Question

The numerator of a fraction is $3$ less than the denominator. If $2$ is added to both the numerator and the denominator, then the sum of the new fraction and the original fraction is $\frac{29}{20}.$ Find the original fraction.

Answer 1

Consider the given fraction.

$\frac{x-3}{x}$

According to the question,

$\frac{x-3}{x}+\frac{x-3+2}{x+2}=\frac{29}{20}$

$\frac{x-3}{x}+\frac{x-1}{x+2}=\frac{29}{20}$

$\Rightarrow \frac{x^2+2x-3x-6+x^2-x}{x^2+2x}=\frac{29}{20}$

$\Rightarrow \frac{2x^2-2x-6}{x^2+2x}=\frac{29}{20}$

$\Rightarrow 40x^2-40x-120=29x^2+58x$

$\Rightarrow 11x^2-98x-120=0$

We know that,

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$x=\frac{-(-98)\pm \sqrt{{(-98)}^2-4\times 11\times (-120)}}{22}$

$x=\frac{98\pm \sqrt{14884}}{22}$

$x=\frac{98\pm \left(122\right)}{22}$

$x=10,\frac{-12}{11}$

When $x=10$

Original fraction is $\frac{x-3}{x}=\frac{10-3}{10}=\frac{x-3}{x}=\frac{7}{10}$

When $x=\frac{-12}{11}$

Original fraction is $\frac{x-3}{x}=\frac{\frac{-12}{11}-3}{\frac{-12}{11}}$

$\Rightarrow \frac{x-3}{x}=\frac{45}{12}$

Hence, original fraction is $\frac{7}{10},\frac{45}{12}$