Divide the sum of 65/12 and 8/3 by their differ-renice.

Question

Divide the sum of \(\frac{65}{12}\) and \(\frac{8}{3}\) by their differ-renice.

Answer 1

According to the question,

\((\frac{65}{12}+\frac{8}{3})\div(\frac{65}{12}-\frac{8}{3})\)

= \((\frac{65\times1+8\times4}{12})\div(\frac{65\times1-8\times4}{12})\)

= \((\frac{65+32}{12})\div(\frac{65-32}{12})\)

= \((\frac{97}{12})\div(\frac{33}{12})\)

= \(\frac{97}{12}\times\frac{12}{33}\)

= \(\frac{97}{33}\)