Problem

If $x=3 a$ and $y=9 b,$ then all of the following…

Problem 9 Hard Difficulty

$$\begin{aligned} y &=x^{2} \\ 2 y+6 &=2(x+3) \end{aligned}$$
If $(x, y)$ is a solution of the system of equations
above and $x>0,$ what is the value of $x y ?$
\begin{equation}
\begin{array}{l}{\text { A) } 1} \\ {\text { B) } 2} \\ {\text { C) } 3} \\ {\text { D) } 9}\end{array}
\end{equation}

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Video Transcript

Okay, So it says if X comma y is a solution of the system, the equations of Buff and X is greater than zero. What is the value of X times? Why? Well, first of all, if we want to know the value of X times why we have to know what x and why actually are. And if we have a solution of that, if we want the solution of assistance of equations, that means that we first have to find a way to combine the two and then solve it up. Well, if y is equal to X squared and we're just going to take that x squared and plug it in right there. So that's gonna give me two X squared plus six to go two times X plus three. Okay. From here, I'm going to distribute out the two to both the ax and the three giving us two X plus six. And that's gonna be equal to X squared plus six. From here we can divide the entire equation by two giving me X squared plus three as he puts X plus three. And then from here we know that we congest cancel out the three and the three from both sides. Because if we do that, it's kind of like subtracting three from the entire equation. And then so from here, I'm going to just have X squared is equal to X, which means that if we divide both sides by X, we would get that s Siegel toe one. But we also know that if it's a square root, we have X squared is equal to X. We could've square square rooted on both sides. To get X is equal to the square root of X, which means it's going to be plus or minus X. So it's going to be not only the positive one, but it's also going to be the Axis negative one. But it does tell us that we wanted, except greater than zero, so we don't actually want to deal with the negative one. So then we know that our exes of one and we have to find her y stole in order to be able to put that into the X Times. Why? Well, if we have the one we know that Y is equal to one squared. So one squared is just one. So why is also one so quick. Currently, we've got that accessing what's one and that wise also equal to the one. So if you do extends, why one times one is just one, so they should be our answer.

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