Question

If $a:b=c:d,$ prove that $(6a+7b)(3c-4d)=(6c+7d)(3a-4b)$

Answer 1

Concept of componendo
Given: $a:b=c:d$
$\Rightarrow \frac{a}{b}=\frac{c}{d}$
Multiply both sides by $\frac{6}{7^{\prime }}$ we get
$\Rightarrow \frac{6a}{7b}=\frac{6c}{7d}$
$\Rightarrow \frac{6a+7b}{7b}=\frac{6c+7d}{7d}$ [using componendo]
$\Rightarrow \frac{6a+7b}{6c+7d}=\frac{7b}{7d}$
$\Rightarrow \frac{6a+7b}{6c+7d}=\frac{b}{d}$ $\cdots \left(i\right)$
Concept of dividendo
$\because \frac{a}{b}=\frac{c}{d}$
Multiply both sides by $\frac{3}{4^{\prime }}$ we get
$\Rightarrow \frac{3a}{4b}=\frac{3c}{4d}$
$\Rightarrow \frac{3a-4b}{4b}=\frac{4c-4d}{4d}$ [using dividendo]
$\Rightarrow \frac{3a-4b}{3c-4d}=\frac{4b}{4d}$
$\Rightarrow \frac{3a-4b}{3c-4d}=\frac{b}{d}\cdots \left(i\right)$
from eq.(i) and (ii),we get
$\Rightarrow \frac{6a+7b}{6c+7d}=\frac{3a-4b}{3c-4d}$
$\therefore (6a+7b)(3c-4d)=(6c+7d)(3a-4b)$
Hence,proved.