# Two years ago, father was three times as old as his son and two years hence, twice his age will be equal to five times that of his son. Find their presents ages.

The given question can be simplified using the linear equations in two variables and the concept of ratios.

## Answer: The present ages of the father and son are 38 years and 14 years respectively.

Let's try to rewrite the given information mathematically.

**Explanation:**

Let the age of the son be x and the age of the father be y.

2 years ago | Present age | 2 years hence | |
---|---|---|---|

Son | (x - 2) | x | (x + 2) |

Father | (y - 2) | y | (y + 2) |

Given that two years ago, the father was three times as old as his son.

⇒ (y - 2) = 3(x - 2)

⇒ y - 2 = 3x - 6

⇒ y - 3x = 2 - 6

⇒ y - 3x = - 4

⇒ **3x - y = 4** -----------------> equation (1)

Two years hence, twice the father's age will be equal to five times that of his son.

⇒ 2(y + 2) = 5(x + 2)

⇒ 2y + 4 = 5x + 10

⇒ 2y - 5x = - 4 + 10

⇒ 2y - 5x = 6

⇒ **5x - 2y = - 6** -----------------> equation (2)

By solving equation (1) & (2) using method of elimination, we get

6x - 2y = 8** **-----------------> (By multiplying [equation (1)] × 3)

5x - 2y = - 6

⇒ x = 14

substituting value of x in eqaution(1):

⇒ 3(14) - y = 4

⇒ 42 - y = 4

⇒ -y = -38

⇒ y = 38

Thus, present age of son: x = 14 years

Present age of father: y = 38 years