Question

6 years ago A man was three times old as his son is 6 years time he will be twice as old as his son find their present ages​

Answer 1

$hello \: mate$

Let the man's age be x and his son's age be y.

Six years before:

x-6=3(y-6)

x-6=3y-18

x-3y= -12-----(1)

After six years:

x+6=2(y+6)

x+6=2y+12

x-2y=6--------(2)

Subtracting (2) from (1),

-y= -18

y=18

Substituting y=18 in (1),

x=42

Hence, their present ages are

Father's age:42

Son's age:18

Answer 2

Their ages before 6 years :

F = 3x

S = x

Their ages after 6 years :

F = 3x + 6

S = x + 6

A.t.q..

$x - 6 = 2x - 6 \\ \\ (3x + 6) = (x + 6)2 \\ \\ 3x + 6 = 2x + 12 \\ \\ 3x - 2x = 12 - 6 \\ \\ x = 6$

Age of son = 6 + 6 =12

Age of father = (6)3+6 = 25