**Question:**

The present age of a father is three years more than three times the age of the son. Three years hence father's age will be 10 years more than twice the age of the son. Determine their present ages.

**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.