Nonlinear equations in one variable and systems of equations in two variables SAT question

Choice C is correct. The second equation in the given system of equations is $y=6x-12$. Substituting $6x-12$ for $y$ in the first equation of the given system yields $6x-12=(x-2)(x+4)$. Factoring $6$ out of the left-hand side of this equation yields $6(x-2)=(x-2)(x+4)$. An expression with a factor of the form $(x-a)$ is equal to zero when $x=a$. Each side of this equation has a factor of $(x-2)$, so each side of the equation is equal to zero when $x=2$. Substituting $2$ for $x$ into the equation $6(x-2)=(x-2)(x+4)$ yields $6(2-2)=(2-2)(2+4)$, or $0=0$, which is true. Substituting $2$ for $x$ into the second equation in the given system of equations yields $y=6(2)-12$, or $y=0$. Therefore, the solution to the system of equations is the ordered pair $(2,0)$.

Choice A is incorrect and may result from switching the order of the solutions for $x$ and $y$.

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

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