Question

A man is 7 times as old as his son now 4 years later he will be 4 times as old as his son find their present ages

Answer 1

Let x be the man's present age and y be the present age of son
A man is 7 times as old as his son
So x = 7y. --------------------------(1)
4 years later he will be 4 times as old as his son
x+4 = 4 (y+4)
x+4 = 4y+16
x = 4y + 16 - 4
x = 4y + 12
x - 12 = 4y ----------------------------(2)
(1) — (2)
x - (x-12) = 7y - 4y
x - x + 12= 3y
3y = 12
y = 12/3= 4
The present age of son = 4
The present age of man = 7y
= 7*4= 28

Answer 2

Let the present age of the boy be x. Then his age after 4 years is x + 4.
His father's present age is 7x. His age after 4 yrs is 7x + 4, which is also 4 times his sons age (given in the question).
So, 
7x + 4= 4×(x+4)
7x + 4 = 4x + 16(distributive law of multiplication over addition)
7x - 4x = 16 - 4
3x = 12
x = 12/3 = 4
∴The son's present age is 4 yrs and the father's age is 7×4, i.e. 28 yrs.