Let the numerator of the fraction to be A and the denominator of the fraction to be B.

So, the required fraction is $A/B$.

ATQ,

Thus, the equation so formed is,

$(A–1)/(B−1)=1/3$

⇒ $3(A–1)=(B–1)$

⇒ $3A–3=B–1$

⇒ $3A–B–2=0$…. (i)

And also it’s given in the question as,

If numerator and denominator are added by $1$, it becomes $1/2$ , so now the expression is

$(A+1)/(B+1)=1/2$

⇒ $2(A+1)=(B+1)$

⇒ $2A+2=B+1$

⇒ $2A–B+1=0$ …….. (ii)

Solving (i) and (ii), to find the fraction

By using cross-multiplication, we will get

$\begin{array}{l}

\frac{A}{{( – 1) \times 1 – ( – 1) \times ( – 2)}} = \frac{{ – B}}{{3 \times 1 – 2 \times ( – 2)}} = \frac{1}{{3 \times ( – 1) – 2 \times ( – 1)}}\\

\frac{A}{{ – 1 – 2}} = \frac{{ – B}}{{3 + 4}} = \frac{1}{{ – 3 + 2}}\\

\frac{A}{{ – 3}} = \frac{{ – B}}{7} = \frac{1}{{ – 1}}\\

\frac{A}{3} = \frac{B}{7} = 1

\end{array}$

⇒ $A=3$, $B=7$

Hence, the fraction is $3/7$.