Question

If $\frac{a}{b}=\frac{7}{3}$ then find the values of the following ratios.
$\frac{3a^2+3b^2}{2a^2-3b^2}$

Answer 1

Given that,

$\frac{a}{b}=\frac{7}{3}⟹\frac{a^2}{b^2}=\frac{49}{9}$

$\therefore \frac{3a^2+3b^2}{2a^2-3b^2}=\frac{a^2(3\times \frac{a^2}{b^2}+3)}{a^2(2\times \frac{a^2}{b^2}-3)}$

$=\frac{3\times \frac{49}{9}+3}{2\times \frac{49}{9}-3}$

$=\frac{147+27}{147-27}$

$=\frac{174}{120}=\frac{29}{20}$(Ans)