Question

If $\frac{a}{b}=\frac{7}{3}$ then find the values of the following ratio.
$\frac{7a+9b}{7a-9b}$

Answer 1

Given that,

$\frac{a}{b}=\frac{7}{3}$

$\therefore \frac{7a+9b}{7a-9b}=\frac{a(7\times \frac{a}{b}+9)}{a(7\times \frac{a}{b}-9)}$

$=\frac{7\times \frac{7}{3}+9}{7\times \frac{7}{3}-9}$

$=\frac{49+27}{49-27}$

$=\frac{76}{22}=\frac{38}{11}$(Ans)