Q3(vi):

Form the pair of linear equations for the following problems and find their solution by substitution method. (vi) 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?

Given:

1. Five years hence (in the future), Jacob's age will be three times that of his son.

2. Five years ago (in the past), Jacob's age was seven times that of his son.

To Find:

The present ages of Jacob and his son.

Step 1: Defining Variables

Let the present age of Jacob be $x$ years.

Let the present age of his son be $y$ years.

Step 2: Formulating the Equations

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: $(x + 5) = 3(y + 5)$

$x + 5 = 3y + 15$

$x - 3y = 10$ --- (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: $(x - 5) = 7(y - 5)$

$x - 5 = 7y - 35$

$x - 7y = -30$ --- (Equation 2)

Step 3: Solving by Substitution Method

From Equation 1, express $x$ in terms of $y$:

$x = 3y + 10$ --- (Equation 3)

Substitute the value of $x$ from Equation 3 into Equation 2:

$(3y + 10) - 7y = -30$

$3y - 7y + 10 = -30$

$-4y = -30 - 10$

$-4y = -40$

$y = \frac{-40}{-4}$

$y = 10$

Step 4: Finding the value of $x$

Substitute $y = 10$ into Equation 3:

$x = 3(10) + 10$

$x = 30 + 10$

$x = 40$

Step 5: Verification

Five years hence: Jacob will be $45$, son will be $15$. $45 = 3 \times 15$ (Correct).

Five years ago: Jacob was $35$, son was $5$. $35 = 7 \times 5$ (Correct).

Final Answer: The present age of Jacob is 40 years and the present age of his son is 10 years.