A father is three times as old as his son. After twelve years, his age will be twice as that of his son then. Find the their present ages.

#### Solution

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

Father is three times as old as his son. Thus, we have

`x=3y`

`⇒ x-3y=0`

After 12 years, father’s age will be `(x+12)` years and son’s age will be `(y +12)` years. Thus using the given information, we have

`x+12=2(y+12)`

`⇒ x+12=2y+24`

`⇒ x-2y -12=0`

So, we have two equations

`x-3y=0`

`x-2y-12=0`

Here *x* and *y* are unknowns. We have to solve the above equations for *x* and *y*.

By using cross-multiplication, we have

`x/((-3)xx(-12)-(-2)xx0)=(-y)/(1xx(-12)-1xx0)=(1)/(1xx(-2)-1xx(-3))`

`⇒ x/(36-0)=(-y)/(-12-0)=1/1`

`⇒ x/36=(-y)/(-12)=1/1`

`⇒ x/36=y/12=1`

`⇒ x= 36 , y = 12`

Hence, the present age of father is 36 years and the present age of son is 12 years.