Chapter 6: Linear Equations in Two Variables (Exercise 6.1)

Chapter 6: Linear Equations in Two Variables (Exercise 6.1)

Question 1:
The cost of a notebook is twice the cost of a pen. Write a linear equation in two variables to represent this statement.

Solution:
* Let the cost of a notebook be x rupees.
* Let the cost of a pen be y rupees.

According to the problem statement:
Cost of notebook = 2 × (Cost of pen)
x = 2y

Expressing in standard form (ax + by + c = 0):
x - 2y = 0

Question 2:
Express the following linear equations in the form ax + by + c = 0 and indicate the values of a, b, and c in each case:
1. 2x + 3y = 9.35
2. x - (y / 5) - 10 = 0
3. -2x + 3y = 6
4. x = 3y

Solution:
1. 2x + 3y - 9.35 = 0
Comparing with ax + by + c = 0:
a = 2, b = 3, c = -9.35

2. x - (1/5)y - 10 = 0
Comparing with ax + by + c = 0:
a = 1, b = -1/5, c = -10

3. -2x + 3y - 6 = 0
Comparing with ax + by + c = 0:
a = -2, b = 3, c = -6

4. x - 3y + 0 = 0
Comparing with ax + by + c = 0:
a = 1, b = -3, c = 0

Question 3:
Find four different solutions for the linear equation x + 2y = 6.

Solution:
Given equation: x = 6 - 2y

* If y = 0 ⇒ x = 6 - 2(0) = 6 ⇒ (6, 0)
* If y = 1 ⇒ x = 6 - 2(1) = 4 ⇒ (4, 1)
* If y = 2 ⇒ x = 6 - 2(2) = 2 ⇒ (2, 2)
* If y = 3 ⇒ x = 6 - 2(3) = 0 ⇒ (0, 3)

The four solutions are (6, 0), (4, 1), (2, 2), and (0, 3).

Question 4:
Draw the graph of the linear equation x + y = 4 in two variables.

Solution:
To draw the graph of x + y = 4, we find two or three points:
* When x = 0, y = 4 ⇒ (0, 4)
* When x = 2, y = 2 ⇒ (2, 2)
* When x = 4, y = 0 ⇒ (4, 0)

Plotting these points on the Cartesian plane and joining them gives a straight line representing x + y = 4.