Question

If ${\displaystyle \frac{15a^2+4b^2}{15a^2-4b^2}=\frac{47}{7}}$ then find the value of the following ratio.

${\displaystyle \frac{7a-3b}{7a+3b}}$

Answer 1

${\displaystyle \frac{15a^2+4b^2}{15a^2-4b^2}=\frac{47}{7}}$

Applying componendo and dividendo, we get

${\displaystyle \frac{(15a^2+4b^2)+(15a^2-4b^2)}{(15a^2+4b^2)-(15a^2-4b^2)}=\frac{47+7}{47-7}}$

⇒ ${\displaystyle \frac{15a^2+4b^2+15a^2-4b^2}{15a^2+4b^2-15a^2+4b^2}}$ =${\displaystyle \frac{54}{40}}$

⇒ ${\displaystyle \frac{30a^2}{8b^2}=\frac{54}{40}}$

⇒ ${\displaystyle \frac{a^2}{b^2}=\frac{54\times 8}{40\times 30}=\frac{9}{25}}$

⇒ ${\displaystyle \frac{a}{b}=\sqrt{\frac{9}{25}}}$

∴ ${\displaystyle \frac{a}{b}=\frac{3}{5}}$

${\displaystyle \therefore \frac{7a-3b}{7a+3b}}$

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

= ${\displaystyle \frac{7\times \frac{3}{5}-3}{7\times \frac{3}{5}+3} ...(\frac{a}{b}=\frac{3}{5})}$

= ${\displaystyle \frac{\frac{21-15}{5}}{\frac{21+15}{5}}}$

= ${\displaystyle \frac{6}{36}}$

= ${\displaystyle \frac{1}{6}}$