Form the pair of linear equations for the following problems and find their solution by substitution method.

A fraction becomes $\dfrac{9}{11}$, if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes $\dfrac{5}{6}$. Find the fraction.

Let numerator be x and denominator be y.

Fraction = $\dfrac{x}{y}$

Given,

If 2 is added to both numerator and denominator, fraction becomes $\dfrac{9}{11}$.

$\Rightarrow \dfrac{x + 2}{y + 2} = \dfrac{9}{11} \\[1em] \Rightarrow 11(x + 2) = 9(y + 2) \\[1em] \Rightarrow 11x + 22 = 9y + 18 \\[1em] \Rightarrow 11x - 9y + 22 - 18 = 0 \\[1em] \Rightarrow 11x - 9y + 4 = 0 \text{ ……….(1)}$

Given,

If 3 is added to both numerator and denominator, fraction becomes $\dfrac{5}{6}$.

$\Rightarrow \dfrac{x + 3}{y + 3} = \dfrac{5}{6} \\[1em] \Rightarrow 6(x + 3) = 5(y + 3) \\[1em] \Rightarrow 6x + 18 = 5y + 15 \\[1em] \Rightarrow 6x - 5y + 18 - 15 = 0 \\[1em] \Rightarrow 6x - 5y + 3 = 0 \text{ ……….(2)}$

Multiplying equation (1) by 5, we get :

⇒ 55x - 45y + 20 = 0 ………….(3)

Multiplying equation (2) by 9, we get :

⇒ 54x - 45y + 27 = 0 ……….(4)

Subtracting equation (4) from (3), we get :

⇒ 55x - 45y + 20 - (54x - 45y + 27) = 0

⇒ 55x - 54x - 45y + 45y + 20 - 27 = 0

⇒ x - 7 = 0

⇒ x = 7.

Substituting value of x in equation (2), we get :

⇒ 6 × 7 - 5y + 3 = 0

⇒ 42 - 5y + 3 = 0

⇒ 45 - 5y = 0

⇒ 5y = 45

⇒ y = 9.

Fraction = $\dfrac{x}{y} = \dfrac{7}{9}$.

Hence, pair of linear equations are 11x - 9y + 4 = 0, 6x - 5y + 3 = 0, where x and y are numerator and denominator and fraction = $\dfrac{7}{9}$.