Question

Before 6 years the father was 3 times older than his son,after 6 years he will be 2 times older.How many years old are the father and the son?

Answer 1

Let $f$ = father's age
Let $s$ = son's age

Before $6$ years :
$f-6 = 3(s-6)\implies f=3s-12$

After $6$ years :
$f+6 = 2(s+6)\implies f=2s+6$

Subtract second equation from first:
$0=s-18\\s=18$

Plug this in second equation and get:
$f=2(18)+6\\f=42$

Therefore the father is $42$ years old and the son is $18$ years old

Answer 2

Let the present (current) ages of father and son be  F  and  S  years.  We need to form two equations in F and S and then solve them.

      Father was 3 times older than his son, 6 years earlier.  Actually this statement should be that  Father was 3 times as old as  the son.  6 years ago, their ages were F-6 years and  S-6 years respectively.

         =>  F - 6 =  3 *  (S - 6)     --- (1)

     After  6 years from now, father will be 2 times as old as his Son. After 6 years their ages are F+6 years and S+6 years respectively.

         =>  F + 6  = 2 * (S + 6)      --- (2)

subtract (2) from (1), we get

                - 12 = S - 30      =>    S = 18 years
 Using equation (1) ,   we get,   F - 6 = 3 * (18-6) = 36

        Hence,   F = father's age = 42 years.    
             and son's age = 18 years.