Question

If $\frac{m+n}{m+3n}=\frac{2}{30}$, find : $\frac{2n^2}{3m^2+m{n}^2}$

Answer 1

$\frac{m+n}{m+3n}=\frac{2}{30}$

$\frac{m+n}{m+3n}=\frac{1}{15}$

Using componendo-dividendo rule,

$\frac{m+n-m-3n}{m+n+m+3n}=\frac{1-15}{1+15}$

$\frac{-2n}{2m+4n}=\frac{-14}{16}$

$\frac{n}{m+2n}=\frac{7}{8}$

$\Rightarrow 8n=7m+14n$

$\Rightarrow 8n-14n=7m$

$\Rightarrow -6n=7m$ or $7m=-6n$

$\Rightarrow \frac{m}{n}=\frac{-6}{7}$ or $\frac{m^2}{n^2}=\frac{36}{49}$

Now,$\frac{2n^2}{3m^2+m{n}^2}$

$=\frac{2\frac{n^2}{n^2}}{3\frac{m^2}{n^2}+m\frac{n^2}{n^2}}$

$=\frac{2}{3\times \frac{36}{49}+m}$

$=\frac{2}{\frac{108}{49}+m}$

$=\frac{98}{108+49m}$