Question

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

Answer 1

Let father age be $x$, his son age be $y$.

Before $2$ years,

$\Rightarrow$$(x-2)=5(y-2)-------\left(1\right)$

After 2 years

$\Rightarrow$$(x+2)=8+3(y+2)--------\left(2\right)$

From (1) $x-5y+8=0$

From (2)

$\Rightarrow$$x+2-3y-6-8=0$

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

Solving them

$\Rightarrow$$-2y+20=0$

$\Rightarrow$$y=10$

When $y=10,x=42$

So the present age of the father is $42$ and the son age is $10.$