Define the operation "$\circ$" by $x\circ y=4x-3y+xy$, for all real numbers $x$ and $y$. For how many real numbers $y$ does $3\circ y=12$?
$\text{(A) } 0\quad \text{(B) } 1\quad \text{(C) } 3\quad \text{(D) } 4\quad \text{(E) more than 4}$
Note that $3 \circ y = 4 \cdot 3 - 3y + 3y = 12$, so $3 \circ y = 12$ is true for all values of $y$. Thus there are more than four solutions, and the answer is $\fbox{E}$