If $x=12$ and $y=-6$, then the value of $\frac{3 x+y}{x-y}$ is
(A) 3
(B) 7
(C) $\frac{5}{3}$
(D) 5
(E) $\frac{7}{3}$
Since $x=12$ and $y=-6$, then

$$
\frac{3 x+y}{x-y}=\frac{3(12)+(-6)}{12-(-6)}=\frac{30}{18}=\frac{5}{3}
$$

ANswer: (C)