One year ago a man was four times as old as his son. After 6 years, his age exceeds twice his son’s age by 9 years. Find their present ages.

Let the present age of man be x years and his son be y years.

Given,

One year ago a man was four times as old as his son,

⇒ x - 1 = 4(y - 1)

⇒ x - 1 = 4y - 4

⇒ x = 4y - 4 + 1

⇒ x = 4y - 3     ….(1)

Given,

After 6 years, the man's age will exceed twice his son's age by 9,

⇒ x + 6 = 2(y + 6) + 9

⇒ x + 6 = 2y + 12 + 9

⇒ x = 2y + 12 + 9 - 6

⇒ x = 2y + 15     …..(2)

From equation (1) in (2), we get :

⇒ 4y - 3 = 2y + 15

⇒ 4y - 2y = 15 + 3

⇒ 2y = 18

⇒ y = $\dfrac{18}{2}$ = 9.

Substituting value of y in equation (1), we get :

⇒ x = 4y - 3

⇒ x = 4(9) - 3

⇒ x = 36 - 3 = 33.

Hence, man's present age = 33 years and son's present age = 9 years.