Looking for accurate SEBA Class 9 Maths Exercise 4.2 (Linear Equations in Two Variables) textual solutions for the 2026–27 academic year? Access complete step-by-step solutions covering infinitely many solutions, finding four different solutions for linear equations, verifying given solutions, and calculating unknown values of k to score top marks in your annual exams.
EXERCISE 4.2
1. Which 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
soln: Correct Option: (iii) infinitely many solutions
Reason:
For every real value of x, there exists a corresponding real value of y, and vice versa, because y = 3x + 5 represents a straight line in two variables.
2. Write four solutions for each of the following equations:
(i) 2x + y = 7
soln: 2x + y = 7 ⟹ y = 7 – 2x
Put x = 0:
y = 7 – 2(0) = 7 ⟹ (0, 7)
Put x = 1:
y = 7 – 2(1) = 5 ⟹ (1, 5)
Put x = 2:
y = 7 – 2(2) = 3 ⟹ (2, 3)
Put x = 3:
y = 7 – 2(3) = 1 ⟹ (3, 1)
Four solutions: (0, 7), (1, 5), (2, 3), (3, 1)
(ii) πx + y = 9
soln: πx + y = 9 ⟹ y = 9 – πx
Put x = 0:
y = 9 – π(0) = 9 ⟹ (0, 9)
Put x = 1:
y = 9 – π(1) = 9 – π ⟹ (1, 9 – π)
Put x = 2:
y = 9 – π(2) = 9 – 2π ⟹ (2, 9 – 2π)
Put x = -1:
y = 9 – π(-1) = 9 + π ⟹ (-1, 9 + π)
Four solutions: (0, 9), (1, 9 – π), (2, 9 – 2π), (-1, 9 + π)
(iii) x = 4y
soln: x = 4y ⟹ y = x/4
Put x = 0:
y = 0/4 = 0 ⟹ (0, 0)
Put x = 4:
y = 4/4 = 1 ⟹ (4, 1)
Put x = 8:
y = 8/4 = 2 ⟹ (8, 2)
Put x = -4:
y = -4/4 = -1 ⟹ (-4, -1)
Four solutions: (0, 0), (4, 1), (8, 2), (-4, -1)
(iv) 2x – 3y = 6
soln: 2x – 3y = 6 ⟹ 2x – 6 = 3y ⟹ y = (2x – 6)/3
Put x = 0:
y = (2(0) – 6)/3 = -2 ⟹ (0, -2)
Put x = 3:
y = (2(3) – 6)/3 = 0 ⟹ (3, 0)
Put x = 6:
y = (2(6) – 6)/3 = 2 ⟹ (6, 2)
Put x = -3:
y = (2(-3) – 6)/3 = -4 ⟹ (-3, -4)
Four solutions: (0, -2), (3, 0), (6, 2), (-3, -4)
SEBA Class 9 Maths Exercise 4.2 Solutions
3. Check which of the following are solutions of the equation x – 2y = 4 and which are not:
(i) (0, 2)
soln: Put x = 0, y = 2:
LHS = 0 – 2(2) = -4
RHS = 4
LHS ≠ RHS
Answer: Not a solution.
(ii) (2, 0)
soln: Put x = 2, y = 0:
LHS = 2 – 2(0) = 2
RHS = 4
LHS ≠ RHS
Answer: Not a solution.
(iii) (4, 0)
soln: Put x = 4, y = 0:
LHS = 4 – 2(0) = 4
RHS = 4
LHS = RHS
Answer: It is a solution.
(iv) (√2, 4√2)
soln: Put x = √2, y = 4√2:
LHS = √2 – 2(4√2) = √2 – 8√2 = -7√2
RHS = 4
LHS ≠ RHS
Answer: Not a solution.
(v) (1, 1)
soln: Put x = 1, y = 1:
LHS = 1 – 2(1) = 1 – 2 = -1
RHS = 4
LHS ≠ RHS
Answer: Not a solution.
4. Find the value of k, if x = 2, y = 1 is a solution of the equation 2x + 3y = k.
soln: Given equation: 2x + 3y = k
Given x = 2, y = 1 is a solution.
Substituting the values:
2(2) + 3(1) = k
4 + 3 = k
k = 7
Answer: k = 7
Class 9 Maths Chapter 4 Exercise 4.2 Question Answer SEBA 2026-27
5. If x = 3k + 2 and y = 2k – 1 is a solution of the equation 4x – 3y + 1 = 0. Find the value of k.
soln: Given equation: 4x – 3y + 1 = 0
Given x = 3k + 2 and y = 2k – 1.
Substituting x and y in the equation:
4(3k + 2) – 3(2k – 1) + 1 = 0
12k + 8 – 6k + 3 + 1 = 0
6k + 12 = 0
6k = -12
k = -12/6
k = -2
Answer: k = -2
Linear Equations in Two Variables Class 9 SEBA Ex 4.2 Solutions
5. For what value of k, x = 2k – 1 and y = k is a solution of the equation 3x – 5y – 7 = 0.
soln: Given equation: 3x – 5y – 7 = 0
Given x = 2k – 1 and y = k.
Substituting x and y in the equation:
3(2k – 1) – 5(k) – 7 = 0
6k – 3 – 5k – 7 = 0
k – 10 = 0
k = 10
Answer: k = 10
6. Choose the correct option (Q. No. 7 to 9)
Each of the points (-2, 2), (0, 0), (3, -3) satisfies the linear equation
(a) x – y = 0
(b) x + y = 0
(c) -x + 2y = 0
(d) x – 3y = 0
soln: For point (-2, 2):
x + y = -2 + 2 = 0
For point (0, 0):
x + y = 0 + 0 = 0
For point (3, -3):
x + y = 3 + (-3) = 0
Correct Option: (b) x + y = 0
SEBA Class 9 Maths Chapter 4 Exercise 4.2 Textbook Answers
7. Statement P: If (2, 0) is a solution of the linear equation 2x + 3y = k then the value of k is 4.
Statement Q: If (3, 4) is a solution of the linear equation 3y = ax + 7 then the value of a is 5/3.
(a) Both the statements P and Q are true
(b) Both the statements P and Q are false
(c) Statement P is true but Q is false
(d) Statement P is false but Q is true
soln: For Statement P:
Put (2, 0) in 2x + 3y = k:
2(2) + 3(0) = k ⟹ k = 4
So, Statement P is True.
For Statement Q:
Put (3, 4) in 3y = ax + 7:
3(4) = a(3) + 7
12 = 3a + 7
3a = 12 – 7
3a = 5 ⟹ a = 5/3
So, Statement Q is True.
Correct Option: (a) Both the statements P and Q are true
8. Consider the linear equation 2x – 3y = 12
(i) It has infinite number of solutions.
(ii) (5, 7) is one a solution of the equation.
(iii) (6, 0) is not a solution of the equation.
(iv) (9, 2) is one of the solutions.
Choose the correct option
(a) (i) and (iv)
(b) (ii) and (iv)
(c) (iii) and (iv)
(d) (i) and (iii)
soln: (i) Every linear equation in two variables has an infinite number of solutions. (True)
(ii) Check (5, 7):
LHS = 2(5) – 3(7) = 10 – 21 = -11 ≠ 12 (False)
(iii) Check (6, 0):
LHS = 2(6) – 3(0) = 12 = RHS
So (6, 0) IS a solution. Statement says “is not a solution”. (False)
(iv) Check (9, 2):
LHS = 2(9) – 3(2) = 18 – 6 = 12 = RHS
So (9, 2) IS a solution. (True)
Hence, statements (i) and (iv) are correct.
Correct Option: (a) (i) and (iv)
Class 9 Maths Exercise 4.2 Find Value of k
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