Question

If $x=\frac{6pq}{p+q}$ then find value of $\frac{x+3p}{x-3p}+\frac{x+3q}{x-3q}$.

Answer 1

$$\begin{gathered} x=\frac{6pq}{p+q}\left(\mathrm{Given}\right) \\ \Rightarrow \frac{x}{p}=\frac{6q}{p+q}...\left(\mathrm{i}\right) \\ \Rightarrow \frac{x}{q}=\frac{6p}{p+q}...\left(\mathrm{ii}\right) \\ \mathrm{We}\mathrm{have}\mathrm{to}\mathrm{find}\mathrm{the}\mathrm{value}\mathrm{of}\frac{x+3p}{x-3p}+\frac{x+3q}{x-3q}. \\ \mathrm{On}\mathrm{applying}\mathrm{componendo}\mathrm{and}\mathrm{dividendo},\mathrm{we}\mathrm{get}: \\ \frac{x+3p+x-3p}{x+3p-x+3p}+\frac{x+3q+x-3q}{x+3q-x+3q} \\ =\frac{x}{3p}+\frac{x}{3q} \\ \mathrm{From}\left(\mathrm{i}\right)\&\left(\mathrm{ii}\right),\mathrm{get}: \\ =\frac{1}{3}\times \frac{6q}{p+q}+\frac{1}{3}\times \frac{6p}{p+q} \\ =\frac{2\mathrm{q}}{\mathrm{p}+\mathrm{q}}+\frac{2\mathrm{p}}{\mathrm{p}+\mathrm{q}} \\ =\frac{2\left(\mathrm{p}+\mathrm{q}\right)}{\mathrm{p}+\mathrm{q}} \\ =2 \end{gathered}$$