3・1 Let $c$ be a constant, the system of equations $\left\{\begin{array}{l}x-y=2, \\ c x+y=3 .\end{array}\right.$. A necessary and sufficient condition for the system to have a solution $(x, y)$ in the first quadrant is
(A) $c=-1$.
(B) $c>-1$.
(C) $c<\frac{3}{2}$.
(D) $0<c<\frac{3}{2}$.
(E) $-1<c<\frac{3}{2}$.
(38th American High School Mathematics Examination, 1987)
[Solution] From the given system of equations, we solve to get $x=\frac{5}{c+1}, y=\frac{3-2 c}{c+1}$. Since $(x, y)$ is in the first quadrant, we have
$$
\left\{\begin{array}{l}
\frac{5}{c+1}>0, \\
\frac{3-2 c}{c+1}>0 .
\end{array} \text { Solving, we get }-1<c<\frac{3}{2} .\right.
$$

Therefore, the answer is $(E)$.