If 2 is added to the numerator of a fraction,

Solution:

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

If 2 is added to the numerator of the fraction, it reduces to $\frac{1}{2}$. Thus, we have

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

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

$\Rightarrow 2 x+4=y$

$\Rightarrow 2 x-y+4=0$

If 1 is subtracted from the denominator, the fraction reduces to $\frac{1}{3}$. Thus, we have

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

$\Rightarrow 3 x=y-1$

$\Rightarrow 3 x-y+1=0$

So, we have two equations

$2 x-y+4=0$

$3 x-y+1=0$

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

By using cross-multiplication, we have

$\frac{x}{(-1) \times 1-(-1) \times 4}=\frac{-y}{2 \times 1-3 \times 4}=\frac{1}{2 \times(-1)-3 \times(-1)}$

$\Rightarrow \frac{x}{-1+4}=\frac{-y}{2-12}=\frac{1}{-2+3}$

$\Rightarrow \frac{x}{3}=\frac{-y}{-10}=\frac{1}{1}$

$\Rightarrow \frac{x}{3}=\frac{y}{10}=1$

$\Rightarrow x=3, y=10$

Hence, the fraction is $\frac{3}{10}$.