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in Linear Equations by (56.3k points)

Six year hence a man’s age will be three times age of his son and three years ago, he was nine times as old as his son. Find their present ages.

1 Answer

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Best answer

Let’s assume the present ages of the father as x years and that of his son’s age as y years.

From the question it’s given that, 

After 6 years, the man’s age will be (x + 6) years and son’s age will be (y + 6) years.

So, the equation formed is 

x + 6 = 3(y + 6) 

x + 6 = 3y + 18 

x – 3y – 12 = 0……. (i) 

Also again from the question it’s given as, 

Before 3 years, the age of the man was (x – 3) years and the age of son’s was (y – 3) years. 

Furthermore, the relation between their 3 years ago is given below 

x – 3 = 9(y – 3) 

x – 3 = 9y – 27 

x – 9y + 24 = 0……. (ii) 

Thus, by solving (i) and (ii), we get the required solution 

Using cross-multiplication, we get

⇒x = 30, y = 6 

Hence, the present age of the man is 30 years and the present age of son is 6 years.

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