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Form the pair of linear equations for the following problems and find their solution by substitution method : 

Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present . ages ?

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Present age of Jacob be ‘x’. 

His son’s present age is ‘y’. 

After 5 years, Age of Jacob will be x + 5 

After 5 years, Age of his son will be y + 5 

At that time Jacob’s age will be three times that of his son. 

x + 5 = 3(y + 5) 

x + 5 = 3y + 15 

x – 3y = 15 – 5 

x – 3y = 10 ………… (i) 

5 years ago Jacob’s age was x – 5 

5 years ago, age of Jacob’s son was y – 5 

At that time Jacob’s age was seven times that of his son. 

∴ x – 5 = 7(y – 5) 

x – 5 – 7y – 35 

x – 7y = -35 + 5 

x – 7y = -30 ………….. (ii) 

From eqn. (i), 

x – 3y = 10 

x = 10 + 3y 

Substituting the value of ‘x’ in eqn. (ii), 

x – 7y = -30 

10 + 3y – 7y = -30 

10 – 4y = -30 

-4y = -30 – 10 

-4y = -40

∴ y = \(\frac{40}{4}\)

∴ y = 10 

Substituting the value of ‘y’ in 

x = 10 + 3y 

= 10 + 3 × 10 

= 10 + 30 

∴ x = 40 

∴ Present age of Jacob is 40. 

Present age of Jacob’s son is 10 .

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