Q2(ii):

Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method : (ii) Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?

Given:

1. Five years ago, Nuri's age was thrice the age of Sonu.

2. Ten years later, Nuri's age will be twice the age of Sonu.

To Find:

The present ages of Nuri and Sonu.

Step 1: Defining Variables

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

Let the present age of Sonu be $y$ years.

Step 2: Formulating the Equations

Condition 1: Five years ago

Nuri's age was $(x - 5)$ and Sonu's age was $(y - 5)$.

According to the problem: $(x - 5) = 3(y - 5)$

$x - 5 = 3y - 15$

$x - 3y = -15 + 5$

$x - 3y = -10$ --- (Equation 1)

Condition 2: Ten years later

Nuri's age will be $(x + 10)$ and Sonu's age will be $(y + 10)$.

According to the problem: $(x + 10) = 2(y + 10)$

$x + 10 = 2y + 20$

$x - 2y = 20 - 10$

$x - 2y = 10$ --- (Equation 2)

Step 3: Solving by Elimination Method

We have the system of equations:

(1) $x - 3y = -10$

(2) $x - 2y = 10$

To eliminate $x$, subtract Equation (1) from Equation (2):

$(x - 2y) - (x - 3y) = 10 - (-10)$

$x - 2y - x + 3y = 10 + 10$

$y = 20$

Step 4: Finding the value of $x$

Substitute $y = 20$ into Equation (2):

$x - 2(20) = 10$

$x - 40 = 10$

$x = 10 + 40$

$x = 50$

Step 5: Verification

Five years ago: Nuri was $50 - 5 = 45$, Sonu was $20 - 5 = 15$. Since $45 = 3 \times 15$, the first condition is satisfied.

Ten years later: Nuri will be $50 + 10 = 60$, Sonu will be $20 + 10 = 30$. Since $60 = 2 \times 30$, the second condition is satisfied.

Final Answer: The present age of Nuri is 50 years and the present age of Sonu is 20 years.