Question

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

Answer 1

Let the age of father be $x$ years and that of the son be $y$ years.

As per the question, $x-2$ = $5(y-2)$
or, $x-5y$ =$-10+2$
or, $x-5y$ = $-8$
or, $x$ = $5y-\mathrm{8...}\left(i\right)$

Also, $x+2$ = $3(y+2)+8$
or, $x-3y$ = $6+8-2$ = $\mathrm{12...}\left(ii\right)$

From (i) and (ii)

$5y-8-3y$ = $12$
$\Rightarrow 2y=12+8$
$\Rightarrow 2y=20$

$\Rightarrow y$ = $\frac{20}{2}$
$\Rightarrow y=10$

So from (i) , $x=5y-8=50-8=42$

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