Answer
$\huge\color{blue}{\frac{7b}{8b + 7a}}$
Work Step by Step
$\large(\frac{8}{7} + \frac{a}{b})^{-1} \ne \frac{7}{8} + \frac{b}{a}$
We need to work the inside of the parentheses first.
$\large (\frac{8}{7} + \frac{a}{b})$$^{-1}$
Lets find the lowest common denominator so we can add the fractions.
$\large ((\frac{8}{\color{blue}{7}})(\frac{\color{green}{b}}{\color{green}{b}}) + (\frac{a}{\color{green}{b}})(\frac{\color{blue}{7}}{\color{blue}{7}}))$$^{-1}$
$\large (\frac{8b}{7b} + \frac{7a}{7b})$$^{-1}$
$\large (\frac{8b + 7a}{7b})$$^{-1}$
Now we can get rid of the $-1$ exponent by flipping the fraction.
$\large (\frac{8b + 7a}{7b})$$^{-1} \large= \frac{7b}{8b + 7a}$
The answer is:
$\huge\color{blue}{\frac{7b}{8b + 7a}}$
