smc

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 $x>0⟹c+1>0⟹c>-1$. 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 $3-2c>0⟹3>2c⟹c<\frac{3}{2}$. Thus we combine the inequalities $c>-1$ and $c<\frac{3}{2}$ to get $-1<c<\frac{3}{2}$, which is answer $\boxed{-1<c<\frac{3}{2}}$.