 In this video, we're going to learn the second of our techniques for solving a system of linear equations known as the elimination method. The elimination method gets its name because we're trying to eliminate variables. If you're really in a grouchy move today, the elimination method is really going to make you feel better because we're just going to kill these variables X and Y in the system here. So in order to start the elimination method, you have to choose a variable to eliminate. That basically means that we're going to choose some variable that we want gone, right? And so this could be done in the following way. We could choose the variable X to eliminate or we could choose the variable Y to eliminate. We can choose either one. It doesn't matter. Just make a choice. Now you might have to adjust the coefficients in order to make this work. We'll do an example of that just very shortly. But for this problem, choosing to eliminate Y is actually a great choice because once you've chosen to eliminate the variable and you adjust the coefficients, you simply then add together the equations and the variable you chose to eliminate will now then be gone, right? If we take these two equations and we add them together, you're going to get 3X plus X, which is a 4X. You're going to get 2Y minus 2Y, which is a 0Y, and then you get 14 plus 2, which is 16. Notice this just becomes 4X. 4X is equal to 16. Divide both sides by 4 and you see that X is equal to 4 itself. So we very quickly found out what the value is by elimination. And so then we're going to say X equals 4. We're going to keep that in check here. So once you've eliminated a variable, you can solve for the other variable pretty quickly. And then we want to figure out what is the other variable, right? We eliminated Y and we found out X. Well, what is Y then? Well, there's two ways to kind of figure out how to finish this thing off. The first one, we're going to call this the hybrid elimination, sort of the path we can take here. And once we figure out what X is, we can find out what Y is, basically the same way we do with the substitution method. You're going to take X equals 4 and plug it into one of these equations. Well, not for Y, for X, something like that. And so then you're going to get 3 times 4 plus 2 times Y equals 14. You'll notice there's no more X, you just have a Y. You get 12 plus 2Y equals 14. Subtract 12 from both sides. You get 2Y equals just a 2, 14 takeaway 12. And so divide by 2, you get Y equals 1. And so you get Y equals 1 over here, or in other words, the solution to the system is 4 comma 1. We're going to record this here. That's how we get this hybrid elimination. You eliminate one variable, find it, and you substitute it back into the other equation to figure out the first variable. That's one option. If we're in a really, really grouchy mood, right, maybe we just want to destroy everything in our path. And so I'm going to clear off the screen right here. And so if one wants to try a pure elimination method, what I mean is you're just going to go back, start the problem over again, but you're going to eliminate the other variables. So we started off by eliminating Y that found X. Now we're just going to eliminate X and we're going to find Y. All right, so how would that do that? Well, in this situation, we actually do have to adjust the coefficients, right? If this time we add together the X's, we're not going to cancel out the X's. We need to adjust the coefficients. And so what we could do is we could times the second equation by negative 3. We're just going to keep the first equation as it is. But we're going to times the second equation on both sides by negative 3. And this will give us negative 3X plus 6Y is equal to negative 6. And so this time when we add them together, you're going to see that because of the adjustment of the coefficients, the 3X minus 3X will actually cancel each other out. That's what we're trying to do. We want to make the coefficients of the X be opposite, but, you know, opposite, but equal, right? So we want an absolute value of 3, but once positive, one's negative. On the other hand, though, when you combine the Y's together, you get 2Y plus 8Y, you get, sorry, 2Y plus 6Y, which is 8Y. They don't cancel out this time. And then 14 takeaway 6 is itself 8. Divide by 8 on both sides, you get Y equals 1, which is what we saw earlier, right? It's the same solution. So we can, we can eliminate one variable and then substitute for the other, or we can just eliminate both variables in two different sets. And that can give us the solution either way. This either technique works for systems of 2 by 2 linear equations. That is two equations, two unknowns. It's a pretty nice technique. And the substitution elimination, there's not really one method that's better than the other. It's kind of what do you prefer to use? There are situations where I think substitutions should be a little bit easier. I think there's some situations where elimination could be a little bit easier. It depends on the coefficients. Like this one, this one was set up for elimination. You want to take out Y. But either one's going to be okay. I would recommend trying out both techniques and see which one you like better. And then if you feel like you gravitate towards one over the other, then use that one to solve these 2 by 2 linear systems.