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

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

To do:To find the fraction.

Solution:Let the numerator of fraction$=x$

And denominator of fraction$=y$

According to the first given condition

$x+y=2y-3$

$x-y=-3\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \dotsc \dotsc \dotsc \dotsc \dotsc \dotsc .\ \ \ ( 1)$

According to the second given condition

If each of the numerator and denominator is decreased by one then the fraction becomes$\frac{1}{2}.$

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

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

$\Rightarrow 2x-y=1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \dotsc \dotsc \dotsc \dotsc \dotsc \dotsc \dotsc .\ ( 2) \ $

Let subtract equation $( 1)$ from equation $( 2)$

$2x-y-( x-y)=1-( -3)$

$\Rightarrow x=4$

And by putting the value of x in equation.

$\Rightarrow x=4$

And by putting the value of x in equation

$4-y=-3$

$\Rightarrow y=7$

By solving equations $( 1)$ and $( 2)$

We get $x=4$ and $y=7$

$\therefore$ The fraction is $\frac{4}{7}$ 
 

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Updated on: 10-Oct-2022

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