The numerator of a fraction is 3 less than denominator . If 2 is added to both den and num , then the sum of

Let n = the denominator of the fraction.
(n-3)/n + (n-3+2)/(n+2） = 29/20
Multiply both sides by n(n+2),
(n-3)(n+2) + (n-1)n = (29/20)n(n+2)
2n^2 - 2n - 6 = (29/20)n(n+2)
40n^2 - 40n - 120 = 29n^2 + 58n
11n^2-98n-120 = 0
(11n+12)(n-10) = 0
n = 10, (n-3)/n = 7/10
or
n = -11/12, (n-3)/n = 47/11

Check:
7/10 + 9/12 = 29/20
47/11 + 49/13 ≠ 29/20

Answer: 7/10

let the original fraction be x/y.
by data x-3=y----->x= y+3--------------------(1)
by 2nd data,adding 2 to both num and denom of original fraction,we get
(x+2/y+2) this is our new fraction.


(x+2/y+2) + x/y = 29/20
take lcm and solve it.
substitute equation (1)

Let d = the denominator.
(d-3)/d + (d-3+2)/(d+2) = 29/20
LCM = d(d+2),
(d-3)(d+2) + (d-1)d = (29/20)d(d+2)
2d² - 2d - 6 = (29/20)d(d+2)
40d² - 40d - 120 = 29d² + 58d
11d²-98d-120 = 0
factorise , to get
(11d+12)(d-10) = 0
either d = 10 or (d-3)/d = 10 - 3 / 10 = 7/10
Answer

[(x - 3) / x] + [(x - 3 + 2) / (x + 2)] = 29/20

[(x - 3)(x + 2) + x(x - 1)] / [x(x + 2)] = 29/20

(x^2 + 2x - 3x - 6 + x^2 - x) / (x^2 + 2x) = 29 / 20

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

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

11x^2 - 98x - 120 = 0


(98 ± √14884) / 22
(98 ± 122) / 22

x = {-12/11, 10}

original fraction is

(x - 3) / x

[(-12/11) - 3] / (-12/11)
[(-12 - 33) / 11] / (-12/11)
(-45/11)(-11/12)
45/12
15/4

or

(10 - 3) / 10
7/10

not sure what u mean by "new and original fraction"

the sum of ... what ?