Solve the following equation by factorisation:

$\dfrac{a}{ax - 1} + \dfrac{b}{bx - 1} = a + b, a + b$ ≠ 0, $ab$ ≠ 0

Given,

$\dfrac{a}{ax - 1} + \dfrac{b}{bx - 1} = a + b \\[1em] \Rightarrow \dfrac{a}{ax - 1} + \dfrac{b}{bx - 1} - a - b = 0 \\[1em] \Rightarrow \big( \dfrac{a}{ax - 1} - b \big) + \big(\dfrac{b}{bx -1} - a \big) = 0 \\[1em] \Rightarrow \dfrac{a - b(ax - 1)}{ax - 1} + \dfrac{b - a(bx - 1)}{bx -1} = 0 \\[1em] \Rightarrow \dfrac{a - abx + b}{ax -1} + \dfrac{b - abx + a}{bx - 1} = 0 \\[1em] \Rightarrow (a - abx + b) \big( \dfrac{1}{ax - 1} + \dfrac{1}{bx - 1}\big) = 0 \text{ (Factorising left side) } \\[1em] \Rightarrow a - abx + b = 0 \text{ or } \dfrac{1}{ax - 1} + \dfrac{1}{bx - 1} = 0 \text{ (Zero-product rule) }\\[1em] \Rightarrow a + b = abx \text{ or } \dfrac{bx - 1 + ax - 1}{(ax - 1)(bx - 1)} = 0 \\[1em] \Rightarrow x = \dfrac{a + b}{ab} \text{ or } ax + bx - 2 = 0 \times (ax - 1)(bx - 1) \\[1em] \Rightarrow x = \dfrac{a + b}{ab} \text{ or } ax + bx - 2 = 0 \\[1em] \Rightarrow x = \dfrac{a + b}{ab} \text{ or } x(a + b) = 2 \\[1em] x = \dfrac{a + b}{ab} \text{ or } x = \dfrac{2}{(a + b)} \\[1em]$

Hence, the roots of given equation are $\dfrac{a + b}{ab} , \dfrac{2}{(a + b)}$.