Systems of equations with elimination (and manipulation)

Let's solve a few more systems of equations using elimination, but in these it won't be kind of a one-step elimination. We're going to have to massage the equations a little bit in order to prepare them for elimination. So let's say that we have an equation, 5x minus 10y is equal to 15. And we have another equation, 3x minus 2y is equal to 3. And I said we want to do this using elimination. Once again, we could use substitution, we could graph both of these lines and figure out where they intersect. But we're going to use elimination. But the first thing you might say, hey, Sal, you know, with elimination, you were subtracting the left-hand side of one equation from another, or adding the two, and then adding the two right-hand sides. And I could do that, because it was essentially adding the same thing to both sides of the equation. But here, it's not obvious that that would be of any help. If we added these two left-hand sides, you would get 8x minus 12y. That wouldn't eliminate any variables. And on the right-hand side, you would just be left with a number. And if you subtracted, that wouldn't eliminate any variables. So how is elimination going to help here? And the answer is, we can multiply both of these equations in such a way that maybe we can get one of these terms to cancel out with one of the others. And you could really pick which term you want to cancel out. Let's say we want to cancel out the y terms. So I'll just rewrite this 5x minus 10y here. 5x minus 10y is equal to 15. Now, is there anything that I can multiply this green equation by so that this negative 2y term becomes a term that will cancel out with the negative 10y? So I essentially want to make this negative 2y into a positive 10y. Right? Because if this is a positive 10y, it'll cancel out when I add the left-hand sides of this equation. So what can I multiply this equation by? Well, if I multiply it by negative 5, negative 5 times negative 2 right here would be positive 10. So let's do that. Let's multiply this equation times negative 5. So you multiply the left-hand side by negative 5, and multiply the right-hand side by negative 5. And what do you get? Remember, we're not fundamentally changing the equation. We're not changing the information in the equation. We're doing the same thing to both sides of it. So the left-hand side of the equation becomes negative 5 times 3x is negative 15x. And then negative 5 times negative 2y is plus 10y, is equal to 3 times negative 5 is negative 15. And now, we're ready to do our elimination. If we add this to the left-hand side of the yellow equation, and we add the negative 15 to the right-hand side of the yellow equation, we are adding the same thing to both sides of the equation. Because this is equal to that. So let's do that. So 5x minus 15y-- we have this little negative sign there, we don't want to lose that-- that's negative 10x. The y's cancel out. Negative 10y plus 10y, that's 0y. That was the whole point behind multiplying this by negative 5. Is going to be equal to-- 15 minus 15 is 0. So negative 10x is equal to 0. Divide both sides by negative 10, and you get x is equal to 0. And now we can substitute back into either of these equations to figure out what y must be equal to. Let's substitute into the top equation. So we get 5 times 0, minus 10y, is equal to 15. Or negative 10y is equal to 15. Let me write that. Negative 10y is equal to 15. Divide both sides by negative 10. And we are left with y is equal to 15/10, is negative 3/2. So if you were to graph it, the point of intersection would be the point 0, negative 3/2. And you can verify that it also satisfies this equation. The original equation over here was 3x minus 2y is equal to 3. 3 times 0, which is 0, minus 2 times negative 3/2 is, this is 0, this is positive 3. Right? These cancel out, these become positive. Plus positive 3 is equal to 3. So this does indeed satisfy both equations. Let's do another one of these where we have to multiply, and to massage the equations, and then we can eliminate one of the variables. Let's do another one. Let's say we have 5x plus 7y is equal to 15. And we have 7-- let me do another color-- 7x minus 3y is equal to 5. Now once again, if you just added or subtracted both the left-hand sides, you're not going to eliminate any variables. These aren't in any way kind of have the same coefficient or the negative of their coefficient. So let's pick a variable to eliminate. Let's say we want to eliminate the x's this time. And you could literally pick on one of the variables or another. It doesn't matter. You can say let's eliminate the y's first. But I'm going to choose to eliminate the x's first. And so what I need to do is massage one or both of these equations in a way that these guys have the same coefficients, or their coefficients are the negatives of each other, so that when I add the left-hand sides, they're going to eliminate each other. Now, there's nothing obvious-- I can multiply this by a fraction to make it equal to negative 5. Or I can multiply this by a fraction to make it equal to negative 7. But even a more fun thing to do is I can try to get both of them to be their least common multiple. I could get both of these to 35. And the way I can do it is by multiplying by each other. So I can multiply this top equation by 7. And I'm picking 7 so that this becomes a 35. And I can multiply this bottom equation by negative 5. And the reason why I'm doing that is so this becomes a negative 35. Remember, my point is I want to eliminate the x's. So if I make this a 35, and if I make this a negative 35, then I'm going to be all set. I can add the left-hand and the right-hand sides of the equations. So this top equation, when you multiply it by 7, it becomes-- let me scroll up a little bit-- we multiply it by 7, it becomes 35x plus 49y is equal to-- let's see, this is 70 plus 35 is equal to 105. Right? 15 and 70, plus 35, is equal to 105. That's what the top equation becomes. This bottom equation becomes negative 5 times 7x, is negative 35x, negative 5 times negative 3y is plus 15y. The negatives cancel out. And then 5-- this isn't a minus 5-- this is times negative 5. 5 times negative 5 is equal to negative 25. Now, we can start with this top equation and add the same thing to both sides, where that same thing is negative 25, which is also equal to this expression. So let's add the left-hand sides and the right-hand sides. Because we're really adding the same thing to both sides of the equation. So the left-hand side, the x's cancel out. 35x minus 35x. That was the whole point. They cancel out, and on the y's, you get 49y plus 15y, that is 64y. 64y is equal to 105 minus 25 is equal to 80. Divide both sides by 64, and you get y is equal to 80/64. And let's see, if you divide the numerator and the denominator by 8-- actually you could probably do 16. 16 would be better. But let's do 8 first, just because we know our 8 times tables. So that becomes 10/8, and then you can divide this by 2, and you get 5/4. If you divided just straight up by 16, you would've gone straight to 5/4. So y is equal to 5/4. Let's figure out what x is. So we can substitute either into one of these equations, or into one of the original equations. Let's substitute into the second of the original equations, where we had 7x minus 3y is equal to 5. That was the original version of the second equation that we later transformed into this. So we get 7x minus 3 times y, times 5/4, is equal to 5. Or 7x minus 15/4 is equal to 5. Let's add 15/4-- Oh, sorry, I didn't do that right. This would be 7x minus 3 times 4-- Oh, sorry, that was right. What am I doing? 3 times is 15/4. Is equal to 5. Let's add 15/4 to both sides. And what do we get? The left-hand side just becomes a 7x. These guys cancel out. And that's going to be equal to 5, is the same thing as 20/4. 20/4 plus 15/4. Or we get that-- let me scroll down a little bit-- 7x is equal to 35/4. We can multiply both sides by 1/7, or we could divide both sides by 7, same thing. Let's multiply both sides by 1/7. The same thing as dividing by 7. So these cancel out and you're left with x is equal to-- Here, if you divide 35 by 7, you get 5. You divide 7 by 7, you get 1. So x is equal to 5/4 as well. So the point of intersection of this right here is both x and y are going to be equal to 5/4. So if you looked at it as a graph, it'd be 5/4 comma 5/4. And let's verify that this satisfies the top equation. And if you take 5 times 5/4, plus 7 times 5/4, what do you get? It should be equal to 15. So this is equal to 25/4, plus-- what is this? This is plus 35/4. Which is equal to 60/4, which is indeed equal to 15. So it does definitely satisfy that top equation. And you could check out this bottom equation for yourself, but it should, because we actually used this bottom equation to figure out that x is equal to 5/4.