If the numerator of a fraction is increased by 2 and denominator is decreased by 1, it becomes $\dfrac{2}{3}$. If the numerator is increased by 1 and denominator is increased by 2, it becomes $\dfrac{1}{3}$. Find the fraction.

Let numerator be x and denominator be y.

Given,

If the numerator of a fraction is increased by 2 and denominator is decreased by 1, it becomes $\dfrac{2}{3}$.

$\therefore \dfrac{x + 2}{y - 1} = \dfrac{2}{3} \\[1em] \Rightarrow 3(x + 2) = 2(y - 1) \\[1em] \Rightarrow 3x + 6 = 2y - 2 \\[1em] \Rightarrow 3x - 2y + 6 + 2 = 0 \\[1em] \Rightarrow 3x - 2y + 8 = 0 \text{ ………(1)}$

Given,

If the numerator is increased by 1 and denominator is increased by 2, it becomes $\dfrac{1}{3}$.

$\Rightarrow \dfrac{x + 1}{y + 2} = \dfrac{1}{3} \\[1em] \Rightarrow 3(x + 1) = 1(y + 2) \\[1em] \Rightarrow 3x + 3 = y + 2 \\[1em] \Rightarrow 3x - y + 3 - 2 = 0 \\[1em] \Rightarrow 3x - y + 1 = 0 \text{ ………(2)}$

Subtracting equation (1) from (2), we get :

⇒ 3x - y + 1 - (3x - 2y + 8) = 0 - 0

⇒ 3x - 3x - y + 2y + 1 - 8 = 0

⇒ y - 7 = 0

⇒ y = 7.

Substituting value of y in equation (1), we get :

⇒ 3x - 2(7) + 8 = 0

⇒ 3x - 14 + 8 = 0

⇒ 3x - 6 = 0

⇒ 3x = 6

⇒ x = $\dfrac{6}{3}$ = 2.

Fraction = $\dfrac{x}{y} = \dfrac{2}{7}$.

Hence, fraction = $\dfrac{2}{7}$.