 In this video, I want to demonstrate how you can use the technique of elimination to help you solve systems of nonlinear equations. And in this video, we're going to use conic sections as our nonlinear equations. So like you see right here in front of you, we have a system with two equations, two unknowns, 3x squared plus 2y equals 26, 5x squared plus 7y equals 3. These are two nonlinear equations. If we look at their graphs, we actually end up with the graphs of some parabolas. So the first one is this parabola for which it's going to be concave downward. The other one also is going to be a parabola that concave downward. We don't necessarily need to know that in order to solve it, but it can be helpful to have these pictures in front of you, right? When it comes to the intersection between two parabolas, you can have anywhere from zero, one, two, three, or four possible intersections. This picture seems to suggest two intersections, but again, we'll be able to determine this completely algebraically. Did you notice if you look at the y's, we have a 2y, we have a 7y. If I adjusted the coefficients, I could cancel those out by elimination, but even better, look at that. We have x-squares in both the first and second equation. If I could cancel out the x-squares by adjusting the coefficients, that would be phenomenal because after all, the x-squares can be much more complicated than the y-squares. So that's actually the direction I want to go in. Let's eliminate the x-squares. So I could do that by adjusting coefficients. If I multiply the first equation by 5, you're going to get 15x squared plus 10y equals 26 times 5, which would be 130. And then we're going to take the second equation and we're going to multiply it by a negative 3. So we get a negative 15x squared, excuse me. We're times everything by negative 3, right? So I'll actually write this on the screen. So we take negative 3 times that one, and then we were doing positive 5 times the first one, like so. And so then we get negative 3 times 7y, which gives a negative 21y, and then we get negative 3 times 3, which is a negative 9. We're then going to add together these equations combining like terms. Now you'll notice, and this is the reason why we adjusted the coefficients the way we did, is that the 15x-squares are going to cancel out. The reason I chose 5 and negative 3 is I was trying to construct a common multiple of coefficients for x-square, and then make sure their coefficients were opposite and signed but equal in value there. So then combining the rest, we're going to end up with 10y minus 21y, which gives us a negative 11y. And then we take 130 minus 9, that's going to give you 121, for which if we divide both sides by negative 11, like so, you end up with y equals negative 11, 121 is 11 squared, right? So we end up with 11, excuse me, y equals negative 11. Now you'll notice here that we only got one y-coordinate, and when you look at the graph here, that kind of makes sense because the two points of intersection actually live on the same horizontal line. This is going to be the line y equals negative 11, right? Both of these functions are actually symmetric with respect to the y-axis, and so the two intersections are going to be reflective across the y-axis right there. So we only got one y-coordinate. That doesn't mean there's only one intersection, just means the two intersections actually have the same y-coordinate. So how do we figure out what the x-coordinates are? Well, we're going to take this y-coordinate and plug it into one of the two equations. It really doesn't matter which one you do for the sake of conversation, I'm just going to pick the first one here. So we're going to take 3x squared plus two times. Well, y, I know is negative 11, is equal to 26, right? This is a quadratic equation that we're going to try to solve. It only involves x squared, so I can actually solve it just by taking square roots here. So we're going to get 3x squared minus 22 equals 26. Let's add 22 to both sides. So we get 3x squared equals 48. Let's divide both sides by 3. We're going to end up with x squared equals 16. And then take the square root of both sides. We're going to end up with x equals plus or minus 4. And so we end up with 2x-coordinates, which is what our picture suggests right here. We have negative 4 and positive 4. So the two points of intersection are going to be 4, negative 11 and negative 4, negative 11. And so we're going to solve this system of equations using elimination. Let's do another example. In this example, what we're going to do is we're going to take the equation x squared plus y squared equals 13. So notice right here, this is going to be a circle. This is a circle centered at the origin and its radius is the square root of 13. All right, we can live with that, I suppose. The other one, x squared minus y equals 7, that's going to be a parabola. It might be a lot easier to see that if we solve for y, you get y equals x squared minus 7. So notice you could use substitution here. You could substitute this in for y and go from there, which would be perfectly fine to do such a thing. I somewhat hesitate to do that here because if you plug that in there, you're going to have to take a y squared, which requires that you actually have to get x to the fourth, right? So that's a degree 4 polynomial. If there's some way of avoiding that, I would like to do that. And that's why I actually think elimination might be a little bit friendly of a technique in this one. Because notice if we eliminated the x-squares, they'll be left with a y squared, a y and some constants, right? That's a quadratic equation. Given the choice between x to the fourth versus y squared, ding, ding, ding, quadratic is the way we want to go. But because x to the fourth is a possibility, I want you to realize that geometrically, if we start intersecting parabolas with circles, we can have up to four intersections like our picture seems to suggest right here. We could also get like two intersections and we have something like this, like the vertex of the parabolas inside the circle. You could have one intersection of the vertex is on the parabola. You could also get three intersections if you have the vertex on one side, but then it concaves into the circle. So you get something like that. You could also get no intersections because maybe the parabola is just not even touching the circle. So those are, geometrically, we anticipate anywhere from zero to four solutions. But let's find them algebraically. I'm going to do this by elimination. So if you just take the second equation and you times it by negative one, I think that'll actually do it for us here. Because that would then give us the first equation, x squared plus y squared equals 13. And then we're going to combine that with negative x squared plus y equals negative 7. If we add those two equations together, you'll see that the x squares cancel out, which is what we wanted. We really can't combine the y squared with the y. I mean, they're not like terms. So we end up with y squared plus y. That's okay. And on the right-hand side, we're going to take 13 minus 13 minus 7, which is going to give us 6. And so this is a quadratic equation. We can set the right-hand side equal to 0. y squared plus y minus 6 equals 0. So we look for factors of 6, negative 6, that add up to be 1. We can take y plus 3 and y minus 2, that equals 0. And so we end up with y being negative 3 and positive 2. So you'll notice that with our equations or with our y-coordinates right here, that does seem to work, right? There is like a horizontal line right here, y equals negative 3, that seems to match up. And there does also appear to be a horizontal line y equals 2 right there. Now we're going to get four solutions. And that's because as we plug in y equals negative 3 and y equals 2, we're going to get two possibilities here. Plug it into one of the two equations. It doesn't matter. I'm going to use this equation right here so I don't have to bother squaring the y-coordinates. It's by no means a big deal. You could use the other equation. But I'm going to use x squared minus y equals 7 here. So doing the first one here, negative 3, we're going to get x squared minus a negative 3 equals 7. And so therefore, x squared plus 3 equals 7. Subtract 3 from both sides. We get x squared equals 4. That is x equals plus or minus 2. This is going to give us two points of intersection. So we're going to have 2, negative 3. And we're going to get negative 2 comma negative 3 as points on the graph. Now we can see those points illustrated right here in our diagram. All right. For the next one, let's try y equals 2. So if x squared minus 2 equals 7, we can add 2 to both sides. We get x squared equals 9. Taking the square, we get x equals plus or minus 3. So again, there's going to be two more solutions here. There's going to be the point 3 comma 2. And there's also going to be the point negative 3 comma 2, which we can see on the graph right here. 3 comma 2 and negative 3 comma 2. So we found these points completely algebraically. But this technique of elimination is able to help us in this nonlinear system here. So a substitution would have worked in this example as well. But like I said, I think elimination works a little bit better because elimination leads to solving a quadratic equation as opposed to elimination. Elimination will lead to a quadratic equation while substitution leads to a quartic equation. And so oftentimes when you come to solving a problem like these nonlinear systems, you have options in front of you. How do you decide what's the best option? Well, you kind of have to look at where is he going to take me? And I want to take the path of least resistance. Some people call that lazy. I call that being efficient. Efficiency is choosing the better path, the one that requires you to do less work.