Question

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

Answer 1

Let the numerator and denominator of the fraction by $x-3$ and $x$ respectively.

$\therefore$ Original fraction $=\frac{x-3}{x}$

Now, $2$ is added to both the numerator and denominator.

$\therefore$ New fraction $=\frac{(x-3+2)}{(x+2)}=\frac{x-1}{x+2}$

According to the question, we have

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

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

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

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

$\Rightarrow 20\times (2x^2-2x-6)=29\times (x^2+2x)$

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

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

Using Quadratic formula, $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ for standard quadratic equation, $a{x}^2+bx+c=0$

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

$=\frac{98\pm \sqrt{9604+5280}}{22}$

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

$=\frac{98\pm 122}{22}$

$\therefore x=\frac{98+122}{22}$ or, $x=\frac{98-122}{22}$

$x=\frac{220}{22}=10$ or, $x=\frac{-24}{22}=\frac{-12}{11}$

Denominator cannot be negative,

$\therefore x=10$

Numerator $=(10-3)=7$
So, the original fraction becomes $\left(\frac{7}{10}\right)$.