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Question

# Solve for x and y: $\frac{10}{x+y}+\frac{2}{x-y}=4\phantom{\rule{0ex}{0ex}}\frac{15}{x+y}-\frac{9}{x-y}=-2$

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Solution

## The given equations are $\frac{10}{x+y}+\frac{2}{x-y}=4.....\left(i\right)\phantom{\rule{0ex}{0ex}}\frac{15}{x+y}-\frac{9}{x-y}=-2.....\left(ii\right)$ Substituting $\frac{1}{x+y}=u\mathrm{and}\frac{1}{x-y}=v$ in (i) and (ii), we get $10u+2v=4.....\left(iii\right)\phantom{\rule{0ex}{0ex}}15u-9v=-2.....\left(iv\right)$ Multiplying (iii) by 9 and (iv) by 2 and adding, we get $90u+30u=36-4\phantom{\rule{0ex}{0ex}}⇒120u=32\phantom{\rule{0ex}{0ex}}⇒u=\frac{32}{120}=\frac{4}{15}\phantom{\rule{0ex}{0ex}}⇒x+y=\frac{15}{4}\left(\because \frac{1}{x+y}=u\right).....\left(v\right)$ Now, substituting $u=\frac{4}{15}$ in (iii), we get $10×\frac{4}{15}+2v=4\phantom{\rule{0ex}{0ex}}\frac{8}{3}+2v=4\phantom{\rule{0ex}{0ex}}⇒2v=4-\frac{8}{3}=\frac{4}{3}\phantom{\rule{0ex}{0ex}}⇒v=\frac{2}{3}\phantom{\rule{0ex}{0ex}}⇒x-y=\frac{3}{2}\left(\because \frac{1}{x-y}=v\right).....\left(vi\right)$ Adding (v) and (vi), we get $2x=\frac{15}{4}+\frac{3}{2}⇒2x=\frac{21}{4}⇒x=\frac{21}{8}$ Substituting $x=\frac{21}{8}$ in (v), we have $\frac{21}{8}+y=\frac{15}{4}⇒y=\frac{15}{4}-\frac{21}{8}=\frac{9}{8}$ Hence, $x=\frac{21}{8}\mathrm{and}y=\frac{9}{8}$.

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