Question

pls solve this question,ASAP,dont spam ,correct ans will be marked as brainliest and I'll thank ur 20 ans

SHOW THE COMPLETE WORKING​

Answer 1

Let the denominator be x

Given:

$Numerator = x-3$

$Denominator = x$

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

If 2 is added to both Numerator and Denominator:

$Numerator = x-3+2 \\ = x - 1$

$Denominator = x+2$

$New Fraction = \frac{x - 1}{x + 2 }$

$Sum \: of \: both \: fractions = \frac{x - 3}{x} + \frac{x - 1}{x + 2} = \frac{29}{20}$

After cross multiplication :

$20((x -3)(x + 2) + x(x - 1)) = 29( {x}^{2} + 2x) \\ 20( {x}^{2} - x - 6 - {x}^{2} - x) = 29 {x}^{2} + 58x \\ 20(2 {x}^{2} - 2x - 6) = 29 {x}^{2} + 58x \\ 40 {x}^{2} - 40x - 120 - 29 {x}^{2} - 58x = 0 \\ 11 {x}^{2} - 98x - 120 = 0 \\ 11 {x}^{2} - 110x + 12x - 120 = 0 \\ 11x(x - 10) + 12(x - 10) = 0 \\ (x - 10)(11x + 12) = 0$

Cancelling the negative value of x, we have x=10.

So the original fraction = 7/10.

Hope it helps!