NCERT Solutions for Exercise 4.2 Class 9 Maths Chapter 4

NCERT Solutions for Exercise 4.2 Class 9 Maths Chapter 4 - Linear Equations in Two Variables

Edited By Vishal kumar | Updated on Apr 24, 2025 10:05 AM IST

Real-world relationships between two quantities are best expressed using linear equations to achieve an easy, predictable representation. Linear equations between two variables are visualised through graphical methods during this exercise. Every linear equation generates straight lines by connecting points which were plotted on the Cartesian plane. This exercise enables us to connect the visual relationship between equation solutions and their corresponding points located on the same straight line. The combined practice of algebra and coordinate geometry becomes easier through this approach.

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NCERT Solutions for Class 9 Maths Chapter 4 – Linear Equations in Two Variables Exercise 4.2
Access Solution of Linear Equations in Two Variables Class 9 Chapter 4 Exercise: 4.2
Topics Covered in Chapter 4 - Linear Equations in Two Variables (Exercise 4.2)
NCERT Solutions of Class 9 Subject Wise
NCERT Exemplar Solutions of Class 9 Subject-Wise

NCERT Solutions for Exercise 4.2 Class 9 Maths Chapter 4 - Linear Equations in Two Variables

Students receive essential guidance about linear equation graphical solutions through the NCERT Solutions for their Class 9 studies. The exercise guides students toward obtaining solutions by providing variable values before they create ordered pairs from those values. Students gain practical experience in graph drawing while they learn about slopes and confirm if points exist on specified lines through this method as specified in the NCERT Books. The successful mastery of this subject creates a solid foundation that enables further mathematics study and problem modelling in real-life scenarios.

NCERT Solutions for Class 9 Maths Chapter 4 – Linear Equations in Two Variables Exercise 4.2

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Access Solution of Linear Equations in Two Variables Class 9 Chapter 4 Exercise: 4.2

Q1 Which one of the following options is true, and why? $y = 3 x + 5$ has

(i) a unique solution,

(ii) only two solutions,

(iii) infinitely many solutions

Answer:

Given equation: $y = 3 x + 5$

Let's put different values of x into the equation and find the corresponding values of y.

Let x = 0, then

y = 3(0) + 5 = 0 + 5 = 5

Let x = 1, then

y = 3(1) + 5 = 3 + 5 = 8

Let x = -1, then

y = 3(-1) + 5 = -3 + 5 = 2

Let x = 2, then

y = 3(2) + 5 = 6 + 5 = 11

Therefore, we can see for each value of x, we get unique value of y.

Thus, the equation has infinitely many solutions.

Q2 Write four solutions for each of the following equations:

(i) $2 x + y = 7$ (ii) $π x + y = 9$ (iii) $x = 4 y$

Answer:

(i) Given equation: $2 x + y = 7$

Let's put 4 different values of x.

Putting x = 0, we have , $y = 7 - 2 \times 0 = 7$ means $(0, 7)$ is a solution.

Putting x = 1, we have , $y = 7 - 2 \times 1 = 5$ means $(1, 5)$ is a solution.

Putting x = 2, we have , $y = 7 - 2 \times 2 = 3$ means $(2, 3)$ is a solution.

Putting x = 3, we have , $y = 7 - 2 \times 3 = 1$ means $(3, 1)$ is a solution.

Therefore, the four solutions are : $(0, 7), (1, 5), (2, 3), (3, 1)$ .

(ii) Given : $π x + y = 9$

Let's put 4 different values of x.

Putting x = 0, we have , $y = 9 - π \times 0 = 9$ means $(0, 9)$ is a solution.

Putting x = 1, we have , $y = 9 - π \times 1 = 9 - π$ means $(1, 9 - π)$ is a solution.

Putting x = 2, we have , $y = 9 - π \times 2 = 9 - 2 π$ means $(2, 9 - 2 π)$ is a solution.

Putting x = 3, we have , $y = 9 - π \times 3 = 9 - 3 π$ means $(3, 9 - 3 π)$ is a solution.

Therefore, the four solutions are : $(0, 9), (1, 9 - π), (2, 9 - 2 π), (3, 9 - 3 π)$ .

(iii) Given : $x = 4 y$

Let's put 4 different values of x.

Putting x = 0, we have , $y = \frac{0}{4} = 0$ means $(0, 0)$ is a solution.

Putting x = 1, we have , $y = \frac{1}{4}$ means $(1, \frac{1}{4})$ is a solution.

Putting x = 2, we have , $y = \frac{2}{4} = \frac{1}{2}$ means $(2, \frac{1}{2})$ is a solution.

Putting x = 3, we have , $y = \frac{3}{4}$ means $(3, \frac{3}{4})$ is a solution.

Therefore, the four solutions are : $(0, 0)$ , $(1, \frac{1}{4})$ , $(2, \frac{1}{2})$ and $(3, \frac{3}{4})$ .

Q3 (i) Check which of the following are solutions of the equation $x - 2 y = 4$ and which are not: ((0,2)

Answer:

(i) Given equation: $x - 2 y = 4$

According to the question, put (0,2), which means x as 0 and y as 2.

Therefore we get , $x - 2 y = 0 - 2 (2) = - 4 \neq 4$

Thus, $(0, 2)$ is not a solution of $x - 2 y = 4$ .

Q3 (ii) Check which of the following are solutions of the equation $x - 2 y = 4$ and which are not: (2,0)

Answer:

Given equation: $x - 2 y = 4$

According to the question, put (2,0), which means x as 2 and y as 0.

Therefore we get , $x - 2 y = 2 - 2 (0) = 2 \neq 4$

Thus, (2,0) is not a solution of $x - 2 y = 4$ .

Q3 (iii) Check which of the following are solutions of the equation $x - 2 y = 4$ and which are not: (4,0)

Answer:

Given equation: $x - 2 y = 4$

According to the question, put (4,0), which means x as 4 and y as 0.

Therefore we get, $x - 2 y = 4 - 2 (0) = 4 = 4$

Thus, (4,0) is a solution of $x - 2 y = 4$ .

Q3 (iv) Check which of the following are solutions of the equation $x - 2 y = 4$ and which are not: $(\sqrt{2}, 4 \sqrt{2})$

Answer:

Given equation: $x - 2 y = 4$

According to the question, put $(\sqrt{2}, 4 \sqrt{2})$ , which means x as $\sqrt{2}$ and y as $4 \sqrt{2}$ .

Therefore we get, $x - 2 y = \sqrt{2} - 2 (4 \sqrt{2}) = \sqrt{2} - 8 \sqrt{2} = - 7 \sqrt{2} \neq 4$

Thus, $(\sqrt{2}, 4 \sqrt{2})$ is not a solution of $x - 2 y = 4$ .

Q3 (v) Check which of the following are solutions of the equation $x - 2 y = 4$ and which are not: (1,1)

Answer:

Given equation: $x - 2 y = 4$

According to the question, put (1,1), which means both x and y as 1.

Therefore we get, $x - 2 y = 1 - 2 (1) = - 1 \neq 4$

Thus, (1,1) is not a solution of $x - 2 y = 4$ .

Q4 Find the value of k , if $x = 2$ , $y = 1$ is a solution of the equation $2 x + 3 y = k$ .

Answer:

Given equation: $2 x + 3 y = k$

According to the question, put x as 2 and y as 1.

Therefore we get, $k = 2 x + 3 y = 2 (2) + 3 (1) = 4 + 3 = 7$

Thus, k = 7 for $2 x + 3 y = k$ putting x = 2 and y = 1.

Also Read:

Linear Equation in Two Variables Exercise 4.1

Topics Covered in Chapter 4 - Linear Equations in Two Variables (Exercise 4.2)

1. Plotting linear equations on the Cartesian plane: The educational method shows students how to visually display equations with two variables through the process of plotting points on the x-y coordinate system.

2. Drawing straight lines using solutions of linear equations: When connecting plotted points to each other a straight line appears which shows all potential solutions of the equation.

3. Finding multiple solutions by assigning variable values: The assignment requires learners to choose different variable values before computing the associated variable values which results in coordinate pairs.

4. Understanding that all points on a line satisfy the equation: The drawn line contains all valid solutions of the linear equation.

5. Verifying solutions through graphical representation: Students use the line to check which points fulfill the equation by determining their position on the line.

6. Linking algebraic and graphical methods: This exercise creates a stronger bond between interpreting algebraic expressions through both geometric visualization of graphs and algebraic understanding.

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NCERT Solutions of Class 9 Subject Wise

Students must check the NCERT solutions for class 9 of Mathematics and Science Subjects.

NCERT Exemplar Solutions of Class 9 Subject-Wise

Students must check the NCERT Exemplar solutions for class 9 of Mathematics and Science Subjects.

Frequently Asked Questions (FAQs)

1. What is the general form of a linear equation in two variables, according to NCERT solutions for Class 9 Maths chapter 4 exercise 4.2?

ax+by+c=0, is the general form of the linear equation in two variables where a, b,and c are real numbers.

2. Check whether the equation xy-5=8 a linear equation in two variables?

Because of the term xy which is of degree 2 , xy-5=8 is not a linear equation in two variables.

3. Check whether the equation 7x+y=8 a linear equation in two variables ?

7x+y=8 is a linear equation in two variables since the degree of the given expression 7x+y=8 is 1

4. Linear equations in one variable have a ________ solution.

Linear equations in one variable have a unique solution.

5. In exercise 4.2 Class 9 Maths , What is the two-variable equation?

The two-variable equation is nothing but an equation that has two different variables and also two different solutions.

6. What are the coefficients of x and y the equation 9x-y = -12?

The coefficient of x is 9 and the coefficient of y is -1.

7. What is the constant of the equation 2x-4y=-3?

The constant of the equation 2x-4y=-3 is -3

8. According to NCERT solutions for Class 9 Maths chapter 4 exercise 4.2, What is the coefficient of y in the equation 2x-4y=-3?

The coefficient of y in the equation 2x-4y=-3 is -4

Also Read

NCERT Solutions for Class 9 Maths Chapter 6 Lines And Angles

NCERT Exemplar Class 9 Science Solutions - Chapter Wise Problems with Solutions

NCERT Solutions for Class 9 Maths Chapter 5 Introduction to Euclids Geometry

NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables

NCERT Solutions for Class 9 Maths Chapter 3 Coordinate Geometry

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials

NCERT Class 9 Maths Chapter 5 Notes Introduction To Euclid’s Geometry - Download PDF

NCERT Class 9 Maths Chapter 1 Notes Number System - Download Free PDF

NCERT Class 9 Maths Chapter 6 Notes Lines and Angles - Download PDF

Triangles Class 9th Notes - Free NCERT Class 9th Maths Chapter 7 Notes - Download PDF

NCERT Class 9 Maths Chapter 8 Notes Quadrilaterals - Download PDF

Heron's Formula Class 9th Notes - Free NCERT Class 9 Maths Chapter 12 Notes - Download PDF

NCERT Exemplar Class 9 Maths Solutions Chapter 2 Polynomials

NCERT Exemplar Class 9 Maths Solutions Chapter 6 Lines and Angles

NCERT Exemplar Class 9 Maths Solutions Chapter 5 Introduction to Euclids Geometry

NCERT Exemplar Class 9 Science Solutions Chapter 15 Improvement in Food Resources

NCERT Exemplar Class 9 Science Solutions Chapter 14 Natural Resources

NCERT Exemplar Class 9 Science Solutions Chapter 5 The Fundamental Unit of Life

NCERT Books for Class 9 Maths 2025-26; Download Free PDF

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