the sum of denominator and numerator …

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

Given, The sum of the numerator and the denominator of the fraction is 3 less than the twice of the denominator.
Thus, we have
{tex}x + y = 2y - 3{/tex}
{tex}\Rightarrow{/tex} {tex}x + y - 2y + 3 = 0{/tex}
{tex}\Rightarrow{/tex} {tex}x - y + 3 = 0{/tex}..............(1)
Also given, If the numerator and the denominator both are decreased by 1, the numerator becomes half the denominator. Thus, we have
{tex}x - 1 = \frac{1}{2}(y - 1){/tex}
{tex}{/tex}
{tex}\Rightarrow{/tex} {tex}2(x - 1) = (y - 1){/tex}
{tex}\Rightarrow{/tex} 2x - 2 = (y - 1)
{tex}\Rightarrow{/tex} 2x - y - 1 = 0.......................(2)
So, we have formed  two linear equations in x & y as following:-
x - y + 3 = 0
2x - y - 1 = 0
Here x and y are unknowns.
We have to solve the above equations for x and y.
By using cross-multiplication method , we have
{tex}\frac{x}{{( - 1) \times ( - 1) - ( - 1) \times 3}}{/tex} {tex}= \frac{{ - y}}{{1 \times ( - 1) - 2 \times 3}}{/tex} {tex}= \frac{1}{{1 \times ( - 1) - 2 \times ( - 1)}}{/tex}
{tex}\Rightarrow \frac{x}{{1 + 3}}{/tex} {tex}= \frac{{ - y}}{{ - 1 - 6}} = \frac{1}{{ - 1 + 2}}{/tex}
{tex}\Rightarrow \frac{x}{4} = \frac{{ - y}}{{ - 7}} = \frac{1}{1}{/tex}
{tex}\Rightarrow \frac{x}{4} = \frac{y}{7} = 1{/tex}

Using Part I & III , we get x = 4

& From part II & III, we get y = 7 {tex}{/tex}
{tex}\Rightarrow{/tex} x = 4, y = 7
Hence, The fraction is {tex}\frac{x}{y}{/tex} = {tex}\frac{4}{7}{/tex}