Using Componendo and Dividendo solve for x :

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

Given,

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

Applying Componendo and Dividendo, we get :

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

Squaring both sides, we get :

$\Rightarrow \Big(\dfrac{\sqrt{2x + 2}}{\sqrt{2x - 1}}\Big)^2 = 2^2 \\[1em] \Rightarrow \Big(\dfrac{2x + 2}{2x - 1}\Big) = 4 \\[1em] \Rightarrow 2x + 2 = 4(2x - 1) \\[1em] \Rightarrow 2x + 2 = 8x - 4 \\[1em] \Rightarrow 8x - 2x = 4 + 2 \\[1em] \Rightarrow 6x = 6 \\[1em] \Rightarrow x = \dfrac{6}{6} = 1.$

Hence, x = 1.