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 Oct 05, 2023 03:11 PM IST

NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations In Two Variables: Exercise 4.2- Download Free PDF

NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations In Two Variables: Exercise 4.2- The notion of linear equations in two variables is covered in NCERT Solutions for Class 9 Maths exercise 4.2 as well as how to solve linear equations. Linear equations in two variables are systems of equations with a single solution, no solutions, or an infinite number of solutions of the kind ax+by+c=0, where a, b, c are real numbers. We must first determine the value of variables in order to find the set of solutions to the linear equations with two variables. The algebraic representations of linear equations in two variables were the topic of Class 9 Maths chapter 4 exercise 4.2. The substitution approach is the simplest way to solve a linear equation.

Scholarship Test: Vidyamandir Intellect Quest (VIQ)

Don't Miss: JEE Main 2027: Narayana Scholarship Test Preparation Kit for Class 9

This Story also Contains

NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations In Two Variables: Exercise 4.2- Download Free PDF
NCERT Solutions for Class 9 Maths Chapter 4 – Linear Equations in Two Variables Exercise 4.2
Linear Equations In Two Variables Class 9 Maths Chapter 4 Exercise: 4.2
More About NCERT Solutions for Class 9 Maths Exercise 4.2:
Benefits of NCERT Solutions for Class 9 Maths Exercise 4.2
Key Features of 9th Class Maths Exercise 4.2 Answers
NCERT Solutions of Class 10 Subject Wise
Subject Wise NCERT Exemplar Solutions

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

The substitution method is defined as substituting the value of one variable in another equation to obtain the value of another variable. We must first determine the value of variables in order to find the set of solutions to the linear equations with two variables. To use the substitution approach to solve a system of linear equations, we must first determine the value of one variable.

9th class maths exercise 4.2 answers are crafted by subject experts at Careers360 in straightforward language. Students can also download the PDF versions of these class 9 maths chapter 4 exercise 4.2 solutions and make use of them. The following activities are included along with Class 9 Maths chapter 4 exercise 4.2.

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

Download PDF

Linear Equations In Two Variables Class 9 Maths Chapter 4 Exercise: 4.2

Q1 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

Answer:

Given : $y = 3x + 5$

This equation is of a line and a line has infinite points on it and each point is a solution Thus, (iii) infinitely many solutions is the correct option.

Q2 Write four solutions for each of the following equations:

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

Answer:

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

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.

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

(ii) Given : $\pi x +y = 9$

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

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

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

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

The four solutions are : $(0,9),(1,9-\pi ),(2,9-2\pi ),(3,9-3\pi )$ .

(iii) Given : $x = 4y$

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.

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 $x - 2y = 4$ and which are not: ((0,2)

Answer:

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

Putting $(0,2)$ ,

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

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

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

Answer:

Given : $x - 2y = 4$

Putting (2,0),

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

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

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

Answer:

Given : $x - 2y = 4$

Putting (4,0),

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

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

Q3 (iv) Check which of the following are solutions of the equation $x - 2y = 4$ and which are not: $(\sqrt2 , 4\sqrt2)$

Answer:

Given : $x - 2y = 4$

Putting $(\sqrt2 , 4\sqrt2)$ ,

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

Therefore, $(\sqrt2 , 4\sqrt2)$ is not a solution of $x - 2y = 4$ .

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

Answer:

Given : $x - 2y = 4$

Putting (1,1) ,

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

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

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

Answer:

Given : $2x + 3y = k$

Putting (2,1),

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

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

Benefits of NCERT Solutions for Class 9 Maths Exercise 4.2

• NCERT solutions for Class 9 Maths exercise 4.2 helps students to get practice in solving a variety of maths problems, and also helps in gaining efficiency and accuracy by solving questions.

• In NCER syllabus exercise 4.2 Class 9 Maths, The substitution method is a simple strategy to solve equations that removes an equation from the problem to make it easier to solve.

• By solving the NCERT solution for Class 9 Maths chapter 4 exercise 4.2 exercises, we can get good scores in Mathematics as they understand the subject better and help to understand the concepts more clearly.

Key Features of 9th Class Maths Exercise 4.2 Answers

Expertly Crafted Solutions: The solution for class 9 maths chapter 4 exercise 4.2 have been prepared by subject matter experts. They ensure accuracy and clarity in explanations.
Comprehensive Coverage: Exercise 4.2 class 9 maths covers various topics related to quadratic equations and their solutions. It includes problems involving quadratic equations in one variable.
Step-by-Step Format: Class 9 maths ex 4.2 Solutions are presented in a step-by-step format, making it easy for students to follow the solution process and understand the concepts.
Conceptual Clarity: The primary objective of ex 4.2 class 9 is to help students develop a clear understanding of quadratic equations and their solutions.
Variety of Problems: This Exercise 4.2 class 9 maths offers a range of problems with different levels of complexity, allowing students to practice and enhance their problem-solving skills.
PDF Availability: Solutions are often available in PDF format, allowing students to download and access them offline for convenient studying.
Free of Charge: These resources are typically provided free of charge, making them accessible to all students.

Also see-

NCERT Solutions of Class 10 Subject Wise

Subject Wise NCERT Exemplar Solutions

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 - Subject-wise PDFs Updated for 2024-25

NCERT Solutions for Class 9 Science

NCERT Solutions for Class 9 Maths

NCERT Syllabus for Class 9 Maths 2023 - Download Syllabus Pdf

NCERT Solutions For Class 9 Science Chapter 1 Matter in Our Surroundings

NCERT Solutions for Exercise 2.5 Class 9 Maths Chapter 2 Polynomials

NCERT Class 9 Maths Notes - Download Chapter Wise PDFs

NCERT Class 9 Maths Chapter 8 Notes Quadrilaterals - Download PDF

NCERT Class 9 Maths Chapter 2 Notes Polynomials - Download PDF

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

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

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

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

NCERT Exemplar Class 9 Solutions - Subject Wise Problems with Solutions

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 13 Why do we fall ill?

NCERT Exemplar Class 9 Science Solutions Chapter 12 Sound

Articles

Simultala Awasiya Vidyalaya Answer Key 2025, SAV Class 6 Answer Sheet Prelims Exam

Nov 01, 2024

MPSOS 10th Result 2024 - Check MPSOS Class 10 December 2024 Exam Result @mpsos.nic.in

Oct 29, 2024

Education Loan for School Students in India – Eligibility, Benefits & Application Process

Oct 29, 2024

How to Track NSP Payment Status – Step-by-Step Guide for Scholarship Updates

Oct 29, 2024

TOSS SSC Time Table 2025, Telangana Open School Class 10 Exam Dates @telanganaopenschool.org

Oct 29, 2024

IB Primary Years Programme (PYP) – Curriculum Overview and Key Benefits

Oct 28, 2024

Importance of Internships in Student Life - 10 Points, 100 Words, 200 Words, 500 Words

Oct 28, 2024

Cambridge Advanced Curriculum: AS and A levels (Grade 11 & 12)

Oct 28, 2024

Upcoming School Exams

Himachal Pradesh Board of Secondary Education 10th Examination

Application Date:09 September,2024 - 14 November,2024

Exam Pattern

Admit Card

Result

Himachal Pradesh Board of Secondary Education 12th Examination

Application Date:09 September,2024 - 14 November,2024

Result

Exam Pattern

National Institute of Open Schooling 12th Examination

Admit Card Date:04 October,2024 - 29 November,2024

Application Process

Exam Pattern

Admit Card

National Institute of Open Schooling 10th examination

Admit Card Date:04 October,2024 - 29 November,2024

Application Process

Exam Pattern

Mock Test

National Means Cum-Merit Scholarship

Application Date:05 October,2024 - 04 November,2024

Eligibility Criteria

Application Process

Exam Pattern

View All School Exams

Get answers from students and experts

Popular Questions

A block of mass 0.50 kg is moving with a speed of 2.00 ms^-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times. Assume that the potential energy lost each time he lowers the mass is dissipated. How much fat will he use up considering the work done only when the weight is lifted up ? Fat supplies 3.8×10⁷J of energy per kg which is converted to mechanical energy with a 20% efficiency rate. Take g = 9.8 ms⁻² :

Option 1)

2.45×10⁻³ kg

Option 2)

6.45×10⁻³ kg

Option 3)

9.89×10⁻³ kg

Option 4)

12.89×10⁻³ kg

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

A particle is projected at 60⁰ to the horizontal with a kinetic energy $K$ . The kinetic energy at the highest point

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

In the reaction,

$2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, $Mg_{3}(PO_{4})_{2}$ will contain 0.25 mole of oxygen atoms?

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1) decrease twice	Option 2) increase two fold
Option 3) remain unchanged	Option 4) be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol^-1) is

Option 1) twice that in 60 g carbon	Option 2) 6.023 × 10²²
Option 3) half that in 8 g He	Option 4) 558.5 × 6.023 × 10²³

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t²) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m² , the number of rotations made by the pulley before its direction of motion if reversed, is