Question

Two years ago a man's was five times his son's age. two years later , his age will be 8 more than three times his son age find their present ages

Answer 1

let f and s be the respective present ages of father and son.

Two years ago:
(f-2) = 5 * (s-2)
f-2 = 5s - 10
f - 5s = -8. **eqn a

Two years later:
(f+2) = 3(s+2) + 8
f+2 = 3s + 6 + 8
f+2 = 3s + 14
f - 3s = 12. **eqn b

eqn b - eqn a:
2s = 20
s = 20/2
= 10

Substituting in eqn b:
f - 3*10 = 12
f - 30 = 12
f = 12 + 30
= 42

Present ages:
Father 42 years old
Son 10 years old