Question: Three years ago the sum of the ages of a father and his son was 48 years and three years hence father's age will be three times that of his son's age. Find the present ages of the father and his son.

Solution:

Let the present ages of the father and son be x years and y years respectively.

According to the question,

Condition I,
Three years ago the sum of the ages of the father and his son was 48 years.
$or, (x -3) + (y-3) = 48$
$or, x -3 + y -3 = 48$
$or, x -6 = 48-y$
$or, x = 54 - y$ - (i)

Condition II,
Three years hence the father's age will be three times the son's age,
$or, (x +3) = 3(y +3)$
$or, x + 3 = 3y +9$ - (ii)

Put value of x from equation (i) in equation (ii), we get,

$or, (54 - y )+3 = 3y +9$
$or, 54 - y = 3y +9-3$
$or, 54 -6 = 3y +y$
$or, 48 = 4y$
$or, y = \frac{48}{4}$
$\therefore y = 12$

Put the value of y in equation (i), we get,

$or, x = 54 -12$
$\therefore x = 42$

So, (x,y) = (42, 12)

Therefore, the required present ages of the father and the son are 42 years and 12 years respectively.