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

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

Applying componendo and dividendo, we get

$$\begin{gathered} \frac{\left(3a+7b\right)+\left(3a-7b\right)}{\left(3a+7b\right)-\left(3a-7b\right)}=\frac{4+3}{4-3} \\ \Rightarrow \frac{3a+7b+3a-7b}{3a+7b-3a+7b}=\frac{7}{1} \\ \Rightarrow \frac{6a}{14b}=7 \\ \Rightarrow \frac{a}{b}=\frac{14\times 7}{6}=\frac{49}{3}.....\left(1\right) \end{gathered}$$
Now,

$$\begin{gathered} \frac{3a^2-7b^2}{3a^2+7b^2} \\ =\frac{{\displaystyle \frac{3a^2}{b^2}}-{\displaystyle \frac{7b^2}{b^2}}}{{\displaystyle \frac{3a^2}{b^2}}+{\displaystyle \frac{7b^2}{b^2}}} \\ =\frac{3\times {\left({\displaystyle \frac{a}{b}}\right)}^2-7}{3\times {\left({\displaystyle \frac{a}{b}}\right)}^2+7} \\ =\frac{3\times {\left({\displaystyle \frac{49}{3}}\right)}^2-7}{3\times {\left({\displaystyle \frac{49}{3}}\right)}^2+7}\left[\mathrm{Using}\left(1\right)\right] \\ =\frac{{\displaystyle \frac{2401-21}{3}}}{{\displaystyle \frac{2401+21}{3}}} \end{gathered}$$
$$\begin{gathered} =\frac{2380}{2422} \\ =\frac{2380÷14}{2422÷14}\left(\mathrm{HCF}\mathrm{of}2380\mathrm{and}2422=14\right) \\ =\frac{170}{173} \end{gathered}$$