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

new and original fraction is 29/20 . Find the fraction . Please solve this question . I couldnt .
Ans : 7/10

a/b + (a + 2) / (b + 2) = 29/20

a = b - 3

(b - 3) / b + (b - 3 + 2) / (b + 2) = 29 / 20
(b - 3) / b + (b - 1) / (b + 2) = 29 / 20
((b - 3) * (b + 2) + b * (b - 1)) / (b * (b + 2)) = 29 / 20
(b^2 - 3b + 2b - 6 + b^2 - b) / (b^2 + 2b) = 29 / 20
(2b^2 - 2b - 6) / (b^2 + 2b) = 29 / 20
20 * (2b^2 - 2b - 6) = 29 * (b^2 + 2b)
40b^2 - 40b - 120 = 29b^2 + 58b
40b^2 - 29b^2 - 40b - 58b - 120 = 0
11b^2 - 98b - 120 = 0
b = (98 +/- sqrt(98^2 - 4 * 11 * (-120))) / (2 * 11)
b = (98 +/- sqrt(9604 + 5280)) / 22
b = (98 +/- sqrt(14884)) / 22
b = (98 +/- 122) / 22
b = 220/22 , -24/22
b = 10 , -12/11

(b - 3) / b =>
(10 - 3) / 10 =>
7/10

or

(-12/11 - 3) / (-12/11) =>
(-12/11 - 33/11) / (-12/11) =>
(-45/11) / (-12/11) =>
(45/12) =>
15/4

So that obviously doesn't fit the conditions of our original equation

7/10 and 9/12 are the fractions

original fraction p/(p+3)
new fraction (p+2)/(p+3+2)

according to question:

(p+2)/(p+3+2) + p/(p+3) = 29/20,

=> (p+2)(p+3) + p(p+3+2) = 29*(p+3+2)(p+3)/20,

=> 40p^2 + 200p + 120 = 29p^2 + 232p + 435,

=> 11p^2 - 32p - 315 = 0,

the roots of the equation ax^2 + bx + c is given by (-b + sqrt(b^2- 4ac))/2a & (-b - sqrt(b^2- 4ac))/2a

now, (-b - sqrt(b^2- 4ac))/2a will give a decimal number but p cannot be decimal as it is the numerator of a fraction which is always an integer.

so the solution is given by (-b + sqrt(b^2- 4ac))/2a i.e p=7.

therefore, the orignal fraction is a/(a+3) = 7/10

Assume the denominator of the fraction to be x, then:

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

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

2x^2 - 2x - 6 = 29/20*x^2 + 29x/10