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in Linear Equations by (56.3k points)

Ten years ago, a father was twelve times as old as his son and ten years hence, he will be twice as old as his son will be then. Find their present ages.

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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 10 years, father’s age will be (x+10) years and son’s age will be (y + 10) years.

So, the equation formed is 

x + 10 = 2(y + 10) 

x – 10 = 2y + 20 

x – 2y – 10 = 0……… (i) 

Also again from the question it’s given as, 

Before 10 years, the age of father was (x – 10) years and the age of son was (y – 10) years. 

Furthermore, the relation between their 10 years ago is given below 

x – 10 = 12(y – 10) 

x – 10 = 12y – 120 

x – 12y + 110 = 0……… (ii) 

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

Using cross-multiplication, we have

⇒ x = 34, y = 12 

Hence, the present age of father is 34 years and the present age of the son is 12 years.

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