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?

Let present age of Jacob be x years and his son's age be y years.

Given,

Five years later, the age of Jacob will be three times that of his son.

⇒ (x + 5) = 3(y + 5)

⇒ x + 5 = 3y + 15

⇒ x - 3y + 5 - 15 = 0

⇒ x - 3y - 10 = 0 ……..(1)

Given,

Five years ago, Jacob's age was seven times that of his son.

⇒ (x - 5) = 7(y - 5)

⇒ x - 5 = 7y - 35

⇒ x - 7y - 5 + 35 = 0

⇒ x - 7y + 30 = 0 ……….(2)

Subtracting equation (2) from (1), we get :

⇒ x - 3y - 10 - (x - 7y + 30) = 0

⇒ x - x - 3y + 7y - 10 - 30 = 0

⇒ 4y - 40 = 0

⇒ 4y = 40

⇒ y = $\dfrac{40}{4}$ = 10.

Substituting value of y in equation (1), we get :

⇒ x - 3 × 10 - 10 = 0

⇒ x - 30 - 10 = 0

⇒ x - 40 = 0

⇒ x = 40.

Hence, pair of linear equations are x - 3y - 10 = 0 and x - 7y + 30 = 0, where x and y are the ages in years of Jacob and his son; x = 40, y = 10.