If $\dfrac{m + n}{m + 3n} = \dfrac{2}{3}$, find : $\dfrac{2n^2}{3m^2 + mn}$.

Given,

$\phantom{\Rightarrow} \dfrac{m + n}{m + 3n} = \dfrac{2}{3} \\[1em] \Rightarrow 3(m + n) = 2(m + 3n) \\[1em] \Rightarrow 3m + 3n = 2m + 6n \\[1em] \Rightarrow 3m - 2m = 6n - 3n \\[1em] \Rightarrow m = 3n.$

Substituting value of m in $\dfrac{2n^2}{3m^2 + mn}$ we get,

$\Rightarrow \dfrac{2n^2}{3(3n)^2 + (3n)n} \\[1em] \Rightarrow \dfrac{2n^2}{3(9n^2) + 3n^2} \\[1em] \Rightarrow \dfrac{2n^2}{27n^2 + 3n^2} \\[1em] \Rightarrow \dfrac{2n^2}{30n^2} \\[1em] \Rightarrow \dfrac{1}{15}.$

Hence, $\dfrac{2n^2}{3m^2 + mn} = \dfrac{1}{15}.$