 This video is going to use elimination to solve systems. The goal of the elimination method is to add the two equations in the system together so that one of the variables is eliminated. There's our word for the day. That is, the term containing the variables add to zero. This process creates an equation which is one variable that enables us to solve for one of the variables in the system. If we look at this system right here, if we add these two equations in the system, then 3x plus a negative 3x is going to be 0x, so they cancel each other out. And y plus three more y is going to give me 4y, and 5 plus 7 is going to give me 12. Now you can see that we've eliminated one variable, so we're down to a one variable equation that we can solve for. And let's go ahead and solve this one. Divide by four. Now we know that y is equal to three. Now you have two equations, and we know y is equal to three. I just have to plug it back into one of them. This type of equation would be nice because this y has a coefficient of one. So let's try 3x plus, and then I'm going to put in my y, equal to five, and my y is three. So that means I want to subtract three from both sides so I can solve for x. 3x is going to be equal to five minus three is two, and if I divide by three, then I find out that x is equal to two-thirds. So my ordered pair is two-thirds and three. This system has opposite coefficients on the y. I can add these right away because they're going to drop out with opposite coefficients. So 3x plus 2x would give me a total of 5x. This cancels out and gives me zero, and eight plus seven is going to be 15. So if I divide by five, then I know that x is equal to three. Now remember, again, it takes an ordered pair, and all we have is the x. So I have to plug it back into one of the other equations. I can plug it into either one. I'm going to choose to do the second equation because it was a positive y, so I won't have to worry about signs. So we have two times our x, which we found, plus y equal to seven, and our x was three. So two times three is six, plus y is equal to seven, and if we subtract the six from both sides, y will be equal to one. So our ordered pair is x is three, and y is one. Now this system doesn't have opposite coefficients, but it does have opposite signs on both the x and the y. If I choose to get rid of x, then I need to get these two things to be the same. I need to multiply one of my equations. Two is the smaller number, so I would have to multiply this equation by five to get rid of the x's. Let's try that. So my original equation for the first one stays, and then this tells me I'm going to distribute five all the way through, even across the equal sign, which will give me negative 10x. Five times six is plus 30y, and five times 16 is 80. And if I add those two equations, 10x plus a negative 10x gives me zero. Negative 3y plus 30y will be 27y, and one plus 80 will be 81. And if I divide everything by 27, I get y, and 81 and 27, I know they have nine in common. Nine times nine is 81, and nine times three is 27, but it must be that they went in perfectly because nine divided by three is actually three. So now we have y equal three. I have to find the x, plug it into either equation. The bottom equation is a little bit smaller, but it has a nice little negative in here, so you can plug it into the top one if you want. In fact, let's do that. So we do that and we get 10 times x minus three times our y, which is three, is equal to one. So 10x minus nine is equal to one. Add nine to both sides. 10x is going to be equal to 10, which means when we divide by 10, we find that x is equal to one. Now I want to do one other thing, but I'm not going to solve the whole problem again because we know what the answer is. But what, let's get rid of this five over here. What if we had wanted to get rid of the y instead of the x? That means I want these three and six to become the same number. Again, three is smaller, and I can make it become a six by multiplying by two. So I would have to multiply by two and distribute it clear through. If you want to set up the system, if you want to solve it on your own later and verify that it's three, one, three is your solution, you could do that. So two times 10x is going to be 20x. Two times negative three y is going to be minus six y. And two times one is going to be equal to two. Don't forget to go across the equal sign. And then the original bottom equation stays. We haven't done anything to it because now when I look at my y's, I have opposite coefficients, and you can add them. I'll add them to get down to unvariable. So 20x plus and negative two x will be 18x. Negative six y plus six y gives you zero, and two plus 16 will be 18. And you can see pretty easily that that x is going to be one. This system doesn't have any opposite coefficients, and it doesn't have any opposite signs anywhere. So we have a little bit more work to do. What we're doing, trying to do, is to get the coefficients to be the same number but opposite signs. I can't make three become four. It's a smaller number, but I can't make it become four. So I'll try my y's, but I can't make two become five either. So this time I have more work than I did before. I'm going to either have to eliminate the x's and find out what three and four have in common, or eliminate the y's and decide what x and two and five have in common. It really doesn't matter. We'll set it up both ways again. If I want to eliminate x, then this top equation, I need four and three are going to have 12 in common. The least common multiple is going to be 12. So I need to multiply this one by three, and that'll give me 12x. But if I multiply that one by three, I need this one to be the opposite, negative 12. So four times three would give me 12, but then I need to make it a negative. And it really doesn't matter which one's negative, but since they were both positive, I had to multiply one equation by a positive and one equation by a negative. So let's set that system up, and then we'll go back and see what it would have been if we eliminated y and then we'll solve. In the top equation, we have 12x plus 6y will be equal to 36, and on the bottom equation, we'll have negative 12x, and negative four times 5y will be minus 20y, and 16 times four, or negative four, is going to be a negative 64. There's that system that we have to add. Now let's go through and figure out if I wanted to get rid of my y's. Well, the LCM in that case is going to be 10. This is for my y's. Let's take care of the negative first this time. To make two become negative 10, I have to multiply by negative five. I have to distribute the negative five in. I have to distribute it everywhere, but especially to that negative, or that 2y. And then to make five become 10, I have to multiply by two, and I'm going to distribute it all the way through. On the top equation, negative five times four is negative 20x, negative five times 2y will be minus 10y, and negative five times 12 is going to be equal to negative 60. And if I take the two times everything, I'll have two times 3x, which will be 6x, plus two times 5y would be 10y, and two times 16 is going to be 32. It doesn't matter which way we did it. You don't have to do it both ways. You're either going to eliminate x, or you're going to eliminate y. And we eliminated x the first time last time, so let's eliminate y the first time here. So we're going to use this system and go to work. Negative 20x plus 6x will be a negative 14x, and my y's cancel out, and negative 60 plus 32 is going to be equal to 28. If I divide by negative 14 on both sides, then I find out that x is equal to negative two. I've got my x, I need a y, I need to go back up to my original equations. And in my original equations, they're about the same kinds of numbers. It really doesn't matter which equation. I'm going to choose to put it into the top equation. Don't ask me why, I just am. So 4 times my x, which I just found, then plus 2y is going to be equal to 12. This is negative 8 plus 2y equal to 12. If I add 8 to both sides, that would give me 2y equals 20. And if I divide by 2, I now find out that y is equal to 10. So my solution is the ordered pair negative 2, 10. Alright, so now for our special cases. We've got a lot to think about here. We've got to think about coefficients. We have to think about signs. This is solving any system. I need to think about coefficients and signs on each variable. So for this x, I don't have opposite signs. And on the y's, I don't have opposite signs. So that's not very helpful. And I don't have the same coefficients. So that's not real helpful either. But I can see that it won't be too bad to get 3 to become 6. And actually I could have 6 become 12 as well. But 3 is a little bit smaller to get to the 6. So I'm going to eliminate my x's. I need this to become a positive 6. Okay, my LCM that I'm going for is 6. And I already have negative 6 down here. So I need to multiply times 2. That'll give me negative 6 though. So if I multiply by a negative 2, now I'll get a positive 6. So if I distribute, that gives me positive 6x. And distribute here, I'll get a positive 12y. And distribute here, I'm going to get a positive 4. This gives me 0x plus 0y. So on this side I have 0. And negative 4 plus 4 is also 0. But remember when it's true, that means we have the same line or infinite solutions. But that's what it looks like when we solve them. We come down with the true statements and all my variables are gone. Come over here to this problem. I can see here that I don't have any opposite signs. And I don't have any coefficients that are the same. But I could make 3 become 9. And I very easily could make this y become a 3y. So again, I this time want to get rid of y. We got rid of x over there, so let's get rid of y this time. If I get rid of y, I'm going for at least common multiple of 3. Well, this one's already a negative 3. So I'm going to take my bottom equation. And I'm going to multiply by, if I only multiply by 3, I still have negative 3y. So I need to multiply by negative 3. Because negative 3 times negative y will be a positive 3y. You're at my problem. This becomes negative 9x. We already said that that was going to be plus 3y. And then when we distribute all the way down to the negative 10, negative 3 times negative 10 is positive 30. Now we add the two equations. And our x's drop out. And our y's drop out. So we have 0 on the left hand side. And 12 plus 30 is 42. And 0 is not equal to 42. So we know that we have no solution. Again, your variables will drop out, but you'll end up with a not true statement with your numbers.