Form the pair of linear equations in the following problem, and find their solutions (if they exist) by the elimination method :

If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes $\dfrac{1}{2}$ if we only add 1 to the denominator. What is the fraction ?

Let x be the numerator and y be the denominator.

Original fraction = $\dfrac{x}{y}$

Given,

If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1.

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

Given,

Fraction becomes $\dfrac{1}{2}$ if we only add 1 to the denominator.

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

Multiplying equation (1) by 2, we get :

2x - 2y + 4 = 0 ……..(3)

Subtracting equation (3) from (2), we get :

⇒ 2x - y - 1 - (2x - 2y + 4) = 0

⇒ 2x - 2x - y + 2y - 1 - 4 = 0

⇒ y - 5 = 0

⇒ y = 5.

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

⇒ x - 5 + 2 = 0

⇒ x - 3 = 0

⇒ x = 3.

Fraction = $\dfrac{x}{y} = \dfrac{3}{5}$.

Hence, pair of linear equations are x - y + 2 = 0, 2x - y - 1 = 0, where x and y are the numerator and denominator of the fraction and fraction = $\dfrac{3}{5}$.