Common Core Algebra II.Unit 3.Lesson 7.Systems of Linear Equations

All right. So in exercise one, what we've got is we've got two systems that we're going to solve by substitution and by elimination. If you think you remember how to do this, pause the video now and take shot at it. All right, let's go through the two methods. Now we're going to be concentrating on elimination today. But we do want to remind ourselves how substitution works. If we take a look at this first, oh, that was interesting. If we take a look at this first system, what we want to do to substitute is we want to substitute one equation into the other in order to eliminate one of the variables. Now, this is most easily done if one of the two equations is solved for. So in other words, there's Y equals or X equals. Now, neither one of those is the case. But we can pretty easily take this equation, move that two X to the other side, where it will become negative. 

All right, and get this equivalent equation. We can now take negative two X -7, and we can substitute it in for Y and we'll get three X plus two times negative two X -7 equals negative 9. And notice kind of like elimination, we only have one variable now. Now of course I can solve this for X by distributing that too using the distributive property. I'll get negative four X -14. Equals negative 9. I can now combine some light terms positive three X negative four X is negative X -14 is negative 9. I can now add 14 to both sides. And get negative X equals positive 5. Now, of course, that's pretty easy to finish. Either multiply or divide both sides by negative one, and we'll get X equals negative 5. Now remember, you are not done solving a system until you've found the values of all the variables that make the equation simultaneously true. In order to do that, we need to now take that negative 5 and either substitute it into this first equation, the second equation, or most easily into this third equation. 

So that's what I'm going to do. I'm going to put that negative 5 into the rule that says Y is negative two times X -7. Of course a negative times a negative is a positive. So I positive ten -7, and then we'll get positive three. It's completely okay to leave your answers that way. A lot of times students will prefer to put them as a coordinate point negative 5 three because that's actually the point where those two lines intersect each other. Okay? Now that's substitution. Let's talk about elimination. The method of elimination which we'll be using for the rest of the day works because you can always add two equations together and get a third valid equation. Now let's see what happens when we add these two equations together. We get three X plus two X all right, no big deal, 5 X, two Y plus one Y, three Y and negative 9 plus negative 7 is negative 16. 

Now that is a completely valid equation and whatever solutions solve the system would also be a solution to that equation. But it didn't really do any good because we didn't eliminate either one of the two variables. So what I'm going to do is I'm going to erase that. And I'm now going to manipulate one of these two equations using the multiplicative property of equality. I'm going to manipulate it so that when I do add the two equations together, something eliminates. Let's do this. Let's number the equations just to keep track of our work. I like to number them and put those numbers in circles, all right? So what I'm going to do is I'm going to take equation two, and I'm going to actually multiply both sides of it by negative two. All right? So I'll do two X times negative two, and I'll get negative four X. I'll get positive Y times negative two, which will be negative two Y and then I'll get negative 7 times negative two, which will be positive 14. 

Now, I'm not going to do anything to equation one other than rewrite it. And the reason I'm not going to now, you can probably see this is that I now have opposites negative two Y and positive two Y so when I add these negative four X positive three X gives me negative one X but negative two Y and positive two Y cancel each other out. 14 plus negative 9 is positive 5. And we're now basically at this point. We can finish off the problem by dividing both sides by negative one and we'll get that X equals negative 5. Now we don't have that really easy convenient one to substitute it in. So I think I'll take this moment to substitute it into the first one. And just see if my Y value continues to turn out to be positive three. I hope it does. Otherwise I made a mistake in the first problem. But we just have to work through this carefully. It's simply going to be a linear equation typically two step equation, but still we can all make mistakes, so let's take our time. 

We'll get two Y is equal to a positive 6, divide both sides by two, and here is a screen full of writing. Y equals three. Awesome. All right. So the method of elimination takes two equations adds them together, eliminates one of the variables, and we end up with one equation with one unknown. Let me just highlight that in red. We added these two equations. And we ended up with this one equation with one unknown. And we were able to easily solve that equation and back substitute to get the value of the other variable. We're going to do something very similar when we have three by three systems. So you really need to understand the two by two first. Pause the video now, take whatever time you need. All right, we're moving on. Let's do a three by three. There it is, three by three system. I've already numbered the equations. 

Notice we've got three variables. We've got X, we've got Y we've got C it's all good. All right. And I'm really going to walk you through this first one. So let's take a look at number letter a the addition property of equality allows us to add two equations together to produce a third valid equation. Create a system by adding equations one and two, so we're going to add these two together. And one and three, so we're going to add these two together. And then we're going to figure out why that was effective. So anyway, let me write down equation one. Two X plus Y plus Z equals 15. And we'll do it in red. 6 X minus three Y minus Z is equal to 35. Now we can always add two equations anytime we want. So I'm just going to add these two two X and 6 X is 8 X positive Y, negative three Y is negative two Y and positive Z and negative Z well they cancel. 15 and 35, that gives us 50. All right, let's take a look at what happens when we add equation one. 

Let me write that down again. I know that's a little bit irritating. Lots of writing in these problems, equation one, with equation three. This takes me a little while to write it all down. There we go. Two X minus four X or two X plus negative four X gives me negative two X Y plus four Y gives me 5 Y and negative Z and C canceling again. 15 and negative 14 is positive one. Now take a look what we now have, let me go back to blue is a new system. And that new system is now a two by two. Two equations with two unknowns. Let's solve it. All right, let's write down 8 X minus two Y is equal to 50. And negative two X plus 5 Y is equal to one. Now, if we were really lucky, we'd be able to just add these two equations together and we'd have another variable eliminate. But that doesn't look to be the case. So what I'm going to do is I'm going to take this equation and I'm going to multiply both sides by four. Okay, let me write that one down. 

Gives me negative 8 X plus 20 Y is equal to positive four. Then I'm going to just write this one down again. That's going to be positive 8 X minus two Y is equal to 50. Now, of course, that was in particularly good strategy. We never really talked about why this was an effective strategy over here. That was because the Z's canceled and left us with just those two variables. But this is now an effective strategy for the same reason. Now when we add the X's are going to cancel each other. And we're going to simply be left with 18 Y equals 54. Now, if everything works out nicely, when we solve this, divide both sides by 18, Y will turn out to be nice, and it does. Looks like I really like Y equals three. But there we go. Now before we back substitute and finish this problem, let me just really explain how all these work. We take a three by three system. We manipulate as we need to. We add together, we eliminate variables, and we get it down to a two by two system. Once we have a two by two system, we do exactly the same thing. 

We manipulate equations we add them together to get yet another variable to cancel until we have, I guess, a one by one system, one equation, one unknown, right? And we solve. Now we have to back substitute, which is a little bit annoying. Okay, and I've got to find room for it. So I'll kind of squeeze it in right here. Now the first thing we're going to do is we're going to back substitute to figure out what X is. And I think I'll just put it right into that first formula right here. 8 X minus two times three equals 50. So I'll get 8 X -6 is equal to 50. 8 X is equal to 56. And X is equal to 7. All right, finally, we take these two, and we pick an equation, any equation from up here, I think I'm just choose the first one. So then I'll do two times 7. Plus three. Plus Z is equal to 15. So we'll get 14 plus three plus Z is equal to 15. 17 plus Z is equal to 15. 

And it looks like Z is equal to negative two. Wow. That's it. Add a couple pairs together to eliminate one of the variables. Then get a two by two, manipulate that, add together to eliminate another variable, gets you down to just one, and then solve and back substitute. All right, that's the whole idea. I mean, I could probably give many of you the homework right now and you'd be fine. But what we want to do is we want to walk you through some additional exercises and have you practice on some yourself. Pause the video now and write down anything you need to. All right, I'm clearing it up. Let's move on to another three by three. All right, so, you know, when you attack a three by three system, there's a lot of choice, a lot of things that you should think about. So what I want you to do is really take a look at this system and then take a look at a, it says which variable will be easiest to eliminate Y and then it says use the multiplicative property of equality and elimination to reduce the system to a two by two. 

So think about this for a moment and even take a little bit of time to try to reduce it to a two by two system. All right. Well, in my mind, the easiest variable to eliminate will be the variable Y and that's because. In equation one and I'm going to label them now. One, two. It's good to label your equations in equation one. The coefficient of Y. Is one. And that's going to be easy to manipulate, right? The fact that this thing's got a coefficient equal to one makes it very, very easy to manipulate. But I'm going to actually erase this so that we have some room to do our work. All right? So if possible, try to find the variable whose coefficient is one. Now, if I want to eliminate Y in the second equation, then what I'm going to do is I'm going to take that first equation, and I'm going to multiply it by negative four. All right? Notice how I show that equation one times negative four. 

That's going to get me negative 16 X minus four Y plus watch out for that 12 Z equals positive 24. Take a look at that again. Negative 16 negative four positive 12 positive 24. And now I'll just write down equation two. All right, and that's going to be negative two X plus four Y there's my opposites plus two C is equal to 38. All right. Well, when we add these two together, we get negative 18 X when we add these two, the Y's eliminate. That's what we wanted. Add these together, we get 14 Z and add these two together, and we get 62. All right. Now, it's really, really important, because this is easy to miss up front with three by three systems. We just eliminated Y we've got to eliminate Y again, okay? We can't eliminate Y and then turn around and eliminate Z I mean, we can, but it just won't get us very far. 

Now, the second manipulation we really don't have to do anything at all. We can just take equation one, four X plus Y minus three Z equals negative 6. And equation three, 5 X minus Y minus whoops. -7 C that almost looks like 7 Z equals negative 19. Add together, and we're going to get 9 X see these two just normally cancel, which is nice. I don't really have to do anything. Negative three C negative 7 C negative ten C and this gives me negative 25. So there we have it, right? Let me just circle those in red. There's now our two by two system. All right, let me write that over here. Negative 18 X plus 14 Z equals 62. And 9 X minus ten Z equals negative 25. Now we have to think a little bit. Which would be the better variable to now eliminate. Think about that for a moment. 

Well, you can certainly eliminate either one. But I think taking the second equation and multiplying both sides by two makes a lot of sense because then we'll have a negative 18 X and a positive 18 X so I'm just writing that first equation down again. I'm going to have a positive 18 X -20 Z equals negative 50. Now when I add these two cancel here, be careful. We'll get a net negative negative 6 Z is going to be equal to positive 12, 62 -50, and of course that means Z is going to be negative two. Once we're at that point, we've really done the bulk of the work. It's all back substitution now. It doesn't mean we won't make a mistake. But it's all back substitution. So let's try to find space to do that. I think I just stick to pattern and I'll put this into this first one. So I'll get negative 18 X plus 14 times negative two equals 62. Just barely snuck it in there. Negative 18 X -28 equals negative 62. Got to add 28 to both sides, right? That's going to give me negative 18 X when did I do? Yeah. Oh, of course that's not the way I solved it. I solved it previously. Let me put this in my calculator. That's going to give me oh. Something's going wrong. What happened? 18 X -20 Z, Z is definitely negative two. So negative 28. One is X, I'm definitely not getting that right answer here. Um. Oh, I see why. Do you see the mistake? That's better. Somehow that negative 62 became a positive 62 there. 

Thinking to myself, boy, this is just not working out right. There it is. Sorry about that. Hey, always good. Always good to make mistakes and check our work. There we go, X equals negative 5. We still need to figure out what Y is. Okay, let's see if I can do this one without a mistake. Give myself a little bit of extra room. All right, let's take that first equation. Let's do four. Times X plus Y minus three times Z it's a good I have my solution sitting next to me. It gives me negative 20 plus Y plus 6, negative three times negative two. Negative 6. So that'll be Y -14 equals negative 6, and then Y equals 8. Well, I stumbled a little bit there, and it's amazing what can happen when you have a positive 62 that suddenly becomes a negative 62 in the next line. That's where you want to have a lot of space to work. All right, but pause the video now and think again about what we did in order to solve this three by three because we'll make it just a little bit harder in the next problem. Okay, I'm going to clear out the text. All right, exercise four. Solve the system of equations shown below. 

Show each step in your solution process. So here's what I'd like you to do. I would like you to pause the video and then I want you to think about what variable is going to be easiest to eliminate. Then I want you to eliminate that variable, bring it down to a two by two system. Once you have it as a two by two, then I want you to eliminate another variable, get it down to a one equation with one unknown and then trot a back substitute after you solve it. Pause the video and take maybe up to ten minutes to solve this problem. Okay? All right, let's go through it. Now again, the key is look for a variable that has a coefficient of one. Now in this problem we actually have two nice choices. We could either get rid of X because it shows up there with a coefficient of one. Or we could get rid of Y, because it shows up there with a coefficient of one. Just to change things up a little bit, I think I'm going to eliminate the X, all right? I'm going to number my equations one, two, three, although not mandatory.