Question

IT'S URGENT. Father is 7 times older than hisson . 2 years ago father was 13 times the son's age . Find their present age.

Answer 1

Let the son's age be x years and the father age be y years.
7x = y
or, 7x - y = 0 ------ [i]
2 years ago, 
13(x-2) =  y-2
or, 13x - 26 = y-2
or, 13x - y = -2+26
or, 13x-y = 24------[ii]
Subtracting both the equations we get,
x = 24/6 = 4
Therefore, the son is 4 years old.
Father's age = 7x = 7*4 = 28 years old. 

Answer : Son's age is 4 years and Father's age is 28 years.

Answer 2

Let Son's present age be x years

So, the present age of Father be 7x years

2-years ago -

Son's age = (x - 2) years

Father's age = (7x - 2) years

According to Question,

(7x - 2) = 13(x - 2)

=> 7x - 2 = 13x - 26

=> - 2 + 26 = 13x - 7x

=> 24 = 6x

=> 24 / 6 = x

=> 4 = x

Required ages -

Son' age = x = 4 years

Father's age = 7x = 7(4) = 28 years

Hence, the present age of Son is 4 years and present age of Father is 28 years