y = x + 9 y = x 2 + 16 x + 63 A solution

Choice A is correct. It's given that $y=x+9$ and $y=x^2+16x+63$; therefore, it follows that $x+9=x^2+16x+63$. This equation can be rewritten as $x+9=\left(x+9\right)\left(x+7\right)$. Subtracting $\left(x+9\right)$ from both sides of this equation yields $0=\left(x+9\right)\left(x+7\right)-\left(x+9\right)$. This equation can be rewritten as $0=\left(x+9\right)\left(\left(x+7\right)-1\right)$, or $0=\left(x+9\right)\left(x+6\right)$. By the zero product property, $x+9=0$ or $x+6=0$. Subtracting $9$ from both sides of the equation $x+9=0$ yields $x=-9$. Subtracting $6$ from both sides of the equation $x+6=0$ yields $x=-6$. Therefore, the given system of equations has solutions, $\left(x,y\right)$, that occur when $x=-9$ and $x=-6$. Since $-6$ is greater than $-9$, the greatest possible value of $x$ is $-6$.

Choice B is incorrect. This is the negative of the greatest possible value of $x$ when $y=0$ for the second equation in the given system of equations.

Choice C is incorrect. This is the value of $y$ when $x=0$ for the first equation in the given system of equations.

Choice D is incorrect. This is the value of $y$ when $x=0$ for the second equation in the given system of equations.