Question

Solve the following question by factorisation:-$\dfrac{x + 1}{x - 1} + \dfrac{x - 2}{x + 2} = 3$​

Answer 1

Given Equation

$\tt \to \: \dfrac{(x + 1)}{(x - 1)} + \dfrac{(x - 2)}{(x + 2)} = 3$

Now Take LCM

$\tt\to \: \dfrac{(x + 1)(x + 2) + (x - 2)(x - 1)}{(x - 1)(x + 2)} = 3$

$\tt \to \: \dfrac{ {x}^{2} + 2x + x + 2 + {x}^{2} - x - 2x + 2 }{ {x}^{2} + 2x - x - 2} = 3$

$\tt \to \: \dfrac{ {x}^{2} + 3x + 2 + {x}^{2} - 3x + 2}{ {x}^{2} + x - 2 } = 3$

$\tt \to \: \dfrac{ 2{x}^{2} + 4}{ {x}^{2} + x - 2} = 3$

Using Cross multiplication

$\tt \to \: 2{x}^{2} + 4 = 3( {x}^{2} + x - 2)$

$\tt \to \: 2{x}^{2} + 4 = 3 {x}^{2} + 3x - 6$

$\tt \to \: 3 {x}^{2} - 2{x}^{2} + 3x - 6 - 4 = 0$

$\tt \to \: {x}^{2} + 3x - 10 = 0$

Now Using Middle term Splitting

$\tt \to \: {x}^{2} + 5x - 2x - 10 = 0$

$\tt \to \: x(x + 5) - 2(x + 5) = 0$

$\tt \to \: (x - 2)(x + 5) = 0$

$\tt \to \: (x - 2) = 0 \: and \: (x + 5) = 0$

$\tt \to \: x = 2 \: and \: x = - 5$

Answer

$\tt \to \: x = 2 \: and \: x = - 5$