If $\;$ $\dfrac{4m + 3n}{4m - 3n} = \dfrac{7}{4}$, $\;$ find
- $m : n$
- $\dfrac{2m^2 + 11n^2}{2m^2 - 11n^2}$
Given: $\;$ $\dfrac{4m + 3n}{4m - 3n} = \dfrac{7}{4}$
- By componendo and dividendo we have,
$\dfrac{4m + 3n + 4m - 3n}{4m + 3n - 4m + 3n} = \dfrac{7 + 4}{7 - 4}$
i.e. $\;$ $\dfrac{8m}{6n} = \dfrac{11}{3}$
i.e. $\;$ $\dfrac{m}{n} = \dfrac{11}{3} \times \dfrac{6}{8}$
i.e. $\;$ $m : n = 11 : 4$
- By componendo and dividendo we have,
$\begin{aligned} \dfrac{2m^2 + 11n^2}{2m^2 - 11n^2} & = \dfrac{2m^2 + 11n^2 + 2m^2 - 11n^2}{2m^2 + 11n^2 - 2m^2 + 11n^2} \\\\ & = \dfrac{4m^2}{22 n^2} \\\\ & = \dfrac{2}{11} \times \left(\dfrac{m}{n}\right)^2 \\\\ & = \dfrac{2}{11} \times \left(\dfrac{11}{4}\right)^2 \;\;\; [\text{from part (1)}] \\\\ & = \dfrac{11}{8} \end{aligned}$