Question

If $$\frac{2a+2b-3c-3d}{2a-2b-3c+3d} = \frac{a+b-4c-4d}{a-b-4c+4d}$$, prove that $$\frac{a}{b} = \frac{c}{d}$$.

\[ \begin{array}{l} \left[ \textbf{Hint :}\ \text{We have}\ \dfrac{(2a - 3c) + (2b - 3d)}{(2a - 3c) - (2b - 3d)} = \dfrac{(a - 4c) + (b - 4d)}{(a - 4c) - (b - 4d)} \right. \\[16pt] \text{Now apply componendo and dividendo.}] \end{array} \]

Answer 1

Given: $$\frac{2a + 2b - 3c - 3d}{2a - 2b - 3c + 3d} = \frac{a + b - 4c - 4d}{a - b - 4c + 4d}$$

To prove: $$\frac{a}{b} = \frac{c}{d}$$

Proof:

1. $$\frac{(2a - 3c) + (2b - 3d)}{(2a - 3c) - (2b - 3d)} = \frac{(a - 4c) + (b - 4d)}{(a - 4c) - (b - 4d)}$$
2. or, $$\frac{[(2a - 3c) + (2b - 3d)] + [(2a - 3c) - (2b - 3d)]}{[(2a - 3c) + (2b - 3d)] - [(2a - 3c) - (2b - 3d)]} = \frac{[(a - 4c) + (b - 4d)] + [(a - 4c) - (b - 4d)]}{[(a - 4c) + (b - 4d)] - [(a - 4c) - (b - 4d)]}$$ [By componendo and dividendo]
3. or, $$\frac{2(2a - 3c)}{2(2b - 3d)} = \frac{2(a - 4c)}{2(b - 4d)}$$
4. or, $$\frac{2a - 3c}{2b - 3d} = \frac{a - 4c}{b - 4d}$$
5. or, $$\frac{2a - 3c}{a - 4c} = \frac{2b - 3d}{b - 4d}$$ [By alternendo]
6. or, $$(2a - 3c)(b - 4d) = (2b - 3d)(a - 4c)$$ [By cross multiplication]
7. or, $$2ab - 8ad - 3bc + 12cd = 2ab - 8bc - 3ad + 12cd$$
8. or, $$-8ad - 3bc = -8bc - 3ad$$
9. or, $$8bc - 3bc = 8ad - 3ad$$
10. or, $$5bc = 5ad$$
11. or, $$bc = ad$$
12. or, $$\frac{a}{b} = \frac{c}{d}$$

Hence proved.