Question

The sum of numerator and denominator of a fraction is $3$ less than twice the denominator.If each of the numerator and denominator is decrease by $1$, the fraction becomes $\frac{1}{2}$.Find the fraction.

Answer 1

Let the fraction be $\frac{x}{y}\Rightarrow x+y+3=2y$

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

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

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

$\Rightarrow 2x-y=1$ _______ (2)

Substituting the value of (1) in (2) we get

$2(y-3)-y=1\Rightarrow 2y-6-y=1$

$\Rightarrow y=7$

$x=y-3=7-3=4$

$\therefore x=4,y=7$

Hence the fraction is $\frac{x}{y}=\frac{4}{7}$