Question

divide the sum of 65/12 and 8/3 by their difference

Answer 1

$$\begin{gathered} \frac{\left({\displaystyle \frac{65}{12}}+{\displaystyle \frac{8}{3}}\right)}{\left(\frac{65}{12}-\frac{8}{3}\right)} \\ =\frac{\left({\displaystyle \frac{65+32}{12}}\right)}{\left(\frac{65-32}{12}\right)} \\ =\frac{\left({\displaystyle \frac{97}{12}}\right)}{\left({\displaystyle \frac{33}{12}}\right)} \\ =\frac{97}{12}\times \frac{12}{33} \\ =\frac{97}{33} \end{gathered}$$