Form the pair of linear equations for the following problem and find their solution by the 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?'

(N/A) Let the present age of Jacob be $x$ years and the present age of his son be $y$ years.
Case $1$: Five years hence,Jacob's age will be $(x + 5)$ and his son's age will be $(y + 5)$.
According to the problem,$(x + 5) = 3(y + 5) \implies x + 5 = 3y + 15 \implies x - 3y = 10$ (Equation $1$).
Case $2$: Five years ago,Jacob's age was $(x - 5)$ and his son's age was $(y - 5)$.
According to the problem,$(x - 5) = 7(y - 5) \implies x - 5 = 7y - 35 \implies x - 7y = -30$ (Equation $2$).
From Equation $1$,$x = 3y + 10$.
Substituting this value into Equation $2$: $(3y + 10) - 7y = -30 \implies -4y = -40 \implies y = 10$.
Substituting $y = 10$ into $x = 3y + 10$: $x = 3(10) + 10 = 40$.
Therefore,the present age of Jacob is $40$ years and the present age of his son is $10$ years.