Using properties of proportion, solve for $x$:

Given,

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

Applying componendo and dividendo,

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

Squaring both sides,

$$\begin{aligned}\Rightarrow \frac{2x + 3}{2x - 3} = 4 \\[1em] \Rightarrow 2x + 3 = 4(2x - 3) \\[1em] \Rightarrow 2x + 3 = 8x - 12 \\[1em] \Rightarrow 3 + 12 = 8x - 2x \\[1em] \Rightarrow 15 = 6x \\[1em] \Rightarrow x = \frac{15}{6} = \frac{5}{2}.\end{aligned}$$

Hence, $\bm{x = \frac{5}{2}}$.