Question

$\frac{3a+7b}{3a-7b}=\frac{4}{3}$ then find the value of the ratio $\frac{3a^2-7b^2}{3a^2+7b^2}$

Answer 1

$Wehave\frac{3a+7b}{3a-7b}=\frac{4}{3}\therefore \frac{3a-7b}{3a+7b}=\frac{3}{4}$

$Now,\frac{3a^2-7b^2}{(3a+7b{)}^2}=\frac{\left(3a+7b\right)\left(3a-7b\right)}{\left(3a+7b\right)\left(3a+7b\right)}=\frac{3a-7b}{3a+7b}=\frac{3}{4}=3:4$

$Hence,requiredratiois3:4.$