Common Core Algebra II.Unit 1.Lesson 2.Solving Linear Equations

Hello and welcome to another common core algebra two video by E math instruction. My name is Kirk weiler, and today we're going to be doing unit one lesson two solving linear equations. Now you've been solving linear equations probably since like 6th or 7th grade. So this is a topic that's been around for quite some time. And yet, no matter how far you go in math, whether or not you become an engineer or somebody who uses a lot of math. All the way into college, you're going to be solving linear equations constantly. So it's an important skill to be fluent with. It's an important skill to understand why it works, how it works, et cetera. So this lesson is going to review all of that. Let's jump right into the first exercise. All right. Well, this first exercise we've got four different linear equations and I guess in a certain sense I've arranged them so that they get more difficult as you go on. Certainly D is the most complicated one of all. What I'd like you to do right now is kind of test to see how ready you are to solve linear equations. By pausing the video now and working through all four of these. It shouldn't take you any more than ten minutes, so go ahead and pause the video and then we'll go through each problem one by one. All right, let's go through the problems. Now, the problem letter a, really, I want to look at this as just undoing what's been done. I've got three times X plus 5 equals 26. So I'm going to subtract 5 from each side to leave me with three X is equal to 21. I'm going to divide both sides by three, and I'm going to be left with X equals 7. All right, that is about as basic a linear equation as you can get. Letter B on the other hand has variables on both sides, and you've got some choice here. We could move all the variables to one side and all the non variables to the other or we could do it in the opposite order. I'm going to actually add this 7 to both sides to get rid of it. So I'll have 8 X is equal to four X here's where we have to be careful negative 5 plus 7 is a positive two. Now I'm going to move this four X to both sides. By doing this. Now I'm not giving you really any kind of justification here. What I'm doing right here is just going through the basic manipulations that you've seen year after year. Now we have to be a little careful here to divide it by four isn't two, right? It simplifies down to X equals one half. Now one could argue that letter C is actually easier than letter B because again, it just has got the single X on the left hand side. But it does have a little division involved and some addition up here. Now, what we need to do though is we need to begin the problem by multiplying by two on both sides. We can't subtract 8. But by multiplying by two on both sides, we get rid of that division by two, and we end up having X plus 8 is equal to negative 12. We can then get rid of the 8 by subtracting 8 from both sides, be careful here. Negative 12 plus negative 8 will be X equals negative 20. All right, finally, letter D has got a lot going on because it's got the distributive property twice it's got variables on both sides. This is about as beastly of a linear equation as you can get. So I'm going to begin by distributing the 6 through the parentheses, 6 X plus 24. I'm then going to distribute a negative two through the parentheses, giving me negative two X and watch out a negative times a negative is a positive. All right. Now, I'm going to use a little bit of rearranging, and I'm going to flip flop that 24 in that negative two X. Some of you, of course, will do this without writing down this line. And that's okay. You know, you get to a certain point in math, and you want to be able to do a lot of the manipulation. I'm going to combine like terms here, 6 X minus two X is four X 24 plus two is 26. On this side, we'll have two X plus 20. And now we're really back to a situation like in letter B, I can certainly subtract a 26 from both sides. Giving me four X is equal to two X again, be careful 20 -26 is negative 6. Subtract a two X from both sides. And we'll get two X is equal to negative 6. We're just about out of room here. Divide both sides by two. I don't know what that little straight mark is. And we'll get X equals negative three. All right. That wasn't a long one. But again, if you can do a problem like letter D, if you got negative three there, awesome. It means that you you're up to speed on your linear equation solving. Pause the video now and write down anything you need to. All right, I'm clearing out this text. Moving on to the next problem. Now, solving linear equations has to be almost automatic for you at this point. But it's still good to understand what justifies each thing we did, especially in a problem like the one that we just had and one letter D where there's a lot going on. Now, in the last lesson, we reviewed the commutative associative and distributive properties of numbers. But there's also properties of equality known as the additive and multiplicative properties of equality. And the properties of equality basically say, look, you can add or subtract a number from both sides of an equation. You can multiply or divide by a number on both sides of the equation. As long as you don't divide by zero, then you're fine. Those properties, the community of associative distributive, and then the properties of equality additive multiplicative. Really justify everything that we do with linear equations. So let's take a look at solving this somewhat complicated linear equation and justify each step. So let's take a look here. We've got two times X plus 7 plus four X equals 44. And we make that into two X plus 14. Plus four X equals 44. Well that is the distributive property, right? We have multiplied that two through the parentheses. I'm not sure why I put down property each time. It always is there. But it's the distributive property we have distributed the two through the parentheses. Now, if you look at this next step, what I've done is I've taken 14 plus four X, and I've made it into four X plus 14. That is the commutative property. The property that allows us to say, hey, 14 plus four X is the same as four X plus 14. Why not? Now the next one, this tends to be the trickiest one. I'm going to do a little bit of a racing here so that you can see it a little bit better. Here, I'm going to take two X plus four X and I'm going to write it as X times two plus four. That is again our friend, the distributive property. A lot of students have a hard time seeing that because it's sort of the distributive property run backwards. Now obviously we just combine the two and the four and get 6. In the next line though, once we start with 6 X plus 14 equals 44, I've subtracted 14 from both sides. Strangely enough, that would be called the additive property of equality, the additive property. Really, in a sense, what we're saying is that we're adding a negative 14 to both sides of this and then of course 44 -14 is 30. Our final step comes when we divide both sides of this equation by 6. That is the multiplicative property. You might say, hey, you didn't multiply there. You divided. But every multiplication problem, every division problem can be phrased the other way. So dividing by 6 is the same as multiplying by one 6th. So we still call it the multiplicative property. Otherwise we'd have to have an additive property and subtractive property of multiplicative property division property, et cetera, et cetera. It's word as is. We don't need to bring any more. All right? So pause the video now and write down anything you need to. All right, let's clear it out. And move on. Exercise three says, try to solve the following equation. State whether the equation is an identity or inconsistent explain. Well, there's many different types of equations. Two that are really rather odd are what are known as identities and ones that are inconsistent. And identity is an equation that is always true. Oh boy. Always true, so all values of X solve it. All values of X solve. That's some people might take a little bit of an issue with that, but that's basically the idea. Every value of X would solve it. And inconsistent equation is one where no value of X will solve it. No value. Of X solves it. All right? Now these are kind of annoying equations, and they don't come up a lot, but they come up enough that you want to be aware of them. And aware of how to interpret strange things that happen when you solve an equation. So watch, let's play around with this equation. Let's distribute the negative two. Let's use the distributive property. Again, let's be very careful because negative two times negative four is negative 8. Let's distribute this three X as well. You know, I mean, at this point you kind of are just moving along going well. I don't see anything odd here. Now we could combine some light terms, 6 X minus two X is four X -8. Here I might have to rearrange these using the commutative property, but ultimately I'd have three X plus X, which would be four X and 6 -5, which would be positive one. Now take a look at this for a moment before we move on. Because I don't really want to go any further. I want to understand structure here. Basically what this says is I'm looking for some number. X is some number. That when I take four times that number and subtract 8, I get the same thing as when I take four times that number and add one. Well, there are no numbers. There are no numbers. There are no numbers for which this is possible. Right? I mean, you can not take four times the number, subtract 8, and get the same thing as taking four times the number, and adding one. So this is an inconsistent equation. No solutions. Now, some students like to keep going a little bit. So let me take it just one step further. Again, I think you should be able to look at it right now and go, that's inconsistent. Some students will choose to subtract a four X from both sides. That's fine. That'll leave you with negative 8 is equal to one. Well, I think that we can all agree that negative 8 is never equal to one. There is no value of X that will ever make negative 8 equal to one. And therefore, this is an inconsistent equation, an inconsistent equation, not equations. There's really only one there. It's an inconsistent equation. All right? So now let's play around a little bit with an identity. Pause the video now and write down anything you need to. Okay, here we go. All right, exercise number four. And identity is an equation that is true for all values of the substitution variable. In other words, X oh, let's go back to blue. Trying to solve them can lead to confusing situations. Consider the equation two X -6 plus X minus one. Equals three times X minus three plus two. Let a racist test the values of X equals 5 and X equals three and this equation show that they're both solutions. Well, let's. Let's try one together and then have your work on the other one by yourself. Let's show that X equals 5 is a solution. Now we didn't really talk about this because it goes to the heart of solution solving, but what it means for a value of X to be a solution of the equation. Is that it will make the equation true. It's like it's like we're not lying, right? So when I put 5 in for X everywhere. The left hand side has to be equal to the right hand side. What we have to be careful with our order of operations here. Make sure we're doing them correctly. But if I just work through nice and slow, ten -6 is four. Plus 5 minus one. Let's do that multiplication. 6 plus two, we have four plus 5, which is 9, 9 minus one is 8. And 8 equals 8. So that's true equation when I put X equals 5. And so that's got to be a solution. What I'd like you to do is pause the video and take about a minute, plug X equals three in and show that it is also a solution. All right, well, let's do it. Let's go with X equals three. So again, just like before everywhere there's an X, I'm now going to replace it with a three. Right? Simple enough. All right, again, just like before I want to be careful with my order of operations. All right. Oh, three minus three is zero. Nice. Here are 6 -6 will be zero plus three minus one. Three times zero is zero. That zero doesn't really matter. Net zero doesn't really matter. Three minus one is two. Two is equal to two. Also a solution. Now that's a bit weird because linear equations, linear equations, one where there's X just to the first. Ten don't only have one solution. They don't have two solutions. I don't tend to. Write. So letter B says attempt to solve the equation until you're sure it's an identity. All right, so now let's try to see what's going on here. By actually solving this equation. All right. So again, it's kind of like if a teacher just slapped this in front of you and said, hey, solve it. What would you do? Well, I might like rearrange this side a little bit if I needed to, you know, and use the commutative property there. I'd probably use the distributive property right away on this side. Maybe I'd clean this up a little bit. I'd get three X and then negative 6 negative one is negative 7. Over here, we would have negative 9 plus two. That's also negative 7. And then I think I would stop. Because this really shows it's an identity. Look, I mean, if I'm trying to find all the numbers where that make three X -7 equal to three X -7, it's going to be everything. Right? I mean, that's basically just saying, look, take a number and do exactly the same thing to it on both sides. You know, you're going to have a quality. So there's how we know it's an identity. Oh, excuse me. I hope it's okay. I'm not going to go back and edit out that sneeze. It's been a long winter. So this will always happen. Always happen. Wow, that apparently my why don't always happen. But this always happens so it's an identity. Always actually, how about always true? Always true. So it's an identity. All right. So we've got identities that will always give you something like 5 X plus one equals 5 X plus one or 7 X minus two equals 7 X minus two, and then you've got these inconsistent equations where the things can't ever be true because negative 8 is an equal to one or something like that. Anyhow we're going to get a nice workout with this in the next exercise. So pause the video now and write down anything you need to. All right, let's clear it out. Okay, exercise 5, which of the following equations are identities, which are inconsistent, and which are neither. All right. Well, the best way to figure this out is to start to solve the equation. Just like you always would. And then C exactly what happens, right? If you get a solution that's just normal, well then it's neither. But if something funny comes along, then you have to be able to figure it out. So let's play around with letter a. Let's do this one together and then have you work on other ones maybe on your own. So in letter a, I'm going to just start solving this. I'll distribute the negative two all distribute the 5, all right? I'll combine some terms. And look at that. 6 X -6 equals 6 X -5. I mean, if you had to take it one farther step, you could subtract a 6 X from both sides, giving you negative 6 equals 5. Well, that that's never true. That's never true. So this is simply an inconsistent equation. No solutions. There's no solutions. No value of X will ever make that true. All right, what I'd like you to do is pause the video. Please look pretty darn confusing these equations. And I want you to try to solve each one of them until either you have a solution or you can conclude that it's inconsistent or it's an identity. Pause the video now and take some time. All right, let's go through it. Letter B wow, so many places that you could start on letter B so many different places. You could use the distributive property right here, right? And you'd get four X divided by two, which would be two X don't forget two divided by two. That would be one. Right? So you have to do that distributive property on the division. But then I'm going to get two X plus 9 equals two X plus 9. Well, that's always true. Always true. It doesn't matter what value of X you put in there. That's going to always be true. So this is an identity. Now, again, just to be very clear what that means is that we have an infinite number of solutions. Every value of X will solve that equation. All right, let's take a look at letter C, find out what you had there. Again, you know, when you first take a look at these, oh, look at this. Let's distribute a negative one, negative X plus 7. Not something that we've done for a little while. Again, let's play around with it two X minus X is X plus 15. Four X -6. Hey, notice at this point, you know, I mean, it feels like I'm just solving a kind of a normal equation. I'm going to do some short cutting here right now because we're talking about algebra two students. Well, so X equals 7. So this is neither. It's neither inconsistent nor an identity, because there's only one value of X that solves the equation. X equals 7. Normal equation. Let's do this last one. Two X plus one. Uh oh. I don't want an equal sign there. That's a little weird. Two X plus one, let's use the distributive property plus two X minus two. Let's use the distributive property again. Four X minus one, right? I've got to divide the 16 by four, the four by the four. Let's see, two X and two X is four X, one minus two is negative one. Four X minus one equals four X minus one. This is again. An identity. Always true. Always true. Oh, there's a lot of writing on that board, which means my head is really tiny right now. Strange for me. So pause the video now, write down anything you need to and then we're going to move on. Okay, clearing it out. Okay, so today we reviewed a very, very important skill. How we solve linear equations. Even though you've been doing it for years and years and 7th grade math, 8th grade math, common core algebra one, common chord geometry, and now back in common core algebra two, this is an essential skill. You've absolutely got to have it. We also looked at some weird linear equations, ones that were inconsistent, had no solutions, and ones that were identities. They had an infinite number of solutions. Spotting these, however, rare they might be, is also pretty important in math. All right. Well, thank you for joining me for another common core algebra two lesson from E math instruction. My name is Kirk weiler, and until next time, keep thinking and keep solving problems.