Question: 3 years later a mother will be 4 times as old as her son. 3 years ago, the mother's age was two times her son's age will be 8 years hence. What are their present ages?

Solution:

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

According to the question,

Condition I,
3 years later a mother will be 4 times as old as her son.
$or, (x +3) = 4(y +3)$
$or, x +3 = 4y +12$
$or, x = 4y +12-3$
$or, x = 4y +9 - (i)

Condition II,
3 years ago, the mother's age was two times her son's age will be 8 years hence.
$or, (x -3) = 2(y +8)$
$or, x -3 = 2y +16$
$or, x = 2y +19$ - (ii)

Put value of x from equation (i) in equation (ii), we get,
$or, 4y +9 = 2y +19$
$or, 4y -2y = 19-9$
$or, 2y = 10$
$or, 2×y = 2×5$
$\therefore y = 5$

Put value of y in equation (i), we get,
$or, x = 4×5 +9$
$or, x = 20+9$
$\therefore x = 29$

So, (x,y) = (29,5)

Therefore, the required present ages of the mother is 29 years and that of the son is 5 years.