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 :

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

Squaring both sides we get :

⇒ 2x + 2 = 4(2x - 1)

⇒ 2x + 2 = 8x - 4

⇒ 8x - 2x = 2 + 4

⇒ 6x = 6

⇒ x = $\dfrac{6}{6}$ = 1.

Hence, x = 1.