 In this video, I'm going to show you how to solve linear systems in two variables by using elimination. This is basically just a different way of solving linear systems. Anyway, let's get started with it. So I'm going to solve using elimination. Now, the reason we call it elimination is because we are simply just eliminating one of the variables. Okay, so now notice how this system is set up. 3x plus 2y is equal to 4, and 4x minus 2y is equal to negative 18. Notice the x's are lined up, the y's are lined up, the equal signs are lined up, and the numbers are lined up. Okay, now that's on purpose. We want that to happen because we want everything to basically, we're going to call what I call add down the columns. Okay, so notice what's going to happen though. When I take these two systems and I add them together, I get 7x. Notice here 2 minus 2 is going to give me 0 equals a negative 14. Now notice this little blank spot here. That's what we mean by using elimination. We're going to take these two equations, we're going to add them together, and we're going to eliminate a variable. So in this case, we eliminated a y. So all I have to do now is divide both sides by 7, and I'm going to get negative 2 for my x. Okay, so what happens is you basically create this little equation right here. You basically create an equation with a single variable, and you can solve that equation for that single variable to figure out what the variable is. In this case, the variable is x minus 2. Okay, now just like substitution, whenever you figure out what one of the answers is, what one of the variables is, you simply just need to plug it back into one of your original equations. In this case, we can plug it back into either one of these, it doesn't really matter, but I would use the one that's going to be easiest. In this case, the top one. The numbers are smaller, I don't have negatives or anything like that, so I'm actually going to use this top one. I'm going to plug this back into the top one here to figure out what my y variable is. So 3x plus 2y is equal to 4, so in this case, 3 times negative 2 plus 2y is equal to 4. Negative 6 plus 2y equals 4, and let's see, add 6 to both sides, 2y is equal to 10. Negative by 2 on both sides, y is equal to 5, there we are. So my solution to this is going to be negative 2, 5, okay, now when I write my solutions, I always write them as coordinates, gives me the idea of where. So in the negative 2, 5, that would be the second quadrant is where these two lines would meet, okay, negative 2, 5, this guy gives me a little bit better understanding of what's going on. Okay, so that's an easy example of solving using elimination. Now what we're going to do is look at a different example, and we're going to do the same thing, we're going to eliminate a variable. Now, as you can see with this second example, we have 3x and 2x, 5y and 3y, and then negative 16 and negative 9. We want to look at these variables over here, now it's not quite apparent which one we're going to get rid of, and that's actually kind of the point of this example is that you can eliminate whatever variable you want to, you really don't, you're not really limited to what variable you have to solve for. You can eliminate the x's, eliminate the y's, it doesn't really matter. In this case, we need the front numbers, we need the coefficients here to be the same. So in this case for the x's, I need to multiply both of these equations so that the x's are 6, or I need to multiply the y's so that these coefficients here are 15, okay, the 6's and the 15 are the least common multiples of either the x's or the y's, okay. So in this case, like I said, it doesn't really matter which one you use, in this case, I'm going to use, I'm going to eliminate the x's. Two reasons why, the first reason is 6 is smaller, so all I have to do is multiply the top one times 2 and the bottom one times 3 to get 6, plus just to show you on the other example that I did, we eliminated the y's, this one I'm just going to eliminate the x's, so two reasons there. Alright, so as I said before, the top one, I'm going to multiply times 2, now the bottom one, I'm actually going to multiply by negative 3, now that might confuse you at first, but you'll see why I'm going to multiply by 3, let me multiply this out and you'll see why I multiply times a negative number, alright, so this one, I'm going to take times 2, now if I want to take it times 2, I have to take the entire equation times 2, okay, so this is going to be 6x plus 10y equals a negative 32, okay, everything needs to be multiplied times 2, same thing here on the bottom, everything is going to be multiplied by this negative 3, okay, so this is going to be negative 6x minus 9y equals a positive 27, okay, now I have two new equations to use, now notice why I multiply by 3, now it's a little bit more apparent, notice the x's here, one's a positive x and one is a negative x, when I add down my columns, when I add the x's together, when I add these x's together, I'm going to get zero and then that is why one of them is positive and I made the other one negative, it just makes the solving just a little bit easier, anyway, add that together to get zero, add down this column to get a positive 1y and then add this together to get a negative, negative 6, nope, not negative 6, negative 5, excuse me, 1, 2, 3, 4, 5, yeah, so negative 5, alright, so y is equal to negative 5 in this case, alright, let's double check in my work, yep, okay, so now what you do just like last time, since you've solved for one of these, what you need to do is plug it back into one of your other equations, now again it doesn't matter which one, I'm going to pick the bottom one to plug it back into, so I'm going to plug this back into the bottom, okay, so 2x plus 3 times negative 5 is equal to negative 9 and 2x minus 15 equals negative 9, okay, just multiply that together, negative 15, so I have to add that over to the other side, 2x equals, add 15 would be 15 minus 9 is 6, I don't know why that took me so long, x is equal to 3, there we go, alright, so that 3, that 3 and negative 5 that is my solution of 3, negative 5, and there we are, alright, and that's solving using elimination, just two quick examples there, the first one was relatively simple, this one though I wanted to show you this one because we have to change one, sometimes you have to change one or both of the equations to match, to match what you want, in this case I had to change both of the equations so that I could get, so I could eliminate one of my variables, in this case I eliminated the x's, yeah, just wanted to show you an example of that, anyway that is solving linear systems using either, the first video that I did was using substitution, and in this video I showed you a couple of examples using elimination.