Systems of two linear equations in two variables SAT question

Choice A is correct. A solution to a system of equations must satisfy each equation in the system. It follows that if an ordered pair $(x,y)$ is a solution to the system, the point $(x,y)$ lies on the graph in the xy-plane of each equation in the system. The only point that lies on each graph of the system of two linear equations shown is their intersection point $(8,2)$.  It follows that if a new graph of three linear equations is created using the system of equations shown and the graph of $x+4y=-16$, this system has either zero solutions or one solution, the point $(8,2)$. Substituting $8$ for $x$ and $2$ for $y$ in the equation $x+4y=-16$ yields $8+4(2)=-16$, or $16=-16$. Since this equation is not true, the point $(8,2)$ does not lie on the graph of $x+4y=-16$. Therefore, $(8,2)$ is not a solution to the system of three equations. It follows that there are zero solutions to this system.

Choice B is incorrect and may result from conceptual or calculation errors.

Choice C is incorrect and may result from conceptual or calculation errors.

Choice D is incorrect and may result from conceptual or calculation errors.