 In this video, I want to solve the nonlinear system of equations or whether it's two equations, two unknowns, where we have 3xy minus 2y squared equals negative 2 and 9x squared plus 4y squared equals 10. Now this example here is going to be challenging but doable. When you look at the second equation, it might be easy to see and maybe it's not so easy, but we could recognize that this second equation is an example of an ellipse. So you're going to get something like this, right? That's the graph of this thing. But the first equation is like, what? You have like x and y and a y squared. It's like, I have no idea how to graph that thing. Well, when it comes to nonlinear systems, you don't actually have to graph them, right? The graph can help intuitively help you solve this thing because we're looking for the intersections of the graphs. But this can be done purely algebraically and the strategies we have will be very effective. So let me show you what you can do. So first of all, I noticed that both equations have a y squared, so that's going to be my target. So we're not going to eliminate y squared, so we're going to be like Arnold Schwarzenegger and, you know, terminate this thing. So how are we going to do that? Well, I could times the first equation by a 2, right? And the signs are already opposite and so that would give us a 6xy minus 4y squared is equal to negative 4 and this will then be added to 9x squared plus 4y squared equals 10. Like so, for which then the y squares will cancel and you can add together the constants, but not everything's going to be like terms here. You're going to end up with a 9x squared plus 6xy is equal to 6 when we combine these things together. I noticed that both sides of the equation are visible by 3, so let's do it to both sides, divide by 3. That gives us 3x squared plus 2xy equals 2, right? We like smaller coefficients with possible. But we still have this equation with an x and a y together, but you'll notice with the y here, there's just a single y. I could solve for y right here. So if you solve for y, we got rid of the y squares, now we're going to solve for y. That would look something like the following. It's attracting negative 3x squared from both sides. So we get 2xy equals 2 minus 3x squared and I could get rid of the coefficients in front of, so I'll just mark down what we did here. We subtracted 3x squared from both sides. We could get rid of the 2x on the left by division and so we end up with y equals 2 minus 3x squared over 3x, like so. And so, okay, we have a y equals this. What do we do with that? Well, whenever you solve for a variable, it makes sense to be like, oh, I'm going to substitute this. We're going to substitute out the y-coordinate. And so I'm going to take this to, let's say, which one should we do? We could do the first equation, the second equation. The second equation seems a little bit cleaner to me. So what if we substituted y equals this into the second equation? Right? What would that look like? So I'm going to bring this down a little bit. So that's, that elliptical equation looked like 9x squared plus 4y squared equals 10. You can't see it on the screen right now. But if we substitute, if we substitute this value, this expression in for y, we end up getting y, our 9x squared plus 4 times y. But instead of y, we're going to get 2 minus 3x squared over 3x, quantity squared, that equals 10 right there. Oh, why did I write 3? That should be a 2, my mistake. My penmanship went down the drain, I suppose. So let's fix that. That should be a 2x. And then 2x. This is often why it's a good idea to show your work because you can detect your errors more often in that regard. So we have a 2x right there. So let's, let's go forward with this thing here. This, this is what we have to do. So I'm going to have to multiply out that fraction. So we're going to get the 9x squared times that by 4. On the top, you're going to have to foil. So you get 2 times 2, which is 4. You get 2 times negative 3x squared, which is a negative 6x. You do that twice. You're going to negative 12x squared. Then you're going to get negative 3x squared times negative 3x squared. You're squaring it. So you get a 9x to the fourth. Let's make sure that looks like a 4. And then on the bottom, we have to also square that thing. So you get a 4x squared. And then the right-hand side equals a 10. So what can we do here? Well, we have a, those 4s cancel out, which is nice enough here. To get rid of this x squared, I actually want to clear the denominators. So I can multiply both sides of the equation by an x squared. For which you're going to have to distribute it onto the two terms. On to the first term, you're going to end up with a 9x to the fourth. On to the second term, they actually will cancel out. And so you're left with just a 4 minus 12x squared plus 9x to the fourth is equal to 10x squared. Alright, so this is a polynomial equation. We're going to want to set the right-hand side equal to zero in factor. So I'm going to subtract 10x squared from both sides to combine like terms. We also have a 9x to the fourth and a 9x to the fourth right there. So combining these things together looks like we have an 18x to the fourth. In terms of the x squared, we're going to have a negative 22x squared. And then we have a plus four equals zero. So again, if I ever see a common coefficient, I just want to get rid of it. Multiply the left-hand side, or divide the left and right-hand side by two. What's good for the goose is good for the gander. So we get 9x to the fourth minus 11x squared plus two is equal to zero. Like so. So how do you work with an equation like this? Well, in a situation like this, it's like we have this degree four polynomial. We could start doing like rational roots and things like that. So we look for factors of two divided by factors of nine. We could look for rational roots in that way and we could do that. The problem, of course, with this one is that all of the solutions, well, not all of them, but some of the solutions are actually irrational, which complicates the issue a little bit. We can go forward with that. But actually, this equation is what I like to call a quadratic-like equation. Oops, assuming I can spell quadratic here. Quadratic-like equation. Did I spell it right? I don't know. I can't spell where I used to abbreviate it. Well, it doesn't even look like here. I'm just going to erase it and try again. 3, 2, 1. This equation is what we call a quadratic-like equation. Because if we make a substitution of variables, like if we say u equals x squared, notice that u squared would equal x squared squared, which is x to the fourth. And so you see that right here. We have an x squared and an x to the fourth. With this substitution in hand, we could actually rewrite this equation as 9u squared minus 11u plus 2. For which then we could try to factor this thing. Factors of 2 times 9, which is 18, that add up to be negative 11. We could factor this thing. I'm just going to take the liberty of factoring this thing. This is going to look like 9u minus 2 times u minus 1 equals 0. Notice if you multiply this thing out again, 9u times u is the 9u squared. Negative 2 times negative 1 is positive 2. And then you're going to get negative 2 minus 9. That's the negative 11 right there. So we get the following factorization, which then tells you u should equal 2 ninths and 1. Great. But u is not what we want to solve for. We want to solve for x. So u is equal to x squared. So this tells us that x squared equals 2 ninths and 1. For which when we take the square root of both sides, we end up with x equals plus or minus the square root of 2 over 3 and plus or minus the square root of 1, which is 1, right there. So we actually get four possible solutions to this thing right here. This gave us the x-coordinates. There's four possible x-coordinates. Come back up to the equation we had for y earlier. Here we go. So we had this thing, y equals 2 minus 3x squared over 2x. Let's bring that down right here. I'm going to copy it down. y equals, what did we say it was? 2 minus 3x squared over 2x. If we plug these four values of x inside of there, what's going to happen? So there's four cases to consider. We're going to mention all of them. So when x equals the square root of 2 over 3, then y is going to equal 2 minus 3 times the square root of 2 over 3 squared over 2 times the square root of 2 over 3. This is going to give us a 2 minus 3 times 2 over 9 over, well, 2 root 2 over 3. And so what we're going to do here is, we'll notice the 3 goes into the 9, leaving the 3 on the bottom. I'm going to times the big fraction by 3 over 3. Kind of clear out the denominators here. This is going to give us the number 6 minus 2 over 2 root 2. For which if we keep on going with that, we end up with 4 over 2 root 2, which gives us 2 over root 2. For which then if we rationalize the denominator, we get the square root of 2 over the square root of 2. This then gives us 2 root 2 over 2, the 2's cancel. And so we get that the value is actually the square root of 2, which is like approximately 1.4 or something like that. So we end up with the square root of 2. Now, if we had inserted instead negative the square root of 2 into this, notice that we get a negative sign right here and a negative sign right here. Well, when you square a negative, it just makes everything positive. So the rest of the numerator is exactly the same. The difference is there's going to be a negative right here. So what we can do here is actually if we put in the sign plus or minus, then we're going to get plus or minus square root of 2 as our solutions in the end. So if we record those, I'm going to keep track of my solutions right here. So we have one, we're going to get the square root of 2 over 3 comma square root of 2. That's one solution. The other one was negative square root of 2 over 3 comma negative square root of 2. We have to also consider what happens when x equals 1. You're going to get y equals 2 minus 3 times 1 squared over 2 times 1, as this is a rational solution and integer solution in fact. It's a lot easier. So we end up with 2 minus 3 over 2. So we end up with negative 1 half as the solution. So we get 1 comma negative 1 half as my solution right there. Now, if I plugged in a negative sign, what happened there? Well, you put in a negative right here. You get a negative right here. The negative on the top would disappear, but they have this negative sign on the bottom. So you get a plus 1 half there. And so that's how the answer would change. So in this situation, if you have a plus or minus 1, your answer is going to be plus or minus negative 1 half. Or some people write this as negative plus 1 half. So that the top goes with the top and the bottom goes with the bottom. So the last solution would look like negative 1 comma 1 fourth. So this one was much more of a doozy than some of the other ones would do. So what you can do to solve these nonlinear systems, sometimes you can combine together substitution and elimination in order to solve the linear system.