Question

Two years ago ,The age of father was five times more than his son's age. after two years the father's age will be 8 more than three times of his son's age.Find the current age of both Father and son?

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Answer 1

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

Before 2 years,

⇒(x−2)=5(y−2)−−−−−−−(1)

After 2 years

⇒(x+2)=8+3(y+2)−−−−−−−−(2)

From (1) x−5y+8=0

From (2)

⇒x+2−3y−6−8=0

⇒x−3y−12=0

Solving them

⇒−2y+20=0

⇒y=10

When y=10,x=42

So the present age of the father is 42 and the son age is 10.

Answer 2

Given :

• Two years ago the age of father was five times more than his son's age.
• After two years the father's age will be 8 more than three times of his son's age

To find :

• The current age of father
• The current age of son

Solution :

★Let:-

• The age of father = x years
• The age of the son = y years

★ Then by the given question,

x - 2 = 5(y - 2)

→ x - 5y = -10 + 2

→ x - 5y = -8

→ x = 5y - 8

x + 2 = 3(y + 2) + 8

→ x - 3y = 6 + 8 - 2

→ 5y - 8 - 3y = 12

→ 2y = 12 + 8

→ y = 20/2

→ y = 10

★ Then:-

x = 5y - 8 = 50 - 8 = 42

•°• The age of father is 42 years. and the age of son is 10 years