**Question:**

Ten years ago, a father was twelve times as old as his son and ten years hence, he will be twice as old as his son will be then. Find their present ages.

**Solution:**

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

After 10 years, father's age will be $(x+10)$ years and son's age will be $(y+10)$ years. Thus using the given information, we have

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

$\Rightarrow x+10=2 y+20$

$\Rightarrow x-2 y-10=0$

Before 10 years, the age of father was $(x-10)$ years and the age of son was $(y-10)$ years. Thus using the given information, we have

$x-10=12(y-10)$

$\Rightarrow x-10=12 y-120$

$\Rightarrow x-12 y+110=0$

So, we have two equations

$x-2 y-10=0$

$x-12 y+110=0$

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

By using cross-multiplication, we have

$\frac{x}{(-2) \times 110-(-12) \times(-10)}=\frac{-y}{1 \times 110-1 \times(-10)}=\frac{1}{1 \times(-12)-1 \times(-2)}$

$\Rightarrow \frac{x}{-220-120}=\frac{-y}{110+10}=\frac{1}{-12+2}$

$\Rightarrow \frac{x}{-340}=\frac{-y}{120}=\frac{1}{-10}$

$\Rightarrow \frac{x}{340}=\frac{y}{120}=\frac{1}{10}$

$\Rightarrow x=\frac{340}{10}, y=\frac{120}{10}$

$\Rightarrow x=34, y=12$

Hence, the present age of father is 34 years and the present age of son is 12 years.