Given, $\dfrac{a}{b} = \dfrac{c}{d}$, prove that :

$\dfrac{3a - 5b}{3a + 5b} = \dfrac{3c - 5d}{3c + 5d}$

Given,

$\dfrac{a}{b} = \dfrac{c}{d}$

Multiplying both sides by $\dfrac{3}{5}$:

$\Rightarrow \dfrac{3a}{5b} = \dfrac{3c}{5d}$

Applying componendo and dividendo:

$\Rightarrow \dfrac{3a + 5b}{3a - 5b} = \dfrac{3c + 5d}{3c - 5d}$

By invertendo: $\Rightarrow \dfrac{3a - 5b}{3a + 5b} = \dfrac{3c - 5d}{3c + 5d}$

Hence, proved that $\dfrac{3a - 5b}{3a + 5b} = \dfrac{3c - 5d}{3c + 5d}$.