Question

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

Answer 1

Answer:

Father's age = 42

Son's age = 10

Step-by-step explanation:

Let father age be x, his son age be y.

Before 2 years

⇒(x − 2) = 5(y − 2) ⇒ x − 5y + 8 = 0

After 2 years

⇒ (x + 2) = 8 + 3(y + 2) ⇒ x − 3y − 12 = 0

Solving,

⇒ −2y + 20 = 0

⇒y = 10

Now substitute for y  in x - 3y - 12 = 0

We get x = 42.

Therefore,

Father's age = 42

Son's age = 10