Question

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

Answer 1

Given: $$\frac{\sqrt{3x} + \sqrt{2x - 1}}{\sqrt{3x} - \sqrt{2x - 1}} = 5$$

To prove: $$x = \frac{3}{2}$$

Proof:

1. $$\frac{\sqrt{3x} + \sqrt{2x - 1}}{\sqrt{3x} - \sqrt{2x - 1}} = \frac{5}{1}$$ [Given]
2. $$\frac{(\sqrt{3x} + \sqrt{2x - 1}) + (\sqrt{3x} - \sqrt{2x - 1})}{(\sqrt{3x} + \sqrt{2x - 1}) - (\sqrt{3x} - \sqrt{2x - 1})} = \frac{5 + 1}{5 - 1}$$ [By Componendo & Dividendo]
3. or, $$\frac{2\sqrt{3x}}{2\sqrt{2x - 1}} = \frac{6}{4}$$
4. or, $$\frac{\sqrt{3x}}{\sqrt{2x - 1}} = \frac{3}{2}$$
5. or, $$\frac{3x}{2x - 1} = \frac{9}{4}$$ [On squaring both sides]
6. or, $$4(3x) = 9(2x - 1)$$ [By cross multiplication]
7. or, $$12x = 18x - 9$$
8. or, $$6x = 9$$
9. or, $$x = \frac{9}{6} = \frac{3}{2}$$

Hence proved.