$a$ and $b$ are real numbers and satisfy $ab^2=\frac{27}{5}$ and $a^2b=135$. Compute $a+5b$.
This problem involves two equations with two unknowns, so I will try to eliminate one of them by using substitution or elimination. I notice that both equations have $ab$ as a factor, so I can divide them to get rid of it. Dividing the first equation by the second, I get $\frac{b}{a}=\frac{1}{25}$. This means I can solve for $b$ in terms of $a$, or vice versa. I will choose to solve for $b$, since it has a lower degree. Multiplying both sides by $a$, I get $b=\frac{a}{25}$. Now I can plug this expression for $b$ into either equation to find $a$. I will use the first one, since it has smaller numbers. Substituting $b=\frac{a}{25}$ into $ab^2=\frac{27}{5}$, I get $a(\frac{a}{25})^2=\frac{27}{5}$. Simplifying, I get $\frac{a^3}{625}=\frac{27}{5}$. Multiplying both sides by 625, I get $a^3=3375$. Taking the cube root of both sides, I get $a=\sqrt[3]{3375}$. I can recognize that 3375 is $15^3$, so $a=15$. Now that I have $a$, I can find $b$ by plugging it into $b=\frac{a}{25}$. I get $b=\frac{15}{25}=\frac{3}{5}$. Finally, I can compute $a+5b$ by plugging in the values of $a$ and $b$. I get $a+5b=15+5(\frac{3}{5})=15+3=18$. 