 In this video, we'll talk about solving systems of equations with elimination. First thing we want to do is to see, understand the concept. So it asks us to add these two equations. So if we add them, 7f plus 3f would be 10f, and then positive g and negative g kind of cancel each other out. So they just give us zero, and 20 plus 20 would be 40. So these cancel each other out. And that's what we want to see with elimination. If we have opposite coefficients, then we know that we're going to be able to eliminate a variable. So in this first example, we can see that our f term has opposite signs on the coefficients. But the coefficients are not the same. But if I multiply 3 by 3, I can get a 9. But if I'm going to multiply this term by 3, I have to multiply the whole equation by 3. So I have to multiply 2g times 3, which will give me 6g. And I have to multiply negative 3f by 3, which will give me negative 9f. And I also have to multiply the negative 20 by 3, which will give me negative 60. The second equation, I don't have to change because it has the coefficient on f that I want, so I just rewrite that equation. And when I add these two equations, I get 6g plus 5g is 11g. And then my 9f and negative 9f cancel each other out. And then negative 60 plus negative 17 would be negative 77. And when I divide by 11, I find out that g is equal to negative 7. Now this is only part of the answer because remember it's an ordered pair that solves the system of equations. So we know what g is, but we don't know what f is. So we have to plug g either into the top equation or into the bottom equation. It doesn't matter which one. We could do both of them and we'd find out the same thing for the f when we solved. I'm going to take the top equation. And the only reason that I'm taking the top equation is because they're smaller numbers and it'll make it, I think, easier to solve. So I come in here and I say 2, but then I've got to put my g in there that I now know is negative 7. And then I write the rest of the problem minus 3f is equal to negative 20. So this is negative 14 minus 3f is equal to negative 20. Adding 14 to both sides, I get negative 3f is equal to negative 6 because the negative 20 is bigger than the positive 14 I added to both sides. And dividing by a negative 3, negative 6 divided by a negative 3 is going to be positive 2. So now I know that my answers are f is 2 and g is negative 7. Now when I look at this example, I look in my x coefficients are both positive and my y coefficients are both negative. So when my signs are the same for each variable, then I need to multiply by a positive or a negative. And then the other thing in this example is that a 4 cannot become 10 very easily and an 8 can't become 12 very easily. So I'm going to have to multiply both equations to get multiplied by something. So one of them will be a positive and one of them will be a negative. So if I look at this, I can either say that x is going to be looking at 4 and 10, they have 20 in common, or I could look at the y's and say that 12 and 8 have 24 in common. And it really doesn't matter which one we get rid of first, I'm going to get rid of my x. We've been getting rid of the middle one the last couple of times. So let's get rid of x. Again, that's the only reason why we could do either one. So if I want to get rid of x, this equation, to make it 20x, I'd have to multiply by 5. And I'm going to make that a negative 5 just so I don't forget to multiply by a negative. And this bottom equation then, if I want it to be 20x, I have to multiply by 2. So let's multiply. Remember, you have to multiply every term. So negative 5 times 4 would be negative 20x. Negative 5 times negative 12 would be plus 60y. And negative 5 times negative 8 would be positive 40. And then the second equation, 2 times 10x would be positive 20x. 2 times negative 8 would be minus 16y and 92. So when I add these two equations then, I'm going to have my x is cancel out. Negative 20 plus 20x is cancel each other out. So I have no x term. And 60 plus a negative 16 would give me 44y. And 40 and 92 will give me 132. And when I divide by 44, I find out that y is equal to 3. Remember now, we know y, we still need to find x. So we plug it into the top or bottom equation, it doesn't matter. Again, I'm going to try the top equation because I like the numbers better, they're smaller. So I come in here and I say 4 times x and then minus 12, I know that y is 3. And then equal to negative 8. So 4x, negative 12 times 3 would be negative 36 is equal to negative 8. And if I take 4x and add 36 to both sides, then I'm going to have a positive 28. And dividing by 4, x is going to be equal to 7. So x equals 7, y equals 3 is the solution that satisfies this system. All right, when we look at this equation, we have opposite signs on our coefficients for m. And opposite signs on our coefficient for n. But I can't make 2 become 7 very easily. I can't make 6 become 21 very easily. So again, I'm going to have to multiply both equations. And they're both going to get to be multiplied by a positive number. Since they're already opposite signs, I don't have to worry about making one of my signs the opposite. So which one do we want to get rid of? One of them is kind of a big number. So I'm going to try to get rid of the m. Could you get rid of n? Yes. But I'm going to look at the m and I'm looking at 2 and negative 7. And I want m to have a coefficient of 14. The top equation to become 14, I'm going to multiply by 7. And the bottom equation, if I want it to become 14 or negative 14 in this case, I'm going to multiply by 2. Distributing, I get 14 m minus 42 and then the n, negative 28. And for the bottom equation, I get negative 14 m plus 42 n and 28. So adding the two equations then, I get my positive 14 m and negative 14 n cancel each other out. And negative 42 n, and that was in positive 42 n, then they also cancel out. So on this side, I have 0. And then negative 28 plus 28 also gives me 0. So this is true. And if you remember from substitution, we talked about this. And if you get a true statement, that means that they are the same line or you have infinite solutions. Infinite solutions. When we look at this equation, I can see that I have all my signs are the same. So I'm going to have to multiply something by a negative. And I can make this k right here become a negative 2k if I just multiply one equation. So we're going to have to multiply by negative 2 on the top equation. So if I multiply here by negative 2, that will give me a negative 15 p minus 2k and negative 6. And then I can recopy the second equation because it has the coefficient on k that I want, a positive 2k. And now I can add the two equations. And when I add those two equations, negative 15 p and 15 p cancel each other out. Negative 2k and positive 2k also cancel out. So on this side, again, I have nothing or 0. And negative 6 plus 7 would give me 1. So this one's a false statement. And again, if you remember from what we did in substitution, that means that there is no solution.