Question

If ${\displaystyle \frac{3x+4y}{3u+4v}=\frac{3x-4y}{3u-4v}}$, then show that ${\displaystyle \frac{x}{y}=\frac{u}{v}}$.

Answer 1

We have
${\displaystyle \frac{3x+4y}{3u+4v}=\frac{3x-4y}{3u-4v} ...\left[\text{Applying alternendo}\right]}$

${\displaystyle \frac{3x+4y}{3u-4v}=\frac{3x+4y}{3u-4v}}$   ...[By componendo and dividendo]
${\displaystyle \frac{3x+4y+3x-4y}{3x+4y-3x+4y}=\frac{3u+4v+3u-4v}{3u+4v-3u+4v}}$
⇒ ${\displaystyle \frac{6x}{8y}=\frac{6u}{8v}}$
⇒  ${\displaystyle \frac{x}{y}=\frac{u}{v}}$.
Hence proved.