The easiest method, which in this case is not very time consuming, is to guess and check and use process of elimination. 
We can immediately rule out (B) since it does not satisfy the third equation. 
The first equation allows us to eliminate (A) and (C) because the bases are different and aren't powers of each other.
Finally, we manually check (D) and (E) and see that our answer is $\fbox{D}$
In the set of equations $z^x = y^{2x},\quad  2^z = 2\cdot4^x, \quad x + y + z = 16$, the integral roots in the order $x,y,z$ are: 
$\textbf{(A) } 3,4,9 \qquad \textbf{(B) } 9,-5,-12 \qquad \textbf{(C) } 12,-5,9 \qquad \textbf{(D) } 4,3,9 \qquad \textbf{(E) } 4,9,3$