A fraction becomes 1/3 if 1 is subtracted from both its numerator and denominator.

Let the numerator and denominator of the fraction be $x$ and $y$ respectively. Then the fraction is $\frac{x}{y}$

If 1 is subtracted from both numerator and the denominator, the fraction becomes $\frac{1}{3}$. Thus, we have

$\frac{x-1}{y-1}=\frac{1}{3}$

$\Rightarrow 3(x-1)=y-1$

$\Rightarrow 3 x-3=y-1$

$\Rightarrow 3 x-y-2=0$

If 1 is added to both numerator and the denominator, the fraction becomes $\frac{1}{2}$. Thus, we have

$\frac{x+1}{y+1}=\frac{1}{2}$

$\Rightarrow 2(x+1)=y+1$

$\Rightarrow 2 x+2=y+1$

$\Rightarrow 2 x-y+1=0$

So, we have two equations

$3 x-y-2=0$

$2 x-y+1=0$

Here x and y are unknowns. We have to solve the above equations for x and y.

By using cross-multiplication, we have

$\frac{x}{(-1) \times 1-(-1) \times(-2)}=\frac{-y}{3 \times 1-2 \times(-2)}=\frac{1}{3 \times(-1)-2 \times(-1)}$

$\Rightarrow \frac{x}{-1-2}=\frac{-y}{3+4}=\frac{1}{-3+2}$

$\Rightarrow \frac{x}{-3}=\frac{-y}{7}=\frac{1}{-1}$

$\Rightarrow \frac{x}{3}=\frac{y}{7}=1$

$\Rightarrow x=3, y=7$

Hence, the fraction is $\frac{3}{7}$.