Transcript for:
Comprehensive Overview of Algebra Concepts

well hello internet and welcome to what i think is gonna be a pretty crazy video i am going to try to teach algebra one and two normally kids take multiple years to learn algebra one and two and i'm going to try to do it in one video so what exactly is this video going to be well i am going to focus on the hard stuff because people often say to me well i know algebra and then i show them a formula and they say maybe i don't know algebra so i am going to really focus on the hard stuff but i'm going to start off by going over the basics real quick and of course if you have any questions this is a live stream at least it's a live stream right now you can go and ask me any questions like if you say i want to see more problems like that say i want to see more problems like that and i'll solve them the way i structured this is i'm basically going to solve every type of algebra problem you could ever see and i'm going to specifically i'm going to um i'm going to target problems where you solve them using one rule and then i'm going to show you the same similar problem that is solved using other rules and i'm hopefully if you go through this video pause when you see the video try to solve it and then watch how i solve it if you do that enough times i think that you basically will master algebra and be able to solve any type of problems well i will what will you cover this isn't an octave video this is an algebra video [Music] and i'm going to basically cover all of algebra 1 and 2 except for matrices i'm gonna cover those in a learn linear algebra in all in one video because i thought that kind of made sense and this video is already gonna be super super long so let's jump in and to the presentation and i am going to be drinking a lot of water so here we are some basic random algebra things on the screen let's start off very basic and it will ramp up very quickly of course a variable is a variable if you don't know what that is is just a letter that represents a currently unknown number um this is going to be targeted towards anybody that wants to learn algebra which can be used of course in programming and i'm mainly making this because i want a complete course on machine learning and data science and a complete understanding of algebra is needed to be able to do that of course knowing algebra also would help dramatically with games okay so a variable is just no problem and believe me by the end of this i'm going to show you pretty much every algebra problem there is okay asides except for matrices i'm going to cover that later okay so we have variables it's just a letter that stands for an unknown number then you have what i guess we're starting to get a little bit more complicated mathematical expressions what them what is a mathematical expression well this is a term right here and a mathematical expression is just 3x or you could go and add on a 5 or you could go and add on to y whatever okay so this is an expression then you have what are called monomial expressions it's just one term okay so this guy right here is called a monomial now binomials have two terms trinomials three terms and thank you very much for subscribing zen uh polynomials cover two three or more terms along with a whole bunch of different math operators then you have equations an equation is how it separates itself from a simple mathematical expression is an equation has an equal sign that's also 7 x is equal to 14. and can you solve that problem that is a equation inequalities they are going to be any time you see a less than symbol a greater than symbol a less than or equal symbol greater than or equal to um you could say not equal to or not equal to whatever you'd like okay so those are inequalities like terms what's a like term well if they have the same variable they are like terms so these are like terms and for example 7z this would be an unlike term in comparison to the other terms that you see on your screen get a little bit more jargon out of the way if you had something like 7 a this guy right here this number is called a coefficient and can you guess what this is called it is called a variable all right and i think that covers a lot of the basics you can basically let's go and erase this so that we can get more stuff here on the screen i'm covering all the basic stuff really quickly so that i can spend a lot of time doing very complicated stuff thank you very much all right i use obviously use algebra constantly all right so we can add like terms so if you had something like 3x plus 7x these could be added to give you 10x now you cannot add 3x plus 4y why because they're unlike terms why are they unlike terms because they use different variables and uh you can also multiply you can also subtract like terms you can't subtract unlike terms you can multiply like uh like terms so this would give you 2x um what else can you do you can multiply a number times a variable so 3 times x is going to be equal to 3x all right so basic basic basic basic stuff but extremely important to be able to understand you can also multiply expressions that have unlike terms and you can also divide expressions that have unlike terms or i should say you can also multiply and divide terms so for example if you had 3x times 2 y this would be equal to 6xy exactly like that um and like i said anything you you can't add or subtract unl um unlike terms but you can multiply and divide them go ahead what's your question hey somebody just said hey i'll just continue on so for example if you had something like three x divided by four x what would happen these would cancel out we can erase them just as if they didn't exist why can they cancel out well x over x is equal to one so this is exactly the same as saying three fourths times one well what's that going to give you it's going to give you three fourths all right so simple stuff um what else give me a bigger eraser here okay so what else can we do well i'm not really gonna cover order of operations because i kind of think that's pretty simple let's start getting into a little bit more complicated stuff and of course if you uh you can subtract yeah and you can't divide by zero we'll get more into those rules we're getting into some rules right now now whenever it comes to what are called properties of numbers there is the commutative property of addition and it just means you can add your values in any order and it doesn't matter so a plus b is exactly the same as b plus a the commutative property of multiplication i know my daughter took algebra and she had to memorize these names i leave that to you to figure out how to memorize them basically the commutative property of multiplication says that you can multiply in any order and get the same exact result you can also based off the associative property of addition you can add a group of numbers together in any order and also get the same results the associative property of multiplication says when multiplying you can also multiply in any order all right and then we get the distributive property of multiplication and that just says that a times b plus c is going to be equal to a times b plus a times c all right so pretty easy now we're going to get into a little bit more complicated we're going to talk about solving equations now how do we solve equations well the very first thing we want to do is we want to simplify each side of the equation what does that mean well it means that we're going to be adding and subtracting like terms and then after we do that we're going to be dividing or multiplying unlike terms so if we had something like 4x plus 2x minus seven plus nine well basically you just wanna look at these as their like term types so those are like terms or maybe i should use a slightly different color here we'll go there's a like term there's a like term and these are like terms all right so there you are and you just add them together so if we have 4x plus 2x plus x what's that going to be equal to what's going to be equal to 7x then you get 9 minus 7 which is going to be equal to plus 2 which is then if you take 5x minus x you get 4x and then you're left with bringing the 4x over how do you do that well you subtract the 4x from this side anything you do on one side you have to do on the other side this leaves you with 3x and then you have to bring the 2 over how do you do that you subtract 2 from this side you subtract 2 from this side and you're left with 3x is equal to negative 2. then you want to get rid of the 3 you can multiply and divide unlike terms so of course the final answer well let's write it out so you need to divide this side by three can i draw three that's legible i think i might be able to and you divide both sides by three well that cancels these out you're left with x is equal to negative 2 over 3. all right so there's an example of just solving a simple equation let's ramp it up a little bit more solving inequalities again remember solving an inequality anytime you see these types of little symbols that means that we are going to be working with what are called inequalities whenever you have the situation in which you have greater than or less than whenever you're drawing these on a number line you're going to draw an unfilled circle in circumstances in which you have an equal sign you're going to draw a filled circle all right so that is the difference between drawing these onto a number line so let's just keep it very simple um let's say that we want to graph x is greater than two well it's not greater than or equal to so what does that mean we're going to use a circle number so we are a number that's not we're going to be using a circle that is not filled so we're just going to go like this and we're going to go like this and there you go you just graphed it so let's go and do something slightly more complicated so we have 2x minus 5x plus 4 is less than or equal to 10. one thing to also understand is anytime you divide by a negative value that you are going to have to flip this guy right here so if you're going to be dividing with a negative value this is going to become greater than or equal to all right and i'm going to show you that in this example right now so we have 2x minus 5x this is going to be equal to negative 3x plus 4 and less than or equal to 10. we want to get rid of this 4 on the left side so we have to go and subtract 4 from both sides that's going to leave us with negative 3x and that is still going to be you can see there's negative values here so we're probably going to be flipping and let's subtract the 4 from the 10 that is going to give us negative 3x is less than or equal to 6. we are going to need to divide by a negative 3 on both sides to get the x by itself remember if you're going to do that you have to flip the inequality and you're going to end up with x is greater than or equal to negative 2. now we have the greater than or equal to that means we're going to be using a filled value so we're going to find the negative 2. we're going to draw it on the number line exactly like this and x is greater than or equal to negative 2. we're just going to go and draw our line maybe nicer than that exactly like this okay and there you go that is how we solve inequalities that brings us to rate of change yeah thank you for following there that's a interesting looking face it's like a kind of falling sleep face and then it is a waking up face okay so rate of change now we're getting into what is called a slope and which do i want to use here let's probably use the x now the rate of change of slope is going to indicate both the direction and the steepness of lines and we're starting to get into more complicated things here now because we this is actually pertains to calculus so what let's say that we just simply have x is equal to can you see that i think you can negative three how would we go and plot that on our little graph here well we just find negative three which is right here and we would just draw a line like this okay so there you are you just graphed that now however let's say that y is equal to negative don't put it there can't draw anything if i put it there y is equal to negative 3. how would that be drawn out well again we find r negative 3. so 1 2 3 which would be right here and we would draw our line in exactly like that and don't worry we're going to get a lot more complicated than this is this a formal class you're streaming on twitch or are you just streaming education for fun i'm just making videos education videos for fun i guess yeah i assume there's some people watching it looks like there are so that's a good thing so positive rate of change means that a line is going uphill so this would be an example of a positive rate of change and this would be an example of a negative rate of change because it is going downhill all right um what would we like to do let's move on here let's go and let's say you had to create a line and you wanted to um do so using just two points how exactly would you figure those out well you would have to use this formula over here remember we were talking about rate of change and slope well this is a formula that is going to allow you to find rate of change or slope i'm just going to call it slope because slope is easier so what do we do we just go and we take our value right here see we're just using this formula right here and we're going to take our 2 that we have right here and minus but we have a negative 8 here so we're going to put a negative 8 and i'm pretty certain you understand that if you subtract a negative number that is just like adding it will you be streaming calculus other variation yes i am going to bang out uh algebra 1 and 2 linear algebra in one video and probably trig and basically the trig video is going to be all the trig you need to know to know calculus and then i am going to probably create calc learn calculus in one video calculus one and then calculus two they'll probably do two separate videos if this goes well if people like it or not okay so now we gotta worry about this guy over here and this guy over here to calculate slope so we're gonna have negative 6 and minus and a negative 3 so there we are and this is going to then give us 10 divided by negative 3. so now we know just with these two points on a graph we now know the slope of our line so what do we need to do now well we need to find what is called the y-intercept again some more jargon that sounds complicated it isn't it is just that point in which our line hits the y-axis of our graph thank you very much for subscribing granny found switch i appreciate it okay so we need to find that y-intercept and the y-intercept is just that point in which our imaginative line that we're trying to discover hits our y-axis and it is a line so it's you know only going to hit the y-axis in one point this is also my pencil okay so how are we going to figure that out well what we're going to do is we are going to go and use this handy-dandy formula right here y is equal to mx plus b and then what we can do is we can go and actually plug in our slope which we just discovered and we can plug in some values so what i want to do is i'm going to plug in this value this 2 up here and this negative 6. if i do so well this is y of course so i'm going to say 2 is equal to plug in our slope that is what the m is in this situation our slope and then this right here the b is are going to be well we're going to figure out the b in this situation it is the y-intercept all right so let's plug in our slope negative 10 over 3 and this is going to be multiplied times what this 6 up here i keep drawing arrows up here hopefully that's not confusing so we're going to multiply that times negative 6 and then plus b b is our y-intercept that we want to discover so we just go and use all the wonderful stuff we just learned before we're going to get 60 if we multiply negative 10 times negative 6 we get 60 and we keep this 3 right here plus b and if we then work that out that's going to end up being equal to 2 is equal to 20 plus b and that means that our y-intercept is going to be negative 18. so we found our slope just with these two points we found our slope we know exactly what the slope is for our line and we also know what our y-intercept is with that information we also know the formula for our line we can just simply say y is going to be equal to our slope negative 10 over 3 x minus 18. and there you go that is how you take two points and you discover a line let's see let's go and also take it to the next level let's say we want to graph a specific line how would we do it what's going on just went to the evidence making twitch account well thank you very much um thank you very much i greatly appreciate the help there granny found twitch today it is super awesome um y equals mx plus c that works whatever it's the same thing bc i don't know i'm sure they would like to put you know i would work whatever it's all math and the symbols are just there just to designate unknown values so let's say somebody goes and they give you this weird thing and they say hey i would like you to show me how to graph this line well what are we going to do well we're going to find our x-intercept just simply by making y equal to zero all right so it's very simple remember we're working with straight lines here as the video continues we'll get into more complicated functions that start doing all kinds of weird things right now we're working with lines everything's easy so we just find the x-intercept by making y equal to zero and if we do that we get 3x is equal to negative 6 and that means that x whenever y is equal to 0 is going to be equal to negative 2. and then we come over here to our little plot x is equal to negative 2 and y is equal to 0 and we put a dot right there what do we want to do next we want to find our y-intercept how do we do that well we can just go in and say negative 4 y is equal to negative 6. again i'm just making x negative this time and this means that y is going to be equal to 3 over 2 or approximately one and a half so what do we do here well remember x is going to be equal to zero that means y is going to be equal to one and a half and that is going to be approximately right there since we know this is a line and we have two points we can just quite simply come in and draw a semi straight line between those two points yes i know i did not perfect that but pretty good stuff as it goes now let's go on let's say uh we wanted to find the equation of a line and we know that our line has this one point right here negative three and one and we also know that it is parallel parallel to this formula right here well by being parallel that just means that it has the same slope so what we're going to do here is um we're going to graph this guy out so i'm going to say that i want this guy right here to be negative i want to calculate the slope for it because i know it's exactly the same as my line let's go and use a different color here so i'm going to say negative 4 y is going to be equal to negative x plus 1 and in that situation we can divide by negative 4 to get rid of that and get y by itself and if we do that we're going to end up getting y is equal to 1 4 x minus one fourth and in that circumstance what does that tell us well we remember this formula right here y equals mx plus and i'll use the english version y equals mx plus c in this situation that means that this guy right here is the slope and it's also the slope of our line so pretty easy stuff so now what i want to do is i want to find my y-intercept how do we find the y-intercept well we just make x equal to zero so if we do this that means that this is b is going to be equal to four over four plus three fourths which is going to be seven over four and it's kind of a mess but what we ultimately know is the formula for our line is going to be one fourth x plus seven over four okay there you are so that's how you can go and create a line if you know a singular point and you know another line that would be parallel that we could use to make that calculation so how do we find the distance between two points this is another formula i taught i used to talk to people all the time i used to tutor people in calculus calculus has been brought up calculus is always brought up every time math is talked about i apologize i have to drink water or i will lose my voice okay so basically the reason why i think i used to tutor kids with calculus and on the first day you know what i always did i brought one book with me was it a calculus book no it was not a calculus book it was an algebra book and by the way i think i can make all these notes available to you for free for downloading if you want if you want that to say you want it and i'll and uh i'll put them somewhere where where you can get it um at the end of the video i'll post a link or something i don't know okay so if you want these notes if you don't want whatever who cares so anyway i always brought an algebra book because the reason why most people i found i want okay i'll i'll make it's a good notes thing and you can just download the whole thing um i just have to upload it somewhere where you can get it i'll probably put it in microsoft onedrive or something like that okay sorry about getting deviated there so um what i found was almost everybody that struggled with calculus really wasn't good in algebra and you're going to see as we get later in this video all the really complicated algebra that everybody sort of struggles with can you recommend a good algebra book for self learning um i like probably my favorite see i like to learn by solving problems and then if i get it wrong i go okay well why did i get it wrong and i take a note and then i try to solve some more problems and then i go back that's kind of what i've tried to do here with this video series um there's so many good algebra books for basic algebra i would say painless algebra is a real if you're like if you're just starting off with algebra i'd say painless algebra is probably a spectacular book and if you are more of a problem-solving learner i would recommend the humongous book of algebra problems i think it's called i think that's what it is what syllabus you're using here i basically just i went and got it algebra 1 and algebra 2 book went through and recreated to make sure that i have by the end of this video that i will have solved every type of algebra problem that you could possibly have based off of different rule changes that are needed depending upon what type of problem you're working with so it's sort of my course i've been making courses on youtube for 14 years so i've made like over a thousand videos i've probably made more courses than just about any human being and it lives so that's what i'm basing everything on okay so how do we find the distance between these two points we just use this nice formula and i got deviated when i was talking about calculus but basically i think learning algebra is more about just learning these formulas writing them down creating a cheat sheet referring to them over and over again and then all of a sudden that's all algebra is it's just a bunch of rules if you memorize the rules then algebra all of a sudden becomes easy i'm going to cover all the rules okay so how do we solve this well we're just going to take 11 minus 6 and we are going to square that plus and 4 and and normally in math if you're squaring values like this it normally means your ultimate goal is to eliminate a negative so anytime you see why are you squaring and then taking the square root well the reason why is we want to have a a value that's positive all right and then if you go and you work all of these out what you're going to end up having is 5 squared and of course if you take the square root of 5 squared they cancel each other out that's another algebra rule and you're going to end up with a value of five all right so that's the distance between those different lines let's go into a little bit more jargon there's something that is called interval notation this is just just so that you can recognize okay so in this circumstance um you're not going to i'm going to get into all of the other different terms as as we continue okay so we have 3 and six okay so how is this going to be or x is less than or greater than or equal to x and less than or equal to six how would we draw that well we would just simply say three and six exactly like that now in the circumstance and that's because we are using the equal to signs if we are using a less than sign or not an equal sign this is going to be drawn out as a parenthesis and three and six again you see the greater than or equal that means you're going to do the square brackets and of course this is going to transpose down here as well so if we have three and six again we're going to use parentheses here onward um well i'm covering algebra today because i'm want to cover i want to cover machine learning next after i just made a ton of data science videos and i've basically made this vid i'm making this video so that i can time code it and anytime somebody asks me a question in my machine learning videos or any of my data science videos i can say refer to this part of this one video it just makes it very easy for me to both teach algebra and also at the same time it makes me very easy it makes it very easy to answer people's questions because i'll be able to refer to a time code for this one video so that's the reason why i'm doing it math is ridiculous like if you don't know math you might as well just give up on machine learning because all machine learning is is math so that's why i'm covering math in a software world okay so here what do you think we need to use parentheses or bracket-wise remember we have two that are less than we do not have less than or equal to we're just going to use parentheses all right so this is just a basic thing it's just mainly covered so that you know how to um be able to you know properly use notation in in mathematics okay so what i want to do here with this equation we have here is convert it into interval notation well that basically means i need to solve this inequality very simple we're just going to say 6 x minus 8 and i say very simple i i apologize for that i don't mean to necessarily mean that it's it's particularly easy so if you do have a question about something hey most people are not good at math so if you have everything or any problems feel free to ask me um this is just what i do constantly for my entire life and so i'm good at it anybody that wants to see me do something really bad watch me play chess because i'm terrible at it okay so um that was the learning chess actually really taught me a lot about uh the difference between people that you learned a skill very young and continued to develop it over numerous years um multiply inequalities by oh yes i covered already about um what happens with flipping the inequality if we're doing any type of division with negative values thank you for bringing that up okay so let's continue we got this um we're gonna then going to quite simply just simplify this this is going to become again we're dividing by a negative so we have to flip the inequality this is going to be um yeah flip that inequality and this is going to be negative 27 over 4. but remember our ultimate goal was to convert this into interval notation and again we're using greater than or equal to so that means we're going to have square brackets and then we're going to put negative 27 over 4 and then this is going to be infinity and what do we do with infinity well we actually use parentheses because it's not defined all right let's graph some more stuff ah i get everybody is joining me i'm glad that this is something people this is ridiculously important but the problem is most people say i know algebra and then you find out they don't okay so this is the slope remember and what is this this is our y intercept so y intercept and of course you know that that is where the line hits the y-axis so um i'm going to actually graph this this time so and graphing a inequality is quite weird because you have to shade underneath the line you have to know where to shade underneath the line so what we're going to do in this circumstance is i'm just going to go and put 0 inside of here less than and negative one over three why did i pick three for so many things i have such trouble yes yes a lot of people have forgotten algebra i agree with that okay so we get this value right here this of course if we decide we want to simplify it is going to give us negative 2 less than or equal to 1 over 3 x and then this is going to be converted we're going to be dividing by negative values or no no not in this circumstance we won't this is going to end up becoming 6 is less than x exactly like that and so we know that y 0 is going to be equal to x 6. i actually probably should put the x first but whatever so whenever x is equal to 6 y is going to be 0 in that situation so what we want to do is we want to come in here and we want to graph these different guys so we're going to put this guy right here and then we can go to three over here and that will go right here and that's basically all the lines we have now one thing uh that is extremely important here is to understand that whenever we are dealing with a a non uh if this was less than or equal it would it would be a straight line but it is it is not it is a less than so in search situations which we have that i have to turn my little drawing device here we're actually going to have to draw a dashed line okay so dashed line exactly like this why because it's not equal to in that circumstance so let's put that right there and then what we need to do is we have to say that where exactly what values are going to be pertaining to our graph well in this situation that is going since y is going to be less than that means we need to shade underneath our line so we can just go like this and whoops hit the wrong thing let's go grab that okay i think i got it this time so we're going to graph underneath our line and this is the weird thing that we do whenever we are working with inequalities and you get it like you know fill in that entire area and there you got it okay so that is the difference between working with multiple different uh what software i'm just using good notes that's that's all on an ipad um nothing really complicated here okay so now we get into system of equations see i thought it's going to get complicated pretty quick now whenever you're trying to solve systems of equations there's multiple different ways of doing it now if your coefficients are opposite of each other like we have here so whoops that's not the right pen these are opposite each other we got positive 4 and a negative 4. in those situations we can solve them using what is called through the process of addition or subtraction so how do we do it it's very easy let's go i keep saying very easy uh we will say x and this is a 4 y is going to be equal to 5 and then x minus 4y see they cancel each other out that's why we can do this this is going to be negative 3 exactly like that and there's a straight line so what does this do well these guys cancel each other out and then we have just quite simply 2x is equal to and 5 minus 3 is equal to 2 so we know that x is going to be equal to 1. all right then what we need to do after we know the value for x we just need to solve for the other variable so we can do that just by simply going 1 plus 4 y equal to 5 and this means that 4 y is going to be equal to 4 so that of course means that y is also equal to 1. and another thing that is extremely important is anytime you get an answer um am i going to oh thank you i'm very happy that you can you can see it i think it's just a matter of practice um i do it all the time so i'm just used to it so okay so we have these two values how do we check to make sure our answer is right well we can go x plus 4 y is equal to 5 and if we do that we can just simply plug in these values and see if they come back correct and that is going to be 4. whoops no i don't need the y there in that situation is equal to 5. okay so 1 plus 4 equals 5. so we know that's right and then we can also go x minus 4 y is equal to negative three and of course this works out to be one minus four is going to be equal to negative three and we know that yes indeed that is also true so we also know that what we did right here is going to work for us okay another way we can solve these types of systems of equations is graphically so what we want to do here is we have y is equal to x plus 3 and y also is equal to negative 2x minus 6. well what do we want to do we've done this calculation a couple times we want to figure out what our y-intercept is well let's just use this first equation right here so if we give x a value of zero well we know our let's do it like this like an underscore like that that means our y-intercept is going to be equal to three pretty simple stuff now let's find the y-intercept for our other line and that other line is going to be again put a zero in there negative six all right pretty easy so in this situation we know that our slope is going to be equal to one over one or just quite simply one and we know that our slope for the second formula we have here is going to be equal to negative 2 over 1 using all of what we just covered before and that means that we are going to have a point at one and four and another point right here at one and negative eight so here it seems that they are going to intersect at negative three and zero so we can say why if what we can actually do is go and graph all this stuff out so let's use like an orange line here so uh this is going to be approximately right here which is plot the intercepts that's all that i'm doing right now and what do i want to do for the other one let's say i want to plot the other one at negative 6 like this and we can just go in here and roughly draw in these lines so i'm going to well let's plug this into the formula as well to give ourselves another point it's quite simple it's just going to go right there so then we can just draw this line in right here and then draw this line in if we have negative six this means this is going to be negative eight uh one two three four five six seven eight i'm just making sure that i'm plugging everything in in the right position so there goes that this line is going to be right like this and that is why we whoops never mind didn't mean to do that that means that these points are going to hit right here in the center area and that means that our intersect is going to be roughly negative three and zero if i drew better so what we can do is we can actually come in here now and verify this intersection so if we do so let's use a different color y is equal to x plus three like that and zero is equal to negative three plus three so we know that yes indeed that is going to work out and likewise we'll do it with the other formula 2 x plus or this would be minus 6. apologize for that minus 6 and if y is equal to 0 well this is going to be negative 2 times negative three minus six and indeed zero is indeed equal to six minus six so this is also true and we have verified this graphically where these two formulas meet all right so pretty neat stuff can also solve with substitution yes the java tutorial is largely still accurate because java hasn't changed as a core language for many many years that's okay to ask questions ask questions about anything i don't care that's why it's a live stream sorry i lose my voice all the time so i have to drink water oh the video's frozen well i apologize for that i'm i'm going to uh upload this to youtube tomorrow so i'll go and edit it and then upload the entire thing to youtube i just haven't figured out how to make um i think the problem is is i'm not a i'm not a twitch partner and that won't allow me to change my resolution on twitch i apologize for that for anybody that's having any issues hopefully we'll correct that someday there's really nothing i can do i just apologize for it okay so what we want to do here is solve this system of equations which just means find the values of x and y um through the use of substitution 1080p60 is that what i'm broadcasting at i thought yeah i am broadcasting at 1080p i sort of wanted to go between 720 and i definitely didn't want 4k so i don't know um that was just the best uh decision i could make because 720p ends up making everything look really blurry and most of the algebra videos are that that are out there are kind of blurry so i just wanted to make a non-blurry algebra video um yeah i apologize for that okay so what we're going to do now is we will try to solve with substitution now if one equation can easily be solved for one variable you can substitute the result into your other equation so in this circumstance what i want to do is i want to solve this guy right here x plus y is equal to 8 so i'm just going to go this is going to become x plus 11 x and minus 16 is equal to eight and we're using all the stuff that i covered earlier in the video just by com adding and subtracting like terms and multiplying and dividing unlike terms so this is going to be 16 of course is equal to 8 12 x is going to be equal to 24 and then of course x is going to be equal to 2. thank you very much for following um when i am i could have put the parentheses in there yes i'm just sort of skipping steps a little bit um then okay so we have our value for x what we want to do now is find y so i'm just going to go and use this guy here instead so in this circumstance this is going to become y is equal to and i just substitute in my known value for x so this is going to be 11 times 2 minus 16 and y is going to be equal to 22 minus 16 and ultimately y is going to be equal to six and if i want to verify that yes indeed these are true i can just go x plus y is equal to eight throw in those two values and indeed two plus six does equal 8 and then do the exact same thing for our other formula that we have here so y is equal to 11x minus 16 and 6 is equal to 11 times 2 minus sixteen and yes indeed six is going to be equal to twenty two minus sixteen okay so there you go and that is how we solve a system equation with substitution another option we have available to us and this is the last one is well actually that's not true you can solve system equations using matrices very very easy but that's more of a linear algebra type of problem um which is the only subject in algebra i'm not going to cover today because i felt that it made more sense to just cover all the linear algebra and all the matrices stuff all in one separate video okay so variable elimination how's this one work well to use this technique the coefficients of either the x or y terms must be opposite from each other and so we have 2x plus y equal to 7 and x minus y is going to be equal to 2. of course i sort of covered this previously 2x plus y is equal to 7 x minus y is going to be equal to 2. this goes and simplifies down so that we have three x plus zero is equal to nine which means that x is going to be equal to three and then we substitute into our x value um where i stack the equations and add or that well that's that's kind of what i'm doing right now with variable elimination but you can only do them in situations in which c how this is opposite this okay y is the opposite of negative y so that is that specific situation where you stack them and then just sort of add through uh so now we'll take x minus y is equal to two now we're using what we had over here and we're plugging in into our other variable this is going to simplify down to three minus y is equal to 2 and of course y ends up being equal to 1 and then we could verify that that indeed is true but i promise you that it is now let's take it up to another level this is solving system of inequalities now it starts getting more complicated because we're going to have shaded areas in multiple different directions good point there barry carter you could most definitely um multiply through the equations to make these different terms the opposite of each other most definitely that is something else that you could most definitely do okay so what i want to do here is i'm going to graph two linear inequalities and then i'm going to go and shade in the region where the two of them overlap each other um so let's just go and solve them so we have y is less than or equal to negative three x one well what are we gonna do here well we wanna find our y-intercept by making x be equal to zero well in that circumstance the y-intercept is going to be equal to one and so we know that we have a point here at one and then one minus three like that and of course that is going to be equal to 1 negative 2 and we could go and plot that out i'm going to plot out everything all at one time though just to keep that more divided up y-intercept on our second is again going to be equal to negative four so we know that we'll have another point here at two and negative five and then of course we could go and we're going to have to go and perform a test to figure out where exactly we need to shade but first what i want to do is actually come in and draw in these individual lines so let's make one of them orange and one of them like a cyan blue or something like that and see if that works out good for our stuff so we just put in our different intercept lines negative two so we're going to have a line like this and in this circumstance we're using less than or equal to signs so we know that we need to make these make these a straight filled line so i'm just going to roughly draw that in right there and then for our second one let's make this what color should i make this i'm not really sure i don't think white necessarily works really well for it uh let's try like a neon type of green here i think that might work a little bit better we got our other difference intercepts here so let's go down to negative four in this ballpark and another one in roughly here now in this circumstance we're dealing with greater than so that means quite simply that we are just going to use a dashed line if i can do this how do i draw this i have to turn this in multiple different ways so it's going to roughly be like this okay so there you have your other dash lines now i have to figure out exactly oh red yeah i could have used red all right and so now what i need to do is actually figure out which way that i need to fill them do i need to fill to the right or to the left well i'm just going to test with using the values of 0 and 0. so this means that we're going to ultimately end up getting 0 is less than or equal to 1 in this situation well in those circumstances i'm going to fill to the left and then in the other one i'm going to test it for 0 and 0 values for x and y and if i do that i get 0 is greater than negative 4. well that obviously is also going to be true so we're going to put this like this and in that circumstance i'm going to fill right okay and now i can go in here and actually fill these in so i'm going to maybe use yellow in this circumstance whoops didn't mean to do that so i'm going to fill this in here roughly you get the point of what i'm trying to do maybe i'll try to do this a little bit neater so that you can see exactly where these two inequalities cross each other so i'll just fill that area in right there and then i will use the green for the other one and let's see how that ends up working that's going to be up in this vicinity so there we are and then you're going to see where exactly the inequalities are going to cover each other and there goes neatness out the window so it started off neat but then we sort of lifted our pen and you can see that where these two inequalities are going to cover each other is going to actually be in this vicinity right here and it makes like a v shape up inside of there okay so that's where our two inequalities that are going to work all right all right so now let's move on to absolute value equations let's go get another pen maybe get another drink of water uh yes i could also say that y less than means lower and y greater means higher good that's a that's also a good way to explain that okay now in the situation in which you're using absolute value equations there are going to actually be two values because the value we take the absolute value of can be both a positive as well as a negative value so that means to solve this we're going to have to solve for both three x minus seven equal to a positive eight and also three x minus 7 equal to a negative 8 because that absolute value cancels them out so this means that we're going to have 3x is equal to 15 and 3x is equal to negative one which means that our value for x can either be five or it could be equal to negative one over three exactly like that so let's try to get a little bit more complicated with our problem solving here so we have 9 minus 3 times the absolute value of x plus 2 is equal to 15. well we could just go and cancel these coefficients we're going to end up with three and absolute value of x plus two is going to be equal to negative six and then further whoops further simplify this down to x plus 2 absolute value is going to be equal to negative 2 and also x plus 2 is equal to 2. and in this situation we find that there is actually no solution because the absolute value can never be negative as you can see right there it just logically doesn't make sense so that means that there is no solution i wanted to try to get every type of answer also in creating this video um all right so let's continue um let's go and graph this guy sorry it's getting a little bit distracting here let's go and this is going to be 2 absolute value of x minus 7 minus 5 is less than or equal to negative 1 and we'll have 2 by simplifying minus 7 is less than or equal to 4 which further simplifies down to x minus seven is less than or equal to two and this is what would be considered to be a compound inequality let me see where i should write this so this is going to be negative two less than or equal to x minus 7 less than or equal to 2 and also 5 less than or equal to x minus less than or equal to 9. and if we go and graph this because we're dealing with equal signs we are going to fill in both these dots and we're going to have a dot right here and then we're also going to have a dot right here and we're actually going to fill in the line between two said dots and there we are okay uh what about yes i i could continue doing a whole bunch of these different things um if you really want me to figure out who solves more problems like with absolute values but i kind of want to move on from this all right another important concept is what is called the fundamental theorem of arithmetic um yes it would be slightly different if you really want me to solve any of those it want me to solve that problem i will i'm just trying to move through because i still have a ton of stuff that that's also going here i can go what did you actually put here i can go and solve that and i'll put it in the notebook that i'll make available it's free the notebook i'm just going to go and solve this problem throw it inside of the notebook and everybody can see that i'll just uh have to put a note here so that i go and solve that all right okay so let's go to the fundamental theorem of arithmetic what this says is that all natural numbers are either going to be um can either become well they either are prime numbers or they can be expressed as a product of prime numbers and an example with this is well we can find the greatest common factor but let's talk about prime factorization basically in doing it the prime values are going to be multiplied together to equal our final values so how can we go and break 20 into a bunch of prime numbers well one thing you can do is just simply say well what can you do with a prime number in regards to 20. you can divide it by two all right that gives you ten what can you do with that well can you do something with two yes let's just throw two in there okay and then finally you end up with five and of course two times two times five is going to be equal to 20. this is a very important concept okay so you can do exactly the same thing with 48 maybe i want to write this out so that we can say it well 48 can i divide it by two yes i tell my daughter this when i was teaching her this exact same thing i'm gonna keep it as simple as humanly possible well you can turn a 4 48 into 2 times 24 okay so what you do is you take this 2 over here and you throw it inside of there and then you can continue to basically divide by 2 as long as you can divide by 2 until you end up with having to multiply times another different number or divide by a different number is what i should actually say and this ultimately is going to end up giving you 2 to the power of 4 times 3 okay and that is how prime factorization works so let's talk about how that can be used to to be able to find what is called the greatest common factor it's the same exact process um all right so what we're going to do with finding the greatest common factor is you just list all the prime factors of each number multiply those factors um that are in common let me just show you an example because it makes more sense this way so i'm going to say that i have 24 so we have 24 and 36 what is the greatest common factor between them so we have 24 and 36. now remember all you need to do is keep it simple can you divide 24 by 2 yes is prime is 2 a prime yes so keep it simple there uh then you end up with 12 can you divide that by two yes you can and you just continue doing this get until you basically have 2 times 2 times 2 times 3 is going to be equal to 12. all right it's very simple all right so again 36 can you divide by 3 i'm going to switch it up this time i'm going to do a different number so you have three and then you just continue dividing and dividing and dividing in with prime values until you don't have any more of a division to go about okay so then after you do all that what you want to do is just highlight those values that are in common so we have 2 2 2 and 3 3 3 2 and 2. well these are in common these are in common this is in common with that so what does that tell you well that tells you that the greatest common factor in this circumstance is going to be equal to 2 times 2 times 3 so greatest common factor between the values of 24 and 36 is going to be equal to 12. all right let's do a more complicated one like 54 and 72 again we can go and just list 54 and of course you're going to ask yourself can you divide by two yes you can and then you're just going to continue dividing by either twos or threes or fives that normally suffices until you're able to create 54 with just a bunch of prime numbers multiplied times each other do the same thing with 72 72 this is going to be 2 times to the power of 36 if i just want to write this out which is going to be equal to 2 and 2 and 18 like this which is going to be equal to 2 and 2 and 2 and 9 like that and then we have to go and recreate the 9 and what we're going to end up with is 2 times 2 times 2 times 3 times 3 and then we go and we find those values that are in common once again so we have 1 2 here we have a 2 here we have a 3 a 3 a 3 a 3 and that is going to give us our greatest common factor for those two values greatest common factor is going to be equal to 18 in that situation all right um no this is not a java design pattern video i made that video series forever ago i'm surprised it's still hot i mean yes i will be getting into um later on no further down the the tutorial i'll be covering things like um a lot more complicated topics like that yes logarithms and euler's number and all that stuff so this brings us to polynomials what are polynomials they are going to be equations in which you have two three or more different terms inside of them so how do you simplify down a polynomial well anytime you're working with exponents like this and you want to combine them all you need to do is take all of the like terms that we have let me use a different color here i usually do the same color every single time but what am i doing all right let's go there we are and you're just going to take all these different values and you're going to add them together and this rule right here tells us that if we have a variable with an exponent and we multiply it times another variable with an exponent with a like variable term we just add the exponent values so this becomes x to the 15th y to the 17th here is another role so these are the sort remember i said algebra is basically just memorizing some rules those are the rules you can probably fit all the algebra rules that exist on the front and the back of one piece of paper so just write all of them down and you got it okay so how are we going to simplify this well it just says if the exponents are the uh you have an exponent in the numerator and denominator with a common variable term well you can simplify them by subtracting those values so if you go and simplify them let's see if i can draw a straight line well this will help me with it there we go this is just simply if we go and we have w uh to the power of 10 w to the power of two well we're going to subtract this value this two here from this one up here and what does that leave us with well it leaves us with w to the power of eight and then we can do the same subtract the four from the x this is going to be x squared in the circumstance which we have a larger exponent on the bottom versus the top we can still subtract this from this guy right here this means we're going to end up having y to the fourth and then likewise seven to the power of two can i subtract just or delete just the end yes i can all right so those are two rules and how they can be used to uh yeah i'm sort of i'm sort of skipping over every possible way of solving problems if i can solve them well enough using just one concept so yeah that's um it's sort of a choice i had to make you know just to keep this this video which is already going to be very long a little bit shorter guess what we got another rule and here's a whole bunch more rules so we got some more rules here and another rule over here and with the power of these rules you're going to be able to solve further and further more complicated problems what does this mean it just means you have a negative exponent you're going to be able to either eliminate it altogether or like we did in this circumstance or you're going to change the exponent to a positive and you're going to flip the fraction so let's say that i want to come in here and using all of these awesome rules go here's like these are all roles also so there's another rule another rule another role another role okay there's not that many rules it just seems like there's a lot of rules because normally whenever you're learning algebra they hit you with all the rules all at one time and you're like oh there's so many rules but then whenever you actually list out all the roles it's not that many it just seems like they normally come in groups so how exactly are we going to go and simplify 4x over 7y to the third to the negative 2 power well we're going to flip the equation and then we are going to square all of the results inside of that and if we do so this is going to end up being converted into 49y to the sixth okay so this guy right here and this right here whenever you put those two exponents together rather than add them together you're going to multiply them all right so this is going to be to this and because we flipped them we flipped the equation this negative sign goes away so it's it's very simple in that regards and then you also square everything else and that is how you simplify that and solve that now you can also go and do the same with cubes and if you do this this is going to end up being equal to x to the power of 6 y to the 18th and straight line 9 x to the negative 2 y to the 10th which is going to then further simplify down to 9y to the 8th over x to the fourth thank you very much trump daddy i greatly appreciate it greatly appreciate it all right let's get into more stuff um radical expressions um what do we want to do here so whenever we're talking about radical expressions this is when things start getting more complicated well one thing you need to understand is the concept of the perfect square and that's just simply something multiplied times itself so it could be something as simple as 36 is equal to 6 times 6. all right x to the fourth is exactly the same as x squared squared um y to the eighth is going to be equal to y to the fourth squared okay very important concept because it's going to dramatically help you in regards to solving problems like this so if i wanted to rewrite this problem right here and this is also this is called a radical as it the arrow already displays this is your index and then your radicand is the value that is inside of there okay so just some more jargon um well thank you very much wow thank you very much trump daddy i i greatly appreciate that that is ridiculously awesome so thank you so much i mean wow thanks um i'm so happy that my design patterns videos have helped so many people that's one of the most i think it's probably the video i did that i a video series that i did that i think people got the most value out of so i greatly appreciate that here let's see i'm gonna move your thing down here all right thank you very much i i mean it i super appreciate it mainly mainly just because i just like like the idea that people are finding value in what i'm doing that's that's all okay so how can we go and solve this using or rewrite this using perfect squares quite simple let's just go and convert this into 36 times two times this is going to end up dividing or breaking up this x to the power of 5 which is going to be x squared oops times x squared times x times y to the fourth times y to the fourth all right and throw in our square root indicator inside of here and then we can go and further break this down to six squared times 2 times x squared x squared times x times y to the power of 4 squared all right see how they're all breaking down and then we can further then simplify anything that's squared like that like that like that can be pulled outside of the radical expression and you get 6 x squared y to the power of four and then these two guys that are left inside of here trapped well you just leave those inside of there exactly like that all right and that is how you can simplify this rather complex radical expression into its simplified form exactly like that um let's go and solve some more uh in this situation we are going to be working with our cubes so again we want to further break down everything into our cube roots let's go and use a different color here just to break this up okay so this is going to be equal to negative 2 and cubed oops oops let's go use so many threes and i can't draw a three there's a three and again for our next value this is going to be x squared cubed see we're just breaking down everything in that x now that you know is going to get trapped inside of here if we can't get rid of it y cubed and here's our additional awesome rules that we need to memorize then we'll have y squared and z squared can't we can't do anything with it and then we just take everything that we were able to cube all of these guys throw them outside to simplify this so you got negative 2 x squared y 2 yeah so this would end up being just simply y and then we're going to have all the stuff that was trapped so x and y squared z squared and like this and then don't forget to put the cube root symbol right underneath of it so awesome stuff okay now let's say that we want to now subtract a radical two radical expressions well in this circumstance are we going to find the cubes of all of these or are we going to find the perfect squares that's the sort of stuff you have to ask yourself well we know based off of what we're looking at that we're going to be to be able to find perfect squares and we also know based off of all these rules over here how we're going to be able to operate with these guys so let's just break it down into perfect squares so this is going to end up being 4 times 4 times 7 x y and like this so we'll draw that in there it looks nice and clean minus and 4 times 7 x y and again there they are i should probably have made a little bit more space here for myself so that i didn't have to write on top of myself let's go move this down here a little bit more and bring this over here this is going to further simplify then to 2 times 2 square root of a 7 x y the two terms we can't work with minus and two and seven x y which is going to further come down to four and square root of uh four yeah seven x y minus two square root of seven x y exactly like that and we are going to because we have like expressions here this is actually going to further be able to simplify down to we could just subtract the 2 from it and get 2 square root of 7 x y so it ended up working out really nicely for us and how can we go and rewrite this radical expression well we can just go and put these cube root of square root let's bring this right here so this guy is actually the same as saying what i'm going to draw right here so this whole entire term squared so anytime you see this you know you can put it into this specific format and then we just need to go through and break it down into cubes so well this this is going to be 5 to the power of 3 times 2 and like this and just keep it very simple and this is going to be squared and this is also of course going to be let's just take the 5 out of there cube root like this and the 2 is left inside of here like this and this is ultimately going to be equal to 25 and cube root of 4. okay and that is because this guy right here is going to go and be multiplied times these different terms to get that final answer okay that's perfectly fine all right uh let's move on to some additional ones what else do we have here all right so how we going to simplify this well it's just simply going to be three times three times three times three which of course means that this is going to be equal to three all right so i broke those down into four individual equal parts and that gives us our final answer of three uh let's go and do the same thing likewise for this guy this is going to end up being equal to negative eight um for this it's going to end up being equal to 10 because of course 10 times 10 times 10 is going to be equal to 1 000 um book smart um then if we go and try to work this guy out we're going to end up getting a similar type of format where we're just going to try to find everything that is a cube put it inside of here and this will be x squared like that and then this is going to end up we can take this out because those two cancel each other out and we'll end up with x squared and the cube root and inside of here everything that's left behind which is just the x squared more complicated concept here let's go and try to solve this guy so this is going to again basically be solved in exactly the same way we're going to break it down into parts so that we can cancel things out so this is going to be x to the fourth and x to the third and then times this is going to be able to simplify down to y to the fifth and four and then of course this is all based off of these other different rules that we have inside of here and this is going to be z squared to the power of 4 times z to the power of 3 and then this is going to once again just be everything that is inside that we have been able to take to the power of four is going to be thrown out of this guy and then we are going to leave everything that is left inside of it and i think you kind of got the point i don't think i need to go and write any more of this stuff out do i i think i don't need to solve that okay so let's move on to another radical expression problem here radical reason why i'm spending so much time with radical expressions is this was one of the concepts that a lot of people really struggle with so if you're wondering why that's a reason why again we're using these rules back here you might want to write them down to solve some of these different problems but basically if we have x to the 15th this is going to be equal to the same as saying um whoops what am i doing here 3 or 15 divided by 3 which is going to give us a value of five and you're going to do that for the all the other different values as well let's move this out of here so that i have a little bit more room let's move it down here i think that might be enough and let's do the same y to the fifth is going to be and let's go three divided by or five divided by that and that's going to be one which is going to give us a remainder of two and z to the sixth is going to be six divided by three which of course is two and what we wanna do here is separate the factors raised to the power of three on the left so let's put this in a different color so this is going to be x to the fifth like this so i'm converting all of these different values this is going to be y to the 1 three times y squared times z squared to the power of three like that and then this is ultimately going to give us the we just take out everything that's different so then i guess i should put in this cube root part right here so there's cube root and if we take out everything that is a cube this gives us a final answer here of x to the power of 5 y z squared and the only value that's going to be left inside of here is our square like that and there we are so that is how we're able to further simplify those down let's do another one in this situation we're going to be working with perfect squares i'm sort of repeating myself uh just to give you more input found it helpful to express radicals as fractional exponents most definitely um i think i cover that also um yeah i thought i covered this also later on as we continue on here um let's let's see i think i kind of remember every single thing i put inside here this is a really big video so in this situation what i want to do is to solve this problem is what i'm working on is to find products of perfect squares again just like we did in the previous example that we have here so we're going to have two squared to the five there are two times five which is going to be a product with a perfect square again x 9 is going to again you're going to divide 2 into 9 and get a get a value of 4 with a remainder of 1 and then y to the third we're going to divide two by three which is going to give us a remainder of one so let's just go and figure it out from this point all right so this is kind of getting a little bit busy let's go and delete all this stuff so that i have a little bit more room to work with everything because i'm sort of scrunching everything together and that is not a good thing all right so let me write out what i was thinking okay so we're going to y to the third and this is going to be 2 divided by 3 and this is going to of course give us a remainder of 1 also and again we're just finding perfect squares here and i'm just writing all that out just so you can see the thought process that i'm taking so this is going to be 2 squared times 5 times x to the power of 4 squared times and we have a leftover x inside of here and this is going to be y squared and y to the power of 1. and then this ultimately if we take out all of our different squared values is going to leave us with 2 the 2 comes out of course and this is coming out and all the other different ones so this gives us 2 x to the power of 4 like that and y the y also is going to end up coming out this y right here because it is also squared and that just leaves us with the values that cannot be put into a squared form which gives us 5x to the y all right all right um another concept that's important to understand is that you can only add and subtract whenever the indices and the radicands are equal so of course this guy right here is just simply going to be simplified down to 7 to the or square root of x and let's go and get rid of this and it will solve our final little problem here um let's grab it and throw it over here can you see whenever i do that oh yeah you can so there we are and what i want to do here is i want to use exponents to eliminate the radical so i'm just going to simplify this down just by converting it into x plus 16 and i'm going to take the square or i'm going to square the outside value okay so if i sometimes it works better to get rid of this we just square the 5. so it becomes 25 and it makes our problems considerably easier and we find that x is going to be equal to 9. all right what else do we got let's continue another concept that's important is the concept of completing this the square now an expression is going to be a perfect square if it is equal to two expressions multiplied by themselves so we have x minus 4 times x minus 4. well what are we going to do this also brings in the concept of foil and what foil says it's just a one of those mnemonics what we're going to do is we're going to get the first the outer the inner and the last so how we're going to solve this is we're going to take the first times the first so x times x is going to be equal to x squared and then we're going to take the outer negative 4 times x is going to be equal to negative 4x negative 4 times x is negative 4x and then the outer which is going to end up being equal to plus 16 which we then can just simplify to x squared minus 8 x plus 16. all right so let's go and do that with all the other ones so we got two this is two y squared this is just the basics of foil negative two y and uh did i do that in right or no i didn't i didn't do it in the right order i just talked about foil and then i messed it up okay so this is going to be 3 y minus 2 y minus 3. like this and then of course this is going to end up being equal to two y squared um plus y minus three all right and let's do one more it doesn't really matter how big these expressions are in this circumstance we're going to take the first and multiply it towards all these other different values so what this is going to do for you is you're going to get 2x squared plus 10 x y minus 2 x minus x y minus 5 y squared plus y and if we further simplify that down we're going to get 2 x squared plus 9 x y just by combining the like terms minus 5 y to the squared minus 2 x plus y all right and there we go now we're getting into factoring trinomials even more complicated what happened in my box here let's move it up here that didn't work oh my box got moved here let me move it back there we are that looks more professional okay so um factoring trinomials you guys should have told me it just said hey your box is missing okay so how do we factor trinomials boy i really combine these a lot you can see here an example of perfect cubes and also a re repeat of the exponent rule in regards to how they can be combined so in regards to factoring the trinomial this guy right here what i want to do is i want to multiply um to what i want to be able to do is i want to be able to multiply values for our center values that are going to be equal to 12 and then that are also going to be able to be added together to give us a value of 8. and that is quite simple also so we got this this is basically here just to give us our x squared plus and what values are we going to be able to multiply toward together to get a value of 12 that are then going to be able to be added together to give us a value of 8. well 2 and 6 work so we're just going to turn this into 2 we're going to turn this into an 8 and that is how we can factor our trinomials now for this 3x squared plus 9x minus 3 you're going to do exactly the same thing for these two values that you're trying to find every single time you want to ask yourself what do i multiply times to get our final value which was this value and then to get this guy right here you say what two values also can i add together that are going to be equal to eight all right you just say that over and over again well in this situation it's actually going to need you're going to need to divide by 3 to simplify this so we'll just convert this into 3 x squared plus 3 x minus 10 of course and then we can go and factor this trinomial on the right side so this is going to end up being equal to 3 times x plus 5 and minus 2 because if we multiply 5 times negative 2 we get negative 10 and if we subtract negative two from five we get three all right hopefully that is useful um i'm gonna do more of these trinomials because they seem to confuse people a lot now for this one let's go and delete this so that i have more room to actually be able to write so things don't get all pushed up next to each other okay so in this situation we can't do that because we don't have a greatest common factor so what are we going to do in those situations well i'm going to um sorry i deleted it do you want me to go back and bring up what i had here um this is going to be equal to yeah oh you know what you are indeed yes this should have been sex i apologize for that little error this is going to be x plus two um plus six yes you are indeed correct thank you for for catching that maybe i shouldn't be deleting these maybe i should just be moving them around on the screen here a little bit you are indeed correct i made that error okay so let's go and figure this one out so we don't have see before we could take the 3 out we can't do that with 3 5 and 12. so how on earth are we going to be able to actually solve this problem it's kind of complicated but what we can do is we can go and convert this value that we have right here into let's go and convert this into oops why did i use so many threes i don't know why but three is like the only number that i have trouble working with we can go and get this x out or the the 3x out of this so this is going to end up being equal to 3 x squared plus 9 x minus 4 x minus 12. see we can break them apart into individual pieces that we can get a greatest common factor for and then we'll be able to just go and get the three x out of this and that leaves us with three our x plus three like that and minus four x plus three and then this is going to further simplify down we take this part and this part throw those between parentheses and get 3 x minus 4 and x plus three all right so multiple different ways if you if you have a greatest common factor and ways of doing it if you do not have a greatest common factor um yeah i really oh i'm just reading an old comment got confused there for a second okay so how do we go and solve the next one this guy right here well in this situation we can use our exponent rule to make things a lot more simple so we can convert this into 3 to the power of 3 x to the third and this will be plus 2 um squared y to the third also like that and this is going to with the exponent rule give us 3 three x to the power of three plus two y to the power of three and then we can go and work these out to then give us 3 x plus 2 y and multiply all that out to give us 9 x squared minus 6 x y plus 4 y squared and there we go another solution is let's just go and move these around a little bit so that i can actually work with these a little bit better so i'm gonna go get this guy and grab it and move it up a little bit so i have more room to write with my final problem here so move this guy up here and grab this okay so for this problem that we're trying to solve what we want to do is create two binomials with a greatest common factor like we did previously and i'm just going to work this out and this is going to give us um 2. again we're going and getting out and trying to simplify this so we're going to take out the common amounts which would be 2x squared and that's going to leave us with x minus 3 x like that minus and take out the common value again which is going to be 5 x minus 3 in this situation and this is going to simplify down further is this can you even see this at this point let's move it over here which is going to be equal to 2 x squared minus five and x minus three x all right so i think that's enough of those why don't we now talk about synthetic division and how that can make our lives considerably better all right i'm back oh it was exhausting it's like taking the world's biggest algebra test and my ipad keeps shutting itself off it's like enough already oh okay so what we're doing here is we're going to be dividing by some polynomials i'm going to show you how to do it the one way and i'm going to show you how to do it using synthetic division with synthetic division you can divide by a linear binomial [Music] let me phrase that differently you can use synthetic division if you are dividing by a linear binomial but i'm going to show you how to do this a long way in a short way so that you know both ways okay what's important here is we're just going to convert this make sure i'm using a good color okay so this is going to be x squared but we do not have an x value so in that situation you just put 0x okay so that solves that plus four okay so we got this right here then we're going to have to do likewise with all of the missing terms for this so we have x to the fourth but we don't have x to the third x is squared and so on so we're gonna have to work those in as well so this is going to be x to the fourth plus zero if you don't have a term you just throw it in there just checking to make sure i still am um you consider my hearing my voice x squared and then plus 7 x minus 2. okay let's go and let's move this out of here all together because this can get kind of complicated so i want to make sure that it's understandable okay so let's just draw that in there hopefully it makes the line straight and it did all right so we got this how are we going to be able to work this out well we just work with the very first term that we have so what do we need to multiply times x squared to give us 2x to the power of 4 well we multiply it times 2 x squared and then we go and multiply this value times all of these individual values to get all our other values so this ends up being equal to two x to the fourth minus zero x to the third minus eight x squared and then we draw a line underneath of it and we work this out and this is ultimately these cancel out the first two terms are going to cancel each other out and this gives us 8 x squared plus and we bring all our other terms down because we have three over here so we have to have three over on the other side so two or seven x and two and what do we need to multiply times x squared to give us a 8x squared well it's going to be a positive 8x squared so we're going to multiply that times negative 8 and this gives us 8 x squared and then if we continue to work on this is going to give us 0x and 4 times the negative 8 is going to end up being equal to positive 32. so based off of all of this we then finally get a remainder of 7x plus 30. and how we would work this out is this is going to end up being equal to 2 x squared minus eight plus and you'll get the seven x squared plus three or seven x plus three divided by x squared plus four okay so that is how we're able to or basically work through this division we have here but like i said you're going to also be able to use synthetic division to solve these problems in a more simplistic form so what we need to do to do that when we have that situation in this other problem we have here i think it's okay to erase this because i think you get the the concept we're just basically what values do we need to multiply to get the said values and then we are just continuing to add and subtract said values what we're going to do is we need to list the coefficients of the dividend and the opposite of our divisor so we're going to convert this into two x plus um well we'll just go let's just go and get rid of this all right whoops i didn't mean to do that okay so what we're going to do is we're going to take this 4 that we have right here we're going to get the opposite of that so it's a negative 4 so we're going to use a positive 4 and we're going to put this inside of a little box here just the way that it's all set up then we're going to take our different coefficients so we've got our 2 r6 and our um negative 8 we're going to then draw a line right like this there it is we're going to take this first value which is 2 we're going to drop that through and then what we need to do is multiply our 4 times r2 which is going to give us a value of 8. if we go and add those through we get a value of 14. once again we're going to take the 4 times that 14 we're going to get a 56 and if we subtract 8 from 56 we get a value of 48 and what this means is if we go and perform this division using synthetic division that this is ultimately going to give us a value of 2x plus 14 plus and we have our remainder which is the last guy we have here we throw x minus 4 right here and the 48 on top and that is synthetic division and how it works quadratic formula here is our quadratic formula that we have here now you're going to be able to solve equations of did i write that down i did not write that down basically you're going to be able to solve equations of the form a x squared plus b x plus c equal to zero using this wonderful formula called the quadratic formula so let's go and just plug all the values into the formula and figure out how it works so this is going to end up being equal to um x1 you're going to have two values here because we're using plus and minus so this is going to be equal to ultimately um do you need me to i'll work it out why not so let's go let's delete this and then let's rewrite this okay so this is going to be negative 19 plus or minus 19 squared and minus 4 times 6 times 15. and let's get the square root of that value and put a line underneath of it then we go and transpose all of those different values into the a b and c spots this means that we're going to get one value x1 is going to be negative 19 plus 1 divided by 12 in that circle or yeah that's going to work and then this is ultimately going to end up being equal to negative 3 over 2 and then we'll have a second version again because of this plus a negative that we have here and that's going to be negative 19 just like we did before minus 1 divided by and again 12 which is going to simplify down to 5 over 3. so fairly simple way of solving those problems as well and this brings us to rational expressions they're called rational expressions because they have a fraction in them so what do we want to do to be able to solve these guys well what we're going to do is we're going to find the least common multiplier of 16 y z because remember we're dealing with fractions so the denominator has to be equal so it's going to be z and 12 x so we're just going to worry about the numbers themselves so we talked about finding the least common multiplier before what are we going to do for 16 that's going to be 2 times 2 times 2 times 2 and for 12 that is going to be equal to 2 times 2 times 3 and we go and combine these terms to find our least common multiplier so that that multiplier is going to be these values and these values all multiplied together so this is going to be 2 times 2 times 2 times 2 times 3 and that's going to be equal to 48 all right and now that we have that we know that our divisor is going to be equal to 48 x y and z can't do anything with those variables other than that and now what we need to do is to get our denominator the same for each one of them we need to figure out what we need to multiply times the numerator to make the denominators equal so let's use a different color we're going to take x squared and 16 y and we're going to have to multiply that times three x z over three x z to give us a final value of three x to the third z whoops there we are and again our common denominator which is going to be 48 x y and z this is starting to get a little bit more complicated isn't it okay so now we need to do the same thing for y over z so y over z is going to to get 48 x y we're actually going to have to or x y z i mean we're going to have to multiply the numerator and the denominator by 48 x y 48 x y and this is going to give us a final value of 48 x y squared [Music] over 48 x and y and z again from over here and we're going to do exactly the same thing for r r5 z [Music] cubed and over to 12 x what do we need to multiply times that to give us 48 x y on the bottom well we need to multiply it times four y z and four y z [Music] and that is going to give us 48 x y z on the top and then we'll get 20 y z to the fourth and now that we have all of those common values across all of them then we can basically just copy and paste these so let's just grab this one and drop it over there and this one and drop it over here so we'll throw that there this is going to be minus and then oh i'll just write it in why not plus 20 y z to the fourth all with the common denominator of 48 x y and z and there you go that is how you simplify that rational expression um and i need to do another one well this is kind of something that's sort of interesting so i'm gonna cover this this problem because you solve it in a different way slightly different way all right so for this what we're going to do again we need to find the least common multiplier for our denominators once again so how are we going to do this well i'm going to take 6 x squared and minus 54. and because i talked about previously about how we can go and break down trinomials into individual pieces this is actually going to convert into x minus three x plus three okay everything's building upon itself and likewise x squared plus x plus oops minus 6 minus 6 is going to convert into x plus 3 x plus or x minus 2 minus 2. so minus 2 exactly like that and what we're going to be able to do now is to combine both of these and create a common uh denominator let's make this in orange instead so we'll get six x minus three x plus three x minus two and that is going to be our denominator in that situation and then what we need to do is multiply the numerators of each of the individual parts to get our final denominator so it gets a little tiny bit complicated x to the fourth and what do we we're going to have 6 x squared minus 54 times to get that value we need to multiply times x squared over x squared to give us a value of x squared minus six x plus eight and i'm not going to write in that that the denominator uh yeah the denominator because it's so extremely big and in the circumstance in which we have 2x minus three i am going to again this will be multiplied x squared plus x minus six and times this will be negative six x and three like this and this is going to give us a final value of negative six two x squared minus nine x plus nine and if we go and add and add these values and that value together we get a final answer of negative 11 x squared plus 48 x minus 46 all over exactly like that this guy that we went and created right up here that i'm going to steal and put down there instead of writing all that out all right um i i say sometimes i say squared when it's supposed to be minus okay sorry about that sorry though hopefully that's not that's not confusing this is just a lot of information to be covering in one video um i'm doing my best okay now we're going to be multiplying so what can we do here well we can go and further simplify these just by going and uh getting the the cubes and multiplying 2 times 2 times 2 and so forth and so on so we're going to get convert this into 8 x to the sixth just to simplify and eliminate those exponents which can become quite confusing and this would then end up because this is negative we need to flip this fraction so this is going to give us z to the fourth and 16 y to the sixth and then we just multiply through so this is going to be eight x squared z to the fourth and y to the third uh 16 y six and then that is going to be able to be simplified further down from that point onwards uh if we wanted to factor out the eight um in regards to four x minus three this is going to be able that is a really complicated one let's go and just move this up here a little bit so that i have more room to write with this um this is going to be able to be solved well this is actually supposed to be four x minus three x minus 8 is equal to 4 x minus 3. i actually copied the problem down wrong so how do we solve a problem like this well what we're going to do is we're going to cross multiply these two individual fractions so this is going to end up being equal to i guess i should draw like an arrow instead because otherwise it'll cause confusion if we cross multiply these values we're going to end up with 4x squared minus 19 x plus 12 is going to be equal to because the rule of cross multiplication is those values would be equal is 4 x squared plus 29 x minus 24 and what we can do here is subtract out our 12 which is going to be 4x squared minus 19 x which is equal to 4 x squared plus 29 x minus 12 and then it's going to be further simplified down to 48 x is equal to negative 36 and if we divide those values and simplify we get a value of for x of being equal to three over four all right wow yes this is all going to be uploaded onto youtube tomorrow i'll edit out the pauses and stuff like that now we're talking about functions we still got logarithms and ranges and all this other stuff after this okay it's just simply an eq a function is just simply an equation that for each one input it's going to provide exactly one output now some of the ideas that people get a little bit confused about is domain codomain and range i think it's better to okay so let's say we have x and we have our function which is just simply x squared this is how range and domain and all those other different things play into it now your domain is just going to be the set of values that a function can accept so it could accept one two three four it can accept any of these values the codomain is going to be what potentially could come out of it so one two three and four and five so what exactly is the range well the range is the set of actual outputs so the range the difference between the codomain and the range is the range or actual values that could come out all right whenever you are squaring the values of x so this is going to be the domain and this is the co domain and then the range are the actual values that confuses some people some time and hopefully that is not confusing anymore oh another drink now remember i said that a function is an equation that for every input there is exactly one output well that is exactly why this guy right here is not a function we have a thing called the vertical line test and what that means is if you can draw a line and touch to a vertical line and you can touch more than one part in your plot for the function that means that it is not a function so this function that we have right here indeed does not pass the vertical line test and hence is not a function so how exactly do we find the range well this is a function we have right here and what we're going to do um is we are just simply going to simplify this equation so let's say i don't know if you guys like the blue color more or the orange color i'm going to stick with the blue so what we have here is i can say x squared is equal to y plus one and that means that x is going to be equal to the square root of y plus one and that means that y is going to be equal to and here is our notation that we learned earlier one two infinity all right and remember if it's undefined you have to use a parenthesis instead of the square bracket and you can't because basically you can't include anything that isn't a real number all right so let's move on to some function problems so is this guy right here a valid function well let's take a look at it we said for every one input there has to be exactly one output well in here we have negative two as an input and it gives one output here we have 4 as an input it gives 2. here we get 4 and we get 3. uh oh that's a problem for the same input we are not getting the same outputs as we can see right there so in that situation you would know that no indeed this does not qualify as a function because you're getting two outputs from that one input let's move on to the next one so if we have a value f x this is just a symbol symbol for a function so we have this guy right here and we have this other function g what how exactly would we model out fg to the 4th well we would just convert this guy this would end up being x squared plus 2 like that times 2 x plus 4 exactly like that and then we could just fill in those values to get our simplified answer which is going to end up being equal to 16 plus 2 times 8 plus 4 and if we work all of those out that's going to give us 18 times 12 and that gives us a final answer of 216. all right what else do we got here getting a little messy well here we have our uh other formula let's say yeah this is best going to be solved with this quadratic equation that we have here and that's the reason why we have our handy-dandy quadratic equation so this is going to be converted down into x squared plus x minus six is equal to zero we'll use our quadratic so we'll say this is x is going to be equal to negative one plus or minus at oh this will be whoops one squared and minus four times one times negative six we're going to go and get the square root of that value this is going to end up being equal to 2 times 1 which is 2 obviously which is going to then give us a value of negative 1 plus or minus and 1 plus 24 over 2 and of course we have two values of x in this circumstance which are going to be negative 1 plus 5 divided by 2 which is going to give us a value of it's going to be 4 divided by 2 which is going to give us a value of 2 and then our second value where we're using a negative value is going to end up being equal to negative one minus five divided by two which is going to be equal to negative six divided by two which is going to give us a value of negative three all right so that's how we solve those ones let's move on okay so if you ever see this symbol here this guy right here what this means is it's read as f composed with g of x and it is let's see basically going to be equal to f g of x exactly like that all right so how exactly do we solve problems like this well this is going to be converted into 2 x plus 6 squared plus four exactly like that and then if we go and work these out this is going to be four x squared plus 12 x plus 12 x plus 36 plus four and of course we can simplify this down to four x squared plus 24 x plus 36 and 40 plus 40. all right another one so yes now we got inverse functions now two functions are going to be inverses um in those situations in which you compose one function ugh let me swallow what i got in my mouth okay if two functions are inverses then composing one function in another is going to return a value of x so we could see here in this situation if that actually is going to work out so let's say we have f let's give it a value of four well we can just go and multiply three times four and plus two this is going to give us a value of 14. now if this is an inverse it's going to give us a value of four so will it let's throw this here let's throw 14 inside of here minus 2 and divide that by 3 and indeed it does give us a 4. let's go and graph some functions okay and i'm just going to do this real quick because it's let's go and actually just work these out because i just want you to understand the basics of what happens whenever you add constants to a function or whenever you multiply a value times a function or and all of the the different changes do you think you'll bridge a series with a category of functional programming i have no idea what i'll ever do with uh video wise i i i just i didn't expect to do this video but i wanted to teach machine learning and to do that i had to teach calculus and linear algebra and to understand calculus on linear algebra you need to understand all of the the basic algebra so it's sort i was like i might as well make this and i might as well make it in one video and then i can just refer to everything i never know what i'm gonna make videos on i know this year i want to make a ton of machine learning videos and the only way i could possibly do it and refer to other previous videos was to do all the math videos and uh after i get through the math videos which actually aren't going to take me that long i'll probably in one month i'll be able to get through all the math videos so it'll be pretty easy okay so what you see here on the what's next week's topic i i might do trig in one video i haven't really decided next week's going to be kind of hard because i have a bunch of medical tests i have to go through you know anything's possible okay so what i have here the graph you see on this graph here is this function right here so what would happen if we would just simply add 1 to that function well adding a constant to a function is just going to shift it up or down so if we add 1 to it it's just going to shift it up by one value so let's go and let's create i'm just going to roughly draw these in it's basically going to create maybe i want to use a different color or something i would just use red okay so what this is basically going to do is for the second one is it's just going to shift this up in like this okay so that's what adding a individual uh constant to a function is going to do okay it's kind of a sloppy drawing bear with me now what would happen if we would multiply in this circumstance a negative value towards our function well multiplying a function by a negative is going to just simply create a reflection of that function so it's going to roughly be let's say like down inside of here and i don't know let's just roughly draw it in i'm going to roughly draw a reflection of what we have here so it's going to sort of come down here like this and it's going to come down like this and do that okay so in this circumstance whenever we are adding a constant we're just going to either be moving it up or down depending upon if the constant's positive or negative and if we're multiplying it times a negative we're going to create a reflection of the value that we have all right so let's get rid of all of that okay so now we'll move on to the next one so what would happen if we would um multiply it times a fraction well in this circumstance it's going to create a function that is going to be half the distance from our x-axis so that is going to put it maybe i should use a different color let's go and use this guy here instead so it's going to just make it go closer to the x-axis so let's go and um i don't know just go underneath of it so it's going to go something like like this and like like this if we would instead go and multiply times a positive value it's going to be twice as far away from the the x axis so it'll be more like this okay so that's just gonna a rough sort of overview of exactly how that works and then if you would use absolute values what you're going to do is convert all the y values positive all right so just a brief overview of exactly what happens with a function whenever you multiply add to it divide and so forth and so on now we move on to whether a function is going to be considered even or odd um start making videos like making your own um i've made i started actually making electronics videos but people didn't seem very interested in them i actually was going to make a a computer just based off of chips i started off making a basic calculator i recreated the game of pong just using transistors and stuff like that and people didn't seem interested in it so i just stopped doing that tutorial um there's so many things on my my youtube channel that most people don't even know exist like i said i have a lot of people know i have crochet videos because i just re-uploaded a a amalgamation of all my crochet videos but i have weird videos on like everything i have a whole entire video series where i cr recreate junk food using healthy alternative food i mean i have like all the really weird stuff yeah yeah basic electronics are kind of complicated okay so what i want to do here is i want to figure out if if um if a function is going to be considered either even odd or neither so if you substitute a minus x for your function and get the same function then in that situation the function is going to be easier yes i don't even know what healthy food is though i'm a particularly healthy eater and i'm sick right right now with i have like this thing called long covet and i said it mainly affects people that are healthy so keep eating junk food if it makes you happy okay so what that means what i just said is uh f x is going to be equal to f negative x in that situation it will be considered even all right so in this situation you can see um basically what that's saying is the functions are just going to be symmetric upon the y axis so we have here if we take f to the negative x this is going to give us negative x squared plus 2 which is going to be negative x oops squared plus 2 and you can see in this situation that x squared plus 2 is indeed equal to x squared plus 2. and in this situation a negative times a negative is always going to be positive all right so there you have an example of a function that is even let's see do i have another one yes i do all right so in this circumstance we're going to do the same type of thing odd functions are going to be symmetric about our origin and to be odd the signs need to be the opposite of whatever the original function was so in this circumstance we have f x or x to the third minus two x if we go and throw a negative value inside of this we're going to get negative x to the third minus 2 and negative x which is going to be equal to whoops negative 3 or x to the power of three plus two x so in this situation we can see that x to the power of three minus two x and this is going to be x to the third plus two x so in this situation we know that this function is going to be odd all right now let's move on let's say do i have another one yes this is going to be a situation in which we have neither and the way this is going to work out is if we go and get and make x negative in this situation we are going to get equal to negative x to the third plus and negative x squared minus two negative x plus two which is going to be equal to x to the third aren't a negative x to the third sorry about that uh plus um x squared plus two x plus two so in this situation we can see let's not rewrite all those out we can see that indeed that change this to an orange color orange color okay so we can see in this situation that this guy right here is not going to be equal to this so it's not going to be even and also we can see that negative x to the third minus x squared plus two x minus two is also not equal to negative x to the third plus x squared uh plus two x minus two okay so in those situations they are not even or odd so they are considered neither all right and this brings us into another concept we're actually very close to covering almost everything i gotta take another break because i need more water sorry for the breaks but otherwise my voice dies and the stream ends early so up next is asymptotes i'll be back in a minute all right i'm back oh i got some tea i tried to find a lollipop or something to to my kids are like cookie or candy monsters any candy in the house just they eat it all boy there's a long stream okay so let's continue now we're getting into the part that not asymptotes but we're right after this is what actually first made me really get fascinated with with mathematics which are logarithms and euler's number and such okay an asymptote is just a uh a line a curve approaches but it it goes off into a into an infinity so here is an example in which we have one what we're going to be able to do is we can find the values of x where there is no value by finding the x value that makes the denominator equal to zero so we got this x plus four so what we're gonna do here is we're going to get to the x squared plus two x minus uh eight and we are going to give it a value of zero and figure it out so very simple we'll just go x plus four we've already worked out how to solve these types of problems earlier in the video is going to be equal to zero and we find out that these are x is going to be negative four and or well we should say or x is going to be equal to two so our vertical um asymptotes are gonna let's just write a is going to be at x equal to negative four oh actually i should write and and two all right so there you go asymptotes and that is how quickly and easily it is to find them all right now we come into logarithms and euler's number and a lot of interesting stuff you can do really cool complica uh complicated calculations very very easily now we have our exponents and what the exponent does it just tells us how many times to multiply a number times itself so if we have for example 2 to the third of course this is 2 times 2 times 2 which is going to be equal to 8. now what makes a logarithm interesting is a logarithm is going to tell us what the exponent is so if would be 2 with a big question mark equal to eight this is going to become our log like this for eight and it is going to give us a number which is going to be equal to 3. all right that just tells us what the exponent is now that might seem rather boring but it gets much more interesting now if we wanted to find n if we have this function right here how would we do it well we would just have n is going to be equal to 2 to the power of 3 and which of course is 2 times 2 times 2 which is also going to be equal to 8. now let's go and solve this this next problem it works in exactly the same way this is just going to be equal to 2. now in our next situation let's go and get 25 and convert this to n this is going to be equal to 5. that is exactly the same as if we would take n and run it like this and equal to 5 2 square root of 25 which is going to be equal to 5 which is the same as log 25 5 and this is going to be equal to one half all right now a base 10 logarithm is that what i have here base 10 is how many times do we use 10 to get to a specific value there's different types of logarithms now in this situation how many times we have to multiply 10 times itself to give us a value of 100 well of course this is going to be equal to 2. a natural logarithm to differentiate itself is going to be how many times we use euler's number which is this guy right here in a multiplication to get a specific result and let's go and let's work this out so this function we have here is going to be equivalent to three x equal to one over nine and three x we're going to use the fraction as a negative exponent in this situation is going to be equal to nine and negative one and if we then go and get three to one like this x this is going to be equal to three squared negative one and this is going to then be equal to three x equal to three to the power of negative two and in this situation we know that x is going to be equal to negative two and i'm going to provide a link to all of my notes so you can download all this stuff for free after the video i'll put it in the description for this video on twitch and then on youtube i'll put it in the description on youtube so you'll be able to download all this stuff and i won't charge you anything i promise okay so then we continue to solve and it's just basically the same exact format again whoops this becomes three of course and then this is 27 so we're basically just juxtaposing around the different values that is all that's going on here and again this is going to end up being x 2 4 over 3 is equal to 16 and this gets to be a little bit more complicated let's go x 4 over 3 like this 3 over 4 is going to be equal to and just run this like this like that 16 and we'll take these to the power of three and this is then going to give us a value for x of four and twenty four two over four and this is going to be to the third and we can bring that two out of there and this becomes two to the third is equal to eight all right natural logarithm now let's move on um logarithm or is also that we got some little rules here we're going to be able to let's just work through this i don't think i need to cover that again basically what we're going to do is just work through these problems and figure them out so this guy right here is going to be exactly the same as this natural logarithm 3 x squared y and let's go and find the difference well that's just going to be this guy right here in this format and let's move on to our next one this again is just going to be this natural logarithm and it'll be x squared y over three and that's about all i got for that all right let's move on to some more interesting stuff because it's basically when you're working with logarithms you're just juxtaposing different values around depending upon those now let's get into euler's number can you see these values i think you can so basically what's awesome about euler's number is it is the rate of growth for all continually growing processes and we're going to be able to do really cool things in regards to organic matter and life and investing and things like that with euler's number um let's go and what we want to do here figure out exactly what we got with my different examples that we have here okay so what this is is e x is going to represent the amount we'll have after starting at one so let's say for example like an investment you make an investment of one dollar how much would you have at the end if it continually grows over x units of town uh x units of time so what we're going to do is just show you an example so that everything makes a lot more sense so let's draw like a little graph here so i'll draw this line here and this line right here and this is going to be on day one and this will be on one so what we want to do is let's assume you found an investment that is going to double your money in one year how is that going to work out so this is day zero the very beginning and then this is going to be at the end of that period well so you're going to start off with an amount which is going to be your investment amount and at the end of the one year remember it's going to double your investment you're going to have your original investment amount plus you're going to have your return on your investment so good stuff um the thing you have to understand however is that your return isn't going to happen all at one time it is going to increase gradually to eventually double your money after six months for example you're gonna have fifty 50 more money than and then at the end of the year you'll have 100 but a thing that's important to understand is your interest is also going to be accumulating additional interest so let's go and change our form here just slightly i'm going to delete this part and i'm just going to delete this whole thing so let's get rid of that and let's start looking at how your investments are going to grow over the halfway mark as well as the final year now at the end or at the six month mark you're going to still have your original investment but you're also going to have a 50 return on that investment because remember it's growing by half to get to the final where it doubles your money so how is it going to work that at the very end of this period of time you're going to have your original investment but you're also going to be able to double your money what would happen you may say if i was if you as an investor were able to constantly double your money well euler's number is going to be able to help you figure that out so let's do another thing um let's use real money because i think that that kind of makes things more understandable so we're going to start off here we're still going to be using the starting point let's say you invest one dollar okay so you have one dollar you invest at this point and then at the end of the year you're gonna have two dollars okay so we know we're going to head there well what are we going to have at the middle point well you're gonna have your original one dollar investment and then you're going to have a 50 cents of additional investments so what we're going to do is really sort of wrap our mind around the whole concept of because we know that interest is going to provide us a return along with the different investments how can we go or can we actually generate an infinite amount of money if we collect interest at different targeted time periods or even better yet what would happen if we were to invest money and we constantly were able to get a return on our money well there's actually a formula for this and the formulated formula for calculating the total over a given number of time periods is going to be this formula so 1 plus 1 over n to the nth where n is the number of time periods and 1 over n is a 100 growth over a certain number of periods now euler's number which i'm going to write it in here again because it was on the previous slide is roughly this it represents the maximum possible result if you were to compound a hundred percent growth over any given period of time and every rate of growth can be written using this basic universal constant which is why it is so fascinating so let me show you some different examples so what if we grow by 50 how exactly would we model that well or how exactly would we create that in an equation well we could go one plus with the formula that i showed you before and you're just going to put 50 that is your 50 percent growth n over n which is going to be equal to 1.648 which just so happens to also be equal to euler's number if the exponent is one half alright now what is the total with a 200 percent growth well to figure that out all you have to do is go and get euler's number to the power of two which is going to be equal to seven three eight nine okay so if you had a two hundred percent growth rate over a certain period and you invested one dollar you would know that that actually is going to grow to seven dollars and 38 cents roughly now we can also use different times like let's say we want to say if we earn a hundred percent in a year how much would we have after six months well that actually works out to being equal to euler's number so what'd we say 100 a year is what at six months okay so if we had 100 at six months that's going to be euler's number and this is just going to be rate times time exactly like that so that's going to work out to being euler's number to 1 times 1 half which is going to be equal to 1.648 so that's going to allow you to do that which is even more interesting let's say we wanted to find out what would be the total at an 8 return after ten years we can do that we can just say euler's number again times well what is our return it's going to be point eight percent times our period of time which is going to be 10 years and if you work out that calculation you're going to get 2.225 so if you got an 8 return on your money for a 10-year period it would be converted into 2. and 23 cents roughly um just keep using basic numbers here just to keep it simple how do i get rid of the n in the first example um if we grow well basically i'm just transitioning from using euler's number in and i've basically just replaced it um that was sort of what i'm saying is half of euler's number is going to tell us if we grow at 50 so it's sort of i'm axing this concept of using this formula and i'm instead choosing to use euler's number because it's going to provide me with exactly the same thing sorry if that was confusing um let's see here let's say i got sort of lost there for a second so i apologize now basically with euler's number it's going to provide a growth over a certain period of time with a natural logarithm what it's going to do is you're going to receive the growth and and returns so what we're going to do here for example let's say you had um like log e 1 well that's going to be equal to 0. and the reason why is basically what you're asking with this equation is how long does it take me to get what i originally already have what takes you exactly zero periods of time it's like saying i have one dollar in my pocket how long until i get an additional one or how long until it becomes one dollar we already have one dollar so it takes you no time in that situation um if you wanted to find out another thing that is important is let's say you have log e negative 1 in this situation it would simply be undefined and the reason why it's undefined is it's not possible to have a negative number using these tools let's see let's say you wanted to find out how long it would take to double your money well what we're going to do is um is just take this natural logarithm and you want to double your money which is the same as saying 2 this is going to end up working out to 6.93 and in time base what that would be equal to is 8.3 months if you were working off that calculation and that would be if you had a 100 interest rate that compounded continuously um what else do we have well you could say something like how long would it take to double your money if you had a eight percent growth rate so again you're going to take the rate times the time which is going to be equal to 6.93 and you would then get 0.693 divided by your interest rate which is going to be 0.08 and that means that it would take 8.66 years all right and there's multiple different ways of playing around with these to get sort of interesting answers for all this but now what i want to do q a section in the stream yeah what you can do is if you have any questions well i'm gonna be uploading this to youtube so you can leave any questions there you could also leave questions in chat and this whole entire uh stream that i have here i have i'm going to make a notepad like a little notepad thing in goodnotes which is in pdf form and i'll make it available for everybody to download i'll put it in the description for this video and on youtube i'll put it in the description on youtube you can get it there and then i have a discord you can ask me any additional questions that you have if you want my discord here i'll give it to you um let me see if i can give you a link to my discord and i'm on there pretty much constantly you can always ask me questions uh why can't i not get a link to my discord sort of there we are all right so get this and invite link expires never and there we are so generate link and where is it all right [Laughter] copy generating a link to my discord you know what i'm just going to grab it off of youtube because discord's acting weird here right now where is it i am not going anywhere i'm just bouncing around in browsers trying to get the discord channel any questions that come up during this video the best place you can ask them is on discord because i'll answer anything that you want on there um just give me one moment to get a link to my discord there it is there it is all right i'll post this inside of the chat and then you can go there so put this right like this boom there it is okay just go there ask questions and i will answer all of them that i get all right so now let's figure out where exactly i am here i'll bring this back up aside here and go back to this all right so now what we want to do is we just want to solve some logarithmic functions to sort of end this tutorial now i guess we're done i guess i've covered all of algebra 1 and 2 except for matrices which i will cover in the linear algebra thing now logarithmic functions are going to allow us to isolate the um variable exponents so in this situation did i write this out i did not okay so let's write this out so what we're going to do here is to solve x squared equal to 4 what we can do is we can find the square root of both sides and we can get x is equal to two well we can also do the same with two x two like this equal to four exactly like that so in this situations x equal to 2 to the power of y is going to be the same as y is equal to the log of 2 x so let's solve this problem right here so all we need to do is go and change it into exponential form which is going to be equal to 5 to the power of x is equal to 25 okay very simplistic also if we wanted to solve 2 to the power of x equal to 16 we can put this in log form like this 16 is equal to x and likewise another thing is log just so you know what the technical terms for everything is if we have 4 16 just to use an example this is called the argument this is called the base and this is called our exponent all right so there's that and let's solve a couple more of these problems this log to the 636 of course this is going to be equal to 2 and if we have 1 8 maybe we want to draw an arrow here well we can go and get 1 8 and this is going to be the same as 1 over 2 to the third which can be converted into two negative three so this is going to give us log of two to the one eighth equal to negative 3 i like that and for this last one that is slightly more complicated we can convert this into log to the fifth and 5 negative 1 over 3 which is ultimately going to give us a value of 1 over 3. and there you go guys that is basically at least one problem on just about every algebra concept you could ever possibly imagine let me see if i can go and give you the link to the notebook right now as while you're on here um do i have it somewhere i am going to put it i'm probably going to throw it on my my one drive is that where i want to put it let's see um i will throw it or maybe i could throw it onto maybe i'll throw it on dropbox that might be best so what i'm doing right now is i am getting getting the entire notebook and i'm going to put it on dropbox so that you can download it i'll give you a link to it if i can remember my dropbox password do i remember it no i do not okay well it's going to go on one drive then well thank you for sticking around it was a lot of information i'm actually kind of surprised i was able to survive through the entire video my voice is completely raw right now [Music] whoops yeah if you're wondering what i'm doing i'm trying to save everything and i accidentally keep hitting my caps lock so that i can give you the the notebook that i have here so where do i want to save this [Music] documents and [Music] you can work through all the the different stuff let's put this into a directory here and i can just throw it right here okay so i'm going to get you the notebook that you can download and you can look at every single thing that's inside of it so i have to locate the file first there it is and i called it all the algebra i think that's what i called the the file yeah all the algebra is what the name of the file is so let's drop it right here and it is uploading and as soon as it uploads i will give you the link and then you can download everything it's written i it's all handwritten so it's really kind of funny and one file all the algebra and i'm gonna try to get you a link i don't know if you ever used onedrive or or anything like it but it's for some reason it takes forever to upload hey no problem to anyone with the link i don't want you to edit it so allow editing i'm going to say no you can download the the file though and do whatever you want with it really specific allow editing set our expiration now that's fine apply and copy link and copy and let me verify that the link actually works before i give it to you and it looks like it did there it is yes this is uh this is uh all my notes from this entire video so a whole bunch of crazy stuff one thing i'd suggest maybe put a dot on the three places you're referring to since we can't see oh yeah i apologize for that um let's go this and this here and i will create a link onedrive and create and what nice name could i give this can i get all the algebra all the algebra as a bitly let's see link's been saved look at that i got that i'm verifying it works looks like it does and i will post it right here boom there you go oh i just yeah that was dumb this would throw all the effort of getting all the algebra and it didn't work here let me get it just just for the heck of it get this and oh okay well thank you very much um i thank you very much for stopping by i greatly appreciate it so let's just copy this link right if i can get it and copy so it's also here it's a lot easier to remember all the algebra at least i think so so there it is all right guys well that was a three hour live stream thank you to everybody that joined me i'm gonna try to do a learn trig in one video next week but i cannot guarantee it i might have to move the time because i have to go to the hospital and get random stupid tests that i can't get moved so whatever uh so thank you to everybody that joined me today this will be on youtube tomorrow so i'll see you next time