Question

six years before the age of mother was numerically equal to the square of son's age three years hence her age will be thrice the age of her son then find the present age of the mother and son ???​

Answer 1

Answer:

x=6 Or x= -3

But age can not be negative.

son's present age = x + 6 = 6 + 6 = 12 years. mother's present age = x2 + 6 = 36 + 6 = 42 years.

Answer 2

Answer:

Mother's age = 42

Son's age = 12

Step-by-step explanation:

Let the present ages of mother and son be y & x  respectively.

1)  y-6 = (x-6)^2

 y- 6 = x^2 - 12x + 36

 y = x^2 -12x + 42            --1

2) y+3 = 3(x+3)

  y+3 = 3x +9

  y = 3x + 6                      --2

Equate 1 and 2 .

we get,

y = x^2 -12x +42 = 3x + 6

x^2+36 = 15x

x^2 -15x+36 = 0

X^2 -12x -3x +36

x(x-12) -3 (x-12) = 0

(x-3)(x-12)

x = 3 & 12

Let x =3

y= 3x + 6 = 3(3) +6 = 9+6 = 15

y=15

Let x =12

y = 3x+6 = 3(12)+6

=36 +6

y= 42

[The ages are mostly 12&42 because the age of mother cannot be 15 with having a child of age 3 years.]

Hence, the answer is 12 & 42 years.