Chapter 4: Linear Equations in Two Variables Exercise & In-Text Questions & Answers
1The cost of a notebook is twice the cost of a pen. Write a linear equation in two variables to represent this statement. (Take the cost of a notebook to be ₹ x and that of a pen to be ₹ y).
Answer & Reference:
Given that cost of notebook = ₹ x and cost of pen = ₹ y.
According to the statement: x = 2y
Therefore, the linear equation is: x - 2y = 0 (Page 57, Exercise 4.1)
2Express the following linear equations in the form ax + by + c = 0 and indicate the values of a, b and c in each case:
Answer & Reference:
Equation
Form ax + by + c = 0
a
b
c
(i) 2x + 3y = 9.35
2x + 3y - 9.35 = 0
2
3
-9.35
(ii) x - 5y - 10 = 0
x - 5y - 10 = 0
1
-5
-10
(iii) -2x + 3y = 6
-2x + 3y - 6 = 0
-2
3
-6
(iv) x = 3y
x - 3y + 0 = 0
1
-3
0
(v) 2x = -5y
2x + 5y + 0 = 0
2
5
0
(vi) 3x + 2 = 0
3x + 0y + 2 = 0
3
0
2
(vii) y - 2 = 0
0x + y - 2 = 0
0
1
-2
(viii) 5 = 2x
2x + 0y - 5 = 0
2
0
-5
(Page 57, Exercise 4.1)
3Which one of the following options is true, and why? y = 3x + 5 has (i) a unique solution, (ii) only two solutions, (iii) infinitely many solutions
Answer & Reference:
The correct option is (iii) infinitely many solutions.
Explanation: A linear equation in two variables has infinitely many solutions because for every value of x, there is a corresponding value of y that satisfies the equation, and vice versa. (Page 58, Section 4.3)
4Write four solutions for each of the following equations: (i) 2x + y = 7 (ii) πx + y = 9 (iii) x = 4y
Answer & Reference: (i) 2x + y = 7
- When x = 0: 2(0) + y = 7 ⇒ y = 7 ⇒ (0, 7)
- When y = 0: 2x + 0 = 7 ⇒ x = 7/2 ⇒ (7/2, 0)
- When x = 1: 2(1) + y = 7 ⇒ y = 5 ⇒ (1, 5)
- When x = 2: 2(2) + y = 7 ⇒ y = 3 ⇒ (2, 3)
(ii) πx + y = 9
- When x = 0: π(0) + y = 9 ⇒ y = 9 ⇒ (0, 9)
- When y = 0: πx + 0 = 9 ⇒ x = 9/π ⇒ (9/π, 0)
- When x = 1: π(1) + y = 9 ⇒ y = 9 - π ⇒ (1, 9 - π)
- When x = 2: π(2) + y = 9 ⇒ y = 9 - 2π ⇒ (2, 9 - 2π)
(iii) x = 4y
- When y = 0: x = 4(0) ⇒ x = 0 ⇒ (0, 0)
- When y = 1: x = 4(1) ⇒ x = 4 ⇒ (4, 1)
- When y = 2: x = 4(2) ⇒ x = 8 ⇒ (8, 2)
- When y = -1: x = 4(-1) ⇒ x = -4 ⇒ (-4, -1)
(Page 59, Exercise 4.2)
5Check which of the following are solutions of the equation x - 2y = 4 and which are not: (i) (0, 2) (ii) (2, 0) (iii) (4, 0) (iv) (√2, 4√2) (v) (1, 1)
Answer & Reference:
For the equation x - 2y = 4:
- (i) (0, 2): 0 - 2(2) = -4 ≠ 4 → Not a solution
- (ii) (2, 0): 2 - 2(0) = 2 ≠ 4 → Not a solution
- (iii) (4, 0): 4 - 2(0) = 4 = 4 → Solution
- (iv) (√2, 4√2): √2 - 2(4√2) = √2 - 8√2 = -7√2 ≠ 4 → Not a solution
- (v) (1, 1): 1 - 2(1) = -1 ≠ 4 → Not a solution
(Page 59, Exercise 4.2)
6Find the value of k, if x = 2, y = 1 is a solution of the equation 2x + 3y = k.
Answer & Reference:
Given equation: 2x + 3y = k
Substitute x = 2 and y = 1:
2(2) + 3(1) = k
4 + 3 = k
k = 7
Therefore, the value of k is 7. (Page 59, Exercise 4.2)
7What are the key points about linear equations in two variables?
Answer & Reference:
Key points about linear equations in two variables (Page 59, Section 4.4):
1. An equation of the form ax + by + c = 0, where a, b and c are real numbers, such that a and b are not both zero, is called a linear equation in two variables.
2. A linear equation in two variables has infinitely many solutions.
3. Every point on the graph of a linear equation in two variables is a solution of the linear equation. Moreover, every solution of the linear equation is a point on the graph of the linear equation.
8Write each of the following equations in the form ax + by + c = 0 and indicate the values of a, b and c: (i) 2x + 3y = 4.37 (ii) x - 4 = 3y (iii) 4 = 5x - 3y (iv) 2x = y
Answer & Reference:
Equation
Form ax + by + c = 0
a
b
c
(i) 2x + 3y = 4.37
2x + 3y - 4.37 = 0
2
3
-4.37
(ii) x - 4 = 3y
x - 3y - 4 = 0
1
-3
-4
(iii) 4 = 5x - 3y
5x - 3y - 4 = 0
5
-3
-4
(iv) 2x = y
2x - y + 0 = 0
2
-1
0
(Page 56-57, Example 1)
9Write each of the following as an equation in two variables: (i) x = -5 (ii) y = 2 (iii) 2x = 3 (iv) 5y = 2
Answer & Reference:
Equation
Equation in two variables
(i) x = -5
1x + 0y + 5 = 0
(ii) y = 2
0x + 1y - 2 = 0
(iii) 2x = 3
2x + 0y - 3 = 0
(iv) 5y = 2
0x + 5y - 2 = 0
(Page 57, Example 2)
10Find four different solutions of the equation x + 2y = 6.
Answer & Reference:
Four different solutions of x + 2y = 6 are:
- (2, 2): 2 + 2(2) = 6
- (0, 3): 0 + 2(3) = 6
- (6, 0): 6 + 2(0) = 6
- (4, 1): 4 + 2(1) = 6
Therefore, the solutions are: (2, 2), (0, 3), (6, 0), and (4, 1). (Page 58, Example 3)
11Find two solutions for each of the following equations: (i) 4x + 3y = 12 (ii) 2x + 5y = 0 (iii) 3y + 4 = 0
Answer & Reference: (i) 4x + 3y = 12
- When x = 0: 3y = 12 ⇒ y = 4 ⇒ (0, 4)
- When y = 0: 4x = 12 ⇒ x = 3 ⇒ (3, 0)
(ii) 2x + 5y = 0
- When x = 0: 5y = 0 ⇒ y = 0 ⇒ (0, 0)
- When x = 1: 2(1) + 5y = 0 ⇒ 5y = -2 ⇒ y = -2/5 ⇒ (1, -2/5)
(iii) 3y + 4 = 0
- This can be written as 0x + 3y + 4 = 0
- For any value of x, y = -4/3
- Two solutions: (0, -4/3) and (1, -4/3)
(Page 58-59, Example 4)