Form the pair of linear equations for the following problem and find their solution by the substitution method.
$A$ fraction becomes $\frac{9}{11}$ if $2$ is added to both the numerator and the denominator. If $3$ is added to both the numerator and the denominator,it becomes $\frac{5}{6}$. Find the fraction.

(A) Let the numerator of the fraction be $x$ and the denominator be $y$. The fraction is $\frac{x}{y}$.
According to the first condition: $\frac{x+2}{y+2} = \frac{9}{11}$.
Cross-multiplying gives: $11(x+2) = 9(y+2) \implies 11x + 22 = 9y + 18 \implies 11x - 9y = -4$ (Equation $1$).
According to the second condition: $\frac{x+3}{y+3} = \frac{5}{6}$.
Cross-multiplying gives: $6(x+3) = 5(y+3) \implies 6x + 18 = 5y + 15 \implies 6x - 5y = -3$ (Equation $2$).
From Equation $2$,$5y = 6x + 3 \implies y = \frac{6x+3}{5}$.
Substituting this into Equation $1$: $11x - 9(\frac{6x+3}{5}) = -4$.
Multiply by $5$: $55x - 9(6x+3) = -20 \implies 55x - 54x - 27 = -20 \implies x = 7$.
Now,find $y$: $y = \frac{6(7)+3}{5} = \frac{42+3}{5} = \frac{45}{5} = 9$.
The fraction is $\frac{7}{9}$.