Question

Solve it -

$\frac{x}{x - y} + \frac{y}{x + y} + \frac{2xy}{ {y}^{2} - {x}^{2} }$

Answer 1

$\bold{\huge{\underline{Answer-}}}$According to the question here we use the a² - b² = (a+b) (a-b)




$\frac{x}{x - y} + \frac{y}{x + y} + \frac{2xy}{ {y}^{2} - {x}^{2} }$




$\bold{\huge{Step-1}}$



$\frac{x(x + y) + y(x - y)}{(x - y)(x + y)} - \frac{2xy}{ {x}^{2} - {y}^{2} }$



$\bold{\huge{Step-2}}$




$\frac{ {x}^{2} + xy + xy - {y}^{2} }{ {x}^{2} - {y}^{2} } - \frac{2xy}{ {x}^{2} - {y}^{2} }$



$\bold{\huge{Step-3}}$




$\frac{ {x}^{2} + 2xy - {y}^{2} - 2xy}{ {x}^{2} - {y}^{2} }$



$\bold{\huge{Step-4}}$




$\frac{ {x}^{2} - {y}^{2} }{ {x}^{2} - {y}^{2} } = 1$





Hence Solved,

Answer 2

$\underline{\underline{\Huge{\textbf{Question:}}}}$

Solve
$\mathbf{\dfrac{x}{x - y} + \dfrac{y}{x + y} + \dfrac{2xy}{ {y}^{2} - {x}^{2} } }$ ———[1]



$\underline{\underline{\Huge{\textbf{Solution:}}}}$

$\underline{\Huge{\textsf{Step-1:}}}}$
Consider the first two terms in the given Expression.

$\mathbf{\dfrac{x}{x - y} + \dfrac{y}{x + y}}$ ———[2]



$\underline{\Huge{\textsf{Step-2:}}}}$
Consider LCM of the denominators of [1]

$\implies LCM(x-y, x+y) = (x-y)(x+y)$



$\underline{\Huge{\textsf{Step-3:}}}}$
Divide LCM by Denominator of first term of [1] & Multiply and Divide the result with first term of [1]

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

Multiply and Divide this value with first term in [1]

$\implies \dfrac{x}{x - y} \times \dfrac{x+y}{x+y}$

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

$\therefore \dfrac{x}{x - y} = \dfrac{x(x+y)}{(x-y)(x+y)}$———[3]



$\underline{\Huge{\textsf{Step-4:}}}}$
Divide LCM by Denominator of second term of [1] & Multiply and Divide the result with second term of [1]

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

Multiply and Divide this value with first term in [1]

$\implies \dfrac{x}{x + y} \times \dfrac{x-y}{x-y}$

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

$\therefore \dfrac{x}{x + y} = \dfrac{x(x-y)}{(x-y)(x+y)}$ ———[4]



$\underline{\Huge{\textsf{Step-5:}}}}$
Rewrite Equation [2] using [3] and [4]

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

$\mathbf{\dfrac{x(x+y) + y(x-y)}{(x + y)(x-y)}}$ ————[5]



$\underline{\Huge{\textsf{Step-6:}}}}$
Simplify and Rewrite [5] using the Identity

$\boxed{ \mathbf{(a+b)(a-b)= a^{2} - b^{2}}}$

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

$\implies \mathbf{\dfrac{x^{2}+2xy-y^{2}}{x^{2}-y^{2}}}$ ————[6]



$\underline{\Huge{\textsf{Step-7:}}}}$
Consider third term of [1]

$\mathbf{\dfrac{2xy}{y^{2}-x^{2}}}$

$\implies \dfrac{2xy}{-(x^{2}-y^{2})}$

$\mathbf{\dfrac{2xy}{y^{2}-x^{2}} =\dfrac{-2xy}{x^{2} - y^{2}}}$ ————[7]



$\underline{\Huge{\textsf{Step-8:}}}}$
Rewrite Eq.[1] using [6] & [7]

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

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

$\implies \dfrac{x^{2}-y^{2}}{x^{2} - y^{2}}$

$\implies \dfrac{1}{1}$

$\implies 1$



$\blacksquare \: \mathbf{\dfrac{x}{x - y} + \dfrac{y}{x + y} + \dfrac{2xy}{ {y}^{2} - {x}^{2} } } = \underline{\underline{\huge{1}}}$