Question

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

Answer 1

Let the denominator be n

$\therefore$ Numerator = n-3

Fraction = $\frac{n-3}{x}$

When 2 is added numerator and denominator

We have fraction as $\frac{(x-1)}{(x+2)}$

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

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

$(x^2-6-x+x^2-x)20=29x^2+58x$

$40x^2-120-40x=29x^3+58x$

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

$11x^2-110x+12x-120=0$

$(11x+12)(x-10)=0$

concelling the negative fraction value of x we have x=10

Then the fraction $\frac{x-3}{x}=\frac{7}{10}$