each x there's one y and for each y there's one x so it's one to one so then we can create this inverse which is g by just interchanging x and y so we have the ordered pair two comma three we swapped everything so this became three comma two so now three is the x value two is the y value then we saw we went from eight comma one to one comma eight and from nine comma four to four comma nine so obviously the domain or the set of allowable x values for f became the range for g because we swapped x and y values and then the range for f became the domain for g again because we just swapped x and y values so the next thing we want to talk about is the notation that's involved with an inverse function so this is a very big source of confusion for students so the inverse of a one-to-one function f is notated as f inverse so that's how that's read it's not f to the power of negative one like it appears to be so this is a big source of confusion if i have a function f of x and it's a one to one function when i notate the inverse i say f inverse like that okay that's what it means so f inverse of x so these two are inverses of each other so these are inverses and basically that just means these two functions will reverse each other now one thing i don't want you to do let's say you see f inverse of x i do not want you to think that this means 1 over f of x okay this is a common mistake again because you see that negative 1 there you're thinking about the rules of exponents this is wrong okay wrong and again i don't know who set this up but it's really really confusing when you look at the notation especially if you don't have somebody there to tell you if you just read it in a book and they don't tell you that you might really think that this is f to the power of negative 1 right it's not so this is f inverse of x again that's the inverse of f of x all right so f inverse consists of all ordered pairs y comma x where x comma y are components of f so we saw that with the simple example we saw that every x value became a y value in the inverse so if this is the original right here my x value here goes into the y position and then my y value here goes into my x position so the x values become y values the y values become x values and we talked about this already the domain of f inverse is the range of f so in other words the x values of the inverse came from the y values of the original and then the range or the y values of the inverse is the domain of f so the y values of the inverse came from the x values of the original function so i know this looks a little complicated in this form but when you break it down like that it's super simple x values of the original function become y values of the inverse and y values of the original function become x values of the inverse all right so you might see this notation in your textbook when you first get started and a lot of students get really confused with this one too so i just want to take a moment and kind of break this down i want you to see an example where you can look and understand how the functions reverse each other when they're inverses okay so if we see f of f inverse of x we say this is equal to x we also see that f inverse of f of x is equal to x so you might say what in the world does that notation mean so let's look at an example and then we'll kind of work through that and then we'll come back in the end we'll explain what this generic notation means let's say you had f of x is equal to 2x plus 3. okay now i haven't taught you yet how to determine if something is one to one we'll get to that later on in the lesson but i just want you to believe me when i say it is a one-to-one function and therefore it will have an inverse now how do we find the inverse and we'll talk more about this later but the simplest method to do this is to replace f of x with y so we know f of x just took the place of y when we started talking about functions y or f of x is nothing more than the dependent variable that's all it is so y is equal to or y is the same as this quantity 2x plus 3. okay now all we need to do is swap x and y so y becomes x and x becomes y all right so now we're going to solve for y again so to do that i would subtract 3 away from each side of this equation and that would give me what i would have x minus 3 on the left is equal to this would cancel so just 2y and if i wanted to solve for y i would divide both sides of the equation by 2 and i would have x minus 3 over 2 is equal to y right this would cancel with this let's write this over here so you have your original function which is y equals 2x plus 3 and then you have your inverse which is y equals x minus 3 over 2. so these two functions are going to reverse each other now the notation for this remember y is nothing more than f of x when we're talking about a function and if these two are inverses and this is the original this is the inverse so we notate it by saying f inverse of x okay very very simple all right so now we have our two inverse functions we have f of x equals 2x plus 3 and we have f inverse of x equals x minus 3 over 2. so let me give you a little bit of insight right now into how these two functions are considered inverses of each other or in other words how they reverse each other so if i start out with the original function f of x equals 2x plus 3 i take x whatever value i choose for that doesn't matter what it is i multiply it by 2 to start so kind of step 1 i multiply by 2. then in step 2 once that's done i'm going to add 3 so i'm going to add 3. so as an example let's say i choose 4 for x so the notation for that with a function is f of 4 just means i'm plugging in a 4 for x so this equals 2 times 4 plus 3. 2 times 4 is 8 8 plus 3 is 11. okay let's think for a second what happens with the inverse so f inverse of x we first subtract away 3 from whatever x is so step one we subtract away 3. now we can see that subtracting away 3 would undo the addition of 3 that we saw in the original function then for step 2 we see that once this numerator is done right we've subtracted away 3 we're going to divide by 2. so we're going to divide by 2 and that undoes the multiplication by 2 that we did in the original step of the first function so you can see how these two would reverse each other so if i took f inverse of 11 of 11 i would reverse the process and i should get back to four so this would be equal to what so let me kind of scroll down a little bit get some room going this would be plug in 11 for x minus 3 over 2 11 minus 3 is 8 so you'd have 8 over 2 which would give me 4. so you can see how the x value of 4 became the y value of 4 in the inverse and the y value of 11 became the x value of 11 in the inverse so in other words in the original function we have the ordered pair 4 comma 11 meaning i plug in a 4 for x and i get an 11 for y or for f of x in the inverse function these two are interchanged so in other words i plug in 11 for x and i get a 4 for y so you can see here 11 started out as y became x 4 started out as x became y so we're just interchanging the x and the y values now let me erase this and i'll show you this generically and then we'll kind of move on to some other topics and we'll swing back to this let's scroll up a little bit and let's see if we can make sense of this notation here so the first thing is that we have f of f inverse of x what does that mean when we say f inverse of x so remember when we think about an equation the two sides are equal to each other meaning they're the same as so you might as well think of this forget about the fact that it's an inverse for a second we could really just say this is a function or an equation where we say we have some dependent variable let's just call it y is equal to x the independent variable minus 3 over 2. so if this right here is just asking me for what is y in the inverse well y equals this guy right here so i could plug that in right here i could plug that in right here and i could say in the original function so f of x minus 3 over 2 should be equal to x let's see if that's the case so this is equal to 2 times i'm plugging this in for x so i'm plugging in an x minus 3 over 2 and then plus three so what happens here is that this two cancels with this two and i'd be left with what i'd have x minus three plus three minus three plus three those two cancel i'm left with x which is exactly what it tells me i would be left with if i think about f inverse of f of x f of x or again you could just think about this as y if it's giving you a mental roadblock this is equal to 2x plus 3 or it's the same as 2x plus 3. so i could just plug this in here i'm saying hey if i have the inverse function and i plug in the y from the original function i should get x back right that's what i would expect because if i take the y from the original function and i plug it into the inverse i know i'm going to get x so this is f inverse of we'll say 2x plus 3 and this should give me x back does that happen well let's go ahead and calculate this so let me kind of scooch this down and let's see what this gives us so we would plug in a 2x plus 3 there so we would have 2x plus 3 minus 3 over 2. we know 3 minus 3 would give us 0. so this would basically be 2x over 2 the 2's would cancel and again i'm just left with x so hopefully this kind of clarifies some things for you i know a lot of students will see this notation just really freak out and say what does it mean again it all goes to this concept where the x and the y values are just interchanged so the x values from one function will become y values in the other and the y values in one function will become x values in the output all right so let's kind of roll through some problems now and i'm going to talk to you a little bit about how to determine if something is a one-to-one function so first let's look at this simple example we have f here and we see that we have nine comma negative two so this is our first ordered pair and seven comma one then negative 4 comma 4 then 2 comma 5. we then see we want f inverse so the first question would be is this a 1 to 1 function i see for my x values i have 9 7 negative 4 and 2. so what you're looking for in a problem like this you don't want any duplicate x values because that would mean the same x value is linked up to more than one y value then you want to look for duplicate y values so negative 2 1 4 and 5 don't have that so you're looking for that to say hey each y value is only linked up to one x value so this function is 1 to 1 so it will have an inverse again you just swap the x and y values so negative 2 becomes the x value here 9 becomes the y value you know so on and so forth so this would be 1 comma 7 4 comma negative 4 and then 5 comma 2. so f inverse contains the ordered pairs negative 2 comma nine one comma seven four comma negative four and five comma two so the domain of f which consists of nine seven negative four and two became the range of f inverse right nine seven negative four and two the range of f which consists of negative 2 1 4 and 5 became the domain of f inverse right negative 2 1 4 and 5. all right let's take a look at g so will g have an inverse so the first thing is is g a one to one function i see have an x value of six negative twelve seven and eleven so no duplicate x values there so we're good so we think about duplicate y values so you have five four one and then 5 again so here this fails the definition of a one to one function so it's a function because each x is associated with only one y but it's not a one to one function because each y value in this particular case 5 is not associated with one unique x value right the y value of 5 is associated with or linked up to 11 and then also 6. so this fails the definition of a one-to-one function and the reason you want something to be one to one think about if i tried to make the inverse of g so let's say i had g inverse like this and you set it up well i reverse everything so this is five comma 6 this is 4 comma negative 12 this is 1 comma 7 and then this is 5 comma 11. so you say hey if x is 5 what's y when we think about the inverse well it could be 6 but it could also be 11. so in other words it's got to be a 1 to 1 correspondence to say hey x is 5 here y is going to be this right in this case x is 5 we don't know whether to say y is 6 or y is 11. so it's not a one to one function and so it will not have an inverse so we'll just put not a one two one function all right so let's talk a little bit about how you would go about determining if you have a one-to-one function or not so you'll recall that when we talked about functions we had this idea of a vertical line test essentially if a vertical line hits the graph in more than one place we do not have a function this would tell us that the same x value corresponds to more than one y value if we take a look at something like a linear equation we know this is a function i could draw as many vertical lines as i want and i'm never going to impact the graph in more than one spot so you see here each vertical line would only touch the graph once so you think about your x value it's associated with one and only one y value there so this is a function here's an example of something that's not a function clearly this would fail the vertical line test so if we look at all these vertical lines you could see that for a given x value we have more than one y value that's associated with it so for each x we have more than one y value associated with it so this is not a function so this is not a function right so now let's talk about the horizontal line test so this is similar in concept to the vertical line test but the horizontal line test is going to tell us if we have a one-to-one function we think about a horizontal line a horizontal line is y equals something so it's the same y value if it impacts the graph in more than one location that same y value is going to correspond to more than one x value and so we don't have a one to one function so we can say here if any horizontal line intersects the graph of a function in no more than one spot the function is one to one right if it does impact the graph in more than one spot the function is not one to one so this is the graph of f of x or you could say y is equal to x squared minus two so obviously this is not a one-to-one function so you look at these horizontal lines and you can see that they impact the graph more than once in each case and so this fails the definition of a one to one function because for the y values you're given they have more than one x value associated with it now you could solve this guy for x and see why that's the case i can add 2 to both sides of the equation that would be y plus 2 is equal to x squared i could take the square root of both sides if i do that what happens i get plus or minus the square root of y plus 2 is equal to x so think about that for a given y value i'm going to now get two associated x values because of this plus or minus operation let's pick something easy from the graph so we know that an x value of 3 would correspond to a y value of 7. so if i plugged in a 7 for y 7 plus 2 would be 9 i would have plus or minus the square root of 9. square root of 9 is 3. so i'd have positive 3 or negative 3 for x so that y value of 7 is associated with an x value of 3 and then also an x value of negative 3. so this is not 1 to 1. so not 1 2 1. all right what about this function right here obviously this is a one-to-one function you can see that each line i drew would only impact the graph in one and only one spot so for every y value here there's only one associated x value all right so the next graph is one that causes a little bit of confusion this is the graph of f of x is equal to x cubed minus 1. so i'm going to tell you right off the bat this is a one-to-one function so i could go through and draw horizontal lines and you can see that all these lines would just impact the graph once where students get confused or where students get tricked here most computer software most graphing calculators will draw this part right here looking semi-flat right so it kind of looks like a horizontal line would impact the graph there in more than one location but that's not the case if you really just think about this for a second when x is 0 okay when x is 0 y or f of x is negative 1. so you have the ordered pair 0 comma negative 1. that exists right there every value to the left of 0 for x as we move towards negative 1 is going to decrease the y value so this guy is going to drop this way it doesn't drop that quickly when you look on the graph but it does drop so if i took something like negative one half and plugged it in for x i would get a smaller value than negative one right this guy's decreasing if i looked at the values between 0 and 1 it's the same thing this guy is increasing here it looks like a horizontal line right there would impact the graph more than once but it actually wouldn't so that's where you got to kind of look at the equation and think about well all right if i plugged in something infinitesimally larger than zero would i still get negative one as a result the answer to that is no so obviously it's not a horizontal line that would touch the graph in more than one location so this is a one to one function so this is a one two one function all right so now let's get to kind of the main thing the thing that you're going to see most often in this section you're going to be asked how can you find the inverse of a function i showed you earlier when we talked about that notation we want to interchange x and y solve for y and then replace with f inverse so i'm going to start out by just replacing f of x with y and then i'm going to say y equals negative 10 plus 3x over 2. now i'm going to interchange y and x so this would be x equals negative 10 plus 3y over 2. so i just interchanged x and y didn't do anything fancy then i'm just going to solve for y so if i do that up here i would multiply both sides of the equation by 2. so you'd have 2x is equal to if i multiply over here by 2 i would cancel that denominator of 2. so i would just have negative 10 plus 3y then i'm solving for y so let me add 10 to both sides of the equation so this is 2x plus 10 is equal to that's cancelled so just 3y the last step is to divide both sides of the equation by 3 and i would get that 2x plus 10 over 3 is equal to y okay so let's erase everything real quick and this guy y i'm going to now replace with f inverse just to notate that hey this guy is the inverse of this guy that's all it is so f inverse of x is equal to 2x plus 10 over 3. so these two are inverses of each other and again if you wanted to think about how you could prove that to yourself well if i plugged in a 4 for x here so if i had an x value of 4 you'd have 3 times 4 which is 12 12 minus 10 would give me 2. so i have 2 over 2 which is 1. so on this function we have an x value of 4 a y value of 1. remember the x and y's are interchanged so if i take a y value of 1 and plug it in for x in the inverse i should get a result of 4. so if i plugged in a 1 there what would i get 2 times 1 is 2 2 plus 10 is 12 12 divided by 3 is 4. so here the ordered pair would be what it would be 1 comma 4. again the y value became x the x value became y so the values here are interchanged all right let's take a look at another example so we have f of x is equal to negative four-fifths x again to do this as simply as we can y equals negative four-fifths x we interchange x and y so we interchange x and y and then we solve for y so i would just multiply both sides by 5 over negative 4 and so this would cancel with this this with this we'd have y is equal to negative 5 4 x or we could say f inverse of x is equal to negative 5 4 x so if you wanted to check this let's just choose something for x let's say we chose i don't know let's say five so f of five would be equal to negative four times five over five the fives would cancel i'd be left with negative four so if i take negative 4 which is the y value here and i plug it in for x here i should get 5 back so if i plug in a negative 4 here this cancels with this and i've got negative 1 negative 5 times negative 1 is 5. right i go right back to that all right let's say we saw f of x equals x cubed minus 2. so again we say y equals x cubed minus 2 and i interchange y and x so this would be x equals y cubed minus two and to solve for y i would add two to both sides of the equation so that's going to give me x plus 2 is equal to y cubed and to solve for y i'm just going to take the cube root of each side and so what's going to happen is i'll have y is equal to the cube root of x plus 2. let me erase all this we can do the proper notation so these two are inverses so we would say f inverse of x is equal to the cube root of x plus 2. and again if you pick a value for x let's say you pick 3. so 3 cubed we know that's 3 times 3 is 9 minus 3 is 27. subtract away 2 you get 25. so f of 3 is equal to 25 so i should be able to take 25 and plug it in for x and get back to 3. right so if i take f inverse of 25 i expect the result to be 3. so the cube root of 25 plus 2 25 plus 2 is 27 the cube root of 27 is 3. all right let's say we saw f of x equals the cube root of x plus 3. again we'll say y equals the cube root of x plus 3 interchange x and y so x equals the cube root of y plus 3. so let's subtract 3 away from each side of the equation we will have x minus 3 is equal to the cube root of y and then i would just cube each side of the equation so i would have y is equal to the quantity x minus 3 cubed so let's erase all this and we'll use the proper notation we'll say f inverse of x is equal to the quantity x minus 3 and this is cubed let's take a look at another one so suppose we see f of x equals 3 over x minus 1 minus 2. so again to find the inverse we'll say y equals 3 over x minus 1 minus 2. swap out x and y so let's just say this is y and this is x so how do we solve for y now we've got to add 2 to both sides of the equation so we'd have x plus 2 is equal to three over y minus one so now let's multiply both sides of the equation by the quantity y minus one so what would that give me well this would cancel over here i would have the quantity y minus 1 times the quantity x plus 2 is equal to 3. so let's do 4 here y times x is x y the outer y times 2 is plus 2y the inner negative 1 times x is minus x and the last negative 1 times 2 is minus 2 this equals 3. so let's keep everything with a y on the left everything without a y on the right so let's add x let's add 2 to both sides of the equation so that would go away and then let's factor out a y so i would have y times the quantity x plus 2 there so we would have y times the quantity x plus two and so what we would do here is just divide both sides of the equation by x plus two and that's going to give us what y is equal to you would have we'll reorder this to x plus five right because we have x plus three plus two three plus two is five over x plus two and again to notate this properly we'll say f inverse of x is equal to x plus five over x plus two and again you can choose a value for x plug it in the original function you'll get a y take that y value plug it in for x in the inverse and you'll get your original x value back so as an easy example let's say x was four so f of four would be what plug in a four there four minus one is three three over three is one so you'd have 1 minus 2 which is negative 1. so if i plug in a negative 1 4x in this inverse negative 1 plus 5 is 4 right so this would be 4 over negative 1 plus 2 is 1. 4 over 1 is 4 so i'm right back to 4. hello and welcome to algebra 2 lesson 69 in this lesson we're going to learn about exponential functions so an exponential function is of the form we see here that we have f of x is equal to a which is our base raised to the power of x so what makes this a little bit different for us is that we've never seen a variable involved in the exponent before so what we'll do here is we'll define a to be any number that is larger than zero and it's not allowed to be equal to one the reason we put these restrictions on the base is so that we can say that x is any real number and you can just think about for a moment why you need those restrictions on a to make this all work out let's say a was allowed to be some negative value just to make it simple let's say it's negative 2. so if i had negative 2 as my base and i raised it to i don't know let's say the number 3. you'd say okay there's no problems there negative 2 cubed or negative 2 to the third power is negative 2 times negative 2 times negative 2 which we know would be negative 8. so that's all fine but what if we said i want negative 2 raised to the power of one-half or 1 4 or something like that well then this is equal to the square root of negative 2 and although we can deal with this using the complex number system the answer is not real right so if we're dealing with just real numbers only we want to restrict a to again be greater than zero and it's not allowed to be equal to one now let's talk about why it can't be zero if you think about having zero as your base if i raise zero to some positive value let's say i square it zero squared we know is just zero so that's all fine but i can't raise zero to the power of zero that's undefined and i can't raise zero to a negative value because if i did something like zero to the power of negative two we know this is one over zero squared and this is 1 over 0 which is undefined undefined so that's why we have these restrictions now let's cover this last one here where we say a does not equal 1. why do you think a the base is not allowed to be equal to one well we would no longer have an exponential function we would have a linear function you might say well why is that the case well if i say f of x let's for to make this really simple instead of f of x let's just say this is y is equal to a i'm just going to replace that with 1 raised to the power of x so just think about some ordered pairs let's just do 2. let's say i picked 2 for x what would y be 1 squared is 1. so y would be 1. let's say i picked i don't know 10 for x 1 to the 10th power is 1 right so y would be 1. you can choose any value you want for x it doesn't matter anything in the real number system when you take one and you raise it to a real number you're going to get one back so what ends up happening is you could basically say this is a horizontal line right it's the horizontal line y equals one or to make it match this format here we could just say f of x is equal to one right this is a horizontal line that would impact the y axis at the point zero comma one so there's kind of two things we're going to talk about in this lesson the first thing is how to graph an exponential function we'll spend a little bit of time on that and then we'll move into solving exponential equations so for graphing an exponential function just a little bit of background here the first thing you would do just like if you were graphing anything else you would find some ordered pairs again enough to show a good picture you'd plot the points like you always do and then you connect the points with a smooth curve again i say this in a lot of videos when you freehand draw things if it's not a line it's usually going to be pretty inaccurate right so you want to stick to just using this for a conceptual tool if you actually need to graph something to look at it you'd probably want to use either computer software or your graphing calculator so let's start out with the graph of you have f of x equals a to the power of x so this kind of generic form that we get presented in our textbook so the first thing is that 0 comma 1 is a point on the graph again this is the x value this is the y value so in other words if i plugged in a 0 for x a to the power of 0 is going to give me 1 because any nonzero number raised to the power of 0 is going to give me 1. and remember we restricted a to be greater than 0 and we also restricted a to not be equal to 1. so following those restrictions we know if we plug in a 0 for x we would get a 1 as a result for y right so this is a very important feature when you deal with exponential functions the graph approaches the x-axis so it comes very very close but it will never touch it it forms something known as an asymptote and we talked about those before when we're dealing with rational expressions so now the domain which is the set of allowable x values consists of all real numbers so again because we restricted a we could say that we can choose any value in the real number system from negative infinity to positive infinity as our input for x then the range or the output the values that we can use for y or for f of x consist of all positive numbers so zero is not involved in that and you can see the way that this interval notation is set up again if you're using a parenthesis next to zero it says zero is not included but anything larger is so anything larger there could be 0.0000001 okay if it's larger than 0 it's acceptable out to positive infinity all right so when a your base is greater than 1 the graph rises from left to right so you're going to see it kind of over to the left as you come from that direction and going to the right you're going to see it just hockey stick itself up and you'll see an example of this in a second then when a is greater than 0 and less than 1 the graph will fall from left to right so in other words it will start out at a higher point and it will decrease as it moves to the right all right so let's take a look at a few graphs now so we want to graph each and again if we're doing this by hand it's going to be just an approximation right we're just doing this to get a little practice on what it would look like it's not going to be 100 accurate so f of x is equal to 2 to the power of x let's choose about 5 values here just enough to get a good picture i'm gonna choose some negative values so i'm gonna go with negative four i'm gonna go with negative one i'm gonna go with zero i'm gonna go with one and i'm gonna go with three all right so 0 comma 1 we know is a point on all of these we talked about that a minute ago so i can go ahead and just fill this out and say it's 0 comma 1. if i plugged in a 0 for x there 2 to the power of 0 by rule is 1 so f of x or y would be 1. now if we look at the other ones we say y instead of f of x is equal to 2 to the power of in this case x is negative 4. so if i plugged in a negative 4 for x two to the power of negative four is what this is one over two to the fourth power which is one over sixteen so we say this is one sixteenth here and then the next scenario is negative one so we know that if we had a negative 1 there it would be 1 over 2 to the power of 1 which is just one-half so this guy would be one-half we've already figured this one out that's 0 comma 1. so now we're dealing with positive exponents so the first one would be one that's easy two to the first power is just two and then the next one if we raise this to the third power or we cubed it two cubed would be eight all right so our ordered pairs here we know we'd have negative 4 comma 1 16. we know we'd have negative 1 comma one half we know we'd have zero comma one we know we'd have one comma two and we know we'd have three comma eight so let's copy these okay so the first ordered pair is negative 4 comma 160. now the way this coordinate plane is drawn the scale is in increments of 1. so obviously i can't accurately portray 1 16 i just have to kind of guess at it so if i go to negative 4 on the x axis that's here 1 16 let's just say it's about right there okay we're just showing a picture again it doesn't have to be 100 accurate then the next point we're looking at is negative one comma one-half so negative one on the x-axis and then one-half on the y-axis so it's about right there then we have zero comma one so that's about right there then we have one comma two so that's about right there and then we have three comma eight so that's going to be three years to the right eight units up that's about right there and let me kind of go over these with a different color all right so as you go to the left here as these numbers become bigger and negative values what happens is if we go back and you look at this guy you have 2 to the power of something if i have a really big negative let's say it's 2 to the power of negative 10. well this is 1 over 2 to the 10th power so as this number gets bigger and bigger the denominator will get bigger and bigger so the 1 is divided by a bigger and bigger number so it becomes a smaller number overall 2 to the 10th power is 1024 so you can say i have one over 1024 this is a small number right it's close to zero but you can make this thing infinitely close to zero as you choose bigger and bigger numbers in the negative realm for x so if i chose negative one million negative one billion negative two trillion something like that you're getting the result for y or the f of x value very very close to zero because it's a very very small number it'll never be zero and it'll never be negative so that's why this guy right here this this asymptote gets formed because it's going to approach 0 but it's never going to quite touch it so it approaches the x-axis a value where y would be 0 but it's never ever going to actually get there so the graph would look something like this you kind of do like that and then we just hockey stick itself up we'll do that let's take a look at one more of these and then we'll kind of look at some exponential equations so we have f of x is equal to one half raised to the power of x so this guy here is between zero and one so we would expect the graph to fall from left to right so it's going to start out high and it's going to come low it's going to approach the x-axis but it's never going to touch it so let's get some points going let's do negative 3 let's do negative 1 let's do zero let's do one and let's do two so i know that zero comma one is a point on this graph again i can always start with that because i plugged in a zero there i know by rule i would get a one right so f of x or y would be one now let's think about the other ones i have f of x or probably better just to put y is equal to one half raised to the power of negative three so how do i do this quickly again i could flip the fraction here and make that exponent positive so i could say this is equal to 2 over 1 raised to the power of 3. so 2 over 1 is just 2 2 cubed is 8. so negative 3 is x y is eight now what if x was negative one what if x was negative one we would again flip the fraction so it'll become two over one and it'll be raised to the first power so that's just two zero comma one we already have if x was one y would just be one half so then if x was two we could say this is one squared which is one over two squared which is four or simply 1 4. all right so let's erase this and we'll just kind of jot these ordered pairs down so we have negative 3 comma 8 we have negative 1 comma 2 we have 0 comma 1 we have 1 comma one half and we have two comma one fourth so let's just copy this so negative three comma eight so negative three comma eight it's gonna be right there negative one comma two so that's gonna be right there and zero comma one is right there one comma one half will be about right there and then two comma one fourth would be about right there so you can see this guy is gonna fall from left to right so in other words it's gonna come from up here and it's going to fall as we move to the right so we're going to go that way we just say that it will again approach the x-axis but it will never ever ever touch it alright so now that we've seen what an exponential function looks like let's talk about solving exponential equations and this is mostly what you're going to do in this section so an exponential equation is just an equation where there's a variable in the exponent so you see something like 2 to the power of x equals 16. without me explaining how to actually conduct this operation you could probably guess that x here is equal to 4 right just because you know that 2 to the 4th power is 16. but if you didn't know that off the top of your head what would you do let's talk a little bit about this so the first thing is just a little bit of a background so i want you to recall from the last section that a horizontal line test is used to graphically determine if we have a one-to-one function we've already looked at the graphs of two exponential functions you can see that a horizontal line would not impact the graph in more than one location it may seem a little confusing because of that area by the x-axis but again if you think about this you're getting closer and closer to the x-axis it's never the actual same value because it just keeps getting closer infinitely closer but it's never actually the same value so a horizontal line would not impact the graph in more than one location so we could say an exponential function is one to one so because of that we can use this kind of rule here if a is raised to the power of x and this is equal to a raised to the power of y then we could say x is equal to y given these restrictions here that a is greater than 0 and a does not equal 1. all right so solving an exponential equation each side must have the same base and again that goes back to this rule where we say if a to the power of x is equal to a to the power of y the base here is the same the base is the same so we say simplify the exponents that's the next step we set the exponents equal to each other we solve the resulting equation and then we're going to check with substitution so it's a very very easy process overall with the first one we have 2 raised to the power of 3x minus 3 this is equal to 64. so the first thing you would do is think about okay 64 is 2 to what power right you want to express these as the same base well you know it's 2 to the sixth power so i would rewrite this and say this is 2 to the power of 3x minus 3 is equal to 2 to the 6th power so now what do i have i have the same base same base so don't worry about the base just take the exponents so this guy and this guy simplify them and set them equal to each other 3x minus 3 is already simplified 6 is already simplified so really all i have to do is say 3x minus 3 is equal to 6. i'm going to solve this very simple linear equation so add 3 to each side of the equation that's going to cancel i'll have 3x is equal to 9 divide both sides by 3 you get x is equal to 3. so the solution set here just contains the element 3. so let's erase all of this so to check this we're going to plug a 3 in for x right there so you just think about this without even writing anything down 3 times 3 is what that would give you 9. then if i subtracted away 3 that would give you 6. so you basically have two to the sixth power is equal to 64. yeah we know that's true right to the sixth power is 64. you would end up with 64 equals 64. all right so for the next one we have 1 9 raised to the power of 3p is equal to 81. so this one's a little bit more challenging we just want to think about how can i get 81 to have the same base as 1 9 so this is my base and i want a base of one knight so what can i do well just think about how you could get 1 9 and change it into 81 just using exponents well i know that 9 squared would give me 81 but that's not going to quite do it because if i square 9 there i'd have 1 over 81. so i've got to think about negative exponents if i use a negative exponent negative 2 would do the trick right so 1 9 raised to the power of negative 2 this would tell me to take the reciprocal of the fraction nine over one and square it which would give me eighty-one so this is what i'm looking for here a base of one-ninth an exponent of negative two so let's rewrite the problem to say it's one-ninth 9 raised to the power of 3p and this is equal to 1 9 raised to the power of negative 2. the exponents are simplified so i could just set them equal to each other so we'd say 3p is equal to negative 2 divide both sides by three we'll say that p is equal to negative two thirds so our solution set here will just contain negative two-thirds all right what about 216 raised to the power of negative three b minus two is equal to thirty-six so in this case if i think about 36 i know that it's 6 squared if i think about 216 i know that it's 6 cubed so i want to just write this as 6 squared i want to write this as 6 cubed so i think about this just very carefully here this 3 here would just multiply this so i would have 6 raised to the power of 3 times in parentheses negative 3b minus 2 this would be equal to 6 squared so if i simplify here 3 times negative 3b is negative 9b so let's write this over here we'd have 6. raised to the power of negative 9b and then 3 times negative 2 would be minus 6 and this equals 6 squared so i have the same base now all i need to do is take my exponents and set them equal to each other so in other words i will have the equation negative nine b minus six is equal to two i will add six to each side of the equation that will give me negative nine b is equal to eight divide both sides of the equation by negative nine this will give me b is equal to negative eight ninths so my solution set here will contain one element just negative eight ninths all right let's take a look at another one so we have 64 raised to the power of 2 minus 2x times 16 raised to the power of negative x minus 1 and this equals 1 4. so the first thing i'm going to do is i'm going to concentrate on simplifying this side i know that 64 is 2 to the 6th power i know that 16 is 2 to the 4th power so let me write these with identical bases let's say this is 2 to the 6th power times this guy right here 2 minus 2x that quantity then times 16 is 2 to the 4th power times negative x minus 1 that quantity and again this equals 1 4. all right so 6 times 2 is 12. so we would have 2 to the power of 12 minus 6 times 2x is 12x and then times we have 2 raised to the we have 4 times negative x that's negative 4x 4 times negative 1 is minus 4 this equals 1 4. so now if we have the same base here remember if we're multiplying we would add exponents so i could add 12 and negative 4 that would give me 8. so this would be 2 to the power of 8 and then negative 12x minus 4x that would be minus 16x so 2 to the power of 8 minus 16x and then if i have this equal to 1 4. let's think about how we could rewrite this to have a base of 2 well i could say that 4 is 2 squared so that's easy enough but i need 2 to be in the denominator not the numerator so i could really rewrite this and say this is 2 raised to the power of negative 2 right because this is 1 over 2 squared which is 1 4. so let's write this as 2 raised to the power of negative 2. and now i'm good to go i know it seems like a lot of work seems kind of tedious but really if you know the steps if you know your rules for exponents it's not too bad all right so two and two same base same base just going to set these guys equal to each other i'll say 8 minus 16 x is equal to negative 2 and i'll just solve for x so i'm going to subtract 8 away from each side of the equation that'll cancel we'll have negative 16x is equal to negative 10. we'll divide both sides of the equation by negative 16. and so that's going to cancel with that we'll have x is equal to positive if we think about 10 it's divisible by 2 divided by 2 you get 5. divide 16 by 2 you get 8. so x equals 5 8. so again x equals 5 8. all right so the next problem is 64 raised to the power of 3p plus 3 times 1 8 raised to the power of 2p this equals 64 raised to the power of negative p minus 3. now we have a 64 a 1 8 and a 64. so let me just use a base of 8. that would be easy so in other words i will take 64 and say this is 8 squared and this is multiplied by the quantity 3p plus 3 up here when we talk about the exponents then times for 1 8 i could really write that as 8 to the power of negative 1 and that's times 2p this equals for 64 again i could write that as 8 squared and the 2 will be multiplied by the quantity negative p minus 3. so we'll have 8 and then two times three p is six p plus two times three is six and then times eight raised to the power of negative one times two p is negative two p and this equals eight raised to the power of two times negative p is negative two p and then two times negative 3 is negative 6 or minus 6. now we have 8 raised to the power of 6p plus 6 times 8 raised to the power of negative 2p if you're multiplying here with the same base you add the exponents so in other words i would just add 6p and negative 2p to give me 4p so this would be 8 raised to the power 4p plus 6 is equal to 8 raised to the power of negative 2p minus 6. so now we have the same base so we can just look at the exponents so we'd have 4p plus 6 is equal to negative 2p minus 6 we can subtract 6 away from each side of the equation that's gone we can add 2p to each side of the equation that's gone so i'll have 6p is equal to negative 6 minus 6 is negative 12. divide both sides of the equation by 6 we get that p equals negative 2. all right let's erase everything so again p equals negative 2. hello and welcome to algebra 2 lesson 70. in this lesson we're going to learn about logarithmic functions so in our last lesson we learned about exponential functions in that lesson we saw f of x equals we had our base a raised to our exponent of x now for most of us this is the first time we ever dealt with a variable in the exponent as we move forward and math this is going to be very very common for us now to make everything work we put some restrictions on the base a we said a had to be greater than zero we also said that a was not allowed to be equal to the number one now in this lesson we're going to talk about the inverse of an exponential function so the inverse of the exponential function is known as the logarithmic function so the first time you kind of look at a logarithm it may look a little bit scary to you right the notation is very foreign something you've never come across before so the first time you read it you'll say f of x equals you have log base a of x so let me kind of write that you'll see log base a of x your book might read that a different way your teacher might tell you a different way it doesn't really matter you're conveying the same information either way now in order to make this guy work we have some restrictions here as well a which is referred to as the base this is referred to as the base the restrictions are the same as for a in the exponential function so a has to be greater than 0 and a is not allowed to be equal to 1. then x which is referred to as the argument okay it's called the argument this has to be greater than zero and i'll explain why that happens in just a minute so where does this come from well usually things that are generic come from something so they're taking this guy right here and then they're finding the inverse of it so in other words if i had let me kind of scroll down get a little room going if i had f of x f of x is equal to a to the power of x and again a is greater than 0 and a is not allowed to be equal to 1. and i said find the inverse well the first thing you would do is you would change f of x to y so this guy would be y equals a to the power of x then you would swap x and y so in other words y would become x and x would become y so we have x equals a to the power of y now you would solve for y in your next step but the problem is we have not developed a method to allow us to do that up to this point so we introduce the logarithm which allows us to solve for the exponent or you could say isolate the exponent on one side of the equation so with a logarithm i can say y is equal to so in other words y is by itself on one side log base a of x so these two right here are conveying the same information this is logarithmic form and this is exponential form and it's very easy to go back and forth between the two now i want to explain why x is going to be greater than zero so that restriction when we get to this part right here comes from the fact that if i have a and a is greater than 0 and a is not allowed to be equal to 1 and i raise it to y it doesn't matter what y is i'm always going to get a result that's greater than 0. let's say i just picked the number 2. if i raise 2 to some positive number let's say i do 2 squared i get a positive number so i get 4. if i take 2 and i raise this to the power of 0 i get 1. if i take 2 and i raise it to a negative let's say i do negative 4 i'm going to get 1 over 2 to the fourth power which is 1 over 16. this is still positive it's still greater than 0 so that's why x here x is greater than 0. all right so i want to just kind of look at this more cleanly so we have our log form here which is y equals log base a of x we have our exponent form here which is x equals a to the power of y now we've got to come up with a way to kind of easily send this back and forth between the two forms now in your textbook you'll see that a is referred to as the base a is the base this is the easy one to remember the base here is the same as the base here so those two very easy to deal with where you kind of get a little bit confused the technical name for x when you're looking at it in log form this is referred to as the argument sometimes you'll hear it called the power i'm just going to call it the result and the reason i'm going to do that is just to make it very very easy to go back and forth it is the result of taking a to the power of y so i'm going to say this is the result you might say it's the answer whatever you want to do we just need to come up with some easy way to go back and forth between the two so this works well for my students now y here is the answer and log form but i'm just going to refer to it as the exponent and the reason i'm going to do that is because the exponent here so i just want to label everything so that we can just go back and forth very easily so to go back and forth i know that i would take a which is my base i would raise it to my exponent of y and i would get a result of x so it's kind of a round robin you can draw this diagram to help you you can just go from a to y to x so if i saw something like 2 cubed is equal to 8 and i said write this in logarithmic form you can easily do that 2 is the base 2 is the base 3 is the exponent and 8 we know is going to be our result so match that up with what i have here the exponent is by itself so 3 will be on one side of the equation by itself so in other words it's solved for the exponent 3 is going to be equal to you have that log word so just put log and then you have your base so the base is the same so 2 is going to be your base there and then you put the result which is 8. again the official name for that eight is called the argument but i'm just calling it the result so we can keep things consistent and we understand what's going on so in other words 2 to the power of 3 would give me 8. all right so one thing that might help you is just to realize that log base a of x is just the exponent to which the base a must be raised to to obtain x i would write that down in my notes or you could write it down with something with numerical information in it so where you can constantly refer to that so where you can think about the fact that this is just asking me for an exponent right it's just asking me to come up with an exponent so log base 3 of 9 is asking me what power does 3 need to be raised to to obtain 9. in other words 3 to the power of question mark equals not 3 squared equals 9. so this question mark is really just a 2. right i could scratch that out and put a 2. so log base 3 of 9 is equal to 2 or in exponential form 3 squared equals 9. again start at the base go to the exponent you'll get the result what about 4 squared equals 16 again it's solved for the exponent so 2 will be on one side of the equation by itself so 2 is equal to log the base is 4. so log base 4 of the result the result is 16. so in other words the base 4 raised to the power of 2 would give me the result of 16. let's try another one we have 5 cubed equals 125. so again 5 is the base 3 is the exponent 125 would be the result so i would have 3 the exponent is equal to log the base is 5. so log base 5 of our result is 125. let's try one more so we have 10 raised to the power of negative 2 equals 1 over 100 so my exponent is negative 2. negative 2 is on one side by itself this is equal to you have log you have a base of 10 so a base of 10 and then of 1 over 100. now you can again say that 10 which is our base raised to the exponent of negative 2 would get me to a result of 1 over 100. we're going to learn later on that if you have a base of 10 you don't need to write it's called the common logarithm but we'll see that in a few lessons all right so at this point you've seen some examples of how to convert between exponential form and logarithmic form it's really easy it's just something you need to practice a lot kind of to get used to it and then it becomes second nature to you so now we just want to talk about how to solve a simple logarithmic equation when we get into the next few sections we're going to see that these get much more complicated but for right now we're going to focus on the easy one so if we have log base a of b and we say this is equal to k we can solve this by putting it in exponential form and just using the rules that we learned in the last section right remember if you have the same base you can set the exponents equal to each other and you can solve so here we would say the base a is raised to k and this is equal to a result of b so a to the power of k equals b all right so for the first example we have 3 is equal to log base 10 of 3x so 10 is raised to the power of 3 and this is equal to a result of 3x so what do i do here in this particular case it's pretty easy we know that 10 cubed is a thousand so let's just start with that so it's a one followed by one two three zeros so that equals three x if i want to isolate x in this case i just divide both sides of the equation by 3 and i'll end up with x is equal to 1000 over 3. let's erase this real quick checking this guy is super simple if i plugged in a 1000 over 3 for x there what would i have i would have 3 is equal to log base 10 of 3 multiplied by 1000 over three and so is this true well this cancels with this and i'm left with one thousand so in other words is ten cubed equal to a thousand yes it is right so we can check this guy off and say x equals one thousand over three is the correct solution all right let's take a look at another so we have log base five of negative four x plus five equals two so we'd have five which is our base raised to the exponent of two and this is equal to our result which is negative four x plus five so another easy one we'd square five that would give us twenty-five this is equal to negative four x plus five we will subtract 5 away from each side of the equation 25 minus 5 is 20 and this equals negative 4x 5 minus 5 is 0. so divide both sides of the equation by negative 4 this will lead us to this cancels with this we'll have 20 divided by negative 4 which is negative 5 this is equal to x so x equals negative 5. again we can check this very easily so if i plug in a negative 5 there negative 4 times negative 5 is 20. 20 plus 5 is 25 so in other words is 5 squared equal to 25 yes it is so this guy right here checks itself out all right for the next one we're looking at log base 27 of the fourth root of 81 this equals x so we'd have 27 to the power of x is equal to the fourth root of 81 we know is 3. now we have one where again if it's an exponential equation we're thinking about solving based on having the same base and letting our exponents be equal to each other so 27 we know is 3 cubed so i can use my power to power rule to kind of set this up i can say this is 3 cubed raised to the power of x equals 3 and for the sake of this problem we're going to write this as 3 to the power of 1 right that x1 is understood to be 1. so transfer this guy up here power power rule if 3 is cubed and it's raised to the power of x 3 stays the same we multiply 3 times x to give us 3x and this equals 3 to the first power the base now is the same so we have the same base and we learned again in the last lesson if the bases were the same you can set the exponents equal to each other you could solve so we would say 3x is equal to 1. so 3x is equal to 1. divide both sides of the equation by 3 you'll get x is equal to 1 3. so x equals one-third and again easy to check this so if i plugged in a one-third there what would that tell me again it's 27 to the power of one-third would give me a result of the fourth root of 81. the fourth root of 81 is 3. so we know that's true right the cube root of 27 which is take something to the power of one-third you're taking the cube root of it the cube root of 27 is 3. so you'd end up with 3 equals 3 so this will check itself out as well all right let's think about some properties of logarithms we'll learn about more properties of logarithms ones that are a little bit more advanced in the next lesson but for now let's look at these basic properties so log base b of b is equal to 1. this should make sense because if you have some base b and you're raising to the power of 1 you would expect to get b back then log base b of 1 is equal to 0. so if i have a base b and i raise the power of 0 i should expect to get 1 back this is given the restrictions that b is greater than 0 and b does not equal 1. so log base 5 of 5 equals again very very easy 5 raised to what power would give me 5 5 raised to the first power would give me 5. and you could do a million examples of this you could say log base 1 320 of 1 320 equals what again this is 1. same number here and here again following those restrictions we gave we know our answer is going to be 1. all right what about log base 105 of 1. again this is going to be 0 because 105 raised to the power of 0 would give me 1. again another one that you could just do a million examples of if i did like log base 1 million of 1 again i know the result is 0. 1 million raised to the power of 0 is 1. okay if i did log base 20 million of 1 again i know the result would be 0. very very easy if you come across those two scenarios so let's kind of wrap up the lesson by looking at some graphing now at this point we know how to convert between exponential form and logarithmic form and we've seen how to solve some very basic logarithmic equations with graphing a logarithmic function the first thing is to graph a log we put it in exponential form now remember that the exponential function and the logarithmic function are going to be inverses of each other so when we looked at the exponential function 0 comma 1 was always on the graph now remember the x and y's are interchanged with inverses so 1 comma 0 is on the graph so this was the y value from the exponential right when we talked about that this was the x value from the exponential so 1 swap places right it went from y to x 0 swap places it went from x to y now we say if a is greater than 1 the graph rises from left to right then if a is greater than 0 and less than 1 the graph falls from left to right now for this one the graph approaches the y axis but does not touch it it forms an asymptote so remember that we had an asymptote when we worked with exponential functions but that guy was when it approached the x-axis think about what it means to approach the y-axis if you're looking at a coordinate plane the y-axis is the value where x is zero we said that if we had y equals log base a of x that x was greater than zero so we're telling you right there hey x will never be zero so the graph will never actually touch the y axis or a value where x would be zero it's going to get infinitely close the reason for this again if i have in exponential form a to the power of y is equal to x well just think about this for a second again if a is restricted to be greater than zero and a is not allowed to be one well then when i take something positive here and i raise it to a positive value okay i'm going to get a positive result i'm going to get something that's greater than 0. if i raise it to 0 i'm going to get 1. if i raise it to something negative i'm going to get a small number but i'm never going to actually get to 0 right so if i chose something like 2 to the power of i don't know some big negative so let's say negative let's say negative 10. this is equal to 1 over 2 to the 10th power that is going to be very close to zero and i can make this guy infinitely close to zero by just making this part right here a bigger and bigger negative let's say i made this two to the power of negative one million that's going to be really close to zero but i can keep getting closer and closer by just choosing bigger and bigger negatives right there so it will approach the y-axis but it will never actually touch it all right so the domain and the range have flipped the domain when we looked at an exponential function was all real numbers now the domain is anything larger than zero okay it can't be zero we just talked about that anything larger than zero out to positive infinity the range is anything from negative infinity to positive infinity so it can be any real number all right so we're going to look at three graphs today we're going to look at f of x equals 2 to the power of x we're going to look at y equals x and we're going to look at the inverse of this guy right here which is going to be a logarithmic function all right so i'm going to start off by just getting some points here so let's just choose some points like negative 3 for x let's do negative 1 for x let's do 0 let's do 1 and let's do 3. so if i plugged in a negative 3 for x what would i get we would have y is equal to 2 to the power of negative 3 this is 1 over 2 cubed which is 1 over 8. so if x is negative 3 y is 1 8 then if x is negative 1 you just think about plugging in a negative 1 there this would be 1 over 2 which is 1 half if x is 0 we know y is 1 right 0 comma 1 is always on an exponential function graph and then if x is 1 y is going to be 2 that one's very very easy if x is 3 right if i plug down a 3 there 2 cubed would be 8. so i've got 5 ordered pairs so 5 ordered pairs let's go ahead and just plot those so we have negative three comma one eighth we have negative one comma one half we have zero comma one we have one comma two and we have three comma eight all right we'll paste those guys let's just graph this guy real quick so negative three comma one eighth so negative three and then one eighth we just got to kind of put a dot right there we don't know exactly where it is because of our scale but let's just say that's pretty close then negative one comma one half so we'll do negative one on the x axis and then about half way up here so let's just say that's right there zero comma one that's easy that's right there one comma two another easy one that's right there and then 3 comma 8 that's easy as well okay so again as the x values get more and more negative the graph approaches the x axis so it gets infinitely close to that guy but it's never going to actually touch it all right so let me label this this is the graph of f of x is equal to 2 to the power of x now the next one we want to graph is y equals x so if i graph y equals x i don't need to really get points going for that so for this guy we know the slope is one right remember that y equals mx plus b so a point on that line would be zero comma zero right if x is zero y zero so from there i can just use my slope of one so to the right one up one to the right one up one to the right one up one i could go down one to the left one down one to the left one down one to the left one let me use a different color here so we'll label this as f of x equals x okay so let's go up to the last guy okay so we have x equals 2 to the power of y where did that come from again if i have the inverse of f of x is equal to 2 to the power of x i would first say f of x is y then i would swap x and y so this would be y and this would be x now i can write this in logarithmic form i can say y equals log base 2 of x but again when we're working with graphing we want this in exponential form so let's get rid of all this now you might say why do you have these ordered pairs from the first guy we work with right when we work with y equals 2 to the power of x the reason is if we have inverses i can just flip the x and y values and i have ordered pairs so in other words the x value of negative 3 would now be a y value of negative 3 and a y value of 1 8 would now be an x value of 1 8. and i could just go down the list and just switch things so in other words this would be one half this would be negative 1. this would be 1 this would be 0. this would be 2 this would be 1 this would be 8 this would be 3. so you can check these if you want but i can assure you that they do work let's just pick 2. so if i took 2 and said this was equal to 2 to the power of 1 that checks itself out you could take something like one-half also so one-half is equal to 2 raised to the power of negative 1 that's true right that would be 1 over 2 to the power of 1 which is one-half all right so now we can just set up our ordered pairs so 1 8 comma negative 3 we have one half comma negative 1 we have 1 comma 0 we have 2 comma 1 and we have 8 comma 3. so let's copy these so 1 8 comma negative 3. so on the x axis 1 8 is let's say about right there negative 3 we drop down 3 units so let's just say the point's right there one half comma negative 1. so one half would be about right there negative 1 to be about right there 1 comma 0 so that's right there then we have 2 comma 1. so 2 comma 1 is right there and then 8 comma 3. so 8 and then 3. so remember for this one we're never going to actually be able to touch the y as the y-values get more and more negative we're going to get closer and closer to the y-axis or an x-value that is zero but we'll never actually be able to touch it remember x is strictly greater than zero so so this guy right here is f of x or y is equal to log base 2 of x so looking at the graph we can see that we do have inverses and again the way you can kind of show this to yourself if you've drawn that line f of x equals x or y equals x and you do this at home you can take your sheet of paper and you can fold the paper exactly on that line f of x equals x and what's going to happen is this graph right here will be exactly on top of this graph right here and that's how you can prove that you have inverses hello and welcome to algebra 2 lesson 71 in this lesson we're going to learn about properties of logarithms so in the last lesson we started talking about logarithms we showed what a logarithm was and we showed how to go back and forth between exponential form and logarithmic form now in this lesson we're going to dig a little bit deeper and we're going to talk about some of the properties that you're going to encounter when you're studying logarithms so the first property you're likely to come across is known as the product rule for logarithms so if you say something like log base b of x y this is equal to log base b of x plus log base b of y so what is this telling us well essentially what it's doing is it's allowing us to write multiplication as addition so in your book you'll probably read the logarithm of a product is the sum of the logarithms of the factors so let's reflect on that for a minute i start out with log base b of x times y so we can think of x as a factor and we think of y as a factor right those two are being multiplied together and we expand this into we say this is equal to log base b so the base is the same of x x is the first factor then plus log base b of y y is the second factor so if i saw something like log base 3 of i don't know let's say 2 times 5. this would be equal to log base 3 of 2 plus log base 3 of 5. okay very very simple you just take your first factor and put it there your second factor goes there very very easy to do this now there are some restrictions we say x y and b so x and y are the factors here and it could be more than x and y you could have three factors four f you could have as many factors you want but all those factors have to be positive real numbers so greater than zero now b the base has to be a positive real number and then in addition to that we have this added restriction b has to be not equal to the number one so it has to be a positive real number and it can't be one all right so let's take a look at a few easy examples here if we saw something like log base 3 of 7 times 4 again all i'm going to do is just expand this so this tells me i could rewrite this as log base 3 of 7 right my first factor plus log base 3 of 4 right my second factor so this guy went here and this guy went there very very easy okay same thing here so log base 7 of 12 times 3 this is log base 7 of 12 then plus you have log base 7 of 3. all right let's take a look at another one so we have log base 2 of 5 times 12. so it's log base 2 of 5 plus log base 2 of 12. very very simple all right now let's try to reverse the process you might be asked on your homework or on a test to condense so using the same property that we just talked about if i saw log base 3 of 8 plus log base 3 of 7 i know i could rewrite that as log base 3 right log base 3 and log base 3 so it's going to be log base 3 of you've got an 8 here and a 7 here so you would have 8 times 7 or you could really multiply those two together and say it's 56 if you wanted to so let's say this is equal to log base 3 of 56 what if you saw log base 4 of 11 plus log base 4 of 5. so log base 4 log base 4 so it's going to be log base 4 and then of you have 11 here and 5 here so those are going to be multiplied together so it would be 11 times 5 which we know is 55 so we could write it as log base 4 of 55. let's take a look at another so if you saw something like log base 8 of 7 plus log base 8 of 5 what would you do you have that log base 8 so that's log base 8 of you've got a 7 here and a 5 here so 7 times 5 will give me 35. so log base 8 of 35. all right so very very easy to deal with the product rule for logarithms the quotient rule for logarithms is just as simple so let's take a look at that so if we see log base b of x over y this is equal to log base b of x minus log base b of y so in other words the x in the numerator is going right here the y in the denominator is going right here so this is subtraction whereas when we talked about the product rule it was addition okay it was addition so we have the same restrictions x y and b are positive real numbers and b is not allowed to be equal to 1. all right for the first one we have log base 3 of u over v so in other words what i'm going to do is i'm going to say we have log base 3 of what's in the numerator that's a u then minus we'll have log base 3 of what's in the denominator that's v it's always numerator here you'll have your minus sign denominator here okay the log base 3 will be the same in each case so you've just got to work out what's going to go in this position and what's going to go in this position again numerator is going to go here denominator is going to go here all right for the next one we have log base 8 of 12 over 11. so this is going to be log base 8 of 12 what's in the numerator minus you'll have log base 8 of 11. what's in the denominator all right now we have log base 5 of 3 over 7. so this is going to be equal to log base 5 of 3 minus log base 5 of 7. all right so now let's talk about reversing this process so if i saw something like log base 9 of 10 minus log base 9 of 7 what would i do well i know that i would have log base 9 then of what well i know it's this guy the first one that appears is 10 that's going to go in my numerator and then it's over the second guy that appears is 7 that's going to go in my denominator so this becomes log base 9 of 10 over 7. all right for the next one we have log base 10 of 7 minus log base 10 of 11. so we know this is going to be log base 10 of you'll have 7 over then you'll have 11. you just want to go in order so 7 is first so it goes in the numerator 11 is second so it goes in the denominator so again log base 10 of 7 over 11. all right for the next one let's look at log base 5 of 3 minus log base 5 of 8. so again very very easy this is going to be log base 5 of the one that appears first is 3 so that goes in my numerator then over the one that appears second is 8 so that goes in my denominator so this is log base 5 of 3 over 8. all right so now we have the power rule for logarithms so you have log base b of x to the power of r and say this is equal to r which was the exponent here multiplied by log base b of x so in other words the exponent just comes down and goes in front okay that's all we're doing there so again x and b are positive real numbers and b again is not allowed to be equal to 1 or as any real number that you'd like so it can be negative it can be positive it can be 0 whatever you'd like it to be so to give you kind of a quick example of why this occurs let's say you saw a log base 7 of 4 cubed we could write this as log base 7 of 4 times 4 times 4. now using our product rule that we learned earlier we can expand this and say this is log base 7 of 4 plus log base 7 of 4 plus log base 7 of 4. so i have 3 of the same thing so i can just say i have 3 times that thing which is log base 7 of 4. so that's all we're doing here we're not doing any magic tricks so if i saw log again base 7 of 4 cubed this guy just goes in front so we just say this is 3 times log base 7 of 4 very very easy all right so what about log base 4 of x to the fourth power again this guy is just going to go out in front so this is equal to 4 times log base 4 of x what about log base 5 of the square root of x so for this guy you want to write it where you have a fractional exponent so you're going to say this is log base 5 of x raised to the power of 1 half and then you just want to bring this guy down so this is going to be equal to one half multiplied by log base 5 of x what about log base 8 of the cube root of x and you want to write this using a fractional exponent so you would say this is log base 8 of x raised to the power of 1 3 this will be equal to again bring this guy down so this will be one third times log base eight of x all right now let's talk about two special properties that you're going to come across in this section and when you first look at them they might look like gibberish but they really save you a lot of time if you can identify them when you're doing your homework or if you're on a test or really any practice that you're doing so if you see something like b raised to the power of log base b of x you should know this is equal to x now the restrictions here is that b has to be greater than zero b is not allowed to be equal to one and x is greater than zero so as long as those criteria are met we're going to be good to go with this kind of formula here and just to demonstrate this i think when you see a numerical example it will make more sense for you if you see something like 3 raised to the power of log base 3 of 7 what would this be equal to well from just looking at this over here you can kind of compare things and say well i know it would be equal to 7 because this guy right here this b is here and here i have a 3 that's here and here right so 3 is kind of in the place where b is in the generic formula then if i know that 7 is here where this x is here and then there's an x here so a 7 must go there so by that process you can kind of figure it out but you really want to understand where we get the answer of 7 from it's a lot easier than just kind of looking at things and saying oh the generic formula says this so let me just copy that think about what this part is right here independent of anything else if i have log base 3 of 7 this is the exponent that 3 must be raised to to obtain 7. well my base here is 3. this is a base of 3. 3 is being raised to this guy right here which again is the exponent that 3 must be raised to to obtain 7. so whatever that value is it doesn't matter three is being raised to it so i'm going to obtain seven right it comes from the fact that the base here and the base here those are the same so that allows us to just take this value right here the argument in the logarithm and just say that's going to be our answer here if i had 7 raised to the power of log base 7 of let's say 30. well again this is really easy 7 is being raised 2 what well log base 7 of 30 is the exponent 7 must be raised to to obtain 30. so 7 is being raised to that value so i'm going to get this right here again the argument in the logarithm the 30 so that's going to be my result all right let's look at another special property this is another one that's going to be a time saver for you if you can identify so you have log base b of b to the x is equal to x the restrictions here would be that b is greater than 0 and b is not allowed to be equal to 1. now for this one it's something you just need to identify you might see something like log base 5 of 5 squared is equal to 1. well again you can kind of pattern match and say okay well there's a b here and a 5 here so this guy right here would be this guy right here and so my answer is this guy right here it's the exponent so the base here and kind of you think about 5 squared 5 is the base there in that exponential expression so those bases are really the same so what i'm saying is that 5 raised to the power of 2 would give me 5 raised to the power of 2 or 5 squared that's all that's really doing it's not very complicated to figure that one out if i had log base 9 of let's say 9 cubed this would be equal to what 9 to what power would give me 9 cubed well 3. it's always going to be the exponent as the answer there because the base here and again when you think about 9 cubed 9 is the base there so those two are the same so really i'm just using the exponent as the answer okay very very simple now where you might have to think a little bit you see this on a test let's say you've got log base 5 of 25 and you want to quickly figure this out well 25 you should know is 5 squared so the trick there is to rewrite this as log base 5 of 5 squared and then you can quickly identify that hey 5 here and 5 here are the same so i know the result here is going to be 2 right 5 to the second power would give me 5 squared or 25. so let's take a look at some examples we have log base 6 of 6 cubed so again this part right here is the same as this part right here so in other words 6 to what power would give me 6 cubed well it would be 3 right 6 to the third power would give me 6 to the third power 6 cubed log base 2 of 2 to the 7th power again this part right here and this part right here is the same so the answer is just the exponent right there it's just going to be the 7. 2 to what power would give me 2 to the 7th power it would just be a 7. what about 3 raised to the power of log base 3 of 8 well this would just be 8. again this logarithm that's the exponent here is the exponent to which 3 must be raised to to obtain 8. so if i raise 3 to that value i'm going to get 8 as a result all right let's take a look at some practice problems so we want to expand each logarithm all right for the first one we're going to look at log base 4 of we have w times the cube root of x times y times z so a lot of ways to kind of attack the problem here but one way you could do it we could say this is log base 4 of w plus remember this is multiplication here so it allows you to use the plus symbol so we'll have log base 4 of this guy right here so i'm going to use a fractional exponent here the cube root of something is like raising it to the one-third power so x times y times z to the power of one-third now i'm going to take this one-third and i'm going to individually raise those factors to that power so in my next step i'm going to say this is log base 4 of w plus log base 4 of x to the power of 1 3 y to the power of 1 3 z to the power of 1 3. okay so now in my next step i'm going to use the multiplication here to kind of further expand this part so we'll say this is log base 4 of w plus log base 4 of x to the power of 1 3 plus log base 4 of y to the power of 1 3 and then plus log base 4 of z to the power of 1 3. now the last thing i want to do here i want to use my power rule and i want to think about these exponents here remember i can write those out in front so what i'm going to do is i'm going to say this is log base 4 of w plus this guy is going to come down so we'll say 1 3 times log base 4 of x plus this guy's going to come down so 1 3 times log base 4 of y then plus this guy's going to come down so i'm going to say 1 3 times log base 4 of z so my answer here is log base 4 of w plus 1 3 log base 4 of x plus 1 3 log base 4 of y plus 1 3 log base 4 of z all right for the next one we have log base 5 of we have 2 over 3 times 7 to the sixth power this is all raised to the fourth power so the very first thing i'm going to do is i'm going to raise everything inside the parenthesis to the fourth power it's going to get rid of this exponent here so this is log base 5 of 2 to the fourth power over 3 to the fourth power times power to power rule here 7 would stay the same 6 times 4 would be 24. now how do i work with this you have a division inside of here so remember the numerator here is going to go first the denominator here has additional multiplication in it so the way you want to think about this you have to use parentheses okay so let me kind of set this up for you you have log base 5 of 2 to the fourth power minus i'm going to use parentheses here you're going to have log base 5 of 3 to the 4th power plus because of that multiplication there log base 5 of 7 to the 24th power now why am i using parentheses or in this case they're actually brackets just i'm referring to any grouping symbols that you're going to use but why am i using brackets here well i'm using that because i want to subtract away this whole thing right this is from my denominator so i want to subtract away the whole thing so i've got to use brackets or parentheses or any grouping symbols that you can kind of think of all right so the next step here you just put equals we have log base five of two to the fourth power then minus you'll have log base five of three to the fourth power then this minus would be distributed to this so you say minus log base 5 of 7 to the 24th power all right so now let's kind of scroll down get a little bit more room going we're going to bring our exponents down in front of each one of these so we'll have 4 times log base 5 of 2 minus 4 times log base 5 of 3 minus 24 times log base 5 of 7. let's take a look at a note so we have log base 8 of 99 times the cube root of 7. so again i'll say this is log base 8 of 99 plus right i have multiplication there log base 8 of this guy right here i would write with a fractional exponent so i would say this is 7 raised to the 1 3 power now i want to bring this guy out in front so let's say this is log base 8 of 99 plus 1 3 times log base 8 of 7. all right let's take a look at a note so we have log base 10 of the fourth root of 5 over x squared y cubed z so again i think about log base 10 of the fourth root of 5 i could write as 5 raised to the 1 4 power then minus i'm subtracting away all this so i've got to use again some grouping symbols parentheses brackets whatever you want to use so let's just do brackets you will have log base 10 of x squared and then you'll have plus let me kind of scooch this down so i'm not going to have enough room so plus log base 10 of y cubed and then plus log base 10 of z and the grouping symbols are just reminding you to distribute that negative to each term so if i keep simplifying here i'll have log base 10 of 5 to the 1 4 power minus right this will go over here log base 10 of x squared minus log base 10 of y cubed minus log base 10 of z okay let's scroll down a little bit get a little room going so let me make that 3 a little bit better and i'm going to bring all these out in front so that's that and that so we'll say this is 1 4 times log base 10 of 5 minus 2 times log base 10 of x minus 3 times log base 10 of y and then minus log base 10 of z all right so now we want to condense each to a single logarithm so we have 2 times log base 2 of 6 plus 4 times log base 2 of 11 plus 4 times log base 2 of 5. so we're kind of undoing these properties we just used to expand things so the first thing is i know that if i have 2 times log base 2 of 6 i could write this as log base 2 of 6 squared right this would just go back up here then plus if i have the 4 out in front i know i could write it as up here so log base 2 of 11 to the 4th power and then plus this guy could go up here so log base 2 of 5 to the fourth power now i have addition so addition you go back to multiplication so everything is log base 2 here right so in other words it's going to be log base 2 of this guy times this guy times this guy so 6 squared times 11 to the 4th power times 5 to the fourth power so in most cases on a test or your homework this answer is good enough i realize that we could multiply all this stuff together it'd be a really big number we're just proving that we can kind of move back and forth if you wanted to do it your answer would be 329 million 422 500. so log base 2 of again 329 million 422 500. but again this answer right here is perfectly acceptable what about 5 times log base 9 of x plus 5 times log base 9 of z minus 15 times log base 9 of y so again these are all going to go back up here so we'll do that first so we'll say this is log base 9 of x to the fifth power and then plus log base 9 of z to the fifth power then minus log base 9 of y to the 15th power all right now the next thing we want to think about is we have log base 9 log base 9 log base 9. so log base 9 of what we know these two because there's addition here this is multiplied by this so it'd be x to the fifth power times z to the fifth power then i have subtraction here so these two are going to be in the numerator this guy right here that's after the subtraction sign is giving me the denominator so this is over y to the 15th power so that's going to be your simplified answer so log base 9 of x to the fifth power times z to the fifth power over y to the 15th power let's take a look at one more problem so we have 1 3 log base 4 of x plus 2 7 log base 4 of y minus 3 log base 4 of z so again i'm just going to take all these back up real quick and we'll start out with that so we'd have log base 4 of x to the 1 3 power plus log base 4 of y to the 2 7 power minus log base 4 of z to the third power so this equals 1. well we'll have log base 4 of i have my addition here so this one times this one so x to the 1 3 power times y to the 2 7 power so we have this subtraction involved here and so we know this is going to be over this z cubed here that's what comes after the subtraction so that's going to come right here so you end up with log base 4 of x to the 1 3 power times y to the 2 7 power this is over z cubed hello and welcome to algebra 2 lesson 72 in this lesson we're going to learn about common and natural logarithms so far when we dealt with logarithms if we came across a logarithm with a base of 10 we listed the base of 10 just like we would for any other logarithm but there's a shortcut involved when you see a logarithm with a base of 10. so a logarithm with a base of 10 is known to us as a common logarithm this is because we have a base 10 number system so when we encounter one of these guys we don't actually have to list the base so something like log base 10 of x could be written as just log of x right so the base here is understood to be 10. so if i saw something like log of 100 this is understood to have a base of 10. so this is really log base 10 of 100 and we all know this is 2 right 10 to the second power will give me 100. or if i saw something like log of let's say 30 we know this is log base 10 of 30. so if you're working with logs and you see log of some number and you're like what is the base again it's just understood to be 10 you've come across a common logarithm now another important logarithm we're going to come across is known as the natural logarithm so the natural logarithm obtained its name because of its place in biology in natural situations where growth or decay is involved so the natural logarithm has a base of e so if you see something like log base e of x this base right here this is what we're talking about in e this number is approximately equal to 2.71 this e here is going to come up a lot okay when we're working with math moving forward and just think of it as kind of another number that you're going to encounter like pi it's this special irrational number again after the 1 here there's an infinite number of digits there's no pattern to it it just goes on and on and on so when we see log base e of x we can abbreviate this and just say this is the same as saying ln which stands for the natural logarithm of x so if i see ln of 5 this is the same as saying we have log with a base of e of the number five okay that these two are the same but typically you're going to see this written and not this written okay so you'll just see ln of the number all right now let's talk a little bit about how we can use a calculator to either get an exact value or an approximate value for a logarithm so the reason it's important to talk about the common logarithm and the natural logarithm these are the two buttons that are going to be available to you on most calculators now there might be some calculators out there that allow you to change the base of a logarithm the calculator i have the calculator that most of you will have gives you just an ln key for the natural log and gives you a log key which is for a base 10 log or a common logarithm so if you are in that situation like i am you need to use a change of base rule in order to plug logarithms into the calculator and get either an exact value or in most cases an approximate value so the change of base rule goes like this for a greater than 0 a not equal to 1 b is greater than 0 b not equal to 1 and x is greater than 0 you have log base a of x this is equal to log base b of x over log base b of a so just notice that the base here and the base here is the same so in most cases we're just going to use the common logarithm so if i had something like log base 2 of 10 and i wanted to change this to a common logarithm or a base 10 logarithm i could just say okay this is log of 10 over log of 2 okay and you could punch this into your calculator and get an approximate value and if i went to three decimal places on this i could say this is approximately equal to three point three two it's one and then a nine so the 9 allows me to take that 1 up to a 2 and then everything after that would be a 0 if i was rounding so let's just say this is approximately 3.322 okay so knowing that i couldn't have typed this into most calculators but i can type this into most calculators and i can get this so that's where the change of base rule comes into play again my base here and my base here are the same i might as well just wrote this is base 10 and base 10 but again if i just write log of something i understand that the base is 10. so that's why i use that little shortcut then this guy right here came from this okay again just follow the pattern this right here your argument in the logarithm is going to go as the argument in the logarithm in the numerator and this right here the base okay the base is going to become the argument in the logarithm in the denominator so just following that pattern you can change the base and you'll be able to key that in to your calculator and you'll get the same result if you use the natural log so if i said ln of 10 over ln of 2 i would get approximately 3.3 as well okay so same thing either way if you want to use the natural log or if you want to use the base 10 log or the common logarithm it's going to be the same answer all right so now we want to use your calculator and the change of base rule to approximate each to three decimal places all right so we have log base 4 of 53 so i'm going to change the base to a base 10 log and so i'm going to say this is log of this guy 53 over log of this guy which is four so what's that going to give me it's approximately what so my calculator the readout gives me 2 2.86396022 7. if i want to approximate each to three decimal places i'm looking at this guy right here so here's where i would look and say okay well the 9 is in the category of 5 or greater so i would round up right so this would turn into a four and then everything after that would be a zero and i could just cut it off so i end up with approximately 2.864 what about log base 5 of 2.9 so this would be equal to log of 2.9 over log of 5. so this would approximately be 1. so my calculator gives me six 0.661 five four one nine seven six one but again if we're approximating this to three decimal places well then you've got one two three so you're looking there so the guy after is a 5 so i can make this into a 2 and then i can cut everything else off so this is approximately 0.662 all right now we're looking at log base 4 of 3.4 let's go ahead and just use the natural logarithm so let's do ln of 3.4 over ln of 4. so what does this give us well this would approximately be equal to you've got point eight eight two seven six seven three seven three two that's the read i get from my calculator so for three decimal places i'm looking at that digit there and so the seven here the digit after is in the category of 5 or greater so i can go ahead and round up and say this is a 3 and then i can cut everything else off so this is approximately equal to 0.883 what about log base 2 of 23 what do we get there so i can say this is log of 23 over log of 2. so this is going to be approximately equal to five two three five six one nine five six so three decimal places right there and i'm looking at this digit here this is in the category of five or greater so i can round up that 3 will become a 4 and i can cut everything else off so this is approximately equal to 4.524 what about log base 7 of 66 so let's say this is ln of 66 over ln of 7. so this will be approximately equal to you'd have 2.153056627 so three decimal places you're looking at this right here the 0 that comes after falls in the category of 4 or less so we would leave the 3 unchanged and we would just cut everything off here so your approximate answer would be 2.153 what about log base 7 of 49 well most of us know that that's equal to 2. but let's just for the sake of doing the example here let's punch this up on a calculator let's go log of 49 over log of 7 and you'll see that you get an answer of 2 exactly what about log base 2 of 64. this is another one that we can get a perfect answer for we know that 2 to the 6th power is 64. so for this one punch up on your calculator ln of 64 over ln of 2 and you're going to find that you get an answer of exactly 6. all right let's look at one that's a little bit challenging so let's say you get log base 2 of negative 1 8 is negative 1 8 allowed to be the argument of a logarithm no it is not remember we say that this guy right here has to be greater than 0. why is that the case think about this for a second 2 to the power of what would produce negative 1 8. you can't do it if i raise 2 to a positive number i get a positive number so like 2 squared is 4 or 2 cubed is 8 so on and so forth if i raise 2 to the power of 0 i get 1. if i raise 2 to a negative number so like negative 4 it's 1 over 2 to the 4th power which is still positive right this is 1 over 16. as this guy gets more and more negative we're going to get a value that's closer and closer to 0 but it will never actually become negative so this right here is not a possibility for an argument in a logarithm and so because of that we can say this guy is undefined right we don't need to do a change of base rule if we see something like this as a trap question we just say it's undefined what if we saw log base 2 of 1 8 well in this scenario 1 8 is positive so we can get an answer here we know that 2 to the power of negative 3 would give me 1 8 but again for the sake of doing this for this exercise let's just go ahead and say this is log of 1 8 over log of 2. punch that up on your calculator and you should get negative 3 as a result hello and welcome to algebra 2 lesson 73 in this lesson we're going to learn about solving exponential and logarithmic equations so although we've seen how to solve an exponential equation before and also we've seen how to solve a logarithmic equation before the problems we're going to deal with here are going to be significantly more difficult so in order to get through them we have to go through a few properties real fast so this property here you've already seen it when we solved exponential equations we say if x equals y then b raised to the power of x is equal to b raised to the power of y or you could also say if b raised to the power of x equals b raised to the power of y then x is equal to y now the restrictions here say that b your base has to be greater than zero so it's got to be positive and also it's not allowed to be equal to one so given that you follow those restrictions if you have something that sets up like this b to the x equals b to the power of y so in other words the base is the same then you can set x equal to y and you can solve right we've already experienced that before we would make our bases the same we would set the exponents equal to each other and we would solve and get our answer now let's talk about some other properties if x equals y then we can say that log base b of x is equal to log base b of y we could also say if log base b of x equals log base b of y then x is equal to y now you have more restrictions here we say that this is for x is greater than zero y is greater than zero b is greater than zero and b is not allowed to be equal to 1. so x and y here represent arguments in the logarithms so here's x here here's y here remember those have to be greater than 0. the base here the b we say has to be greater than 0 and again it's not allowed to be equal to 1. so we're going to use these properties you might want to pause the video and just write them down we're going to use them as we start to solve these equations all right so we're going to start off by just looking at some exponential equations and the first guy we're going to look at is 19 to the power of x equals 28. now a few lessons ago we looked at easier examples where we could just make the bases the same we could set the exponents equal to each other and then we would solve that's not going to work for us here so we need to use a different technique in this case we're going to take the log of each side and remember if i have log base 10 of something i don't need to write the base of 10 it's implied so i can say this is log of 19 raised to the power of x this is equal to log of 28. so i really didn't do anything special here no magic tricks involved i just took the log of both sides now what does this allow us to do remember that power rule for logarithms that we talked about a few lessons ago that allows me to take this guy right here the exponent and bring it down here so what's going to happen is i'm going to have x times log of 19 and this will be equal to log of 28. now if i want to solve for x i just want to isolate it so what's multiplying x here well it's just log of 19. so all i need to do is just divide both sides of the equation by log of 19 and i'm done so it's very very easy to solve a scenario like this this guy right here will cancel with this guy and i'll have my answer it's x is equal to log of 28 over log of 19. now this is an exact value here if you want an approximation you can type that into a calculator and depending on how accurate you want it to be you can round or you can trunk it whatever you want to do just make sure you use an approximation symbol versus an equal symbol to say hey this is approximately this versus saying it's exactly this another way we could write this remember our change of base rule so we could say this is x is equal to log base 19 of 28 okay so that's two different ways you can write this i would probably stick to this way right here so in my solution set i will write the element as log of 28 over log of 19 and i keep it in that format because it's easier to type it into a calculator that way so this is what i would report as my answer if i was on a test but again if you needed to give an approximation you could just punch it into a calculator and give that all right let's take a look at a note so we have 3 to the power of x equals 40. so we're going to do this the exact same way instead of using the common logarithm let's kind of use the natural logarithm we can use either one both buttons are accessible on pretty much every calculator so i'm going to say this is ln of 3 to the power of x is equal to ln of 40. and again this is the exact same thing as if i would have done log of 3 to the power of x is equal to log of 40. it's going to enable me to do the same thing so this guy right here is going to come down and so i'm going to have x times ln of 3 is equal to ln of 40. i divide both sides of the equation by ln of 3. again that's what's multiplying x and i'm going to have my answer this cancels with this i'll have x is equal to ln of 40 over ln of 3. all right so for my solution set i'm going to put ln of 40 over ln of 3. all right so now we're going to look at one that involves the number e so remember this is a special universal constant something like pi so when e is involved you always want to use the natural logarithm you might say y well if i take e to the power of x equals 88 and i say ln of e to the power of x equals ln of 88 what's going to happen is this right here is going to cancel itself out and become 1. so i'm going to have x equals ln of 88 and i'm done now you might say why in the world does ln of e equal 1 well think about what ln actually means remember ln is a log with a base of e so really what this was was it was log base e of e to the power of x now i can bring this guy down here so really i'm saying this is x times log base e of e so what is this part equal to it's equal to one e to what power gives me e 1 right e to the first power would give me back e so when you have ln of e that automatically equals 1. so when i see something like that i always want to use the natural logarithm you could do this using the common logarithm if you want but it's just going to be more work for you so you want to stick to this technique all right so x equals ln of 88 or in solution set notation i'll just put ln of 88 like that all right let's take a look at another one so we have e to the x equals 75. again i'm going to use ln so i'm going to use my natural log so ln of e to the x is equal to ln of 75 and again this guy would come down here so you'd have x times ln of e we know that this guy equals one right i just showed you that so this would be equal to ln of 75 so there's your answer you would have x equals ln of 75. so let's put this in solution set notation ln of 75. all right let's try a few that are a little bit more tedious and then we'll move on and look at some logarithmic equations so nothing you're going to see in this section super hard it's just a lot of things that you have to remember from all the things throughout the chapter where you're talking about exponentials and logarithms so here we have 2 raised to the power of x plus 3 equals 5 raised to the power of x so i'm going to use my normal technique i'm going to say this is log of 2 raised to the power of x plus 3 equals log of 5 to the power of x so what i'm going to do is i'm going to take these down and so i'll have x plus 3 that quantity multiplied by log of 2 and this equals x multiplied by log of 5. now in a lot of cases when people see this they get stuck at this point they're like what do i do how do i get x by itself i've got x on this side i've got x on this side but it's wrapped in some parentheses just realize that you're going to use your distributive property here this is a number if i had a number multiplied by the quantity x plus 3 i would just distribute it so you're not going to use any different technique simply because you have a logarithm involved it's just a number so i'm going to say this times this plus this times this so we'll have x times log of 2 plus 3 times log of 2 this is equal to x times log of 5. now what do i want to do next well what i want to do is i want to get all the terms with x in it on one side of the equation everything else on the other this is no different than what we've been doing forever so let me subtract x times log of 5 from both sides of the equation let me also subtract 3 times log of 2 away from each side of the equation so this would cancel with this and this would cancel with this so what i'd be left with on the left is x times log of 2 and then we'd have minus x times log of 5 and this equals negative 3 times log of 2. now you might be saying how in the world am i going to simplify this you're going to factor so remember you have this here and this here i could factor the x out i could factor the x out and i would have x times inside of parentheses you'd have log of 2 minus log of 5 and then this is equal to negative 3 times log of 2. so at this point it's pretty easy to see that x is multiplying this guy right here and so i would just divide both sides of the equation by that and i have my solution let me kind of scooch this down a little bit get some room so i would divide both sides of the equation by this log of 2 minus log of 5 so over log of 2 minus log of 5. and this can get pretty tedious depending on what you're working with but in the end it's no more difficult than solving anything else just got to take things one step at a time so our answer here will be that x is equal to negative 3 times log of 2 this is over you have log of 2 minus log of 5. and in solution set notation i'll just put negative 3 times log of 2 over log of 2 minus log of 5. what about 5 times e raised to the power of 10x plus 2 plus 9 equals 93. so in this scenario you want to kind of clean things up first you want to try to isolate this part right here so what i'm going to do in order to do that is just subtract 9 away from each side of the equation first so that will cancel i'll have 5 times e raised to the power of 10x plus 2 and this is equal to 93 minus 9 is going to give me 84. now again i want this part by itself so let me divide both sides of the equation by 5. and so this will cancel with this and now what i'm going to have is something easy to work with so i'm going to have e raised to the power of 10x plus 2 this is equal to 84 fifths the reason i want this by itself is because now i can just say ln of this side equals ln of this side right so now i know that this e part and the ln will cancel right they'll become 1. so i'm just left with this part right here in the exponent because that would have come down so i would have 10x plus 2 is equal to ln of 84 fifths now i've got some more work to do to solve for x let me kind of drag this up here what i'll have is 10x plus 2 equals ln of 84 fifths i will subtract 2 away from each side of the equation so that's going to cancel i'll have 10x is equal to ln of 84 fifths minus 2. i can divide both sides of the equation by 10. over here what i'm going to do is i'm just going to multiply the whole thing by 1 10 to make it a little easier on myself so this will cancel i'll have x is equal to you'll have ln of 84 fifths multiplied by 1 10 then minus two times one-tenth two times one-tenth is one-fifth so this would be my solution so i could write it in solutions that notation just say this is ln of eighty-four-fifths you can put times one-tenth or you can divide it by ten there doesn't matter i'm just going to put times one-tenth i think that looks a little cleaner than minus one-fifth and again that's your answer let's take a look at one more of these and then we'll kind of move on and look at some logarithmic equations so we have 6 raised to the power of 6n minus 1 plus 4 equals 80. so i want to kind of clean it up i want this by itself so let me subtract 4 away from each side of the equation to start i'll have 6 raised to the power of 6n minus 1 is equal to 76 i'm going to take the log of both sides so log of 6 raised to the power of 6n minus 1 is equal to log of 76. so what this is going to again allow me to do is bring this down so i'll have 6n minus 1 multiplied by log of 6 this is equal to log of 76. scroll down get a little bit of room going so at this point i can either use the distributive property or i can outright just divide by log of six so this would cancel so i'd be left with 6n minus 1 is equal to log of 76 over log of 6. okay so what do we do now we want this part y itself right we want the n by itself so i need to add 1 to both sides so that will cancel so we'll have 6n is equal to log of 76 over log of 6. and then plus 1. so the last thing i need to do is divide both sides of the equation by 6. again if i'm going to do that i'm just going to multiply this guy by 1 6 and that will allow me to get n by itself all right let's scroll down get a little bit more room going so we'll say n is equal to we will have log of 76 over log of 6 then times 1 6 or you can put divided by 6 it doesn't really matter then plus 1 6 times 1 would just be 1 6 so plus 1 6 over here so in solution set notation i'm just going to say log of 76 over log of 6 multiplied by 1 6 then plus 1 6 and that's going to be your solution all right now let's take a look at some logarithmic equations so let's say we saw log of x minus log of 7 equals 1. so what would we do we would use our properties of logarithms to rewrite the left side as one single logarithm so if i have subtraction here remember i'm allowed to condense this subtraction is related to division when we think about logarithms so this would be log of the first guy right here will go on the numerator and then you have your subtraction so this is divided by this guy right here the second guy is going to go in your denominator and then of course this is equal to 1. so now that we have this set up in this manner how can we solve this well we can write this using exponential form and that's easy to solve so remember this is log base 10 of x over 7 equals 1. so in other words this is 10 to the first power is equal to x over 7. pretty simple equation to solve i would simply multiply both sides of the equation by 7 and what's going to happen is i get 70 is equal to x or my solution here is just 70. now if you plug that back in you can easily see that makes sense if i had a 70 here 70 over 7 is 10 so i'd have log base 10 of 10 is equal to 1. that's true 10 to the first power does give me 10. all right let's take a look at another so we have log of 9 plus log of x equals 1. so again what i want to do is use my properties of logarithms to condense this to 1 log on the left side of the equation so because this is addition i'm going to condense using multiplication so this is going to be log of this guy times this guy so 9 times x is 9x this equals 1. and again all i want to do at this point is just use exponential form so i know the base here is implied to be 10 so 10 to the first power equals 9x so 10 to the first power equals 9x i can divide both sides of the equation by 9 and i'll get that x is equal to 10 9. all right what about log base 4 of negative 4x minus log base 4 of 5 equals 1. so again i have subtraction here so i'm going to condense using division so i'll have log base 4 of this guy right here which is negative 4x divided by this guy right here which is 5 this is equal to 1. so i want to convert this into exponential form so it's four to the first power is equal to negative four x over five so four to the first power equals negative four x over five if i wanted to solve this i'd multiply both sides by the reciprocal of negative four fifths which is negative five-fourths so this part would cancel and i would have x is equal to fours would cancel so just negative five now you gotta be careful here remember your argument for the logarithm has to be greater than zero so if you end up with some answer that makes it negative or zero you got to reject it okay so that's what you're looking for if i plugged in a negative 5 there negative 4 times negative 5 would be 20. 20 is a positive number so you're good to go this would be log base 4 of 20 minus log base 4 of 5 equals 1. who would that work out to be if we wanted to check it so that would be log base 4 of 20 over 5 is 4 so log base 4 of 4 equals 1 so that does check itself out and then in solution set notation i'll just put negative 5 inside of set braces all right what about log base 2 of 5 minus log base 2 of 5 x equals 2. again i'm going to condense this left side this will be log base 2 of i have some fractions so that's going to mean division so 5 this guy right here divided by 5x that guy right there this will be equal to 2. so 2 to the second power is going to give me 5 over 5x 2 to the second power equals 5 over 5x okay so let's go ahead and solve this guy we know that 2 squared would be 4 so we would get 4 is equal to 5 over 5x so i know these fives would cancel and so this would turn into one over x is equal to four if i wanted to solve this an easy way to think about this is that this is x to the power of negative one equals four i could raise both sides of the equation to the power of negative 1 and i'd have my solution so negative 1 times negative 1 would be 1 so this would be x now this is equal to 4 raised to the power of negative 1 is 1 4. so x here would equal 1 4. and if you wanted to check think about plugging in a 1 4 there what would that do you'd have 5 times 1 4 which would be 5 4. so you'd have log base 2 of five over five fourths so five over five fourths you can think about that as five over one times four over five the fives would cancel and you'd be left with 4. so really this is 4 right here so log base 2 of 4 is equal to 2. is that true well is 2 squared equal to 4 yes it is so this does check out let's take a look at log of 3x squared plus 8 minus log of 2 is equal to 2. so again i'm going to condense this using division because i have subtraction involved there so i'll have log of it'll be 3 x squared plus 8 that'll be in the numerator over 2 which will be in the denominator this is equal to 2. so in exponential form i would have 10 raised to the power of two or you could say a hundred is equal to three x squared plus eight over two now to get rid of the denominator there i can multiply both sides by two so if i multiply this side by two and this side by 2 i would have 200 is equal to 3x squared plus 8 i would subtract 8 away from each side of the equation so that would give me 192 is equal to 3x squared i would divide both sides of the equation by 3. it's kind of a race over here real quick so 192 divided by 3 would give me 64. so i would have this canceling with this x squared is equal to 64. now i'm going to take the square root of each side and you'll have x is equal to plus or minus the square root of 64 is 8. now you do want to check to make sure that nothing you plug in here is going to make this guy zero or negative remember the argument's got to be positive this one right here you're not going to change it's set as 2. you just have to worry about that one so if i plugged in a negative 8 there it would be squared right so if i square that it'd be 64. 64 times 3 would give me 192. 192 plus 8 would give me 200. so essentially you would have 200 divided by 2 which would be 100 so that does work itself out in other words you're saying 10 to the second power gives you 100 so that works whether i plugged in an 8 there or a negative 8 i would not end up with a negative argument because of the squaring operation there so i'm good to go right either way i ended up with 100 right there so this answer here we'll go ahead and put in set braces we'll put plus or minus 8. all right let's take a look at one more problem so we have log of 3x minus 2 minus log of 6 equals log of 11. so i'm going to condense the left side so we're going to say this is log of 3x minus 2. i've got subtraction so it's divided by that's 6 there this equals log of 11. now i could go through and solve this a different way but the easiest way to solve this is to realize i have the same base here so we already talked about the properties if you had log in this case is base 10 so log base 10 of something equals log base 10 of something else the two arguments here are going to be equal to each other okay so i can just set 3x minus 2 over 6 equal to 11 and i can solve so let's start by multiplying both sides of the equation by 6. this will cancel with this i'll have 3x minus 2 is equal to 66. i can add 2 to each side of the equation that'll cancel 3x is equal to 68 is what i have now divide both sides by 3 and let me kind of erase all this so 68 is not divisible by 3 it's 17 times 2 times 2 so we can just leave it as x equals we'll say this is x equals 68 thirds so in solution set notation we'll say 68 thirds and if you wanted to check that you could you could plug in a 68 thirds there and see what you get if you plug in a 68 thirds the threes would cancel you so you'd have 68 minus 2 which is 66 6 over 6 is 11 so you'd have log of 11 equals log of 11. that's true so this does end up working itself out hello and welcome to algebra 2 lesson 74. in this lesson we're going to look at some additional graphs of functions so a few lessons ago we learned about graphing parabolas now in that lesson we talked about the quadratic function f of x equals x squared and then we compared other functions to that and we talked about the fact that we could have a vertical a horizontal or both a vertical and a horizontal shift so what we're going to do today is we're going to look at some other commonly occurring functions things that you'll see in algebra 1 and algebra 2 and we're going to talk about what creates this vertical and horizontal shift and we'll talk about how you can determine if you have a vertical or horizontal shift without actually looking at the graph or pulling out a t table or anything like that there's a very easy formula to follow so the first one we're going to look at is the absolute value function we have f of x equals the absolute value of x when we talk about this function the domain is any real number so i can plug anything i'd like in for x there i can do a negative a positive zero whatever i'd like the range is limited though remember the absolute value operation returns a positive if i plugged in a negative so if i did something like the absolute value of negative 5 that would give me an answer of positive 5. if i did 0 i would get 0 back if i did a positive i would get a positive back so your range is limited the smallest it can be a zero the largest it can be is positive infinity so basically zero or any positive real number now one interesting thing with the absolute value operation every number except for zero has an opposite so five and negative five three and negative three two million and negative two million opposites have the exact same absolute value because again they're the same distance from zero on the number line so what happens is you end up with a v shape when you graph this guy in other words if i was to look for points here for everything i plug in so if i plugged in a negative 6 there i know my y value once i evaluate the absolute value of negative 6 i would get positive 6. well then i could do an x value of 6 there and i would also get a y value of 6. so if i look at my coordinate plane for an x value of negative 6 i'd have a y value of 6 but also for an x value of positive 6 that have a y value of 6. so this guy corresponds to here and here and what's going to happen is if i did an x value of 5 i would get a y value of 5 but an x value of negative 5 will give me a y value of 5. so this keeps going down like this and it creates a v shape as it comes back up here if you plugged in a 2 for x absolute value of 2 would give me 2. but if i plugged in a negative 2 for x the absolute value of negative 2 would give me 2. so let's graph this guy okay so this is f of x is equal to the absolute value of x all right let's talk a little bit about horizontal shifts and i'll give you a little shortcut here on how to tell if you have a horizontal shift and what the amount of the shift is in which direction and then in the example we'll get to in a second i'll show you why it actually happens so if you see f of x equals the absolute value of x and then you come across f of x equals the absolute value of x minus h the first thing you need to understand is that this minus h here happens inside of the absolute value bars so we would say it happens inside of the function so you want to write down in your notes that if it happens inside of the function you will have a horizontal shift the way you can tell your operation there is the absolute value operation so if it happens inside of that you're going to end up with a horizontal shift if it happens outside of it so if you had f of x equals the absolute value of x let's say plus 2 this is outside of it so this will be a vertical shift but in this case it's happening inside so inside of the function one way to tell for sure is to ask yourself can i take this guy right here and plug it in for x in the original function the one i'm comparing it to would i get the same thing the answer is yes i would end up with this guy exactly so it is happening inside the function we've established that now if you have something in this format the horizontal shift will be as follows we say if h is greater than 0 we would shift h units to the right so if i saw f of x is equal to the absolute value of x minus 7 this would mean that i am shifting 7 units right all right the other scenario if we have f of x equals the absolute value of x plus h here if h is greater than zero we're going to shift h units left so f of x is equal to the absolute value of x plus seven in this case here seven is greater than zero so we would be shifting seven units left let's think about why this is the case let's say again compared to f of x equals the absolute value of x we look at f of x equals the absolute value of x minus 3. the shift according to what i just told you because this is a minus 3 here inside the absolute value operation would be 3 units to the right so this shifts 3 units right now why is that the case again most of you will probably say it's a minus three on the x-axis if we're thinking about that a minus three would be three units to the left so why is this guy saying three units to the right what ends up happening is for a given y-value a given vertical value on the coordinate plane x now has to be three units larger to produce that so let me show you with the t table let's say in the original function i choose an x value of negative three i choose an x value of zero and i choose an x value of positive three so the y value here would be three the y value here will be zero and the y value here will be three how can i get the same y values over here let's just choose them already so we're going to say 3 0 and 3 is what we want well i would have to plug in an x value that's 3 units larger than these over here to compensate for this minus three here so in other words if i had a zero which is three units larger than negative three the absolute value of negative three would give me three so you see how once i take away 3 i'm right back to this guy right here and then i'm taking my absolute value so i'm just kind of thinking about what amount would i have to put in there to kind of undo this operation of subtracting away 3 and that would give me the same result as i would have gotten if there was nothing done to it okay so in this case i have to add three to undo the minus three that's all i'm doing so in this case right here if we have 0 we'd add 3 this would be 3. 3 minus 3 would give me 0 absolute value of 0 is 0. and then in this case if it's 3 it would go to 6. 6 minus 3 is 3 absolute value of 3 is 3. so that's all we're really doing we're adding 3 to each x value to undo the effect of this minus 3 here so if we looked at a graph so we did f of x equals the absolute value of x minus 3 some points we looked at so far were zero comma three three comma zero and six comma three so zero comma three is going to be right here then 3 comma 0 is going to be right here and 6 comma 3 is going to be right here so all we're really doing is shifting everything to the right 3 units so this would go 3 units to the right that's my point i could take this point right here and go one two three to the right that would be my point i could take this guy one two three to the right that's my point one two three to the right i already have it one two three is to the right you know so on and so forth it's the same thing with these points over here so i could take this one and go one two three units to the right this one one two three is to the right you know so on and so forth so this guy let me label it this is f of x equals the absolute value of x minus 3. again for a given y value the x value now has to be three units larger and so that creates a graph that's just going to shift to the right by three units so hopefully that makes some sense for you again a lot of people think that minus three would cause you to go to the left by three but it's the opposite of that you wanna go to the right because you're trying to undo the -3 so you basically have to think about it as i'm adding 3 so i'm going to the right by 3 units all right so what if i saw f of x equals the absolute value of x and compared to that i have f of x equals the absolute value of x plus 4. well in this case i'm adding 4. so again if i just think about it simply what do i need to do to undo the addition of 4 i would subtract 4 so that means i'm shifting 4 units to the left so i shift or i could say shifts four units left right pretty easy so i have this guy drawn already for us here's the original and then here's the other guy where we have f of x plus 4. you can see every point has just moved 4 units to the left and again that's to compensate for the fact that we're adding 4 to x so for example if i have f of x equals the absolute value of x plus 4 to get a y value of 0 i now have to plug in an x value of negative 4 to undo that plus 4 there negative 4 plus 4 would be 0 absolute value will give me 0. so to get a y value of 0 now my x value has to be negative 4 whereas before my x value was 0. so everything just shifts 4 units to the left to compensate for that plus 4 that's inside the absolute value operation all right so the next thing we want to talk about would be a vertical shift so if we're looking again at f of x equals the absolute value of x and we see f of x equals the absolute value of x plus k well now what's going to happen is for a given x value i plug something in for x i take the absolute value everything is the same but then i have this plus k that gets tacked on to the operation so my f of x or my y value will change based on this so this is going to directly affect y and when we affect y we're affecting the vertical axis so we're going to shift up or we're going to shift down so if k is greater than 0 meaning i had something like f x equals the absolute value of x plus let's say three well in this case k is greater than zero we would just say that we're shifting up by three units in the case where k is negative so let's say this was plus negative three well in this case we're going to shift the absolute value of k units down so the absolute value of negative three is three we'd say we'd shift three units down now no one would ever write it like this this is just how it is presented in your textbook in most cases you just see -3 here so you would just kind of adapt and say okay i know if i'm subtracting away 3 i go down 3. if i add 3 i know i'm going up by 3. all right let's take a look at an example so we have f of x equals the absolute value of x minus 5 and we have f of x equals the absolute value of x so we can say for a given x value the y value that will be produced will be 5 units less and this will create a vertical shift down by five units so again we can go with negative three zero and three this would give me three zero and three for this guy if we have negative three zero and three again i plug in a negative 3 here i take the absolute value i get this result but then i just get minus 5. so i can basically take these values and subtract away 5 and give myself the y value from over here so 3 minus five would be negative two zero minus five would be negative five three minus five would be again negative two so i can take these points and plot them so we have f of x equals the absolute value of x minus five our first point we have negative three comma negative two we have zero comma negative five and then we had three comma negative two so negative three negative two it's right there we have zero comma negative five and we have three comma negative two so three negative two now i can just take points from this original guy and shift them down five units if i want more so i can take this point go down one two three four five and take this point go down one two three four five and you can see where the line is going this right here would be the lowest point on the graph we know that right because if x is zero that's the lowest i can make this guy because the absolute value operation if you plug in a zero you get zero if i plug in something negative remember i get a positive so the lowest i can get is if x is zero that gives me a y value of negative 5. over here if i shift some of these down if i take this guy i go down one two three four five over here one two three four five and i have enough points to get a good graph going okay so this would be my graph for f of x equals the absolute value of x minus five so again every point on the original just shifted down five units so this went down one two three four five this one went down one two three four five this one went down one two three four five you know so on and so forth so in this next example we see f of x equals the absolute value of x plus 2. so now i'm adding 2 outside of the absolute value operation so again i'm just shifting the graph up by 2 units so we can say this shifts two units up so here's all three of these guys here's the original here's the guy that was shifted five units down and here's the guy that shifted two units up right so if i plugged in a zero for x obviously you would get 2 for y right because you have f of x equals the absolute value of x plus 2 so you get a 0 there 0 plus 2 would give you 2. so everything is just shifted up 2 units as we would expect so what happens if they throw both of these on you so in other words you get a horizontal and a vertical shift this part right here happens inside the absolute value operation this part right here happens outside so this will create a horizontal shift and this will create a vertical shift you just have to remember inside the function or inside the operation is giving you a horizontal shift outside is giving you a vertical so with a vertical shift it's intuitive so if you're adding something you're going up with a horizontal shift it's counter-intuitive if i'm subtracting i'm going to the right so if i have x minus 1 here again i've got to think about what can i do to counteract that minus 1. how can i get this back to where it was i've got to add 1 so i'm shifting to the right so this shifts right by one unit and it's going to go up by 3 up by 3 units so here's a graph of again the original here's a graph of the guy that shifted so it went up one two three and to the right one so there's your new lowest point you can take any given point and go up three units and to the right one you'll be on the new graph so you'll go up 1 2 3 to the right 1. you take this point and go up 1 2 3 to the right 1. so on and so forth alright the next function we're going to talk about is the square root function so we have f of x equals the square root of x the domain and the range for this guy are both limited and the reason for that is simple if i plug something in for x here i could plug in a 0 square root of 0 is 0. i can plug in a positive square root of positive will give me a positive but remember i can't take the square root of a negative number without involving the imaginary unit or the complex number system because we're limiting this to the real number system we would say our domain is restricted to zero or anything larger and the same thing for the range zero or anything larger all right so if the function we're comparing everything to is again f of x equals the square root of x for f of x equals the square root of x minus three this is happening inside the function inside the square root operation so if it happens inside the function we have a horizontal shift and again it's going to be counter-intuitive so what do i need to do to cancel out this minus 3 here i would add 3. so you think about like that this is going to shift or you could say shifts 3 units right because again for a given y value my x value now has to be 3 units larger to compensate for this minus three here as an example let's say we looked at one ordered pair let's say i plug in a zero for x i know i would get a zero for y well here to get that y value of zero what would x have to be i would have to increase it by 3 units i would now have to plug in a 3 for x 3 minus 3 would give me 0 square root of 0 would give me 0. so a 3 for x would now give me a y value of 0 whereas before an x value of 0 will give me that same y value of 0. so i had to increase x by 3 units and it would be the same thing everywhere i'm just shifting the original graph 3 units to the right so if we look at the graph here's the original here's the one that we looked at after so everything is just moving three units to the right here's another point here this goes three units to the right you know so on and so forth all right let's take a look at some more here so we have f of x equals the square root of x plus 5 then plus 2. so you have two different things going on you have something inside the function and something outside the function so inside the function you have plus 5. so again to undo that adding of 5 i've got to subtract 5 away so that means i'm going to shift to the left by 5 units so this shifts left by 5 units and then i have plus 2 that's happening outside the function that would just drive up my y value by 2 for a given x value so this would cause it to shift up by two units so and up by two units then we have f of x equals the square root of x minus seven this is happening outside the function outside the square root operation so we can just say this shifts down by seven units so this shifts down by seven units so here's all your graphs drawn if you want to look at them here's your original here's the one that was shifted down by seven units here's the one that was shifted to the right three units and this guy had both changes right it was shifting horizontally and vertically so it went up one two units and it went to the left 1 2 3 4 5 units all right the next function we're going to look at is known as the reciprocal function so we have f of x equals 1 over x the domain and range i haven't listed here but essentially the domain is anything but zero i can't plug in a zero because i don't want to divide by zero and my range is also anything but zero so any number in the real number system you choose other than zero is allowed so the first thing we'll compare to it we have f of x equals 1 over x plus 3. this is happening outside of the function so if it happens outside the function it's a vertical shift in this case we're just going to move up by three units so we'll say this shifts up by 3 units for a given x value you can see the y value would just increase by 3. if in this one i chose an x value of 1 to make it really easy my y value would be 1 right plug in a 1 there 1 over 1 is 1 so y is 1. in this case if i chose an x value of 1 my y value according to this should be 4. 1 over 1 would be 1 one plus three gives me four so that's what i would have so every point just shifts up by three units so here's a graph of these guys you can see if you pick a point like one comma one on the original you could just shift it up by three units and you get to it on the other so everything is just moved up again by three units what about f of x equals one over x plus two again compared to f of x equals one over x so this part right here is happening inside the function i can take x plus 2 and plug it in for x so what happens is that's going to give me a horizontal shift so if it's plus 2 here you just think about the fact that i would need to subtract 2 away to bring it back to where it was i'm thinking about how i can counteract that plus 2 there so for a given y value my x value needs to be 2 units smaller to get to it so i would say this shifts two units left and to prove that to you again let's do a little t table i know in this guy i could do one comma one as a point over here according to what i just said the x value would now need to be two units less so that would give me an x value of negative one that would produce a y value of positive one and let's try it if i plugged in a negative one there negative one plus two would give me one one over one is one right so here's the two graphs drawn for you this is the original you can see that point right there one comma one and again it's just shifted two units to the left so there's your new point where you have negative one comma one so every point on that original would just shift two units over to give you the new function graph all right so the last function i want to cover today is called the greatest integer function so this guy f of x equals the greatest integer of x you might also hear this called the floor function now the notation for this guy varies greatly you might see brackets around x like this you might see kind of l-shaped brackets like that you might also see something that looks like this this is very common you probably see that in your textbook it's in the textbook that i'm using to write this lesson so no matter what you see it all means the same thing in your calculator you'll probably see int and then x like that all means the same thing what this denotes is the largest integer that is less than less than or equal to x you got to think about this as it rounds down to the nearest integer so if you're at an integer value you stay there if you're not in an integer value you're rounding down to the nearest integer so in the case where you have f of x equals the greatest integer of x if x is greater than or equal to negative 1 but less than 0 what's going to happen is your value is going to be negative 1. so let's say i'm on the number line and i'm somewhere like right here let's say i'm at negative 0.5 if i round down to the nearest integer i'm going to the left so going to the left from that point puts me here that's going to be the largest integer that's less than or equal to negative 0.5 okay so it's very very tricky if you don't reread the definition over and over and over again if x is greater than or equal to 0 but less than 1 so let's say it's 0.5 now well if i round down i'm going to get to 0. okay so in this case i'm at 0. and so on and so forth if x is greater than or equal to 1 and less than 2 so let's say i was at 1.5 1.5 well then i'm rounding down to 1. so i'm always going to round down to the nearest integer if i'm at an integer itself so let's say i was at 1. let's say i plug that in well it's just going to be the integer it can be equal to that but if i'm anything less than 1 let's say i'm at you know 0.9998 well in that case i've got to round down to 0 right i'm in this category where x is greater than or equal to 0 but less than 1 if i'm not at 1 exactly i've got to round down so i'm going to end up at 0. so for the domain and range for the domain i can plug anything i want in for x there's no restrictions right all real numbers but for my range remember if i plug something in here the output is going to be the largest integer that's less than or equal to the value i plugged in so i'm limited to just getting integers back so my range will be the set of integers so we sometimes refer to the greatest integer function as a step function because of the shape of the graph and the way it looks kind of like that is because for a range of x values you now have the same y value so for an x value that is greater than or equal to negative 3 but less than negative 2 we know y would be equal to negative 3. if x was greater than or equal to negative 2 but less than negative one we know y would be negative two if x was greater than or equal to negative one but less than zero y would be negative one if x was greater than or equal to zero but less than one we know y would be zero you know so on and so forth you'll see the pattern a minute let me do one more of these and then you'll be able to draw as many as you want so if x was greater than or equal to one but less than two y would be one so let's just think about this on the coordinate plane here so again if x was greater than or equal to negative three but less than negative two y would be negative three so let's find a y value of negative three let's go over to an x value of negative three i put a bracket here to say this is included here and i'm going to draw a horizontal line across and a negative 2 i'll put a parenthesis to say it's not included then i'm just going to step up to the next integer for x and i'm going to do the same thing i'm going to put a bracket to say negative 2 is included i'm going to draw a horizontal line to negative 1. negative 1 is not included i'm going to step up again put a bracket here go over to the next integer put a parenthesis step up again bracket parenthesis step up again brackets parenthesis step up again bracket parenthesis step up again so again for a given range of x values there's a set y value so if x is between negative 3 and negative 2 where negative 3 is included y is negative 3. if x is between negative 2 and negative 1 negative two included y is negative two so on and so forth as you go up or as you go down now just as before we can still have vertical and horizontal shifts if i have f of x equals the greatest integer of x minus three this happens outside the function so i would just be shifting down by 3 units so this shifts down 3 units for this one f of x equals the greatest integer of x minus 5 this happens inside the function i could plug x minus 5 in for x and this is what i get so this is going to shift to the right by 5 units so this shifts right by five units hello and welcome to algebra 2 lesson 75 in this lesson we're going to learn about the composition of functions so at some point in your algebra 2 course or this might even happen to in an algebra 1 course you're going to come across this topic known as function composition or again you could say the composition of functions so what this really involves is some heavy substituting and some simplifying it's a very very easy topic once you understand the notation that you're going to be working with so the first thing you're probably going to see in your textbook if f and g are functions this is read g of f of x and you could also write this like this you could say g of f of x and then another way to write it a way that you've probably never seen before you'll have g of f of x this way right here is the way that confuses most students it's a foreign notation people always look at that and think it's a multiplication symbol it's not you can see that it's not filled in this is not g multiplied by f in any way shape or form so this is g composed with f okay that's what this is basically telling us but it's just a fancy way of saying hey i'm going to take f of x and i'm going to plug it in for x in g of x and we'll see that when we get to the examples in a second then this guy right here we've kind of flipped the order around so here this is f of g of x and we could also write this like this we could say f of g of x essentially all i'm doing here is i'm taking g of x and i'm plugging it in for x in the function f of x so this section is all going to be about just substituting and simplifying so before we kind of jump in and look at the first example i want to make sure you understand the most basic thing when it comes to function notation if i take this f of x here and in this case it's 2x minus 3. so that's our function if i ask you what is f of 2 what am i asking for 2 is in the place of x so what i'm asking for is what is the function's value when i plug in a 2 for x all i'm asking for so this is 2 multiplied by plug in a 2 for x then minus 3. 2 times 2 is obviously 4 4 minus 3 is 1. so f of 2 or the function's value when the independent variable x is 2 is 1. right so you can say f of 2 is 1. now i'm not limited to just plugging in numbers for my variable i can take something more complex and plug it in so what if i said i want f of i don't know let's say x minus 1. well all i'm going to do is i'm going to plug in an x minus 1 for x and i'm going to evaluate so in this case i would have 2 multiplied by 4x i'm plugging in an x minus 1. so i'm going to use parentheses so x minus 1 that quantity then minus 3. so i'm going to use my distributive property 2 times x is 2x then minus 2 times 1 is 2 then minus 3. you have 2x which you can't do anything with negative 2 minus 3 is minus 5. so f of x minus 1 is equal to 2x minus 5. now i didn't get a numerical value there but i still found what the function would be if i plugged in an x minus 1 for the independent variable x that's all i'm asking for so that's all we're really going to be doing here we're just plugging things in and simplifying so for the first example we're given f of x which equals 2x minus 3 and we're given g of x which is equal to 4x plus 4. what we want to find is f of g of 1 plus x now if i saw g of 1 plus x what would that mean it would mean that i'm taking a 1 plus x and i'm plugging it in for x in that function g of x so you would start by finding this part first because whatever this is that's what i'm plugging in for x and f of x so you're starting on the inside and you're working your way out so let's find g of one plus x or you can flip that around and say this is g of x plus one whatever you want to do so this equals four multiplied by again i'm plugging in for x there what am i plugging in i'm plugging in an x plus one so x plus one and then plus four so four times x is four x then plus 4 times 1 is 4 then plus 4. so this gives me what 4x can't do anything with that then plus 4 plus 4 is 8. so i get 4x plus 8 as the result there so now let's go back up here i'm just gonna write that g of x plus one is equal to four x plus eight and i can erase this so now that i know that g of one plus x or g of x plus 1 however you want to say that is 4x plus 8. i can really say that this problem is f of this right here is 4x plus 8 so f of 4x plus 8. and then i can just plug a 4x plus 8 in for x in this guy right here so this would be what it would be 2 multiplied by the quantity 4x plus 8 and then minus 3. so 2 times 4x is 8x and then 2 times 8 is 16. so plus 16 and then minus 3. so 8x there's nothing i can do with that 16 minus 3 is obviously 13 so you get 8x plus 13 as your answer so the result here f of g of 1 plus x is equal to 8x plus 13. so nothing really super complicated here it's just a lot of substituting and simplifying all right let's take a look at another example so we have f of x equals negative x minus 1. we have g of x equals x plus 5 and we want to find f of g of negative 2x so again i'm going to start out with this guy right here what is g of negative 2x well again all i'm going to do is plug a negative 2x in for x and g of x so this would be what i'm plugging in a negative 2x there so just be negative 2x and then plus 5. super super simple so now all this would be is f of i know that this part right here is this so f of negative 2x plus 5 would give me what well i'm plugging this in for this right here be careful of that negative out in front you want to account for that so i'm going to have a negative times the quantity negative 2x plus 5 that's what i'm plugging in for x right there and then minus 1. so i'm going to distribute that negative to each term so in other words you can think about this as a negative one multiplying each term i'm just going to change the sign so instead of negative 2x i'll have 2x instead of positive 5 i'll have minus 5 and then i'll have minus 1. so you'll have 2x you can just say negative 5 minus 1 is negative 6 so this ends up being 2x minus 6. so f of g of negative 2x ends up being 2x minus 6. all right for the next one we'll look at g of x which equals x squared minus 1 and h of x which equals 3x plus 4 we're trying to find g of h of x over 3. alright so again i want to start with the inside right here so we want to find h of x over 3. so what is that going to be well here's h of x so we want h of x over 3 so this equals i'm just going to plug this in for x in my function so we'd have three multiplied by x over three then plus four what's going to happen is this will cancel with this and i'm left with x so this ends up being x plus four so this part right here is nothing more than x plus 4. so this would be equal to g of x plus 4 and that would be what i'm taking x plus 4 that quantity and i'm plugging it in for x in g of x so i would have the quantity x plus 4 that would be squared and then i'm subtracting away 1. so i'm going to use my special products formula there i can square x plus 4 that quantity pretty quickly this would be x squared plus 2 times x times 4. 2 times 4 is 8 8 times x is 8x so this would be plus 8x and then lastly plus 4 squared is 16 and then minus 1. so you'd have x squared plus 8x and then 16 minus 1 is 15. so plus 15. so we can say g of h of x over 3 is equal to x squared plus 8x plus 15. all right let's take a look at another so we have g of x is equal to x plus 1 h of x is equal to negative 4x minus 1. so we want g of h of negative 3x so all i want to do is start out by finding h of negative 3x so h of negative 3x just plug a negative 3x in for x here what would that give me you'd have negative 4 multiplied by negative 3x plugging in for x there then minus 1. negative 4 times negative 3 is 12 so you'd have 12x and then minus 1. so this guy right here is 12x minus 1. so this would become g of 12x minus 1 which equals what plug in a 12x minus 1 there so you would have 12x minus 1 then plus 1. pretty simple negative 1 plus 1 is 0. so this just turns into 12x so g of h of negative 3x is just 12x all right what about f of x equals 4x plus 5 and then g of x equals x cubed minus 2x so here we want f of g of 3x so i start with g of 3x so g of 3x this would be what plug in a 3x there so you'd have 3x cubed minus 2 times i'm plugging in a 3x there as well and what does that give me 3x cubed would be what 3 cubed is 27 right 3 times 3 is 993 is 27 so 27 and then x cubed and then minus 2 times 3 is 6 6 times x is 6 x so you get 27 x cubed minus 6 x and i want to take this guy and i want to plug it in for x here so in other words i want to say that i have f of 27 x cubed minus 6 x and this would give me 1. it would give me 4 multiplied by 27 x cubed minus 6x again just plugging this in for x there then you have plus 5. all right so what does this give us we would have 4 times 27 which is 108 then times x cubed then minus 4 times 6 is 24 and times x then plus 5. so we can say f of g of 3x is equal to 108 x cubed minus 24x plus 5. all right let's take a look at another one so we have g of x equals 2x minus 4. we have f of x equals 4x so we want g of f of x plus 1. so if i had f of x plus 1 this part right here what would that give me this is 4 multiplied by this gets plugged in for x so 4 times the quantity x plus 1 this gives me 4x plus 4. so that's what's going right here so in other words this is g of 4x plus 4 so this would be equal to i have a 2 multiplied by plug in a 4x plus 4 in for x so 2 times the quantity 4x plus 4 and then you have minus 4. use the distributive property 2 times 4x is 8x then plus 2 times 4 is 8 then minus 4. so you'd have 8 x 8 minus 4 is 4 so 8 x plus 4. so g of f of x plus 1 gives us 8 x plus 4. let's take a look at one more again it's a very easy concept as you can see you're just taking one function and you're using it as an input in another function so we have f of x equals 4x plus 2 we have g of x equals 2x plus 5. so what is f of g of x plus 3. so again i start out with g of x plus 3 so g of x plus 3 gives me 1. this would be 2 times this quantity right i'm plugging that in for x so 2 times the quantity x plus 3 and then plus 5. 2 times x would be 2x plus 2 times 3 is 6 plus 5. so this gives me 2x plus 6 plus 5 is 11. so that's this part right here so in other words you would have f of 2x plus 11. so this gives me what you would have four times in place of x i'm plugging in a 2x plus 11. so 4 times the quantity 2x plus 11 and then plus 2 so use my distributive property 4 times 2x is 8x and then plus 4 times 11 is 44 and then you have plus 2 8x would stay the same nothing i can do with that 44 plus 2 is just 46. so we end up with f of g of x plus 3 as 8x plus 46. hello and welcome to algebra 2 lesson 76 in this lesson we're going to learn about the circle so at some point in your algebra 2 course you will begin talking about conic sections so these include parabolas circles ellipses and hyperbolas now we already discussed how to work with parabolas before now we're just going to tackle the subject of circles so the first thing is let's just get a brief definition going so a circle is the set of all points in a plane that lie a fixed distance from the center so the next part of that is to say that the fixed distance from the center to each point is known as the radius this is something you want to write down because we're going to use it in a minute when we talk about the distance formula so the fixed distance from the center so from the center to each point on that circle is again known as the radius now the first thing you're going to come across when you get into the topic of circles is going to be the equation of the circle now this is given to you in standard form otherwise known as center radius form and it allows you to look at the equation and very quickly identify the key information so i have here we can use the distance formula to find the equation of a circle so how do we go about doing that let's think about what the distance formula does for just a second i want you to remember that the distance formula allows us to take two points on a coordinate plane and find the distance between them so if i had two generic points let's just say x sub 1 y sub 1 and x sub 2 y sub 2 okay these could be any two points well the distance between them is given by this formula the d or the distance equals we would have the square root of you have x sub 2 minus x sub 1 this quantity squared then plus you'd have y sub 2 minus y sub 1 this quantity is squared now how do we use this information to obtain the equation for a circle well if i'm given the center of the circle which is not actually on the circle it's in the center and i'm given the radius what i can say is that the radius gives me a distance from the center to any point on that circle so generically i could say that the distance is 2 the distance is 2 and this equals the square root of so let me just replace this point right here this x sub 1 y sub 1 with a known point of 1 comma 3 and let's plug into the distance formula so the distance between any point instead of calling it x sub 2 y sub 2 i'm going to call it x comma y so any x comma y point that you come up with because again we know that the distance between them would be equal to 2 because our radius is 2. so following this x sub 2 minus x sub 1 squared so x sub 2 now here i've replaced with x so i would say x minus for x sub 1 to replace with 1. so x minus one that quantity squared then plus we would have y sub two which now i replaced with y minus y sub one i replace that with three so then this is squared so so far it's not actually where we need it to be we need to do one additional step which is to square both sides so if i square both sides 2 squared is 4 and this equals this cancels with this so i'm just left with what is underneath the square root symbol which is the quantity x minus 1 squared plus the quantity y minus 3 squared now for the purposes of what you normally see i'm just going to take this 4 that's over here on the left i'm going to put equals 4 over here on the right because this is how you're normally showing this equation so this is your center radius form this is your center radius form and you might hear this called standard form for a circle those two can go back and forth interchangeably but the key thing here is understand that your center will be given by this guy and this guy so as long as it's in the format of x minus something and y minus something you just take whatever the number is here and here and that's how you're going to find your center so your center here is 1 comma 3. we already knew that because we were given that information and the radius comes by taking this guy and taking the principal square root of it so in other words this right here represents the radius squared so taking the principal square root gets me back to the radius take the square root of 4 i get back to 2. now some people will get a little confused and say well why don't you take the positive and the negative the reason we don't say that the radius is also negative two is because the fact that a distance has to be non-negative right you can't say that hey i'm gonna get in my car and drive negative two miles to work you're either gonna go zero miles and not go anywhere sit at your house or you're going to go some positive amount right but you can't just go a negative amount of miles somewhere all right so here is the official formula which you'll see in your textbook so the center radius form you have x minus some value h squared plus y minus some value k squared this equals the radius squared so the center is h comma k and the radius we find again by taking the principal square root of what we see here so if i know my center and my radius i can very quickly graph something so to graph a circle i start at the center which is h comma k i make a nice little clean point then i move by the amount of the radius so up down left and right so you end up just making four nice clean points and then you draw a smooth curve through the four points so let's look at a few examples these are things you would see in your textbook pretty much right away so we want to identify the center and radius then sketch the graph so we have x plus 4 that quantity squared plus we have y minus 3 that quantity squared and this equals 4. so again to match the format we had x minus h this quantity squared then plus y minus k this quantity squared this equals the radius squared now here's where you've got to pay close attention this is a minus sign here and a minus sign here if it's not presented in that exact format you have to make an adjustment to what you're looking at this is a plus here so what i would do is i would say okay i can write that as x minus a negative four so in other words i can say this is x minus a negative four now i've completely matched this format here then plus this guy right here is fine it's y minus some value so i'm going to leave that alone y minus 3 that quantity squared and then this guy right here this 4 i would just take the principal square root so the square root of 4 would be 2. so i'm just going to go ahead and write this as 2 squared just to match everything up so now i can just quickly look at the equation and pull the information out that i want if i want the center again i match things up h here h here k here k here so the center occurs at the point negative 4 that's my x value and then positive 3 that's my y value then the radius the radius again when i take the principal square root of 4 i get 2 that's why i wrote it as 2 squared so my radius is 2. now i can quickly graph this guy with this information so i've already pre-drawn the circle as you can see the reason i did that it's kind of hard to freehand draw a circle especially if you're using a computer a lot of people will graph this on their calculator or if you freehand draw this on a piece of paper you can kind of stick your wrist down and kind of rotate the page around and make a perfect circle but if you really freehand draw it it's kind of difficult to do but again if you wanted to do this we're just going to do the exercise the center is at negative four comma three so on the coordinate plane i would go four units to the left and three units up so that's why i have that point right there that's your center so this is the center and i can just highlight that and then after that i'm gonna make four points i'm gonna go up by the radius so i'm gonna go up by two units so that's right here that's why there's a point there i'm gonna go to the right by two units i'm gonna go down by 2 units and i'm going to go to the left by 2 units so that's where those 4 points came from now once i've done that i can just sketch the graph of the circle i'm going to try to do this to the best of my ability and that's about as good as i can personally do i know a lot of people are really good at drawing circles freehand but personally i am not good at so that's why i made a graph to kind of start off with it's just that easy just identify the center usually i just make a point and then just move up down to the left and to the right by the radius plot those points and then draw a smooth curve to connect those four points all right let's take a look at another problem so we have the quantity x minus 3 squared plus the quantity y plus 2 squared this equals 4. again match this up to we have x minus h that quantity squared plus we have y minus k that quantity squared this equals the radius squared so i can see that this is minus and this is minus so my h here will be 3. so if i'm thinking about the center the x coordinate will be 3. now for the y coordinate i have a plus here so i've got to think about rewriting that so if it's a plus this is going to be minus a negative 2 to get it to match that format so in other words i can say this is y minus a negative 2 this quantity squared then we can see that k would be negative 2 here so my y coordinate would be negative 2 and then what about the radius again you just take this guy here and you take the principal square root so i can really think about this instead of being 4 you say this is 2 squared right so the radius is just 2. it's just this part right here the base of the exponential expression you set up so my radius is two all right let's take this information and graph it so again the center occurs at three comma negative two and the radius is two so i go three units to the right and two units down this is my center so again this is the center and then i would make four points so i'd use my radius which is two to go up by two units to go to the right by two units to go down by two units and go to the left by two units once i have those four points again i can draw a smooth curve or i can attempt to not always going to be perfectly smooth but let's try all right again that's about as good as i can personally do all right let's take a look at one more of these easy ones and then i'll show you a harder scenario where the information is not given to you in this format and so you have to do a little bit of extra work so here we have x plus 2 that quantity squared then plus we have the quantity y minus 3 squared and this equals 9. so again the format that we're looking at is x minus h this quantity squared plus we have y minus k this quantity squared this equals our radius squared so what matches up and what doesn't this is minus k this is minus 3. so 3 here is k the h here is not 2 because again i have this plus sign here so i've got to account for the formula if i see a plus sign i can just cross that out and put minus a negative so i know h is represented by negative 2. so for the center for the center we can say this occurs at the point negative 2 comma 3. now for the radius again i could take 9 i could just write it as 3 squared or i could take the principal square root of and come up with 3. however you want to do that but a lot of people like to just match the formula up exactly so again if you set this up as an exponential expression 3 squared you know the radius is the base 3. so i'll put my radius here is equal to 3. so let's copy this all right so again our center occurs at negative 2 comma 3. so i go 2 units to the left and three units up so this is our center this is our center and then the radius is three so again to make those four points from the center i'd go up by one two three i would go down by one two three i would go to the left by one two three and i'll go to the right by one two three so very easy once i have that let me just change colors here and i'll again attempt to draw the circle i'm just gonna connect those four outer points with a smooth curve all right so there is our circle all right so in some cases we are not given the equation of the circle in standard form again otherwise known as center radius form so when this occurs we can obtain standard form by completing the square so this is actually quite an easy process if you remember solving quadratic equations by completing the square it's kind of the same thing but it's a lot easier so essentially all i'm going to do here if i look at this first equation we have x squared plus y squared minus 4x minus 4y minus 8 equals 0. the first thing i would do is group my x terms together and my y terms together and i would take the constant and send it to the other side so in other words i would say this is x squared minus 4x i would group those two together and then plus i would take the y term so y squared minus 4y group those two together and then i would add 8 to both sides of the equation so this equals 8. now if you remember from again that lesson where we completed the square to solve quadratic equations the steps would be to look at the coefficient on the first degree term so in other words the variable that is raised to the first power you take that coefficient and you cut it in half meaning you multiply it by a half and then you square the result this is what you're going to add to both sides of the equation to complete the square so very very easy to do this so we have x squared minus 4x to complete the square here or again get a perfect square trinomial something i can factor into a binomial squared i would take this coefficient right here so it's negative four i would cut it in half so i'd multiply it by half and then i would square it negative four times a half is negative two negative two squared is four so i would add four here now this gives me a perfect square trinomial this would factor into x minus 2 that quantity squared here's the issue though i added 4 to this side of the equation so i can't just do that i have to balance it by adding 4 to this side of the equation as well as long as i do that everything is okay so then plus we're going to do the same thing over here so we have y squared minus 4y now this coefficient here is the same as the coefficient here right i had a negative 4 here and i have a negative 4 here so without even doing the work of saying negative 4 times a half that amount squared which gives me 4 i can just look here and know that i would add 4 to this side over here and again doing that i have to also add 4 over here just to make sure it's balanced so this is plus we know this would factor into y minus 2 that quantity squared over here 8 plus 4 is 12 12 plus 4 is 16 so this equals 16. now you have this in center radius form or standard form however you want to think about it so we can look at again i can match this up with x minus h that quantity squared plus y minus k that quantity squared equals the radius squared so in this case it's basically perfect the only thing i would do is i could write 16 as 4 squared to match up the h with 2 the k with 2 and the r with 4. so my center my center would occur at the point two comma two and my radius my radius would be four okay very very simple let's take this down to the coordinate plane and we'll graph this guy real quick all right so again the center here is at two comma two so two units to the right and two units up that puts me right there again this is my center and then the points we would generate here the radius is four so i could go up one two three four there's a point down one two three four there's a point to the right one two three four there's a point and then to the left one two three four there's a point again i just want to draw a smooth curve through the points just going to try to trace this guy that i've already pre-drawn all right let's take a look at one more problem here so we have 2x squared plus 2y squared minus 12x plus 8y plus 24 equals 0. the first thing you would notice before you do anything here is that everything is divisible by 2. so if i went through and divided everything by 2 what would i get this would be x squared plus this would be y squared then minus we'd have 6x plus we'd have 4y plus we'd have 12 and this equals 0 over 2 is 0. now again to complete the square the first thing i'd want to do is i want to grab all my x terms and group those together all my y terms and group those together and i want to put the constant on the other side so i would say this is x squared minus 6x group those together then plus we'd have y squared plus 4y group those together and i would subtract 12 away from each side of the equation so this equals negative 12. okay so now we're going to complete the square again i'm going to choose something that takes that binomial there makes it into a trinomial that can then be factored into a binomial squared again perfect square trinomial means it's a binomial squared if we factored it all right so what do i need to do here so this is x squared minus 6x what's the missing part there again i take the coefficient for the variable that's raised to the first power in this case that's going to be negative 6. i cut it in half so i take negative 6 and i multiply it by half and then i square the result negative six times one-half is going to be negative three then negative three squared would be nine so i'm gonna add nine over here but again to make this legal i've gotta add nine over here so that part's done we'll close the parentheses here and then we can go ahead and factor because we know this would factor into a binomial squared this would factor into x minus 3 that quantity squared all right so then plus over here we have y squared plus 4y and then you have something that's missing here so again if i cut 4 and half in other words if i multiply 4 by half i'd get 2 square 2 i get 4. so i'm going to add 4 over here i'm going to add 4 over here so this makes this into a binomial squared which is going to be y plus 2 that quantity squared okay then this equals negative 12 plus 9 is going to be negative 3 negative 3 plus 4 is 1. so to put this into our format i'm going to write this as x minus 3 that quantity squared this has a minus sign here so we're good to go then plus over here i'm going to change this to minus a negative so y minus a negative 2. so now we're good to go there that's squared and this equals you could say this is 1 squared right because 1 to any power is just going to give you 1. so i know the center is going to occur at the point 3 comma negative 2 and the radius the radius will be equal to 1. that guy right there so let's take this information down to the coordinate plane all right so the center occurs at 3 comma negative 2. so i would go three units to the right and two units down so there's my center right there so we'll say this is the center and the radius is one so again i could go from the center up one i could go to the right one i could go down one and go to the left one and then i just want to connect the points on the outside with a smooth curve so don't worry too much about having a perfect graph again the whole idea here is just to get the concept down identify the center identify the four points on the outside by using the radius and sketch a graph through it and you'll have a good idea of what the circle looks like hello and welcome to algebra 2 lesson 77 in this lesson we're going to learn about graphing ellipses so in the last lesson we learned how to graph a circle so it turns out that a circle is just a special case ellipse where the distance from the center to any point is the same remember we had a special name for this distance from the center to any point we call this the radius so kind of looking back on that lesson we saw that if we had a circle that was centered at the origin it followed the equation x squared plus y squared equals r which stands for the radius squared so if i looked at x squared plus y squared equals 4 i know this is the radius squared so if i wanted the actual radius i would take the principal square root of that and that would give me 2 right square root of 4 is 2. so my radius is equal to 2. now i already told you this is centered at the origin so we can say our center is at 0 comma 0. but the way you can tell it's centered at the origin if you just have x squared and you just have y squared so you don't have x minus something or x plus something and that squared you don't have y minus something or y plus something squared when you just see x squared on y squared this guy is centered at the origin so to graph the circle we learned that we started at the center in this case that's at zero comma zero so that's right there then remember the radius is a distance from the center to any given point on that circle what we can do is start at the center and move to the right by the radius and make a point we can move up by the radius and make a point we can move to the left by the radius and make a point and we can move down by the radius and make a point so once we have those four points there we can sketch a smooth curve through the points and we've just graphed our circle then as we move into circles a little bit more we see that not every circle is centered at the origin so if we deal with a circle that's been shifted then what happens is we have the equation x minus h this quantity squared plus we have y minus k that quantity squared and this equals the radius squared so to think about where the center would have moved to if it follows this format we have a minus some h and then you have a minus some k the center will occur at h comma k so in this case where we have the quantity x minus 3 squared plus the quantity x minus 2 squared equals 4 where's my center going to be my center will have moved to 3 comma 2. again when it matches this format i have a minus here so i'm just going to take this 3 and that's going to be my x value here i have the same format so minus something here i'm going to take that 2 and that's going to be my y value if you run into the scenario where you have a plus sign you need to convert it into minus some negative value that way it matches that formula okay so you have to do that but this guy it has the same radius as the last one we looked at if i take the principal square root of 4 i get 2 so the radius is equal to 2. so essentially everything is the same it's just been shifted 3 units to the right and 2 units up my center was right here at the origin but it moved 1 2 3 units to the right and 1 2 units up so this is my center now and all the other points did the same thing but again when you're graphing this guy you just want to look at the center which is three comma two and then again you make your points with your radius so i go two units to the right make a point two units up make a point two units to the left make a point two units down make a point and then i sketch my smooth curve through the points now i just want to show you this one so that you can see how it shifted so given the fact that they both have the same radius of two this guy right here again has been shifted three years to the right and two units up so my center went from here to here three units right two units up you could take any given point so you can take this point right here and say i went one two three inches to the right one two units up the impact here again is that for a given point that was on that original graph that we looked at x squared plus y squared equals four it's now been moved three units to the right and two units up all right so now that we understand the basis of how to graph a circle we want to move into the ellipse so graphing an ellipse is not any more difficult than graphing a circle once you learn how to graph a circle graphing the ellipse is pretty easy so when we see an ellipse we generally think of a squished circle so it may be tall and skinny in the case of a vertical ellipse or it may be short and fat in the case of a horizontal ellipse so the first ellipse we're going to look at is x squared over 9 plus y squared over 25 equals 1. so this type of ellipse is known as a vertical ellipse you can see that it's tall and skinny and you'll see an example in a minute where it's going to be short and fat but this guy is a vertical ellipse i want you to notice something this is what makes this a vertical ellipse this guy right here this 25 the denominator under the y squared is larger than this 9 here the denominator under the x squared think about the fact that y is the vertical axis okay it's the vertical axis so when this guy is bigger you'll get a vertical ellipse i want you to notice something real quick let's say that i told you we wanted to find the y-intercepts here how would we do that we'd plug in a 0 for x so this would go away i'd have y squared over 25 equals 1. that's easy to solve y squared over 25 equals 1. i can multiply both sides of the equation by 25 and get y squared equals 25 and i'll take the square root of this side and then plus or minus the square to this side so y is equal to plus or minus 5. now with the y intercept we know that x is 0. so that's what leads to the points 0 comma 5 and 0 comma negative 5. okay those are right here for the x intercepts i would do the same thing i would just plug in a 0 for y and i would solve for x you'll find that you would get 3 comma 0 and you would get negative 3 comma 0. so here and here so what's creating this shape is the fact that the distance from the center to this 5 up here this is 5 units this right here is 5 units then your distance from this guy right here to here well this is only three units this is only three units the fact that when we graph this guy we're drawing a curve through these points we have to go further in the vertical direction than we do in the horizontal direction and so that's what creates this shape this vertical ellipse that is tall and skinny now let's take a look at a horizontal ellipse so in this case i've just switched the denominators around now the 25 is underneath the x squared the 9 is underneath the y squared so the intercepts kind of just changed around so now the x-intercepts would be at five comma zero and negative five comma zero the y-intercepts would be at zero comma three and zero comma negative three so this guy changed from being tall and skinny to shortened fat in other words the distance from here to here is now 5 units whereas the distance from here to here is now only three units so as this guy gets graphed again we draw our curve it's a further distance horizontally when this curve is being drawn than it is vertically right so it creates the shape of being short and fat so when you start working with ellipses you're going to see two different standard forms this can be a little bit confusing at first but it's something you will get used to when a vertical ellipse is centered at the origin it has the equation x squared over b squared plus y squared over a squared equals one now we talked about the fact that a vertical ellipse has a bigger value here underneath y squared than here and the reason for that is we always want to have a being greater than b okay so if the denominator under y squared is bigger you're going to use this formula here this is for a vertical ellipse in the case where we have a horizontal ellipse well when a horizontal ellipse is centered at the origin it has the equation x squared over a squared plus y squared over b squared equals one in this case we said with a horizontal ellipse this is greater than this again you've got to set this up to where a is greater than b this has to do with a formula you're going to deal with down the road when you really start talking about ellipses you're going to see for the purposes of subtraction a is going to need to be greater than b so that's why these formulas are going to kind of switch back and forth all right so how do we graph an ellipse let's start out with the easy scenario where it's centered at the origin so we have graphing ellipses centered at the origin we first want to find and plot the x and y-intercepts then we just sketch the graph draw a smooth curve through the points so just like we saw when we graphed a circle we'll end up with four points and we just draw a smooth curve through the points all right so let's start out with this first example we have x squared over nine plus y squared over 49 and this equals one so the first thing is how do we find our x-intercepts and let me make that a little better and how do we find our y-intercepts well this concept has not changed from when we first learned it when we were working with linear equations in two variables essentially if i want the x intercepts i want to make y equal to zero i want to make y equal to zero and i want to solve for x so i know that it's something comma zero in each case right because y is zero so if this is 0 this is gone right 0 squared is 0 0 49 is 0 it's gone so we have x squared over 9 is equal to 1. i multiply both sides of the equation by 9 and we know that this would cancel with this and i have x squared is equal to 9. now to solve this for x i take the square root of this side i take plus or minus the square root of this side and i get that x is equal to plus or minus 3. so the x values when y is 0 would be positive 3 and negative 3. now what's the shortcut here because i don't need to go through this every single time what ends up happening is if this guy right here is a 1 which is always going to be in the case of an ellipse sometimes you'll get a equation that you have to put into this form but once it's put into this form this is what you're going to see so what happens is if this is always gone let's just say this is x squared over q it doesn't matter what q is it's just a placeholder for now and this equals 1. well i'm always going to multiply both sides of the equation by q so i'll have x squared equals q and to solve this i'm going to take the principal and negative square root of q every time right so x would equal the positive and negative square root of q so to get the x values for the x intercept i just need to look at this denominator here and take the principal and the negative square root of that i don't need to go through plugging in 0 and doing all this work because in every situation that's what's going to happen the same thing goes for finding the y values for the y-intercepts i already know that the x values are zero i don't need to guess on that so if i went through the whole thing i'd plug in a zero there this would be gone i have y squared over 49 equals one i multiply both sides of the equation by 49 that cancels i get y squared is equal to 49. again i take the square root of this side plus or minus the square root of this side so we get y is equal to plus or minus seven so zero comma seven and then zero comma negative seven but again the shortcut would be to just take this guy right here and say okay for the y values for my y intercepts i just take the principal and the negative square root and that's how i get them right positive or negative square root of 49 would be plus or minus 7. so once we have the intercepts the process is very very easy it's just like graphing a circle we just plot the points and then we sketch a smooth curve through the points so let's take these points down to the coordinate plane all right so we have everything kind of pre-drawn just to make it a lot easier for us so here's our points we have the x-intercepts that occur at 3 comma 0 and negative 3 comma 0. so here's 3 comma 0 here here's negative 3 comma 0. so those are two points then my y-intercepts occur at 0 comma 7 and 0 comma negative 7. so here's 0 comma 7 and here's 0 comma negative seven and then i just need to sketch the curve around those points so you can see the computer did a perfect job i'm going to trace over what they did and i will attempt to do a mediocre job all right let's take a look at another one so we have x squared plus y squared over 16 equals one so the first thing is if i don't have a visible denominator under x squared or y squared i'm going to put this guy over 1. the reason for that is i'm going to need to use these denominators to again get my x and y-intercepts so for the x-intercepts again we learned the shortcut for that so they would occur where we know the y values are 0 in each case what are the x values again i look at this denominator under x squared i take the principal and negative square root of that so plus or minus the square root of one is plus or minus one so i get positive one and negative 1 right as my x values then for the y intercepts we would get what well we know that the x values would be 0 in each case and again i would just look at my denominator here in this case it's 16 i take the principal and negative square root of that guy this equals plus or minus 4 and so the y values here would be 4 and a negative 4. so again very very easy to get these points going i just take them down to the coordinate plane now all right so we would start out by plotting the point one comma zero so one comma zero is here and then negative one comma zero is here and then 0 comma 4 is here and then 0 comma negative 4 is here so we have our intercepts plotted and then again we're just going to graph a smooth curve through the points again the computer did a perfect job already and i am going to again attempt to do a mediocre job by tracing over what the computer did so let's go ahead and do that okay so again not perfect but you get the general idea of what this guy would look like and that's what you're looking to do so just like when we saw circles that weren't centered at the origin you're also going to see ellipses that aren't centered at the origin as well so generally speaking and this will change up based on if you have a horizontal ellipse or a vertical ellipse but let's just use the formula for a horizontal ellipse just to kind of think about this we have the quantity x minus h now squared over a squared plus we have the quantity y minus k squared over b squared this equals 1. now if you had a vertical ellipse the difference would be that this guy right here would be b squared and this guy right here would be a squared but again to keep things simple let's just work with the horizontal ellipse for right this second the center will occur at h comma k okay so whatever's right here for h and whatever's right here for k but that's given that you have a minus sign here and a minus sign here you've got to pay attention to that because if you end up with a plus something you've got to make an adjustment to match the formula so what we can say is that this guy shifts h units to the right if h is greater than zero if h was less than zero then what would happen is it's shifting h units to the left then we could also say it shifts k units up if k is greater than zero or it would shift k units down if k was less than 0. all right so let's take a look at x minus 3 that quantity squared over 9 plus we have y plus 1 that quantity squared over 36 and this equals 1. so the first thing is this guy will be a vertical ellipse we know that because the 36 is larger than the 9 and 36 is underneath where the y variable is now that we've gotten kind of that out of the way we think about how could we graph this guy well if i was graphing x squared over 9 plus y squared over 36 equals 1. what would i do well i know the center i know the center is at 0 0 i know my x-intercepts are where again i would look at this guy and i'm going to go plus or minus the square root of that plus or minus the square root of 9 is plus or minus 3. so 3 comma 0 and then negative 3 comma 0. and then my y-intercepts my y-intercepts would occur where again i'd look at this plus or minus the square root of 36 is plus or minus six so zero comma six and a zero comma negative six okay so i could plot those four points i could sketch my curve around the four points and i'd be done what happens is this guy we're looking at here has been shifted three units to the right and one unit down so the center will now be at three comma negative one you might say well where did you get that again the formula is x minus h that quantity squared over in this case we have a vertical ellipse so this would be b squared but we take whatever's right here after the minus sign so in this case that's 3 that is the x value for the center then we have plus we have y minus k that value squared and then again because this is a vertical ellipse this would be a squared here this equals one but whatever this guy right here if this was minus we're taking that now it's not a minus i have a plus here so i want to rewrite this as minus on negative one so it matches the formula and i can just take negative 1 and plug it in as my y coordinate for the center so now i know let me just kind of erase this if i scooch this up i know that these points here have just shifted three units to the right and one unit down now they're no longer the x and y intercepts they're just going to be in the case of the x-intercepts they just became the furthest points to the right and to the left of the center on the ellipse for the y-intercepts it's the furthest point above the center and below the center on the ellipse so these are just four additional points we're going to get so if i take three comma zero and i shift it three units to the right and one unit down then i would get six comma negative one if i take negative three comma zero and i shifted three inches to the right and one unit down i would get zero comma negative one if i take zero comma six and again i shifted three inches to the right and one unit down i would get three comma five and lastly if i take zero comma negative six and i shift it three units to the right and one unit down i would get three comma negative seven now each time i do this i don't wanna go through this comparison so the easy way to kind of do this is to just find the center based on the formula we know we're looking for h comma k we know how to find those values that's my center for points where we move horizontally from the center i'm looking at this value here so plus or minus the square root of that would be the horizontal distance from the center again to the furthest point to the right and to the left on that ellipse so starting at 3 comma negative 1 i could add 3 to my x value and get 6 comma negative 1 and i could subtract 3 from my x value and get 0 comma negative 1. there's two points there then i take this guy right here and now i'm thinking about getting a vertical distance from the center to the highest and the lowest point on the ellipse then plus or minus the square of 36 would give me that so that'd be plus or minus six with negative one i go up by six i get to five with negative one i go down by 6 and i go to negative 7. so that's how you could quickly get those 4 points you grab the points let's start out with 6 comma negative 1 so that's going to be right here we would have 0 comma negative 1 which is right here so again from the center this is moving one two three to the right and one two three is to the left then we have three comma five so that's right here and then we have three comma negative seven that's right here again from the center i'm moving one two three four five six units up and one two three four five six units down so that's how we get my four points and again i'm just going to draw a smooth curve through the points okay again not perfect but you get the general idea of what this guy would look like all right let's take a look at one final problem now so we have x squared over 49 plus you have the quantity y plus 4 squared over 9 this equals 1. so this guy we want to start by again finding the center and in this case we have what the 49 is bigger than the 9 so this would be a horizontal ellipse so the center occurs where it's at h comma k i just have x squared here i don't have x minus something i don't have x plus something that quantity i don't have that squared so the center will occur here the x value hasn't shifted it's just 0. and to really write this in that format you could say this is x minus 0 that quantity squared over 49. so this guy right here is your x value which is 0. now for the y value again i want to write this as minus a negative 4 so negative 4 will be the y value there and then to get my additional points again all i'm looking to do to get some horizontal points from the center i just take this guy right here i take the principal and negative square root of it principle of negative square root of 49 is 7. so i'm just going to move by 7 units to the right of the center and to the left of the center so add 7 to 0 you get 7 then comma negative 4. subtract 7 from 0 you get negative 7 comma negative 4. then over here again i'm going to do the same thing for this guy plus or minus the square root of 9 would be plus or minus 3. so from the center i'm now going to look for points that are vertically above the center and below the center so my x value stays the same negative 4 plus 3 would be negative 1 and then again my x value stays the same negative 4 minus 3 would be negative 7. so there's your four points there so again we're going to plot these four points you have 7 comma negative 4 that's right here and you have negative 7 comma negative four so that's right here again in each case from the center we just moved seven units to the right and seven units to the left okay so that's how we got there then we have zero comma negative one and we have zero comma negative seven again from the center we moved three units up and three units down so let's go ahead and sketch the graph here okay so again not perfect but again you're just giving a general idea of what the ellipse would look like hello and welcome to algebra 2 lesson 78 in this lesson we're going to learn about graphing hyperbolas so up to this point we've learned how to graph parabolas how to graph circles and how to graph ellipses so the next thing we're going to come across is how to graph the hyperbola so these are a little bit more challenging to draw but again when you're working with these things especially if you're freehand drawing something the expectation is you're just giving a general idea of what it looks like so with hyperbolas the equation looks similar to the ellipse the difference is the minus sign so i want you to recall that when you worked with ellipses you had a horizontal ellipse that was short and fat and you had a vertical ellipse that was tall and skinny when we look at hyperbolas you're going to have two different scenarios that occur you'll have a horizontal hyperbola and you'll have a vertical hyperbola so with a horizontal hyperbola you get this equation in your textbook you'll see x squared over a squared minus y squared over b squared equals 1. now let's stop for a minute let's go back to when we learned about ellipses the horizontal ellipse had the following equation it was x squared over a squared plus it was y squared over b squared and this was equal to 1. the difference between this equation and this equation is the minus sign here that's how you know that you have a hyperbola versus an ellipse once you figure that out you have to be able to establish do i have a horizontal hyperbola do i have a vertical hyperbola and this comes from looking at which term is positive and which term is negative so in other words i could think about this term right here i could say this is plus negative y squared over b squared so if this term is negative the term that's positive if that contains the x variable we're going to have a horizontal hyperbola again think about the x-axis as being the horizontal axis the y-axis is being the vertical axis so in this case you can think about it as let's go back to the original this guy right here the x squared over a squared is larger or more dominant than this guy right here the y squared over b squared i might say well how do you know that this guy minus this guy equals this guy if this guy minus this guy equals one one is a positive number so for that to be true whatever i started with the x squared over a squared had to be larger than y squared over b squared because i've got a positive amount left just think about if you're spending money out of your checking account if the result of paying a bill leaves your checking account with a positive balance well the bill was smaller than what you started with okay so it's the same thought process here so however you want to remember that if you want to think about the fact that this term is bigger than this term or you want to say well this x squared over a squared is positive and this guy's negative however you remember it this guy right here the positive one the one that's larger that's what's going to dictate if we have a horizontal hyperbola or a vertical hyperbola in the next one we see a vertical hyperbola now y squared over a squared this is the larger one now so it's a vertical hyperbola y is on the vertical axis so we have y squared over a squared minus x squared over b squared equals one and again when you think about the vertical ellipse what was our formula there it was x squared over b squared plus y squared over a squared equals 1. now i could change the order of this around and put y squared over a squared first because there's addition here but i just want you to notice that the difference again is the plus sign here became a minus sign here that's how you can tell if you have a hyperbola now for today we're just going to look at graphing a hyperbola that's centered at the origin the first thing you need to do is locate the intercepts if you have a horizontal hyperbola you will have x-intercepts and no y-intercepts if you have a vertical hyperbola you will have y-intercepts and no x-intercepts and when we get to a problem i'll show you why that's the case the next thing you're going to do is find this fundamental rectangle the points for this i'll show you how to get those you'll sketch the asymptotes these are going to be lines that are extended from the fundamental rectangle or you can actually get the equation for them from the hyperbola you can graph them that way then lastly we will sketch the graph so each branch of your hyperbola goes through an intercept and approaches but doesn't touch the asymptotes so let's take a look at the first example we have y squared over 9 minus x squared over 9 equals 1. so in this case this term right here is larger so it involves the y variable so i know i have a vertical hyperbola so i know this guy is vertical so what i told you is that if you have a vertical hyperbola you will have y-intercepts and no x-intercepts why is that the case we know how to find our x-intercepts and y-intercepts if i want the y-intercept i plug in a 0 for x and i solve for y we know that this would just go away and we know from our last lesson that the shortcut is i could just take the positive and negative square root of that denominator there plus or minus the square root of 9 is plus or minus 3. so my y-intercepts my y-intercepts would occur at 0 comma 3 and 0 comma negative 3. and you might say where did that come from again if i plug in a 0 here that term's gone so you would have y squared over 9 is equal to 1. to get y by itself i would multiply both sides of the equation by 9. because in standard form this guy always is equal to 1 whatever the denominator was here it always ends up on that side of the equation so i end up with y squared is equal to whatever that denominator was now the end result to get y by itself is to take the square root and over here i have to go plus or minus the square root of that guy so the shortcut is always to look at the denominator and say plus or minus the square root of that that would give me the y values for the y-intercepts and again you could do the same thing if you're working with x-intercepts the concept is the same so this gives me y is equal to plus or minus 3. of course we already know that 0 comma 3 and 0 comma negative 3 are the y-intercepts now you might say why don't we have x-intercepts why does that occur well because of this minus sign here you're going to run into some problems if i plugged in a 0 here and this term was gone you'd have negative x squared over 9 is equal to 1. multiply both sides of the equation by 9 you'd have negative x squared is equal to 9. now let's say i divide both sides of the equation by negative 1 to get rid of that negative so let's say we have x squared now is equal to negative 9. is there a real solution that i can get there no there's not because if i take the square root of this side and i do plus or minus the square root of this side the solution there is a complex solution although we can do it when we think about just involving real numbers we're not going to have a solution so that's why we won't have any x-intercepts here so let's erase this and a similar thing will happen when you have a horizontal hyperbola now we have our y-intercepts the next thing i told you you want to do is find this fundamental rectangle so you'll see a formula for this in your book the reason i don't teach this with the formula is because things are switching back and forth all the time the a is different the b is different it's all over the place and students have a hard time remembering if you just remember that the first point you want to get is x comma y where does x come from where does y come from for x you take the principal square root of that number right there whatever is under x squared so the principal square root of 9 is 3. so it's 30. now for y you take the principal square root of whatever is underneath the y squared variable so in this case that's three also then you're going to do a variation of this so there's four points total to make this rectangle so now i'll make this guy negative and this guy will stay positive then i'll make this guy positive and this guy negative and then i'll make them both negative getting the fundamental rectangle is very very easy if you just follow that formula there again just think about x and y in that format now to get the asymptotes they extend from the diagonals of this rectangle so that's the easiest way to do it if you want the formula for the asymptotes all you'd have to do is say y is equal to plus or minus you'll have a slope here then times x now where do i get this guy right here again i want you to think about slope as rise over run so there's going to be a vertical value that you move in a horizontal value that you move again think about rise over run so where do i get the rise from again you take the principal square root of this guy right here whatever is underneath y squared so that's going to be 3. where do i get the run you go underneath the x squared variable take the principal square root of that that's going to be 3. in this particular case we have the same denominator in each case so it's really really easy right we know that it would be three over three or just one so this is one x and you're going to have two scenarios so one line would be y is equal to x the other line will be y is equal to negative x so let's take all of this information to the coordinate plane and i'll show you how to sketch the graph so let's start with the y-intercepts they occurred at 0 comma 3 and also 0 comma negative three so zero comma three is right here zero comma negative three is right here then let's think about that rectangle so we know the endpoints were at three comma three they were at three comma negative three then we had negative three comma positive three and then we had negative three comma negative three so three comma three is right here three comma negative three is right here negative 3 comma 3 is right here and then negative 3 comma negative 3 is right here so now we connect those endpoints now once you have the rectangle drawn you can extend from the diagonals your asymptotes so you can either again extend these from the diagonals there or you could just find the asymptotes on your own again we found that it was y equals plus or minus 3 over 3 which is 1 times x so this is y equals x or y equals negative x so if you did y equals x that would give you this guy right here this line right here this is y equals x so if i start at the origin i go up one i go to the right one up one to the right one you know so on and so forth y equals negative x that's going to be this guy right here okay that's y equals negative x now once you have all that drawn you would think about the actual hyperbola so i've had the computer draw this for us just so it's accurate but i'm going to trace over it and again when you draw this each branch is going to go through an intercept and approach but not touch the asymptotes so let's kind of start here and we'll come down and again we'll go through this intercept here and we will come back up and again we approach the asymptote but we don't touch it do the same thing down here okay so we'll take this through the intercept again we approach the asymptote but we do not touch it okay and again not perfect but you're giving a general idea of what this guy looks like let's take a look at another one so we have x squared over 4 minus y squared over 9 and this equals 1. so in this case this guy right here is the dominant one we know this is larger because this minus the y squared over nine is equal to positive one so again this will tell me that if the x variable is involved in the larger term or the term that's positive however you want to think about that it's a horizontal hyperbola with a horizontal hyperbola i will have x-intercepts and no y-intercepts so how do i get my x-intercepts the shortcut is to take that guy right there and take the principal and negative square root of that so plus or minus the square root of 4 is plus or minus 2. so my x-intercepts my x-intercepts occur at two comma zero and negative two comma zero all right let's talk about the fundamental rectangle so our rectangle again how do we get these endpoints i'm just going to say x comma y with the formulas you get in your textbooks again they go back and forth based on what's a what's b it's kind of all over the place and it takes a lot to really memorize that stuff if you just think about the fact that you're going to have an x value and a y value it's very easy to get this so for the rectangle think about the first point that you would have so it's x comma y where does x come from take the square root of whatever's under x squared the square root of 4 is 2. where does y come from take the square root of whatever's under y squared square root of 9 is 3. so it's 2 comma 3 and then i get variations of that so i end up with negative 2 comma 3 i end up with 2 comma negative 3 and i end up with negative 2 comma negative 3. very easy once you think about it that way you never have to remember any kind of crazy formulas none of that you get your rectangle right away and some people don't even want the rectangle they just want the asymptotes you can do that too the asymptotes are easy to find also so y equals you have your slope times x right your slope again what is slope it's rise over run well where do i get the rise from again rise is vertical we're thinking about the y variable again take the principal square root of nine that's three then for the run take the principal square root of 4 that's 2. so you would have y equals plus or minus 3 halves x plus or minus 3 halves x okay that would be your two equations for the asymptotes so let's take this information to the coordinate plane again we saw that the x-intercepts are going to occur at 2 comma 0 and then negative 2 comma 0. so 2 comma 0 is here negative 2 comma 0 is here we found that the fundamental rectangle endpoints were what we had 2 comma 3 we had negative two comma three we had two comma negative three and then we had negative two comma negative three so two comma three is right here negative two comma three is right here 2 comma negative 3 is right here and negative 2 comma negative 3 is right here let's go ahead and make this rectangle okay and again once this is drawn i can extend my asymptotes from the diagonals so from here and here i could make a line going through that and then from here and here i can make a line going through that or again i can just find the equation we found that it was y equals plus or minus 3 halves x so y equals 3 has x and y equals negative 3 halves x so if i started at the origin if i went up 3 units so 1 2 3 and i went to the right 2 that would put me at that point there if i went from the origin down 3 units 1 2 3 and to the left 2 that puts me right there so you can see that would be the line so let's go ahead and sketch that guy so let's graph the hyperbola now so i would again approach the asymptotes but i wouldn't touch them and this guy is going to go through the intercept and coming back around i'm approaching the asymptote but i'm not going to touch it again pretty hard to do this freehand and be accurate but that's why i pre-drew everything just to make it a little bit better again through the intercept and then approaching the asymptote but not touching let's take a look at one more so we have x squared over 16 minus y squared over 16 equals one again this guy right here you think about this as being the positive one or the larger one the more dominant one in the end x squared over 16 minus y squared over 16 is one so this guy has to be bigger so this is going to be a horizontal hyperbola so again we'll have x intercepts and no y intercepts so those would occur at 4 comma 0 and negative 4 comma 0. i got that by taking the principal and negative square root of 16 right plus or minus the square root of 16 is plus or minus 4. that's how i get the x values there and there getting the fundamental rectangle here is very easy because i have the same denominator in each case so i know it would be 4 comma 4 to kind of get things started and then i would do negative 4 comma 4 i would do 4 comma negative 4 and then i would do negative 4 comma negative 4. and again this is all you really need you can extend your asymptotes from the diagonals of the rectangle or you can just figure out what the asymptotes are in this case again you have 4 and 4 if i think about the square root of 16 and the square root of 16. so the asymptotes would just be y equals plus or minus 4 over 4 is 1 so y equals plus or minus x all right so let's take this information and let's graph this guy real quick so again the x intercepts occurred at 4 comma 0 and you had negative 4 comma 0. so 4 comma 0 is here negative 4 comma 0 is here then for the rectangle again we have the points 4 comma 4 you had negative 4 comma 4 you had 4 comma negative 4 and you had negative 4 comma negative 4. so for 4 comma 4 that's here for negative 4 comma 4 that's here for 4 comma negative 4 that's here and then for negative 4 comma negative 4 that's here okay so the rectangle is drawn again if i want to i can just extend the diagonals there and create my asymptotes or again i know it's y equals x or y equals negative x so starting here i would go up one and to the right one up one to the right one you know so on and so forth until i got to that point this point this point or going down it's the same thing for the other guy if it's y equals negative x i would go down one and to the right one down one into the right one you know so on and so forth and have points down here it also make points going up here okay so let's draw the asymptotes so i have my asymptotes drawn again it goes through the intercept again it goes through the intercept okay again not perfect but you get the general idea of what the shape is and that's all we're looking for when you're asked to do this free hand hello and welcome to algebra 2 lesson 79 in this lesson we're going to learn about solving non-linear systems of equations so at this point in algebra 2 you should be very comfortable with solving a system of linear equations in two variables or three variables using substitution elimination graphing or matrix methods here what we're going to do is just introduce something new the concept is very much the same though the only difference here is you just have to be very selective about what method you're going to use based on the situation that you're in so the first thing that we're going to look at we have a second degree equation here we have x squared plus y squared equals 25 we know this is the equation of a circle then we have a line so we have a linear equation here as well so the definition of a non-linear system just has to have one non-linear equation involved so if we wanted to solve this type of scenario we're going to turn to substitution and you can see with this linear equation what happens is i have a y here that has a coefficient of 1. so very easily i can solve this guy for y and i could substitute into this kind of top equation so the first thing i'm going to do is i'm going to label this equation up here as equation 1 and this guy right here is equation 2. so i'm going to solve equation 2 for y so let's start with that so i have negative 2x plus y equals 5. very easy i'm just going to add 2x to both sides of the equation and this is going to cancel so what i'll have is y is equal to we'll have 2x plus 5. let me erase this now with substitution we remember that we substitute in for the variable so i'm going to substitute in for y and i'm going to substitute in here right i'm not going to substitute in from where i took it from so i'm going to take this 2x plus 5 and i'm going to plug that in right there for y so that would give me x squared plus i want this whole thing to be squared i'm plugging in a 2x plus 5 there so the quantity 2x plus 5 the whole thing needs to be squared and this is equal to 25. now that we have that out of the way once you've got that concept down it's just pretty much some simplification from that point so we have x squared plus i'm going to square this guy i can use my special products formula i know that i would take this first guy this 2x that term and i would square it 2 squared is 4 x squared is just x squared then plus 2 times this guy times this guy 2 times 2x is 4x 4x times 5 is 20x then plus this last guy 5 squared which is 25 and then this equals 25. so the first thing we can do is subtract 25 away from each side of the equation so that will go away and that will go away i'll have a zero over here and what i'll have is x squared plus 4x squared which is 5x squared then plus 20x and this equals 0. so let me erase everything to this point don't need any of this now how can i solve this for x well we have a few different things we can do we could use our quadratic formula but kind of the quickest way to do this here is to factor it i could factor out a 5x and i could write this as x plus 4 inside the parentheses i could say this is equal to 0. so now i can use my zero factor property or you could say the zero product property and i could set each factor equal to zero so i could say 5x equals 0 or the quantity x plus 4 equals 0. if i solve this i divide both sides by 5 and i get x equals 0 over here if i solve this i subtract 4 away from each side and i get x equals negative 4 over here so i'm not done i have found out two solutions for x but i don't know the corresponding y values so one x value would be x equals zero the other x value would be x equals negative 4. so let's erase this all right so if x equals 0 what is y so i'm going to plug a 0 in for x in one of these equations doesn't matter which one i want to pick the linear equation because it's easier to work with so i would say negative 2 times 0 plus y equals 5 and so this is going to go away i'll have y equals 5. so if x is 0 y is 5. now the other scenario occurs when x equals negative four so we'd have negative two times negative four plus y equals five negative two times negative four is positive eight so eight plus y equals five we will subtract eight away from each side of the equation this cancels we'll have y equals 5 minus 8 is negative 3. so if x is negative 4 y is equal to negative 3. now i'm not going to do this in every video but i do want you to understand that you should check these solutions and make sure they work in each equation so if x equals 0 y equals 5 that's the ordered pair 0 comma 5. then if x equals negative 4 y equals negative 3 that's the ordered pair negative 4 comma negative 3. so in solution set notation put 0 comma 5 and negative 4 comma negative 3. all right so let's check these guys real quick so i'm going to check 0 comma 5 in each one real fast so 0 squared plus 5 squared equals 25 we know that works then what about here if i had 0 here and 5 here we know that works as well right this would go away you'd have 5 equals 5. all right so this guy is a solution for both this one checks out what about negative 4 comma negative 3. so negative 4 squared plus negative 3 squared equals 25 negative 4 squared is 16 negative 3 squared is 9 16 plus 9 is 25 so it does work there all right let's check it in the last one so negative 2 times negative 4 would give me 8. so i would have 8 plus negative 3 equals 5 we know that's true so this does check out in each one as well all right so i'm not going to check each one moving forward i've already checked these it's something you can do on your own for practice i just really recommend you check things especially when you're on a test and you're trying to get all the credit that you can now i want to show you something before we kind of move on i've graphed the circle and i've graphed the line on the coordinate plane so you can see that there's an intersection between the circle and the line at these two points all right so we can see we have our circle here and again this represents x squared plus y squared equals 25 we see we have our line again this is negative 2x plus y is equal to 5. and we can see that this point right here 0 comma 5 exists on both of the graphs right it exists on the graph of the line and it exists on the graph of the circle so we can see that it satisfies both equations the same thing happens here negative 4 comma negative 3 this point is on the graph of the circle it's also on the graph of the line all right for the next example we'll see the exact same concept we have a second degree equation and a linear equation so we have x squared plus y squared plus 18 x minus 2y minus 18 equals 0. we have 2x plus y minus 3 equals 0. so this is equation 1 and this is equation 2. now what i'm going to do again is solve the linear equation equation 2 for one of the variables in this case i'm just going to solve for y again because there's a nice easy coefficient of 1 there by y so i would subtract 2x away from each side and i would add 3 to each side and that would give me y is equal to negative 2x plus 3. so once i've done that i can then plug in for y in this equation up here so i'm going to plug in here but now also here so this problem is going to be a little bit more tedious because there's more stuff to do so i'll end up with what i'll have x squared plus again plugging in for y i'll have this quantity negative 2x plus three this is squared plus 18 x then minus 2 times again i'm plugging in for y so this negative 2x plus 3 and then minus 18. okay and then this equals 0. all right so now we have that out of the way let me erase this and then i'll just slide this up all right so now we just have some simplifying to do so we have x squared plus i used my special products formula there i would square negative 2x that would be 4x squared and then i would do 2 times negative 2x times 3. if you think about 2 times negative 2 that's negative 4 negative 4 times 3 is negative 12. so you'd have negative 12 times x or minus 12x and then i'd have 3 squared which would be 9 so plus 9. then i have this plus 18x and then i'm going to distribute this negative 2 to each term here so negative 2 times negative 2x is plus 4x and then negative 2 times 3 would be minus 6 and then we have minus 18 and then this equals 0. all right so what can we combine here well these are like terms these are like terms and then these are like terms there all right so x squared plus 4x squared would be 5x squared negative 12x plus 18x is 6x and then 6x plus 4x is 10x so plus 10x then we have 9 minus 6 which is 3 and then 3 minus 18 would be negative 15 and then this equals 0. all right so let me erase everything let's slide this up so most of you will notice that you could divide both sides of the equation by 5 here and that will make the numbers you're working with a little small so you'll have x squared plus 10 over 5 is 2 so 2x and then minus 3 and this equals 0. so i can solve this by factoring i know the left part here would factor into what so give me two integers that sum to 2 and multiply together give me negative 3. well i could do plus 3 and minus 1. 3 minus 1 is 2 3 times negative 1 is negative 3. so x plus 3 and x minus 1 and this equals 0. all right let's erase all this and we'll bring this up so i'm going to set these two factors equal to 0 and i'm going to solve i'll have x plus 3 is equal to 0 or x minus 1 is equal to 0. of course these are very easy to solve i just subtract 3 away from each side of the equation here i get x equals negative 3. over here i add 1 each side of the equation i get x is equal to 1. so i know that we have x equals negative 3 and then we have x equals 1. so let's erase everything and again i just want to plug in to one of the equations and see what my value is for y for each given x value now the linear equation is much easier to work with if you want to plug into equation 1 you can i recommend you end up doing that just as a check with the solutions that you get so if x is negative 3 i'd have 2 times negative 3 plus y minus 3 equals 0. this would be negative 6 plus y minus 3 equals 0. negative 6 minus 3 would be negative 9 so i would have y minus 9 equals 0. add 9 to each side of the equation i get y is equal to 9. so if x is negative 3 y is equal to 9. all right let's try the other scenario now so what about when x is 1. so 2 times 1 is 2 so you'd have 2 minus 3 which would give me negative 1. so you'd have y minus 1 equals 0 add 1 to each side of the equation you get y is equal to 1. so as our solution set goes we'll have the ordered pair negative three comma nine and we'll have the ordered pair one comma one now i'm not going to go through and check these but i recommend that you do i already know they work because when i set up these problems i did check them all right let's look at a different scenario now so when you see two second degree equations what you want to use now is elimination so let's say we see something like 2x squared plus 2y squared minus 5x plus 20 y equals 0 and then 7x squared minus 2y squared minus 49x minus 20y equals 0. so i'm going to call this equation 1 and this equation 2. now most of the problems you're going to see in algebra 2 course will set up to where you can just add the equations right away and you'll have one of the variables be eliminated so if i add equation 1 and equation 2 together that will give me equation 3. so 2x squared plus 7x squared would give me 9x squared 2y squared minus 2y squared would give me 0y squared or we could just say 0 and then negative 5x minus 49x we'll say that's minus 54x and then 20y minus 20y let's say this is plus 0 y or again this is 0 and then this equals 0. so what i'm left with now is 9x squared minus 54x and this equals 0. so this is something i can solve very quickly using factoring i can factor out a 9x and i'll have x minus 6 inside the parentheses this equals 0. let me erase all of this we don't need any of this information anymore and i will just scooch this up 9x is a factor so i'm going to set that equal to zero then or the quantity x minus six that's a factor i'll set that equal to zero very very easy equations to solve divide both sides of the equation by nine here we see x is equal to 0. here i can add 6 to each side of the equation i'll see that x is equal to 6. so let me erase everything and the substitution part here is a little bit more tedious than it was before because you don't have a linear equation anymore all right so let's see what happens when x equals 0. so i'm going to go back and label this as 1 and 2 and i'm going to plug in a 0 for x in equation 1. so 2 times 0 squared plus 2y squared minus 5 times 0 plus 20y is equal to 0. all right so this is gone and this is gone so basically what i have is 2y squared plus 20y is equal to 0. now what i can do is i can divide each side of the equation by 2. i can make those numbers smaller so that cancels i'll have y squared plus this cancels with this and i have 10. so 10y and this equals 0. let me erase all this and we'll slide this up let's get some room i can solve this by factoring if i factor out a y i would have y plus 10 inside the parentheses this equals zero so one of the solutions here would be that y is equal to zero the other solution would be y plus 10 equals 0. we know we could subtract 10 away from each side of the equation and that would give me that y is equal to negative 10. so the solution here when x equals 0 there's two different y values that's kind of partnered up with that so in other words you'd have the ordered pair 0 comma 0 but you'd also have the ordered pair 0 comma negative 10. so let me kind of write this over here x equals 0 y equals 0 and you also have x equals 0 y equals negative 10. now let's see what happens when x equals 6. so if x equals 6 again i'm just going to plug into equation 1. the numbers are a little smaller so it's easier to work with so we'd have 2 times 6 squared plus 2 y squared minus 5 times 6 plus 20 y equals 0. so 6 squared is 36. so 2 times 36 is 72 plus 2 y squared negative 5 times 6 is negative 30. so minus 30 then plus 20 y and this equals 0. so what is 72 minus 30 it's going to give us 42. so i can just get rid of this and just put this as 42 over here then i can subtract 42 away from each side of the equation so this is going to go away i'll have 2 y squared plus 20 y is equal to negative 42. all right so let's erase all this and i'll bring this back up all right so i'm going to divide each side of the equation by 2. so this will cancel with this and i'll have y squared then plus this will cancel with this and i have 10. so plus 10y this equals this cancels with this so i know it's negative 42 divided by 2 is 21. so y squared plus 10y equals negative 21. so what i can do there i can add 21 to each side of the equation i'll have y squared plus 10y y plus 21 equals zero now this is factorable so i can solve it that way just slide this up all right so what does this factor into on the left side give me two integers whose sum is 10 and whose product is 21. it's easy seven and three so y plus seven and y plus three so we know the solutions here we would set y plus seven equal to zero then we would also set y plus three equal to zero over here i would subtract seven away from each side of the equation i would get y is equal to negative seven or over here i subtract three away from each side of the equation and i'd find that y is equal to negative 3. so if x is 6 y could be negative 7 or if x is 6 y could be negative 3. so you have x equals 6 y equals negative 7 you have x equals six y equals negative three all right so let's erase all this all right so our solution set here we're gonna have the ordered pair zero comma zero we have the ordered pair zero comma negative ten we have the ordered pair 6 comma negative 7 and then we'll have the ordered pair 6 comma negative 3. all right let's take a look at one more of these so we have 4x squared plus 2y squared minus 29x minus 8y plus 6 equals 0. we have 7x squared minus 2y squared minus 37x plus 8y minus 6 equals 0. so let's call this equation 1 and this equation 2. two second degree equations and again in your book or whatever you're using to kind of study this especially in algebra 2 they give you these easier problems where it's kind of set up where these are opposites and these are opposites so i can add the two equations right now and i can basically eliminate y so we're going to use elimination 4x squared plus 7x squared for equation 3 would give me 11x squared these would be eliminated these would be eliminated negative 29x plus negative 37x would be negative 66x and then 6 minus 6 that's gone so this just equals 0. so the first thing is i can divide both sides of this equation by 11 and make everything smaller this would be x squared minus six x equals zero so let me erase this and i'll slide this over of course we can solve this guy by factoring if i factor out an x i'll have x minus 6 inside the parentheses we set this equal to 0 and we say that x could be 0 or x could be what if i set x minus 6 equal to 0 i know the solution would be positive 6. so x could be 0 or x could be 6. all right so let's erase everything and i'll keep this here for now kind of slide it over just a little bit so what happens when x is 0. let's just pick i don't know equation 1 looks a little bit easier let's pick equation 1 to work with so if x is 0 we'll have 4 times 0 squared plus 2 y squared minus 29 times 0 minus 8y plus 6 equals 0. so we know that this is going to go away and this is going to go away i'll basically have 2y squared minus 8y plus 6 equals 0. now i can divide each side of the equation by 2 and make that a little easier so y squared minus 4y plus 3 equals 0. i can factor this y and y so give me two integers whose sum is negative four and whose product is three you could do negative three and negative one so minus three and minus one so i could set each factor here equal to zero so you'd have y minus 3 equals 0 and then you'd have y minus 1 equals 0. of course the solution to this one would be that y equals 3. the solution to this one would be that y equals 1. if x is 0 y is equal to 3 or it's equal to one so we're going to set this up over here and i'll just say we have the ordered pair zero comma three we also have the ordered pair zero comma one all right so let's get rid of that all right now let's see what happens when x equals six so i'm going to go into equation one again and we'll say four times you'll have six squared plus two y squared minus 29 times 6 minus 8y plus 6 and this equals 0. so 6 squared we know is 36. 4 times 36 is 144. so let me erase this and just put 144. 29 times 6 is 174. so let me put minus 174. and we know that 144 minus 174 is negative 30. so let me just put negative 30 and then we're adding 6 so that's negative 24. so i'll put negative 24 back here so what i'm left with is 2y squared minus 8y minus 24 equals 0. so let me erase this and we'll kind of scooch this down now one thing i can do is i can divide everything by two so this would cancel i'd have y squared this would cancel with this and give me negative 4 so i'd have minus 4y this would cancel with this and give me negative 12 so minus 12 and this would equal 0. now can i solve this by factoring yes i can let me erase this real quick and i'll slide this guy up so i can factor the left side here this is why this is y give me two integers that sum to negative four and multiply together to be negative twelve well of course you could do negative six and positive two so we know the solutions here would be that y equals six or y equals negative two right if i set each one of these factors equal to 0. so y equals 6 or y is going to equal negative 2. so let's erase this and so our ordered pairs would be 6 comma 6 and then also six comma negative two so in solution set notation we'll say we have zero comma three as a solution zero comma one as a solution six comma six as a solution and 6 common negative 2 as a solution hello and welcome to algebra 2 lesson 80. in this lesson we're going to learn about nonlinear inequalities so hopefully at this point everybody knows how to solve a linear inequality in two variables and also how to solve a system of linear inequalities in two variables i just want to take a few minutes here and review how to do this type of problem pretty much everything we're going to do in this problem is going to carry forward into solving a non-linear inequality and then also the same concepts are going to be used when we solve a system of nonlinear inequalities if we have a problem like 6x plus 2y is greater than 12 what's the first thing we want to do we want to find the boundary we want to find the boundary and in this case it's going to be a boundary line but later on if we're working with non-linear inequalities we won't have a boundary line the boundary will be a different shape but the boundary is an important concept to get down when you're on the coordinate plane the boundary is going to separate the coordinate plane into two regions one region is the solution region so all the points there will satisfy the inequality the other region on the other side of the line is going to be in the non-solution region so every point there will not work or will not satisfy the inequality so you're either in the solution region or you're in the non-solution region it's one or the other now the boundary line there's two scenarios there's one where the boundary line is included in the solution set so when it's included is when we have a non-strict inequality so basically a greater than or equal to or less than or equal to and for this we get a solid boundary line now if it's a different shape you're working with maybe a circle or an ellipse or whatever it is we wouldn't say solid line we'd say solid circle or solid ellipse or solid hyperbola you know so on and so forth then the other scenario is if you have a strict inequality so if you have something that is strictly greater than or strictly less than you get a dashed or broken up line to show that the boundary line is not part of the solution so how do we get the boundary line or the boundary in general essentially we look at the inequality symbol and we replace it with equals so i would have 6x plus 2y is equal to 12. i can easily solve this guy for y and graph the resulting line so let's subtract 6x away from each side of the equation that will give me 2y is equal to negative 6x plus 12 let me divide both sides of the equation by 2 and we'll see that this will cancel with this and i'll have y by itself this is equal to negative 6 over 2 is negative 3 then times x then plus 12 over 2 is 6. so this line right here is very easy to graph it has a y-intercept of 0 comma 6 with a slope of negative 3. so the first thing we would do is graph this line and we would graph it as a dashed line because we have a strict inequality here strictly greater than and so let's go to the coordinate plane and i've pre-drawn everything today just to make it a little bit quicker for us but again the y-intercept occurs at 0 comma 6 which is right here and the slope is negative 3. so down 3 to the right 1 down three to the right one down three to the right one so on and so forth we graph our line again this guy is dashed because it's not included in the solution now for the next part you basically had two options you could solve this guy for y and you could shade based on the direction of the inequality symbol or you could use a test point so for today we're going to go back to this test point method and with a test point i can choose any point on the coordinate plane that is not on the boundary line so if i go back down i like to use 0 0 because it's easy to work with it's not on the boundary so let's go back up so our test point is 0 0. so essentially i'm going to plug in a 0 for x and a 0 for y and what i want to know is do i have a true statement this would go away this would go away you'd have 0 is greater than 12. is that true no it is not this is false since 0 0 does not work as a solution we know it lies in the non-solution region so you can think about this area right here that's beneath or below the line as the non-solution region so this is the non-solution region so everything on this side of the line does not work as a solution to the inequality anything on the other side of the line or above the line you could say works as a solution so i would shade everything this way this is my solution region and on each example here i included a computer graph so you can see it did it perfectly it shaded everything above the line and it kept the line dashed and the reason it does that again is because the line is not part of the solution but anything above the line works as a solution to the inequality and so it's part of our solution set all right now let's jump in and look at our first example where we have a non-linear inequality so we have x squared plus y squared is less than or equal to 9. again the first thing i want to do is graph the boundary so now i'm not going to have a boundary line in this case if i replace the inequality symbol with equals i'm going to have a circle so the circle is going to be the boundary so let's do x squared plus y squared equals 9. so this is a circle that has a center at the origin and that has a radius of 3 right the square root of 9 is 3. so the center is obviously at 0 0 and the radius is 3. this guy right here remember this is a non-strict inequality so i'm going to have a solid boundary i go to my center and from the center i go three units to the right make a point three units down make a point three years to the left make a point three units up to make a point so i have my circle i would obviously sketch the graph around those points and so again it's a solid graph because it's included in the solution now the next thing we want to do is use a test point let's use 0 zero it's not on the circle it's the center of the circle but it's not on the circle so zero comma zero let's go back up so if i plugged in a zero for x and a zero for y what would i get i would have zero is less than or equal to nine which is true right 0 is less than 9. so that tells me 0 0 lies in the solution region so the solution region here is anything inside of the circle remember the circle is the boundary so you're either on the inside of the circle or you're on the outside of the circle in this case since it's on the inside of the circle i would shade everything inside of the circle all of those points inside the circle would satisfy the inequality again for a nice computer drawn graph you can see we have our circle again it's drawn as a solid circle because the circle itself is part of the solution set and then the shaded area is everything inside of the circle so this is the solution region right everything inside that circle would work as a solution for the inequality and then anything outside the circle would not work right that's the non-solution region all right let's take a look at another one so we have y is greater than x squared plus x minus 6. so again if i want the boundary the boundary what do we get we replace this with equals we'd have y is equal to x squared plus x minus six this is going to be a parabola now it's been a while since we work with parabolas it was a few sections ago but we should kind of remember how to graph a parabola it's one of those things that you really don't forget the first thing you want to do is find the vertex and basically the formula for the vertex in case you forgot it occurs at the point negative b over 2a comma f of negative b over 2a so what is a and what is b well in case you forgot a is going to be the coefficient right here for x squared so this is a and then b is going to be the coefficient for x to the first power so that's b so in each case they're one so i'd have negative b which is negative one over two times one which is two so basically negative b over two a is negative one half so negative one half now i want f of negative b over 2a i know negative b over 2a is negative one-half so i want f of negative one-half what is that it's basically where i just plug in a negative one-half for x and i evaluate so this would be negative one-half squared plus negative one-half minus six so let's see what we get with this so negative one-half squared would be one over four or one-fourth then you'd have plus or you could say minus one-half i want to get a common denominator going now so let's multiply this by two over two we could say this is minus two fourths then lastly we have minus six but again let's get a common denominator going we'll multiply six by four over four and we'll have 24 fourths so 1 minus 2 is negative 1 and then negative 1 minus 24 is negative 25 this is over the common denominator of 4. so let's replace this with negative 25 4 we can erase all this so now generally when we graph a parabola we want to get three points going so we have our vertex in this case because the parabola opens up we know the vertex is going to be the lowest point so it would look something like this so we want to try to find the x-intercepts if we have those in some cases we don't we want to try to find the y-intercept so let's take a look for the y-intercept i can just replace x with zero what would i have zero squared is zero plus zero is zero so you basically have zero minus six which is negative six so if x is zero y is negative 6 this is your y intercept what about the x intercepts so the x intercepts well in this case you're going to have two of them if i replace y with 0 and i solve the resulting equation i get my x-intercepts so you can factor this right side here i'm going to kind of flip it around and say this is x here and x here and i'll set it equal to 0 over here and give me two integers whose sum is one and whose product is negative six well you could do positive three and negative two so if i have the quantity x plus three times the quantity x minus two and this is equal to zero i can set each factor equal to zero so x plus three equals zero the solution to that would be x equals negative three so one x intercept would be negative three comma zero the other one if i set x minus 2 equal to 0 the solution will be x equals 2. so the other x intercept would be 2 comma 0. all right so now that we have those let's go to our coordinate plane all right so the first thing i will do is identify the vertex which is going to be right here basically what it is it's negative one-half negative one-half which is about right here comma negative twenty-five-fourths which is about right here right you think about twenty-four fourths is negative six so you add another fourth to that in the negative direction so you're going down by another fourth so again that's about right there so this is about our vertex right there then our y-intercept occurred right here at 0 comma negative 6 and our x-intercepts occurred at 2 comma 0 and negative 3 comma 0. now we graph the parabola and again notice that this guy is graphed as a dashed parabola that's because our inequality is a strict inequality right so it's not part of the solution now we want to grab a test point something that's not on the parabola zero comma zero is available so let's check it so our test point is zero comma zero so i plug in a zero for each x and a zero for y and i would get that 0 is greater than this is all gone negative 6. 0 is greater than negative 6. this is true so that tells me that 0 comma 0 lies in the solution region so when we think about the two regions here one would be inside the parabola and one would be outside the parabola remember the parabola is the boundary it's separating the solution region from the non-solution region so everything inside the parabola would work as a solution for my inequality and then here's a computer generated image once again you have a dashed parabola to show it's not part of the solution and then everything inside of the parabola is shaded to indicate that this is the solution region right anything outside the parabola will be in the non-solution region all right so let's take a look at one more of these and then we'll look at some systems so we have x squared over 16 minus y squared over 9 is greater than 1. so again the first thing we're looking for is the boundary so if i had x squared over 16 minus y squared over 9 is equal to 1 i'd be looking at a hyperbola again because of the minus sign there and then again because this is the positive one we know this is a horizontal hyperbola and specifically it's a horizontal hyperbola that is centered at the origin so i know in this case i won't have any y-intercepts i'll just have x-intercepts so to find those it's very very easy we know y is 0 in each case and the shortcut is just to take this denominator here and take the principal and the negative square root of that plus or minus the square root of 16 is plus or minus four so the x-intercepts would occur at four comma zero and negative four comma zero so now the next thing you wanna do you can either find your asymptotes and then graph it that way or you can find the rectangle and then from the rectangle you can extend from the diagonals to find your asymptotes so however you want to do it i'm just going to do everything so you have a complete understanding the rectangle there's four endpoints and i want you to remember how we go about getting those so everything is looking at the denominators right i'm messing with the square roots so it's x comma y so i look underneath x squared i have 16 take the square root of that i get 4. i look underneath y squared i have 9 take the square root of that i get 3. that's my x comma y to kind of get things started then i do variations of that so i'm going to do negative 4 comma 3 i'm going to do 4 comma negative 3 and then negative 4 comma negative 3. so that's my fundamental rectangle again i can extend from the diagonals and get my asymptotes if i want the equations for the asymptotes it's y equals plus or minus your slope is rise over run for the rise take the square root of what's underneath y squared the square root of 9 is 3. for the run take the square root of what's underneath x squared the square root of 16 is 4. so y equals plus or minus 3 4 x so we have everything pre-drawn but again you can kind of follow along we know that the intercepts occurred at 4 comma 0 and negative 4 comma 0. we know that the points for the fundamental rectangle were 4 comma 3 negative 4 comma 3 4 comma negative 3 and negative 4 comma negative 3. again if we wanted to we could extend the diagonals here and get our asymptotes so those would be your two lines or you could graph it with the equations we found y equals plus or minus 3 4 x right so if i start at the origin if i went up 3 and to the right 4 i'd have that point or if i went down 3 and to the left 4 i'd have that point so that's one line if i start at the origin and i went up 3 and to the left 4 that'd be that point or down 3 and to the right 4 that'd be that point so however you want to do it you end up graphing the asymptotes and so you have everything you need each branch of the hyperbola is going to go through the intercept and it's going to approach the asymptote but not touch it now these guys again you see that they're dashed and that's because we have a strict inequality now zero comma zero is available it's not on either branch of the hyperbola so if i use zero comma zero as a test point plug in a zero that would be gone this would be gone you basically would have zero is greater than one which is false so that tells me that zero comma zero is in the non-solution region so what we would shade here is everything to the right of this branch right here and everything to the left of this branch here that would give you the solution region for this inequality again just to show you a computer generated graph you see this is the solution region along with this side over here anything outside of those areas that are shaded will not work as a solution in the inequality all right now let's look at a system of nonlinear inequalities the concept here is the same all we're doing is graphing stuff that's a little bit more tedious so that's why it takes more time than when we work with just linear inequalities or systems of linear inequalities so we have x squared plus y squared is less than 49. let's start out with this one we know that the boundary here would be a circle and we know the boundary would be dashed because this guy is strict so it's x squared plus y squared is equal to square root of 49 is 7. so i might as well say this is 7 squared so this is a circle that is centered at the origin with a radius of 7. so let's go down here so centered at the origin a radius of seven means i'd go up seven and make a point to the right seven to make a point down seven to make a point to the left seven to make a point again graph my circle as a dash circle because no point on that circle is part of the solution set for the test point let's choose zero comma zero and if i had a zero for x and a zero for y i would have zero is less than forty nine which is true so that tells me that the solution for that first inequality would be that i would shade everything inside of the circle but i'm not done because it has to satisfy both the first inequality and the second inequality so let's go back up and let's think about this second inequality here and i'm just going to label this so i can refer to stuff this is inequality 1 and this is inequality 2. so let's work on inequality 2 now so this guy will be a line when we make the boundary x plus 5y is equal to 10 and again it's going to be dashed because we have a strict inequality so let's solve this for y this would be 5y is equal to i'm going to subtract x away from each side of the equation and then divide both sides of the equation by 5 that would give me y is equal to negative 1 5 x plus 2. so my y-intercept occurs at zero comma two and my slope is negative one-fifth so my y-intercept let me just erase this real quick my y-intercept occurs at zero comma two right here my slope is negative one-fifth so down one to the right five or i can go up one into the left five so that's where that line came from again it's dashed because it's not part of the solution now where is our solution region for inequality number two take zero comma zero you can use that as a test point and if i plugged in a zero for x and a zero for y essentially i would have zero is greater than ten which is false so 0 0 is not in the solution region so i would want to shade everything above the line so let's think about this we need the overlap of the two graphs with inequality 1 it was anything inside of the circle with inequality 2 it's anything above the line so essentially the overlap of the two graphs or the area of the coordinate plane that satisfies both would be anything above the line that is inside of the circle so this section right here and just to let you see what a computer does you can see this graph is basically perfect you have a dash circle because it's not included also a dashed line because it's not included anything above the line that's inside of the circle is in the solution region all right let's take a look at one more so we have y is less than three we also have y is greater than or equal to x squared minus x minus two so let's call this inequality one and this inequality two for y is less than three it's a strict inequality so the boundary line will be broken but we're basically graphing as a boundary line y equals three now let's take a look y equals three is a horizontal line so really it's just this line here that crosses through zero comma three on the y-axis right but for every x value y is equal to three now in terms of where we wanna shade y is less than three so any value of y that's less than 3 works so we would be shading below the line so anything below the line works as a solution to that first inequality now let's take a look at the second inequality this guy is a non-strict inequality so the boundary will be solid right and we're graphing a parabola here you'd have y is equal to x squared minus x minus 2 this is a parabola that opens up so we would want the vertex to start again this occurs at negative b over 2a comma f of negative b over 2a so again negative b b here is negative one and a here is one so the negative of negative one is one over two a two times one is 2 so you'd have one half for negative b over 2a then we want f of negative b over 2a so we want f of one half let's erase all this i'm going to plug in a one-half there and also there so f of one-half is equal to you'd have one-half squared which would be one-fourth then minus one times a half is a half let's get a common denominator multiply this by two over two you'd have two fourths there then minus two i can multiply two by four over four that would be minus eight fourths so one minus two is negative 1 negative 1 minus 8 is negative 9. so this would be negative 9 4. this right here is negative 9 4. all right so we have the vertex what would our y-intercept be if i plugged in a 0 for x i would get y equals negative 2. what our x-intercepts be well we have 1 2 or 0. i'd plug in a 0 for y so i'd have x squared minus x minus 2 equals 0. i can factor this and solve it so we would have x here and x here so give me two integers whose sum is negative one and whose product is negative two so you could do negative two and positive one so if i solve each one of these factors x minus two equals zero has a solution of x equals 2 so 2 comma 0 will be 1 x intercept and then x plus 1 equals 0 has a solution of x equals negative 1 so negative 1 comma 0 is another x intercept so let's erase this and let's graph this guy i'm just going to erase what we've done already so the vertex again is at a half comma negative nine fourths so right there then our x-intercepts again we have two comma zero and then we have negative one comma zero and our y intercept was at zero comma negative two all right so we have enough to sketch the graph which is already done for us and again this is a solid parabola because it's part of the solution now let's think about a test point let's use zero comma zero i plugged in a zero for x and a zero for y i would have zero is greater than or equal to negative two this is true so that tells me that zero comma zero lies in the solution region so in other words anything inside of the parabola would work as a solution for that inequality now let's think about the overlap we know in the first inequality it was anything below the line so essentially it's going to be anything inside of the parabola that's also below the line so this section right here would be your solution region and again to see it with a computer we have anything that is below the line but also that's inside of the parabola is part of our solution set so this would be your solution region and then also don't forget that this part right here satisfies both of the inequalities and so it's part of your solution region for the system as well but just the part that is underneath that line anything above the line no longer satisfies both so this would be your solution region these are points on the coordinate plane that satisfy both inequalities and so they satisfy the system