sat-dog

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

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.