A father is three times as old as his son. After twelve years,

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

$x=3 y$

$\Rightarrow x-3 y=0$

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

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

$\Rightarrow x+12=2 y+24$

$\Rightarrow x-2 y-12=0$

So, we have two equations

$x-3 y=0$

 

$x-2 y-12=0$

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

By using cross-multiplication, we have

$\frac{x}{(-3) \times(-12)-(-2) \times 0}=\frac{-y}{1 \times(-12)-1 \times 0}=\frac{1}{1 \times(-2)-1 \times(-3)}$

$\Rightarrow \frac{x}{36-0}=\frac{-y}{-12-0}=\frac{1}{-2+3}$

$\Rightarrow \frac{x}{36}=\frac{-y}{-12}=\frac{1}{1}$

$\Rightarrow \frac{x}{36}=\frac{y}{12}=1$

$\Rightarrow x=36, y=12$

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