Question
Divide the sum of $\frac{65}{12}$ and $\frac{8}{3}$ by their difference.
✓
Answer
$\Big(\frac{65}{12}+\frac{8}{3}\Big)\div\Big(\frac{65}{12}-\frac{8}{3}\Big)$
$=\Big(\frac{65}{12}+\frac{8\times4}{3\times4}\Big)\div\Big(\frac{65}{12}-\frac{8\times4}{3\times4}\Big)$
$=\Big(\frac{65}{12}+\frac{32}{12}\Big)\div\Big(\frac{65}{12}-\frac{32}{12}\Big)$
$=\Big(\frac{65+32}{132}\Big)\div\Big(\frac{65-32}{12}\Big)$
$=\Big(\frac{65+32}{132}\Big)\div\Big(\frac{12}{65-32}\Big)$
$=\frac{65+32}{65-32}$
$=\frac{97}{33}$
$=\Big(\frac{65}{12}+\frac{8\times4}{3\times4}\Big)\div\Big(\frac{65}{12}-\frac{8\times4}{3\times4}\Big)$
$=\Big(\frac{65}{12}+\frac{32}{12}\Big)\div\Big(\frac{65}{12}-\frac{32}{12}\Big)$
$=\Big(\frac{65+32}{132}\Big)\div\Big(\frac{65-32}{12}\Big)$
$=\Big(\frac{65+32}{132}\Big)\div\Big(\frac{12}{65-32}\Big)$
$=\frac{65+32}{65-32}$
$=\frac{97}{33}$
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