Question

If $$x = \frac{6ab}{a + b}$$, prove that $$\left(\frac{x + 3a}{x - 3a} + \frac{x + 3b}{x - 3b}\right) = 2$$.

[Hint : We have $$\frac{x}{3a} = \frac{2b}{a + b}$$ and $$\frac{x}{3b} = \frac{2a}{a + b}$$. Apply Componendo & Dividendo on each.]

Answer 1

Given: $$x = \frac{6ab}{a + b}$$

To prove: $$\frac{x + 3a}{x - 3a} + \frac{x + 3b}{x - 3b} = 2$$

Proof:

1. $$x = \frac{6ab}{a + b}$$ [Given]
2. $$\frac{x}{3a} = \frac{2b}{a + b}$$ [Dividing both sides by $$3a$$]
3. $$\frac{x + 3a}{x - 3a} = \frac{2b + (a + b)}{2b - (a + b)}$$ [By Componendo & Dividendo]
4. or, $$\frac{x + 3a}{x - 3a} = \frac{a + 3b}{b - a} = -\frac{a + 3b}{a - b} \qquad (1)$$
5. Also, $$\frac{x}{3b} = \frac{2a}{a + b}$$ [Dividing both sides of $$x = \frac{6ab}{a + b}$$ by $$3b$$]
6. $$\frac{x + 3b}{x - 3b} = \frac{2a + (a + b)}{2a - (a + b)}$$ [By Componendo & Dividendo]
7. or, $$\frac{x + 3b}{x - 3b} = \frac{3a + b}{a - b} \qquad (2)$$
8. Adding (1) and (2), $$\frac{x + 3a}{x - 3a} + \frac{x + 3b}{x - 3b} = -\frac{a + 3b}{a - b} + \frac{3a + b}{a - b}$$
9. or, $$\frac{x + 3a}{x - 3a} + \frac{x + 3b}{x - 3b} = \frac{(3a + b) - (a + 3b)}{a - b}$$
10. or, $$\frac{x + 3a}{x - 3a} + \frac{x + 3b}{x - 3b} = \frac{2a - 2b}{a - b}$$
11. or, $$\frac{x + 3a}{x - 3a} + \frac{x + 3b}{x - 3b} = \frac{2(a - b)}{a - b} = 2$$

Hence proved.