Form the pair of linear equations in the following problems and find their solutions (if they exist) by any algebraic method. $A$ fraction becomes $\frac{1}{3}$ when $1$ is subtracted from the numerator and it becomes $\frac{1}{4}$ when $8$ is added to its denominator. Find the fraction.

(A) Let the fraction be $\frac{x}{y}$.
According to the given information:
$\frac{x-1}{y} = \frac{1}{3} \implies 3x - y = 3$ $...(1)$
$\frac{x}{y+8} = \frac{1}{4} \implies 4x - y = 8$ $...(2)$
Subtracting equation $(1)$ from equation $(2)$,we obtain:
$(4x - y) - (3x - y) = 8 - 3$
$x = 5$
Substituting the value of $x$ in equation $(1)$:
$3(5) - y = 3$
$15 - y = 3$
$y = 12$
Hence,the fraction is $\frac{5}{12}$.