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.
Linear Equations
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Question

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.

KnowledgeBoat · Accepted Answer

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 = = 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.

Mathematics

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.

Linear Equations

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