Question

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

Answer 1

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

$\frac{3a^2-7b^2}{3a^2+7b^2}=?$

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

$\frac{a(3+7\frac{b}{a})}{a(3-7\frac{b}{a})}$ $=\frac{4}{3}$

$9+21\left(\frac{b}{a}\right)=12-28$ $\left(\frac{b}{a}\right)$

$\Rightarrow 49\left(\frac{b}{a}\right)=3$

$\Rightarrow \left(\frac{b}{a}\right)=\frac{3}{49}$

$\Rightarrow \frac{b^2}{a^2}=\frac{9}{49\times 49}$

$\frac{3a^2-7b^2}{3a^2+7b^2}=\frac{a^2(3-7\frac{b^2}{a^2})}{a^2(3+7\frac{b^2}{a^2})}=\frac{3-7\left(\frac{9}{49\times 49}\right)}{63+7\times \left(\frac{9}{49\times 49}\right)}$

$=\frac{3[1-\frac{3}{7\times 49}]}{3[1+\frac{3}{7\times 49}]}$

$=\frac{(7\times 49-3)}{(7\times 49+3)}$

$=\frac{343-3}{343+3}$

$\Rightarrow \frac{340}{346}$

$\Rightarrow \frac{170}{173}$

$\frac{3a^2-7b^2}{3a^2+7b^2}=\frac{170}{173}$.