Question: 2 years ago, father's age was nine times the son's age but 3 years later it will be 5 times only. Find the present age of the father and the son.

Solution:

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

According to the question,

Condition I,
2 years ago, father's age was nine times the son's age.
or, (x -2) = 9(y-2)
or, x -2 = 9y -18
or, x = 9y -18+2
or, x = 9y -16 - (i)

Condition II,
3 years later, father's age will be five times the son's age.
or, (x +3) = 5(y+3)
or, x +3 = 5y +15 - (ii)

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

or, (9y -16) +3 = 5y +15
or, 9y -16 = 5y +15 -3
or, 9y -5y = 12 +16
or, 4y = 28
or, y = \frac{28}{4}
\therefore y = 7

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

or, x = 9×7 - 16
or, x = 63 - 16
\therefore x = 47

So, (x,y) = (47,7)

Therefore, the required present ages of the father and the son are 47 years and 7 years respectively.