Common Core Algebra I.Unit 5.Lesson 4.The Elimination Method.by eMathInstruction

All right, let's get into it. Now, really, we laid the basis for this lesson in the last one. And I really hope that you did watch lesson number three. Because in that lesson, what we saw was that solutions to systems remain solutions if two things happen. One, if properties of equality are used to rewrite either of these equations, and that's primarily the multiplication property of equality, right? Multiplying both sides of the equation by whatever we want. And number two, the equations are added or subtracted or any rewrite is added or subtracted. Really, in a certain sense, number two is also a property of equality. 

You might say it's the additive property of equality, right? We're just adding the same thing to both sides. Now because of that, we can come up with a very slick method. And this is truly the first algebraic method that you're seeing in common core algebra one that you didn't learn in 8th grade common core. All right? So we contrast this with solving systems graphically and solving systems by substitution. Now we saw them using the elimination method. So let's jump in and do that. All right. Consider the system shown below. Solve the system in two ways by eliminating X and a, and by eliminating Y and B so let's jump in. Now, if I want to eliminate X in this problem, then really what I want to do is add the two equations together and have the X's cancel. But the only way that's going to happen is if we have additive opposites. 

So right now we have a four X here and we have a negative two X there. So what I'm going to do is I'm going to use a property of equality and I'm going to multiply both sides of this equation by two. Now what that's going to do is it's not going to do anything to the first equation I haven't changed that at all. But it's going to make the second equation into negative four X plus two Y is equal to 16. Now the beautiful thing about that is that what I then add these together, these two terms cancel. 5 Y and two Y is 7 Y, 12 and 16 is 28. The reason that it's called the method of elimination is we've now eliminated X, making it so easy to solve for Y in fact we get Y is equal to four. Now what we also need is our X value, so we can take that form. We can substitute it into either one of the equations, but I think I'm going to put it into this one. And I'll get four X plus 5 times four is equal to 12. Four X plus 20 is equal to 12. And now we've just got a very typical linear equation solving, four X is equal to negative 8. 

Divide both sides by four and X is equal to negative two. So there's our solution, right? Negative two comma four. Now, the great thing about the illumination method is it really doesn't matter which variable you eliminate first. So let's eliminate Y let's look at these two equations. Here we have a 5 Y here we have a positive one Y what I'd really love is a 5 Y and a negative 5 Y we can do that easily this time by multiplying both sides of the equation by negative 5. Yet again, I'm going to take that first equation and I'm going to leave it alone. The second equation, again, got to be careful make sure to distribute negative 5 times negative two X as positive ten X negative 5 times. Positive Y's negative 5 Y and on the other side we get negative 40. When we add these two together, these eliminate, and we end up having 14 X is equal to negative 28. Divide both sides by 14. And of course, negative 28 divided by positive 14 is a negative two. I would then substitute that negative two into here or here and solve for Y I'm not going to do that. We already know that Y is equal to four. Okay. So I'm not going to go there. 

Now letter C says that show that the point that you found in a and B it's the same point, right? It's the point negative two comma four is a solution to this system of equations. So let's do that really quickly. We've done that quite a few times in recent lessons. So just a little review, and something that's always a good idea, right? Always a great idea. To make sure that the answer that we found is correct. So substitute it in. I'll get negative 8 plus 20 is equal to 12. 12 is equal to 12. You bet it is. Substitute into the other equation negative two X plus Y is equal to 8. A little bit more careful here because we're dealing with negatives. All right, negative times a negative is a positive. And then I think we know that 8 SQL to 8. All right, so because both equations are true, it's got to be a solution. That's it though. It's kind of neat, isn't it? The whole idea of elimination is to add the two equations together so that one of the two variables eliminates. If that doesn't just naturally happen, which most of the time it won't, then often you have to multiply one and sometimes both of the equations by some constant to get additive opposites to a negative two, 5, and negative 5, four, negative four. 

All right? I'm going to clear this out. So pause the video now and write down anything you need to. All right, here it goes. Exercise two. Solve this following system of equations by elimination and check your aunts that your answer is a solution to the system. So let's take a look at these two. Now, the first thing I should always consider is can I just add them together and eliminate something? Well, if I add these together 5 X and two X is 7 X, no good there. Negative two Y and 7 Y no good there. All right? So then the question is, well, what should I eliminate the X's or the Y's? Now, in neither case, is it going to be really, really easy to just magically get something that works out nice? It just isn't, right? So what I think I'm going to do is I think I'm going to try to eliminate the X's. That's me. You could totally eliminate the Y's as well, but I'm going to eliminate the X's. All right. I'm going to do that by trying to make this and this opposite of each other. I want them to be opposites. And the easiest thing to turn them both into is a ten. Well, one of them to attend, and one of them into a negative ten. I think just arbitrarily, what I'm going to do is take this first one and I'm going to multiply by two on both sides. 

That's going to give me ten X minus four Y is equal to 20. On the other hand, this equation I'm going to multiply both sides by negative 5. All right? Negative 5 times two X is negative ten X there are my opposites. Negative 5 times positive 7 is negative 35. And 43 times negative 5 is 200 15. It's negative 215. Good. Just had to look at my cheat sheet. When I add these two together now, these two eliminate, watch out, you're adding these two. So that's going to be a negative 39 Y of course use your calculator if you need to and then 20 plus negative two 15 is negative one 95. The X's have now been eliminated. We'll just divide both sides by negative 39 and Y will be equal to 5. All right, we can now take that 5. I think I'm just going to substitute it into the first equation, but not there. I'm going to substitute it there. All right, I'm going to get 5 X minus two times Y, which is 5. It's equal to ten. 5 X minus ten is equal to ten. Additive property of equality add ten to both sides. 5 X is equal to 20, divide both sides by 5. And X is equal to four. So it looks like the .45 is our solution. 

Let's check. All right. So we check very easily by substituting into both equations. Let's see 5 times four. Minus two times 5 equals ten. 20 minus ten equals ten. Ten equals ten. Great. Let's check it in this one, two X plus 7 Y equals 43. Two times four. Plus 7 times 5 equals 43. 8 plus 35 equals 43. And 43 is 43. So yeah, it works there too. All right? Now, honestly, exercise two is to a certain extent as bad as it ever gets. The worst case scenario when you do a method of elimination is that you have to multiply both equations in order to get opposites. So that was it. It doesn't get any harder than that. So let's get some more practice. All right? And we're going to get some practice and kind of unique ways in ways that I think you might find kind of cool. So pause the video now and clear and write down anything you need to before I clear it out. All right, here we go. One of the coolest things I think that you can do with the method of elimination. Is to find the equation of a line. Now we're fine equation of a line. 

All right? Now we've done this before. But we're going to find the equation of the line now. By doing it a little bit differently. All right, we're going to take Y equals MX plus B and we're going to find the equation of the line that passes through these two points. And here's what we're going to do. Because the line passes through these two points, we're going to substitute those two points in to this equation. All right? That gives me negative 11 equals negative two M plus B and then if I put the 14 and the three in, I'll get 14 equals M times three plus B, which gives me 14 equals 3M plus B so now this is my system of the equations. And what we're going to do is we're going to solve the thing. I think I'll rewrite it like this. Negative two M plus B equals negative 11. And 3M plus B is equal to 14. All right, if I added these two together, I'd get negative two positive three, which would be a positive one M so M's wouldn't be eliminated. B plus B to B, that wouldn't work. All right. But one thing we could do is we could take this equation and we could multiply both sides by negative one. 

Now why would we do that? What we would do that, because suddenly, we would have a negative B. And given that we have a positive B in the other one, when we add these two together, those with cancel, 5 M would be 15, divide both sides by 5, and our slope would be three. Now to find my Y intercept, I can take my slope three, substitute it in here, three times three plus B is equal to 14, 9 plus B is 14, subtract 9 from both sides. And B is 5. Don't forget to actually give the equation of the line Y equals three X plus 5. So a different way of actually coming up with the equation of a line, right? We set up a system, but that system then allows us to solve for the M and the B it's kind of any press procedure, I think, combines two very different things. Solving a system and finding the equation of the line. All right. I'm going to clear out this text, salute complicated. So really think about what we just did here. 

Pause the video and write down what you need to. All right, here we go. Let's try another one like that. All right, here we've got a table. Find the equation of the linear function and Y equals MX plus B form, shown in the table below. Well, really, we just need two points. Which two points, any two points. It doesn't matter. I think I'll pick the first two. Negative two comma three and two comma 5. Now remember, what I'm going to do is I'm going to take these sub points and substitute them into this equation. This generic equations and the three is going to go in there. The negative two is going to go in there. All right, it's always a little bit obnoxious that's written this way to begin. But that's one equation. Now let's take this one. And boom, and I'll get 5 equals M times two plus B so maybe I'll write it down here. 5 equals two M plus B now, one of the things that's really great about the way this one's set up is that because I've got a negative two M and a positive to them, I can just add the two together. They're going to eliminate each other. These two are going to go away and B plus B is two B dividing both sides by two. I'll get the Y intercept is four. 

Now to figure out what my M is, I can take my B equals four, maybe put it in here. 5 equals two M plus four. Subtract four from both sides. We'll get one equals two M divide both sides by two. And we get M equals one half. So one half X plus four. What's cool about this problem is you could check it, right? So let me check. Check with ten 9, right? So if I'm right, and this is the correct equation, then this will be a true statement. One half ten is 5. And 9 equals 9. Great. So I know it's a little bit of a funny way of finding the equation of a line, right? But it's a good way to know how to do it, and it's a nice application of the method of elimination. So pause the video now if you need to and write down whatever you have to. All right, let's move on. Do our final problem. I think it's our final problem. Yes. Final problem. Okay. Now the next couple of lessons, we're going to spend a lot of time just modeling with systems of equation and solving them many times we're going to solve them with the method of elimination because it's new to us this year. But here is our first shot at kind of a word problem involving a system. Two numbers have the following properties. The sum of the larger and twice the smaller is equal to 13. Twice their positive difference is equal to 8. What are the two numbers? 

All right. So we got a larger number. We have a smaller number. I'm going to let X be the smaller number. I'm going to let Y equal to the larger number. All right, given that, let's sum this, or let's model. The sum of the larger and twice the smaller, okay? The larger plus twice the smaller is 13. Good enough. Twice their positive difference is equal to 8. They're positive difference. You see, because I can subtract in two ways. The positive way of subtracting will be taking the larger and subtracting the smaller. But then they say that it's twice that. So twice that difference is equal to 8. Now before I go ahead and do anything, it might make sense to distribute that to I'm not doing it yet. I'm just rewriting this equation. But distribute this two through this parentheses. Not over here, that two is not on this side. No two. So what do I get I'll get two Y minus two X is equal to 8. And I want to solve this. Now the beautiful thing is we can solve it by elimination very easily. Because these two are opposite. 

And in fact, when we add them together, they cancel. And we get three Y is equal to 21. But now that's simple enough to solve, divide both sides by three, and Y will 7. So that is our larger number. Now to get our smaller number, I think I'll take the Y equals 7 and I'll probably put it in here, so I'll get 7 plus two X is equal to 13. Subtract 7 from both sides will get two X is equal to 6. Divide both sides by two, and we'll find our smaller number. Is through. And that's it. The two numbers are three and 7. We're going to work a lot more with word problems involving the method of elimination and the method of substitution in the coming lessons. Pause the video now though, copy down anything you need to. All right, and let's move on. Let's finish up this lesson. Okay. So in this lesson, we saw the method of elimination, and it worked so beautifully. 

It basically works on the concept that when you add two equations together, their solutions still remain their solutions. Now, that's a good thing because if we can add two equations together and eliminate one of the two variables, then we have only the other one left to solve for. And typically, it's very easy to solve. So we're going to work more with this in the coming lessons as I've mentioned. For now, I want to thank you for joining me for another common core algebra one lesson by E math instruction. Until next time, my name is Kirk Weiler. Keep thinking and keep solving problems.