Question:
Divide the sum of $\frac{65}{12}$ and $\frac{8}{3}$ by their difference.
Solution:
$\left(\frac{65}{12}+\frac{8}{3}\right) \div\left(\frac{65}{12}-\frac{8}{3}\right)$
$=\left(\frac{65}{12}+\frac{32}{12}\right) \div\left(\frac{65}{12}-\frac{32}{12}\right)$
$=\left(\frac{97}{12}\right) \div\left(\frac{33}{12}\right)$
$=\frac{97}{12} \times \frac{12}{33}$
$=\frac{97}{33}$