Question

The numerator of a fraction is 3 less than its denominator. If 2 is added to both numerator and denominator, then the sum of the new fraction and original fraction is 29/20. Find the original fraction.

Answer 1

numerator=x-3
denominator=x

then
numerator=x-1
denominator=x+2

(x-3/x)/x+(x-1/x+2)=29/20
by solving
this equation will come
11x^2-98x-120
use formula( -b+-√b^2-4ac)/2a
you will get the value of x
x=10 and -12/11

original fraction will be
10/13 and -12/21
plzz check the answer I am not sure

Answer 2

$\huge\mathfrak\red{Solution} \\ \\ \\ \tt{Let\:the\:fraction\:be \: \frac{x - 3}{x}} \\ \\ \tt{By\:the\:given\: condition,\:new\:fraction= \frac{x - 3 + 2}{x + 2}} \\\\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \tt={\frac{x - 1}{x + 2}} \\ \\ \tt{\frac{x - 3}{x} + \frac{x - 1}{x + 2} = \frac{29}{20}} \\ \\ \tt{ \implies \: 20[(x - 3)(x + 2) + x(x - 1)] = 29(x^{2} + 2x)} \\ \\ \tt{ \implies \: 20[ x^{2} - x - 6 + x^{2} - x] = 29x^{2} + 58x} \\ \\ \tt{ \implies \: 20[2x^{2} - 2x - 6 ] = 29x^{2} + 58x} \\ \\ \tt{ \implies \: 40x^{2} - 40x - 120 - 29x^{2} - 58x = 0} \\ \\ \tt{ \implies \: 11x^{2} - 98x - 120 = 0} \\ \\ \tt{ \implies \: 11x^{2} - 110x + 12x - 120 = 0} \\ \\ \tt{ \implies \: 11x(x - 10) + 12(x - 10) = 0} \\ \\ \tt{ \implies \:(x - 10) (11x + 12 )= 0} \\ \\ \tt{ \implies \:x = 10 \: \: \: \: \: \therefore \: x(i.e. \: denominator) > 0} \\ \\ \tt{ \therefore \: x = \frac{ - 12}{11} \: is \: rejected. } \\ \\ \tt{ \therefore \: the \: fration \: is \frac{7}{10}. } \\ \\ \\ \mathfrak \green{hope \: \purple{it} \: helps}$