The present age of a father is three years more than three times the age of the son.

Solution:

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

The present age of father is three years more than three times the age of the son. Thus, we have

$x=3 y+3$

$\Rightarrow x-3 y-3=0$

After 3 years, father's age will be $(x+3)$ years and son's age will be $(y+3)$ years.

Thus using the given information, we have

$x+3=2(y+3)+10$

$\Rightarrow x+3=2 y+6+10$

$\Rightarrow x-2 y-13=0$

So, we have two equations

$x-3 y-3=0$

$x-2 y-13=0$

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

By using cross-multiplication, we have

$\frac{x}{(-3) \times(-13)-(-2) \times(-3)}=\frac{-y}{1 \times(-13)-1 \times(-3)}=\frac{1}{1 \times(-2)-1 \times(-3)}$

$\Rightarrow \frac{x}{39-6}=\frac{-y}{-13+3}=\frac{1}{-2+3}$

$\Rightarrow \frac{x}{33}=\frac{-y}{-10}=\frac{1}{1}$

$\Rightarrow \frac{x}{33}=\frac{y}{10}=1$

$\Rightarrow x=33, y=10$

Hence, the present age of father is 33 years and the present age of son is 10 years.