Question

solve:x-a/x+b=a+b/a-b​

Answer 1

$\red{ :\implies \tt \: Solve \: \dfrac{x - a}{x + b} = \dfrac{a + b}{a - b} }$

$\tt \: :\implies \: \dfrac{x - a}{x + b} = \dfrac{a + b}{a - b}$

On Subtracting 1 from each term, we get

$\tt \: :\implies \: \dfrac{x - a}{x + b} - 1 = \dfrac{a + b}{a - b} - 1$

$\tt \: :\implies \: \dfrac{x - a - x - b}{x + b} = \dfrac{a + b - a + b}{a - b}$

$\tt \: :\implies \: \dfrac{ - b - a}{x + b} = \dfrac{2 b}{a - b}$

$\tt \: :\implies \: 2b(x + b) = - (b + a)(a - b)$

$\tt \: :\implies \: 2bx + {2b}^{2} = (b + a)(b - a)$

$\tt \: :\implies \: 2bx = {b}^{2} - {a}^{2} - {2b}^{2}$

$\tt \: :\implies \: 2bx = - {b}^{2} - {a}^{2}$

$\tt \: :\implies \: x = \dfrac{ - ( {b}^{2} + {a}^{2})}{2b}$

Verification :-

Consider LHS,

We have,

$\tt \: :\implies \: \dfrac{x - a}{x + b}$

$\tt \: :\implies \: on \: substituting \: x = \dfrac{ - ( {b}^{2} + {a}^{2})}{2b} \: we \: get \:$

$\tt \: :\implies \: \dfrac{\dfrac{ - ( {b}^{2} + {a}^{2})}{2b} - a}{\dfrac{ - ( {b}^{2} + {a}^{2})}{2b} + b}$

$\tt \: :\implies \: \dfrac{\dfrac{ - ( {b}^{2} + {a}^{2}) - 2ab}{2b}}{\dfrac{ - ( {b}^{2} + {a}^{2}) + {2b}^{2} }{2b}}$

$\tt \: :\implies \: \dfrac{ - ( {b}^{2} + {a}^{2} + 2ab) }{ - {a}^{2} - {b}^{2} + {2b}^{2} }$

$\tt \: :\implies \: - \dfrac{ {(b + a)}^{2} }{ {b}^{2} - {a}^{2} }$

$\tt \: :\implies \: - \dfrac{ {(b + a)}^{2} }{(b + a)(b - a)}$

$\tt \: :\implies \: - \dfrac{b + a}{b - a}$

$\tt \: :\implies \: \dfrac{a + b}{a - b}$

➦ LHS = RHS

$\large{\boxed{\boxed{\bf{Hence, verified}}}}$