If $\dfrac{\sqrt{3x} + \sqrt{2x - 1}}{\sqrt{3x} - \sqrt{2x - 1}} = 5$, prove that x = $\dfrac{3}{2}$.

(i) Given,

$\dfrac{\sqrt{3x} + \sqrt{2x - 1}}{\sqrt{3x} - \sqrt{2x - 1}} = 5$

Applying Componendo and Dividendo, we get :

$\Rightarrow \dfrac{\sqrt{3x} + \sqrt{2x - 1} + \sqrt{3x} - \sqrt{2x - 1}}{\sqrt{3x} + \sqrt{2x - 1} - (\sqrt{3x} - \sqrt{2x - 1})} = \dfrac{5 + 1}{5 - 1} \\[1em] \Rightarrow \dfrac{2\sqrt{3x}}{\sqrt{3x} + \sqrt{2x - 1} - \sqrt{3x} + \sqrt{2x - 1}} = \dfrac{6}{4} \\[1em] \Rightarrow \dfrac{2\sqrt{3x}}{2 \sqrt{2x - 1}} = \dfrac{3}{2} \\[1em] \Rightarrow \dfrac{\sqrt{3x}}{ \sqrt{2x - 1}} = \dfrac{3}{2}$

Squaring both sides, we get :

$\Rightarrow \Big(\dfrac{\sqrt{3x}}{ \sqrt{2x - 1}}\Big)^2 = \Big(\dfrac{3}{2}\Big)^2 \\[1em] \Rightarrow \Big(\dfrac{3x}{2x - 1}\Big) = \Big(\dfrac{9}{4}\Big) \\[1em] \Rightarrow 4(3x) = 9(2x - 1) \\[1em] \Rightarrow 12x = 18x - 9 \\[1em] \Rightarrow 18x - 12x = 9 \\[1em] \Rightarrow 6x = 9 \\[1em] \Rightarrow x = \dfrac{9}{6} = \dfrac{3}{2}$

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