Use the properties of proportionality to solve : $\dfrac{\sqrt{12x + 1} + \sqrt{2x - 3}}{\sqrt{12x + 1} - \sqrt{2x - 3}} = \dfrac{3}{2}$.

Given,

Equation : $\dfrac{\sqrt{12x + 1} + \sqrt{2x - 3}}{\sqrt{12x + 1} - \sqrt{2x - 3}} = \dfrac{3}{2}$.

Applying componendo and dividendo we get :

$\Rightarrow \dfrac{\sqrt{12x + 1} + \sqrt{2x - 3} + (\sqrt{12x + 1} - \sqrt{2x - 3})}{\sqrt{12x + 1} + \sqrt{2x - 3} - (\sqrt{12x + 1} - \sqrt{2x - 3})} = \dfrac{3 + 2}{3 - 2} \\[1em] \Rightarrow \dfrac{2\sqrt{12x + 1}}{2\sqrt{2x - 3}} = 5 \\[1em] \Rightarrow \dfrac{\sqrt{12x + 1}}{\sqrt{2x - 3}} = 5$

Squaring both sides we get :

$\Rightarrow \Big(\dfrac{\sqrt{12x + 1}}{\sqrt{2x - 3}}\Big)^2 = 5^2 \\[1em] \Rightarrow \dfrac{12x + 1}{2x - 3} = 25 \\[1em] \Rightarrow 12x + 1 = 25(2x - 3) \\[1em] \Rightarrow 12x + 1 = 50x - 75 \\[1em] \Rightarrow 50x - 12x = 75 + 1 \\[1em] \Rightarrow 38x = 76 \\[1em] \Rightarrow x = \dfrac{76}{38} = 2.$

Hence, x = 2.