# Two years ago, a father was five times as old as his son.

Question:

Two years ago, a father was five times as old as his son. Two year later, his age will be 8 more than three times the age of the son. Find the present ages of father and son.

Solution:

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

After 2 years, father's age will be $(x+2)$ years and the age of son will be $(y+2)$ years. Thus using the given information, we have

$x+2=3(y+2)+8$

$\Rightarrow x+2=3 y+6+8$

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

Before 2 years, the age of father was $(x-2)$ years and the age of son was $(y-2)$ years. Thus using the given information, we have

$x-2=5(y-2)$

$\Rightarrow x-2=5 y-10$

$\Rightarrow x-5 y+8=0$

So, we have two equations

$x-3 y-12=0$

$x-5 y+8=0$

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

By using cross-multiplication, we have

$\frac{x}{(-3) \times 8-(-5) \times(-12)}=\frac{-y}{1 \times 8-1 \times(-12)}=\frac{1}{1 \times(-5)-1 \times(-3)}$

$\Rightarrow \frac{x}{-24-60}=\frac{-y}{8+12}=\frac{1}{-5+3}$

$\Rightarrow \frac{x}{-84}=\frac{-y}{20}=\frac{1}{-2}$

$\Rightarrow \frac{x}{84}=\frac{y}{20}=\frac{1}{2}$

$\Rightarrow x=\frac{84}{2}, y=\frac{20}{2}$

$\Rightarrow x=42, y=10$

Hence, the present age of father is 42 years and the present age of son is 10 years.