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.

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Solution

Let the denominator be n
$∴$ Numerator = n-3
Fraction = $\frac{n - 3}{x}$
When 2 is added numerator and denominator
We have fraction as $\frac{(x - 1)}{(x + 2)}$
$\frac{x - 3}{x} + \frac{x - 1}{x + 2} = \frac{29}{20}$
$(x - 3) (x + 2) + x (x - 1) = x (x + 2) \times \frac{29}{20}$
$(x^{2} - 6 - x + x^{2} - x) 20 = 29 x^{2} + 58 x$
$40 x^{2} - 120 - 40 x = 29 x^{3} + 58 x$
$11 x^{2} - 98 x - 120 = 0$
$11 x^{2} - 110 x + 12 x - 120 = 0$
$(11 x + 12) (x - 10) = 0$
concelling the negative fraction value of x we have x=10
Then the fraction $\frac{x - 3}{x} = \frac{7}{10}$

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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 2920. Find original fraction.

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.