Let $c$ be a constant. The simultaneous equations
\begin{align*}x-y = &\ 2 \\cx+y = &\ 3 \\\end{align*}
have a solution $(x, y)$ inside Quadrant I if and only if
$\textbf{(A)}\ c=-1 \qquad \textbf{(B)}\ c>-1 \qquad \textbf{(C)}\ c<\frac{3}{2} \qquad \textbf{(D)}\ 0<c<\frac{3}{2}\\ \qquad \textbf{(E)}\ -1<c<\frac{3}{2}$
We can easily solve the equations algebraically to deduce $x = \frac{5}{c+1}$ and $y = \frac{3-2c}{c+1}$. Thus we firstly need $x > 0 \implies c + 1 > 0 \implies c > -1$. Now $y > 0$ implies $\frac{3-2c}{c+1} > 0$, and since we now know that $c+1$ must be $>0$, the inequality simply becomes $3-2c > 0 \implies 3 > 2c \implies c  -1$ and $c < \frac{3}{2}$ to get $-1 < c < \frac{3}{2}$, which is answer $\boxed{\text{E}}$.