The sum of numerator and denominator …

Suppose the numerator of the fraction be x 

Denominator of the fraction be y 
{tex}\therefore{/tex} the fraction is {tex}\frac{x}{y}{/tex}
According to the question,

 The sum of the numerator and denominator of the fraction is 12.
{tex}\Rightarrow x + y = 12{/tex}
{tex}\Rightarrow{/tex} {tex}x + y - 12 = 0{/tex}
If the denominator is increased by 3, the fraction becomes {tex}\frac{1}{2}{/tex}.
{tex}\Rightarrow \frac{x}{{y + 3}} = \frac{1}{2}{/tex}
{tex}\Rightarrow{/tex} {tex}2x = (y + 3){/tex}
{tex}\Rightarrow{/tex} {tex}2x - y - 3 = 0{/tex}
So, we have two equations
{tex}x + y - 12 = 0{/tex}
{tex}2x - y - 3 = 0{/tex}
Here x and y are unknowns.
We have to solve the above equations for x and y.
By using cross-multiplication, 
{tex}\Rightarrow \frac{x}{{(1) \times ( - 3) - ( - 1) \times - 12}}{/tex} {tex} = \frac{{ - y}}{{1 \times ( - 3) - 2 \times - 12}}{/tex} {tex}= \frac{1}{{1 \times ( - 1) - 2 \times (1)}}{/tex}
{tex}\Rightarrow \frac{x}{{ - 3 - 12}}{/tex} {tex}= \frac{{ - y}}{{ - 3 + 24}} = \frac{1}{{ - 1 - 2}}{/tex}
{tex}\Rightarrow \frac{x}{{ - 15}} = \frac{{ - y}}{{21}} = \frac{1}{{ - 3}}{/tex}
{tex}\Rightarrow \frac{x}{{15}} = \frac{y}{{21}} = \frac{1}{3}{/tex}
{tex}\Rightarrow x = \frac{{15}}{3},\;y = \frac{{21}}{3}{/tex}
{tex}\Rightarrow{/tex} {tex}x = 5, y = 7{/tex}
The fraction is {tex}\frac{5}{7}{/tex}