Question

HOTS.. Fraction sum.. Don't spam please.. ​

Answer 1

Answer:..solve this... may be

Step-by-step explanation:

Answer 2

Answer:

Let the required fraction be x/y

ATQ

x = y - 3 ......(1)

and

new fraction is (x+2)/(y +2)

new fraction + original fraction = 29/30

(x+2)/(y+2) + x/y = 29/30

from ....(1) put the value of x in this equation

we get

$\frac{(y-3 + 2)}{(y+2)} + \frac{ (y-3)}{y} = \frac{29}{30}$

$\frac{y - 1}{y + 2} + \frac{y - 3}{y} = \frac{29}{30}$

$\frac{((y - 1)(y) + (y - 3)(y + 2)}{(y + 2)(y)} = \frac{29}{30}$

$\frac{ {y}^{2} - y + {y}^{2} + 2y - 3y - 6 }{ {y}^{2} + 2y } = \frac{29}{30}$

$\frac{{2y}^{2} - 2y - 6}{ {y}^{2} + 2y } = \frac{29}{30}$

$60{y}^{2} - 60y - 180 = 29 {y}^{2} + 58y$

$31{y}^{2} - 118y \: - 180 = 0$

it can't be solved further.