Question

(vi) The sum of the numerator and the denominator of fraction is 15. The denominator is 3 more than twice the numerator. Find the fraction.

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{the}\mathrm{numerator}\mathrm{and}\mathrm{denominator}\mathrm{of}\text{the}\mathrm{fraction}\mathrm{be}x\mathrm{and}y,\mathrm{respectively}. \\ \mathrm{Sum}\mathrm{of}\mathrm{numerator}\mathrm{and}\mathrm{denominator}=15 \\ \therefore x+y=15 \\ \Rightarrow x=15-y...\left(1\right) \\ \mathrm{According}\mathrm{to}\mathrm{the}\mathrm{question}, \\ y=3+2x...\left(2\right) \\ \mathrm{Substituting}\mathrm{eq}\left(1\right)\mathrm{in}\mathrm{eq}\left(2\right)\mathrm{we}\mathrm{get}: \\ y=3+2\times \left(15-y\right) \\ \Rightarrow y=3+30-2y \\ \Rightarrow 3y=33 \\ \Rightarrow y=11 \\ \mathrm{Substituting}y=11\mathrm{in}\mathrm{eq}\left(1\right)\mathrm{we}\mathrm{get}: \\ x=15-11 \\ \Rightarrow x=4 \\ \mathrm{Hence},\mathrm{the}\mathrm{fraction}\mathrm{would}\mathrm{be}\frac{4}{11}. \end{gathered}$$