problem: $a$ and $b$ are real numbers and satisfy $ab^2=\frac{27}{5}$ and $a^2b=135$. Compute $a+5b$.
solution: Rearranging the first equation, we have that $a=\frac{27}{5b^2}$. If we substitute this into the original equation, we get $\frac{729}{25b^4}b=135$; multiplying each side by $\frac{b^3}{135}$ yields $b^3=\frac{27}{125}$. Taking the cube root, we see that $b=\frac{3}{5}$. Substituting $b$ into the first equation, we get that $\frac{9}{25}a=\frac{27}{5}$ or $a=15$. Thus, $a+5b=15+3=\boxed{18}$.