Solve the following equation by factorisation:

$\dfrac{x + 1}{x - 1} + \dfrac{x - 2}{x + 2} = 3$

Given,

$\dfrac{x + 1}{x - 1} + \dfrac{x - 2}{x + 2} = 3 \\[1em] \Rightarrow \dfrac{(x + 1)(x + 2) + (x - 2)(x - 1)}{(x - 1)(x + 2)} = 3 \\[1em] \Rightarrow \dfrac{x^2 + 2x + x + 2 + x^2 - x - 2x + 2}{(x - 1)(x + 2)} = 3 \\[1em] \Rightarrow x^2 + 2x + x + 2 + x^2 - x - 2x + 2 = 3(x - 1)(x + 2) \\[1em] \Rightarrow x^2 + x^2 + 3x - 3x + 2 + 2 = 3(x^2 + 2x - x - 2) \\[1em] \Rightarrow 2x^2 + 4 = 3(x^2 + x - 2) \\[1em] \Rightarrow 2x^2 + 4 = 3x^2 + 3x - 6 \\[1em] \Rightarrow 2x^2 - 3x^2 - 3x + 4 + 6 = 0 \\[1em] \Rightarrow -x^2 - 3x + 10 = 0 \text{ (Writing as } ax^2 + bx + c = 0) \\[1em] \Rightarrow x^2 + 3x - 10 = 0 \text{ (Multiplying the equation by -1) } \\[1em] \Rightarrow x^2 + 5x - 2x - 10 = 0 \\[1em] \Rightarrow x(x + 5) - 2(x + 5) = 0 \\[1em] \Rightarrow (x - 2)(x + 5) = 0 \text{ (Factorising left side) } \\[1em] \Rightarrow x - 2 = 0 \text{ or } x + 5 = 0 \text{ (Zero-product rule) } \\[1em] x = 2 \text{ or } x = -5 \\[1em]$

Hence, the roots of given equation are 2 , -5.