Form the pair of linear equations for 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 $\frac{1}{2}$ if we only add $1$ to the denominator. What is the fraction?

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