Question
If a : b = c : d, prove that: (6a + 7b) (3c – 4d) = (6c + 7d) (3a – 4b).
✓
Answer
Given $\frac{a}{b}=\frac{c}{d}$
$\Rightarrow \frac{6 a}{7 b}=\frac{6 c}{7 d} \quad$ (Multiplying each side by $6 / 7$ )
$\Rightarrow \frac{6 a+7 b}{7 b}=\frac{6 c+7 d}{7 d}$ (By componendo)
$\Rightarrow \frac{6 a+7 b}{6 c+7 d}=\frac{7 b}{7 d}=\frac{b}{d} \ldots . .(1)$
Also $\frac{a}{b}=\frac{c}{d}$
$\Rightarrow \frac{3 a}{4 b}=\frac{3 c}{4 d} \quad$ (Mutipling each side by $3 / 4$ )
$\Rightarrow \frac{3 a-4 b}{4 b}=\frac{3 c-4 d}{4 d} \quad$ (By dividendo)
$\Rightarrow \frac{3 a-4 b}{3 c-4 d}=\frac{4 b}{4 d}=\frac{b}{d}$ ...... (2)
From (1) and (2)
$\frac{6 a+7 b}{6 c+7 d}=\frac{3 a-4 b}{3 c-4 c}$
(6a+ 7b)(3c - 4d) = (6c + 7d)(3a - 4b)
$\Rightarrow \frac{6 a}{7 b}=\frac{6 c}{7 d} \quad$ (Multiplying each side by $6 / 7$ )
$\Rightarrow \frac{6 a+7 b}{7 b}=\frac{6 c+7 d}{7 d}$ (By componendo)
$\Rightarrow \frac{6 a+7 b}{6 c+7 d}=\frac{7 b}{7 d}=\frac{b}{d} \ldots . .(1)$
Also $\frac{a}{b}=\frac{c}{d}$
$\Rightarrow \frac{3 a}{4 b}=\frac{3 c}{4 d} \quad$ (Mutipling each side by $3 / 4$ )
$\Rightarrow \frac{3 a-4 b}{4 b}=\frac{3 c-4 d}{4 d} \quad$ (By dividendo)
$\Rightarrow \frac{3 a-4 b}{3 c-4 d}=\frac{4 b}{4 d}=\frac{b}{d}$ ...... (2)
From (1) and (2)
$\frac{6 a+7 b}{6 c+7 d}=\frac{3 a-4 b}{3 c-4 c}$
(6a+ 7b)(3c - 4d) = (6c + 7d)(3a - 4b)
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