1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

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

Open in App
Solution

## $x=\frac{6pq}{p+q}\left(\mathrm{Given}\right)\phantom{\rule{0ex}{0ex}}⇒\frac{x}{p}=\frac{6q}{p+q}...\left(\mathrm{i}\right)\phantom{\rule{0ex}{0ex}}⇒\frac{x}{q}=\frac{6p}{p+q}...\left(\mathrm{ii}\right)\phantom{\rule{0ex}{0ex}}\mathrm{We}\mathrm{have}\mathrm{to}\mathrm{find}\mathrm{the}\mathrm{value}\mathrm{of}\frac{x+3p}{x-3p}+\frac{x+3q}{x-3q}.\phantom{\rule{0ex}{0ex}}\mathrm{On}\mathrm{applying}\mathrm{componendo}\mathrm{and}\mathrm{dividendo},\mathrm{we}\mathrm{get}:\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\frac{x+3p+x-3p}{x+3p-x+3p}+\frac{x+3q+x-3q}{x+3q-x+3q}\phantom{\rule{0ex}{0ex}}=\frac{x}{3p}+\frac{x}{3q}\phantom{\rule{0ex}{0ex}}\mathrm{From}\left(\mathrm{i}\right)&\left(\mathrm{ii}\right),\mathrm{get}:\phantom{\rule{0ex}{0ex}}=\frac{1}{3}×\frac{6q}{p+q}+\frac{1}{3}×\frac{6p}{p+q}\phantom{\rule{0ex}{0ex}}=\frac{2\mathrm{q}}{\mathrm{p}+\mathrm{q}}+\frac{2\mathrm{p}}{\mathrm{p}+\mathrm{q}}\phantom{\rule{0ex}{0ex}}=\frac{2\left(\mathrm{p}+\mathrm{q}\right)}{\mathrm{p}+\mathrm{q}}\phantom{\rule{0ex}{0ex}}=2\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

Suggest Corrections
2
Join BYJU'S Learning Program
Related Videos
MATHEMATICS
Watch in App
Join BYJU'S Learning Program