SAT Question 7dbd46d9: Nonlinear Equations In One Variable And Systems Of Equations In Two Variables answer and explanation with Desmos solution

Choice A is correct. It's given that the graphs of the equations in the given system of equations intersect at the point $\left(x,y\right)$. Therefore, this intersection point is a solution to the given system. The solution can be found by isolating $y$ in each equation. The given equation $8x+y=-11$ can be rewritten to isolate $y$ by subtracting $8x$ from both sides of the equation, which gives $y=-8x-11$. The given equation $2x^2=y+341$ can be rewritten to isolate $y$ by subtracting $341$ from both sides of the equation, which gives $2x^2-341=y$. With each equation solved for $y$, the value of $y$ from one equation can be substituted into the other, which gives $2x^2-341=-8x-11$. Adding $8x$ and $11$ to both sides of this equation results in $2x^2+8x-330=0$. Dividing both sides of this equation by $2$ results in $x^2+4x-165=0$. This equation can be rewritten by factoring the left-hand side, which yields $\left(x+15\right)\left(x-11\right)=0$. By the zero-product property, if $\left(x+15\right)\left(x-11\right)=0$, then $\left(x+15\right)=0$, or $\left(x-11\right)=0$. It follows that $x=-15$, or $x=11$. Since only $-15$ is given as a choice, a possible value of $x$ is $-15$.