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Linear and quadratic systems — Harder example

Video transcript

- [Instructor] If x comma y is a solution to the system of equations shown below, what is the product of the x-coordinates of the solution? This is a non-linear system. The first equation is non-linear. The second one is linear. But we wanna find the solution, the x comma y pair that satisfies both of these equations. Well, lucky for us, both of these have been solved explicitly for y. So this thing is equal to y, and this thing is equal to y. So for the x and y where we get the solution, this thing must be equal to this thing right over here because they're going to be equal to the same y. So let's do that. So we could either, we could view this as a substitution, that this is equal to y. So where we see y here, we can substitute it with two x plus seven. So if we do that, on the left hand side you get x squared plus x plus five is equal to, instead of y, since y is equal to two x plus seven, we could write is equal to two x plus seven. And now let's see if we can solve for x. So we can subtract two x from both sides. I'm doing that to clear out the right hand side. And we can subtract seven from both sides. And we are going to be left with, on the left hand side, we're gonna be left with x squared. X minus two x is negative x. Negative x. Positive five minus seven is negative two, is equal to, well, two x minus two x is zero. Seven minus seven is zero. That was by design. It's going to be equal to zero. Let's see, can I factor this quadrati expression? And if what I'm about to do looks like voodoo, (laughs) I encourage you to watch the videos on Khan Academy on factoring quadratic expressions. But to factor or at least attempt to factor, I have to think, well, are there two numbers whose product is negative two, but if I were to add them, they add to negative one? This negative x, the coefficient here, you can think of it as a negative one. So let's see, well, negative two and one, negative two and one, their sum is negative one. Their product is negative two. So I can factor this as x minus two times x plus one is equal to zero. And you can verify that if you take the product here, you're gonna get that over there. And so what are the x's that would make this, that would be solutions to this equation? Well, it could be the x that makes this expression equal to zero, which would be x is equal to two. Or it could be the x that makes this expression equal to zero, 'cause if either of these are equal to zero, the product is going to be equal to zero. So x is equal to negative one. So these are the x-coordinates of the two solutions. So one solution's gonna be two comma something, and then the other solution is going to be, so the other solution's going to be negative one and then something else. Now, if they wanted the solutions to the systems, you could take these x values and then substitute 'em back and figure out the corresponding y values. But for this problem, they don't even have to do that. They just say, "What is the product "of the x-coordinates of the solution?" So it's just gonna be two times negative one. So two times negative one is just negative two. That's our answer.