If we add 1 to numerator …

Suppose the numerator and denominator of the fraction be x and y respectively.
Then the fraction is {tex}\frac{x}{y}{/tex}.
If 1 is added to the numerator and 1 is subtracted from the denominator, the fraction becomes 1.
Thus, we have {tex}\frac{{x + 1}}{{y - 1}} = 1{/tex}
{tex}\Rightarrow{/tex} {tex}(x + 1) = (y - 1){/tex}
{tex}\Rightarrow{/tex} {tex}x + 1 - y + 1 = 0{/tex}
{tex}\Rightarrow{/tex} {tex}x - y + 2 = 0{/tex}
If 1 is added to the denominator, the fraction becomes {tex}\frac{1}{2}{/tex}.
Thus, we have {tex}\frac{x}{{y + 1}} = \frac{1}{2}{/tex}
{tex}\Rightarrow{/tex} {tex}2x = (y + 1){/tex}
{tex}\Rightarrow{/tex} {tex}2x - y - 1 = 0{/tex}
We have two equations
{tex}x - y + 2 = 0{/tex}
{tex}2x - y - 1 = 0{/tex}
By using cross-multiplication, we have
{tex}\frac{x}{{( - 1) \times ( - 1) - ( - 1) \times 2}}{/tex} {tex} = \frac{{ - y}}{{1 \times ( - 1) - 2 \times 2}}{/tex} {tex} = \frac{1}{{1 \times ( - 1) - 2 \times ( - 1)}}{/tex}
{tex}\Rightarrow \frac{x}{{1 + 2}}{/tex} {tex}=\frac{{ - y}}{{ - 1 - 4}} = \frac{1}{{ - 1 + 2}}{/tex}
{tex}\Rightarrow \frac{x}{3} = \frac{{ - y}}{{ - 5}} = \frac{1}{1}{/tex}
{tex}\Rightarrow \frac{x}{3} = \frac{y}{5} = 1{/tex}
So, {tex}x = 3\ and\ y = 5.{/tex}
The fraction is {tex}\frac{3}{5}{/tex}.