**Question:**

Six years hence a man's age will be three times the age of his son and three years ago he was nine times as old as his son. Find their present ages.

**Solution:**

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

After 6 years, the man's age will be $(x+6)$ years and son's age will be $(y+6)$ years. Thus using the given information, we have

$x+6=3(y+6)$

$\Rightarrow x+6=3 y+18$

$\Rightarrow x-3 y-12=0$

Before 3 years, the age of the man was $(x-3)$ years and the age of son's was $(y-3)$ years. Thus using the given information, we have

$x-3=9(y-3)$

$\Rightarrow x-3=9 y-27$

$\Rightarrow x-9 y+24=0$

So, we have two equations

$x-3 y-12=0$

$x-9 y+24=0$

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

By using cross-multiplication, we have

$\frac{x}{(-3) \times 24-(-9) \times(-12)}=\frac{-y}{1 \times 24-1 \times(-12)}=\frac{1}{1 \times(-9)-1 \times(-3)}$

$\Rightarrow \frac{x}{-72-108}=\frac{-y}{24+12}=\frac{1}{-9+3}$

$\Rightarrow \frac{x}{-180}=\frac{-y}{36}=\frac{1}{-6}$

$\Rightarrow \frac{x}{180}=\frac{y}{36}=\frac{1}{6}$

$\Rightarrow x=\frac{180}{6}, y=\frac{36}{6}$

Hence, the present age of the man is 30 years and the present age of son is 6 years.