Solve :

$\Big(\dfrac{2x - 3}{x - 1}\Big) - 4\Big(\dfrac{x - 1}{2x -3}\Big) = 3$

Let $\dfrac{2x - 3}{x - 1} =$ a …….(i)

$\Rightarrow a - \dfrac{4}{a} = 3 \\[1em] \Rightarrow \dfrac{a^2 - 4}{a} = 3 \\[1em] \Rightarrow a^2 - 4 = 3a \\[1em] \Rightarrow a^2 - 3a - 4 = 0 \\[1em] \Rightarrow a^2 - 4a + a - 4 = 0 \\[1em] \Rightarrow a(a - 4) + 1(a - 4) = 0 \\[1em] \Rightarrow (a + 1)(a - 4) = 0 \\[1em] \Rightarrow a = -1 \text{ or } a = 4.$

Substituting value of a = -1 in (i) we get,

$\Rightarrow \dfrac{2x - 3}{x - 1} = -1 \\[1em] \Rightarrow 2x - 3 = -1(x - 1) \\[1em] \Rightarrow 2x - 3 = -x + 1 \\[1em] \Rightarrow 2x + x = 1 + 3 \\[1em] \Rightarrow 3x = 4 \\[1em] \Rightarrow x = \dfrac{4}{3} \\[1em] \Rightarrow x = 1\dfrac{1}{3}$

Substituting value of a = 4 in (i) we get,

$\Rightarrow \dfrac{2x - 3}{x - 1} = 4 \\[1em] \Rightarrow 2x - 3 = 4(x - 1) \\[1em] \Rightarrow 2x - 3 = 4x - 4 \\[1em] \Rightarrow 4x - 2x = -3 + 4 \\[1em] \Rightarrow 2x = 1 \\[1em] \Rightarrow x = \dfrac{1}{2}.$

Hence, x = $1\dfrac{1}{3}, \dfrac{1}{2}$.