If the numerator of a certain fraction is increased by 2 and the denominator by 1, the fraction becomes equal to $\dfrac{5}{8}$ and if the numerator and denominator are each diminished by 1, the fraction becomes equal to $\dfrac{1}{2}$; find the fraction.

Let numerator be x and denominator be y.

Given, if the numerator of a certain fraction is increased by 2 and the denominator by 1, the fraction becomes equal to $\dfrac{5}{8}$

$\therefore \dfrac{x + 2}{y + 1} = \dfrac{5}{8}$

⇒ 8(x + 2) = 5(y + 1)

⇒ 8x + 16 = 5y + 5

⇒ 5y - 8x = 11 …….(i)

Given, if the numerator and denominator are each diminished by 1, the fraction becomes equal to $\dfrac{1}{2}$

$\therefore \dfrac{x - 1}{y - 1} = \dfrac{1}{2}$

⇒ 2(x - 1) = y - 1

⇒ 2x - 2 = y - 1

⇒ y = 2x - 1 ……(ii)

Substituting value of y from (ii) in eq. (i) we get,

⇒ 5(2x - 1) - 8x = 11

⇒ 10x - 5 - 8x = 11

⇒ 2x = 16

⇒ x = 8.

Substituting value of x in eq. (ii) we get,

y = 2x - 1 = 2(8) - 1 = 16 - 1 = 15.

Original fraction = $\dfrac{x}{y} = \dfrac{8}{15}.$

Hence, fraction = $\dfrac{8}{15}$.