Question

Q88. Simplify:

$\sf{\dfrac{x^{-1}}{x^{-1}+y^{-1}}+\dfrac{x^{-1}}{x^{-1}-y^{-1}}}$​

Answer 1

Solution

To simplify the fraction, we should multiply $\dfrac{xy}{xy}$ on both fractions.

Given: $\dfrac{x^{-1}}{x^{-1}+y^{-1}} +\dfrac{x^{-1}}{x^{-1}-y^{-1}}$

$=\dfrac{xy(x^{-1})}{xy(x^{-1}+y^{-1})} +\dfrac{xy(x^{-1})}{xy(x^{-1}-y^{-1})}$

$=\dfrac{y}{y+x} +\dfrac{y}{y-x}$

Taking LCM of the denominator,

$=\dfrac{y(y-x)}{(y+x)(y-x)} +\dfrac{y(y+x)}{(y+x)(y-x)}$

$=\dfrac{y^2-xy+y^2+xy}{y^2-x^2}$

$=\dfrac{2y^2}{y^2-x^2}=\boxed{-\dfrac{2y^2}{x^2-y^2} }$

Answer 2

${\large{\underline{\underline{\pmb{\frak{Given:-}}}}}}$

$\sf \bullet \;\; \dfrac{x^{-1}}{x^{-1}+y^{-1}} + \dfrac{x^{-1}}{x^{-1}-y^{-1}}$

${\large{\underline{\underline{\pmb{\frak{solution:-}}}}}}$

$\sf : \;\implies \dfrac{x^{-1}}{x^{-1}+y^{-1}} + \dfrac{x^{-1}}{x^{-1}-y^{-1}}$

Solving first fraction :-

$\sf : \implies \dfrac{1}{ \bigg\{1+\dfrac{1}{y} \; x \bigg\} \times \bigg\{ \dfrac{1}{x} \bigg\} \;x }$

$\sf : \; \implies \dfrac{1}{1+\dfrac{1}{y} \; x}$

$\sf : \; \implies \dfrac{1}{1+\dfrac{x}{y}}$

$\sf : \; \implies \dfrac{1}{\dfrac{y}{y} + \dfrac{x}{y}}$

$\sf : \; \implies \dfrac{1}{\dfrac{y+x}{y}}$

$\sf : \; \implies \dfrac{y}{y+x}$

Solving second fraction :-

$\sf : \; \implies \dfrac{1}{ \bigg\{ - \dfrac{1}{y}x + 1 \bigg\} \times \bigg\{ \dfrac{1}{x} \bigg\} x }$

$\sf : \; \implies \dfrac{1}{ - \dfrac{1}{y} \; x + y }$

$\sf : \; \implies \dfrac{1}{ - \dfrac{x}{y} + \dfrac{y}{y}}$

$\sf : \; \implies \dfrac{1}{ \dfrac{-x+y}{y}}$

$\sf : \; \implies \dfrac{y}{-x+y}$

Now, solving both fractions together :-

$\sf : \; \implies \dfrac{y}{y+x} + \dfrac{y}{-x+y}$

$\sf : \; \implies \dfrac{y(-x+y)}{(x+y)(-x+y)} + \dfrac{y(x+y)}{(x+y)(-x+y)}$

$\sf : \; \implies \dfrac{y(-x+y)+y(x+y)}{(x+y)(-x+y)}$

$\sf : \; \implies \dfrac{ -yx + y^{2} + yx + y^{2}}{(x+y)(-x+y)}$

$\sf : \; \implies \dfrac{2y^{2}}{(x+y)(-x+y)}$

$\boxed{\bf{ \leadsto \;\; \dfrac{2y^{2}}{-x^{2}+y^{2}}}}$