$15 \cdot 32$ For all real numbers $x$ and $y$, the operation “$\circ$” is defined as
$$
x \circ y=4 x-3 y+x y \text {. }
$$

The number of real numbers $y$ such that $3 \circ y=12$ is
(A) 0.
(B) 1.
(C) 3.
(D) 4.
(E) more than 4.
(44th American High School Mathematics Examination, 1993)
[Solution] Substitute 3 for $x$ in the expression, we get
$$
3 \circ y=12-3 y+3 y=12 \text{. }
$$

Therefore, $3 \circ y=12$ holds true for all real numbers $y$. Hence, the answer is $(E)$.