Question

two years ago, father was three times as old as his son and two years hence twice his age will be equal to five times that of his son. find their present ages (linear equation in one variable)

Answer 1

Let age of son be x
father's age  two yeas back was  3(x-2) and now is 3(x-2)+2
Fathers age  two years hence = 5(x+2)/2 and now is 5(x+2) -2
  Hence            3x-6+2=((5x+10)/2) -2
                       ⇒ 6x - 8=5x+10-4
                       ⇒ x=14
son's age is 14  and fathers age is 38