Given $2^x=8^{y+1}$ and $9^y=3^{x-9}$, find the value of $x+y$
$\textbf{(A)}\ 18 \qquad \textbf{(B)}\ 21 \qquad \textbf{(C)}\ 24 \qquad \textbf{(D)}\ 27 \qquad \textbf{(E)}\ 30$
Since $8^{y + 1} = 2^{3(y+1)}$ and $9^y = 3^{2y}$, we have:
$2^x = 2^{3(y+1)}$ and $3^{2y} = 3^{x - 9}$

Note that if $a^b = a^c$, then $b=c$.  Setting the exponents equal gives $x = 3y + 3$ and $2y = x - 9$.  Plugging the first equation into the second equation gives:
$2y = (3y + 3) - 9$
$2y = 3y - 6$
$0 = y - 6$
$y = 6$
Plugging that back in to $x = 3y + 3$ gives $x = 3(6) + 3$, or $x = 21$.  Thus, $x+y = 21 + 6$, or $x+y=27$, which is option $\boxed{\textbf{(D)}}$