If $\frac{4^x}{2^{x+y}}=8$ and $\frac{9^{x+y}}{3^{5y}}=243$, $x$ and $y$ real numbers, then $xy$ equals:
$\text{(A) } \frac{12}{5} \quad \text{(B) } 4 \quad \text{(C) } 6 \quad \text{(D)} 12 \quad \text{(E) } -4$
$\frac{4^x}{2^{x+y}}=8\implies 4^x = 2^{x+y+3}\implies 2x=x+y+3 \implies x = y+3$.
$\frac{9^{x+y}}{3^{5y}}=243\implies 9^{x+y}=3^{5y+5}\implies 2x+2y=5y+5\implies 2x = 3y +5$.
So, $2y+6=3y+5\implies y = 1 \implies x = 4$. Therefore, $xy = 4$ or $\fbox{B}$