smc

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 $\boxed{27}$