Using componendo and dividendo solve for x : $\frac{\sqrt{2x + 2} + \sqrt{2x - 1}}{\sqrt{2x + 2} - \sqrt{2x - 1}} = 3$.

Given : $\frac{\sqrt{2x + 2} + \sqrt{2x - 1}}{\sqrt{2x + 2} - \sqrt{2x - 1}} = 3$

Applying componendo and dividendo, we get :

$$\begin{aligned}\Rightarrow \frac{\sqrt{2x + 2} + \sqrt{2x - 1} + \sqrt{2x + 2} - \sqrt{2x - 1}}{\sqrt{2x + 2} + \sqrt{2x - 1} - (\sqrt{2x + 2} - \sqrt{2x - 1})} = \frac{3 + 1}{3 - 1} \\[1em] \Rightarrow \frac{2\sqrt{2x + 2}}{2\sqrt{2x - 1}} = \frac{4}{2} \\[1em] \Rightarrow \frac{\sqrt{2x + 2}}{\sqrt{2x - 1}} = 2 \\[1em] \Rightarrow \sqrt{2x + 2} = 2\sqrt{2x - 1}\end{aligned}$$

Squaring both sides we get :

⇒ 2x + 2 = 4(2x - 1)

⇒ 2x + 2 = 8x - 4

⇒ 8x - 2x = 2 + 4

⇒ 6x = 6

⇒ x = $\frac{6}{6}$ = 1.

Hence, x = 1.