case study questions on linear equation class 9

Class 9 Maths Case Study Questions Chapter 4 Linear Equations in two variables

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Case study Questions in Class 9 Mathematics Chapter 4  are very important to solve for your exam. Class 9 Maths Chapter 4 Case Study Questions have been prepared for the latest exam pattern. You can check your knowledge by solving  Class 9 Maths Case Study Questions Chapter 4 Linear Equations in two variables

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In CBSE Class 9 Maths Paper, Students will have to answer some questions based on Assertion and Reason. There will be a few questions based on case studies and passage-based as well. In that, a paragraph will be given, and then the MCQ questions based on it will be asked.

Linear Equations in two variables Case Study Questions With Answers

Here, we have provided case-based/passage-based questions for Class 9 Maths Chapter 4 Linear Equations in two variables

Case Study/Passage-Based Questions

Case Study 1: Deepak bought 3 notebooks and 2 pens for Rs. 80. His friend Ram said that the price of each notebook could be Rs. 25. Then three notebooks would cost Rs.75, the two pens would cost Rs.5 and each pen could be Rs. 2.50. Another friend Ajay felt that Rs. 2.50 for one pen was too little. It should be at least Rs. 16. Then the price of each notebook would also be Rs.16

Lohith also bought the same types of notebooks and pens as Aditya. He paid 110 for 4 notebooks and 3 pens. Later, Deepak guesses the cost of one pen is Rs. 10 and Lohith guess the cost of one notebook is Rs. 30.

(i) Form the pair of linear equations in two variables from this situation by taking the cost of one notebook as Rs. x and the cost of one pen as Rs. y. (a) 3x + 2y = 80 and 4x + 3y = 110 (b) 2x + 3y = 80 and 3x + 4y = 110 (c) x + y = 80 and x + y = 110 (d) 3x + 2y = 110 and 4x + 3y = 80

Answer: (a) 3x + 2y = 80 and 4x + 3y = 110

(ii) Which is the solution satisfying both the equations formed in (i)? (a) x = 10, y = 20 (b) x = 20, y = 10 (c) x = 15, y = 15 (d) none of these

Answer: (b) x = 20, y = 10

(iii) Find the cost of one pen? (a) Rs. 20 (b) Rs. 10 (c) Rs. 5 (d) Rs. 15

Answer: (b) Rs. 10

(iv) Find the total cost if they will purchase the same type of 15 notebooks and 12 pens. (a) Rs. 400 (b) Rs. 350 (c) Rs. 450 (d) Rs. 420

Answer: (d) Rs. 420

(v) Find whose estimation is correct in the given statement. (a) Deepak (b) Lohith (c) Ram (d) Ajay

Answer: (a) Deepak

Case Study 2: In the below given layout, the design and measurements have been made such that area of two bedrooms and Kitchen together is 95 sq. m.

(i) The area of two bedrooms and kitchen are respectively equal to (a) 5x, 5y (b) 10x, 5y (c) 5x, 10y (c) x, y

(ii) Find the length of the outer boundary of the layout. (a) 27 m (b) 15 m (c) 50 m (d) 54 m

(iii) The pair of linear equations in two variables formed from the statements are (a) x + y = 13, x + y = 9 (b) 2x + y = 13, x + y = 9 (c) x + y = 13, 2x + y = 9 (d) None of the above

(iv) Which is the solution satisfying both the equations formed in (iii)? (a) x = 7, y = 6 (b) x = 8, y = 5 (c) x = 6, y = 7 (d) x = 5, y = 8

(v) Find the area of each bedroom. (a) 30 sq. m (b) 35 sq. m (c) 65 sq. m (d) 42 sq. m

Hope the information shed above regarding Case Study and Passage Based Questions for Class 9 Mathematics Chapter 4 Linear Equations in two variables with Answers Pdf free download has been useful to an extent. If you have any other queries about CBSE Class 9 Maths Linear Equations in two variables Case Study and Passage Based Questions with Answers, feel free to comment below so that we can revert back to us at the earliest possible By Team Study Rate

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Class 9th Maths - Linear Equations in Two Variables Case Study Questions and Answers 2022 - 2023

QB365 provides a detailed and simple solution for every Possible Case Study Questions in Class 9th Maths Subject - Linear Equations in Two Variables, CBSE. It will help Students to get more practice questions, Students can Practice these question papers in addition to score best marks.

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Linear equations in two variables case study questions with answer key.

9th Standard CBSE

Final Semester - June 2015

Mathematics

(ii) Find the length of the outer boundary of the layout.

(iii) The pair of linear equation in two variables formed from the statements are (a) x + y = 13, x + y = 9 (b) 2x + y = 13, x + y = 9 (c) x + y = 13, 2x + y = 9 (d) None of the above (iv) Which is the solution satisfying both the equations formed in (iii)?

(v) Find the area of each bedroom.

(iii) Find the cost of one pen?

(iv) Find the total cost if they will purchase the same type of 15 notebooks and 12 pens.

(v) Find whose estimation is correct in the given statement.

(b) How to represent the above situation in linear equations in two variables ?

(c) If Sita contributed Rs. 76, then how much was contributed by Gita ?

(d) If both contributed equally, then how much is contributed by each?

(e) Which is the standard form of linear equations x = – 5 ?

(ii) Which is the solution of the equations formed in (i)?

(c) If the cost of one notebook is Rs. 15 and cost of one pen is 10, then find the total amount.

(d) If the cost of one notebook is twice the cost of one pen, then find the cost of one pen?

(e) Which is the standard form of linear equations y = 4 ?

(b) If the number of children is 15, then find the number of adults?

(c)  If the number of adults is 12, then find the number of children?

(d) Find the value of b, if x = 5, y = 0 is a solution of the equation 3x + 5y = b.

(e) Which is the standard form of linear equations in two variables: y - x = 5?

(b) If the cost of chocolates A is 5, then find the cost of chocolates B?

(c) Which of the follwing point lies on the line x + y = 7?

(d) The point where the line x + y = 7 intersect y-axis is 

(e) For what value of k, x = 2 and y = -1 is a soluation of x + 3y -k = 0.

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CBSE Case Study Questions for Class 9 Maths Linear Equations Free PDF

Mere Bacchon, you must practice the CBSE Case Study Questions Class 9 Maths Linear Equations  in order to fully complete your preparation . They are very very important from exam point of view. These tricky Case Study Based Questions can act as a villain in your heroic exams!

I have made sure the questions (along with the solutions) prepare you fully for the upcoming exams. To download the latest CBSE Case Study Questions , just click ‘ Download PDF ’.

CBSE Case Study Questions for Class 9 Maths Linear Equations in Two Variables PDF

Checkout our case study questions for other chapters.

• Chapter 2 Polynomials Case Study Questions
• Chapter 3 Coordinate Geometry Case Study Questions
• Chapter 5 Euclids Geometry Case Study Questions
• Chapter 6 Lines and Angles Case Study Questions

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Solve every question of NCERT by hand, without looking at the solution.

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CBSE Class 9 Mathematics Case Study Questions

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Significance of Mathematics in Class 9

Mathematics is an important subject for students of all ages. It helps students to develop problem-solving and critical-thinking skills, and to think logically and creatively. In addition, mathematics is essential for understanding and using many other subjects, such as science, engineering, and finance.

CBSE Class 9 is an important year for students, as it is the foundation year for the Class 10 board exams. In Class 9, students learn many important concepts in mathematics that will help them to succeed in their board exams and in their future studies. Therefore, it is essential for students to understand and master the concepts taught in Class 9 Mathematics .

Case studies in Class 9 Mathematics

A case study in mathematics is a detailed analysis of a particular mathematical problem or situation. Case studies are often used to examine the relationship between theory and practice, and to explore the connections between different areas of mathematics. Often, a case study will focus on a single problem or situation and will use a variety of methods to examine it. These methods may include algebraic, geometric, and/or statistical analysis.

Example of Case study questions in Class 9 Mathematics

The Central Board of Secondary Education (CBSE) has included case study questions in the Class 9 Mathematics paper. This means that Class 9 Mathematics students will have to solve questions based on real-life scenarios. This is a departure from the usual theoretical questions that are asked in Class 9 Mathematics exams.

The following are some examples of case study questions from Class 9 Mathematics:

Class 9 Mathematics Case study question 1

There is a square park ABCD in the middle of Saket colony in Delhi. Four children Deepak, Ashok, Arjun and Deepa went to play with their balls. The colour of the ball of Ashok, Deepak,  Arjun and Deepa are red, blue, yellow and green respectively. All four children roll their ball from centre point O in the direction of   XOY, X’OY, X’OY’ and XOY’ . Their balls stopped as shown in the above image.

Answer the following questions:

Answer Key:

Class 9 Mathematics Case study question 2

• Now he told Raju to draw another line CD as in the figure
• The teacher told Ajay to mark  ∠ AOD  as 2z
• Suraj was told to mark  ∠ AOC as 4y
• Clive Made and angle  ∠ COE = 60°
• Peter marked  ∠ BOE and  ∠ BOD as y and x respectively

Now answer the following questions:

• 2y + z = 90°
• 2y + z = 180°
• 4y + 2z = 120°
• (a) 2y + z = 90°

Class 9 Mathematics Case study question 3

• (a) 31.6 m²
• (c) 513.3 m³
• (b) 422.4 m²

Class 9 Mathematics Case study question 4

How to Answer Class 9 Mathematics Case study questions

To crack case study questions, Class 9 Mathematics students need to apply their mathematical knowledge to real-life situations. They should first read the question carefully and identify the key information. They should then identify the relevant mathematical concepts that can be applied to solve the question. Once they have done this, they can start solving the Class 9 Mathematics case study question.

Students need to be careful while solving the Class 9 Mathematics case study questions. They should not make any assumptions and should always check their answers. If they are stuck on a question, they should take a break and come back to it later. With some practice, the Class 9 Mathematics students will be able to crack case study questions with ease.

Class 9 Mathematics Curriculum at Glance

At the secondary level, the curriculum focuses on improving students’ ability to use Mathematics to solve real-world problems and to study the subject as a separate discipline. Students are expected to learn how to solve issues using algebraic approaches and how to apply their understanding of simple trigonometry to height and distance problems. Experimenting with numbers and geometric forms, making hypotheses, and validating them with more observations are all part of Math learning at this level.

The suggested curriculum covers number systems, algebra, geometry, trigonometry, mensuration, statistics, graphing, and coordinate geometry, among other topics. Math should be taught through activities that include the use of concrete materials, models, patterns, charts, photographs, posters, and other visual aids.

CBSE Class 9 Mathematics (Code No. 041)

Class 9 Mathematics question paper design

The CBSE Class 9 mathematics question paper design is intended to measure students’ grasp of the subject’s fundamental ideas. The paper will put their problem-solving and analytical skills to the test. Class 9 mathematics students are advised to go through the question paper pattern thoroughly before they start preparing for their examinations. This will help them understand the paper better and enable them to score maximum marks. Refer to the given Class 9 Mathematics question paper design.

QUESTION PAPER DESIGN (CLASS 9 MATHEMATICS)

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Class 9 is an important milestone in a student’s life. It is the last year of high school and the last chance to score well in the CBSE board exams. myCBSEguide is the perfect platform for students to get started on their preparations for Class 9 Mathematics. myCBSEguide provides comprehensive study material for all subjects, including practice questions, sample papers, case study questions and mock tests. It also offers tips and tricks on how to score well in exams. myCBSEguide is the perfect door to enter for class 9 CBSE preparations.

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14 thoughts on “CBSE Class 9 Mathematics Case Study Questions”

This method is not easy for me

aarti and rashika are two classmates. due to exams approaching in some days both decided to study together. during revision hour both find difficulties and they solved each other’s problems. aarti explains simplification of 2+ ?2 by rationalising the denominator and rashika explains 4+ ?2 simplification of (v10-?5)(v10+ ?5) by using the identity (a – b)(a+b). based on above information, answer the following questions: 1) what is the rationalising factor of the denominator of 2+ ?2 a) 2-?2 b) 2?2 c) 2+ ?2 by rationalising the denominator of aarti got the answer d) a) 4+3?2 b) 3+?2 c) 3-?2 4+ ?2 2+ ?2 d) 2-?3 the identity applied to solve (?10-?5) (v10+ ?5) is a) (a+b)(a – b) = (a – b)² c) (a – b)(a+b) = a² – b² d) (a-b)(a+b)=2(a² + b²) ii) b) (a+b)(a – b) = (a + b

MATHS PAAGAL HAI

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NCERT Solutions Class 9 Maths Chapter 4 Linear Equations In Two Variables

The linear equations in two variables illustrate the relationship between two variables. These equations explain the connection between two variables and their graphical representation. The linear equations in two variables have two solutions, and their solutions form a line. NCERT solutions for class 9 maths Chapter 4 linear equations in two variables comprise exercises based on writing linear equations, finding solutions of linear equations, graphing linear equations, and graphical representation of equations of lines parallel to x and y axes.

The linear equations in two variables are significantly important to implement various algebraic concepts studied in higher grades. It is also important to understand many real-life calculations like calculating the profits, estimating values, etc. Therefore students must build a strong foundational knowledge of this topic. NCERT solutions class 9 maths chapter 4 linear equations in two variables, explains all the core concepts efficiently so that the students can imbibe the knowledge easily. To study with the class 9 maths NCERT solutions chapter 4 linear equations in two variables, check the pdf files in the links below and also find some of these in the exercises given below.

• NCERT Solutions Class 9 Maths Chapter 4 Ex 4.1
• NCERT Solutions Class 9 Maths Chapter 4 Ex 4.2
• NCERT Solutions Class 9 Maths Chapter 4 Ex 4.3
• NCERT Solutions Class 9 Maths Chapter 4 Ex 4.4

NCERT Solutions for Class 9 Maths Chapter 4 PDF

NCERT solutions maths for class 9 offer the optimal coverage to all the topics as well as subtopics covered in this chapter. To exercise wise prepare this topic, you can download the pdf files as given below.

☛ Download Class 9 Maths NCERT Solutions Chapter 4

NCERT Class 9 Maths Chapter 4   Download PDF

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

Solving and graphing equations is a foundational skill required for the math foundation of every student. Studying these resources will enable the student to interpret any complex algebraic equation easily. The practice of questions provided in the above exercises will give proper guidance to students about the exam patterns and important questions. To learn and practice with NCERT Solutions Class 9 Maths Chapter 4 linear equations in two variables, try the detailed exercise-wise questions provided in the links given below.

• Class 9 Maths Chapter 4 Ex 4.1 - 2 Questions
• Class 9 Maths Chapter 4 Ex 4.2 - 4 Questions
• Class 9 Maths Chapter 4 Ex 4.3 - 8 Questions
• Class 9 Maths Chapter 4 Ex 4.4 - 2 Questions

☛ Download Class 9 Maths Chapter 4 NCERT Book

Topics Covered: The important topics covered in class 9 maths NCERT solutions Chapter 4 are a basic introduction to linear equations in two variables, solution of linear equation in two variables, graphing linear equations in two variables, and equations of lines parallel to x-axes and y-axes.

Total Questions: Class 9 Maths Chapter 4 linear equations in two variables consists of a total of 16 questions, which are mostly categorized as easy and moderate with sub-questions that facilitate the sufficient practice of important concepts.

List of Formulas in NCERT Solutions Class 9 Maths Chapter 4

NCERT solutions class 9 maths Chapter 4 are competent resources covering all the fundamental concepts based on linear equations and graphing. These solutions are helpful to form the basic math foundation in students to understand complex geometry concepts in higher grades. The most important concepts covered in these NCERT solutions for class 9 maths chapter 4 are based on solving linear equations. While solving a linear equation, we must always remember the following two points:

• If the same number is added or subtracted from both sides of the equation, the solution of a linear equation remains the same.
• If both sides of the equation are multiplied or divided by the same number( non-zero number), the solutions remain the same.

Important Questions for Class 9 Maths NCERT Solutions Chapter 4

Video solutions for class 9 maths ncert chapter 4, faqs on ncert solutions class 9 maths chapter 4, what is the importance of ncert solutions for class 9 maths chapter 4 linear equations in two variables.

NCERT Solutions Class 9 Maths are well-researched resources created by experts to form a solid mathematical foundation in students. These resources are based on NCERT textbooks that the CBSE has recommended to study and prepare for exams. These solutions are devised to impart complex knowledge in an easy manner. This makes them a highly reliant exam guide studded with multiple examples and sample problems framed to yield positive results.

Do I Need to Practice all Questions Provided in Class 9 Maths NCERT Solutions linear equations in two variables?

By practicing all the questions in the NCERT Solutions Class 9 Maths Chapter 4 Linear Equations in Two Variables, students will be able to ensure the complete preparation required for gaining an in-depth understanding of this topic. When students have complete knowledge of this topic, they will be able to understand more advanced concepts easily. Solving different types of questions will also enable them to prepare well for facing any competitive exams with confidence and a positive mindset.

What are the Important Topics Covered in Class 9 Maths NCERT Solutions Chapter 4?

NCERT Solutions Class 9 Maths Chapter 4 covers a basic introduction to linear equations in two variables, solution of linear equations, the graph of linear equations, equations of lines parallel to x-axes and y-axes. NCERT solutions class 9 Maths Chapter 4 Linear Equations in Two Variables efficiently cover the complete concepts, formulas , and exercises with suitable examples and sample problems.

How Many Questions are there in NCERT Solutions for Class 9 Maths linear equations in two variables?

NCERT Class 9 Maths Chapter 4 Linear Equations in Two Variables has a total of 16 questions in four exercises. All these questions are gradually paced to provide in-depth learning of linear equations, their solutions, and graphing. These 16 questions can be sub-categorized as long answers, short level, or easy ones to plan a time-based practice and preparation.

What are the Important Formulas in NCERT Solutions Class 9 Maths Chapter 4?

The important concepts covered in the NCERT Solutions Class 9 Maths Chapter 4 are based on finding solutions of linear equations to graph lines. These concepts are also important for students to acquire a clear understanding of applying linear equations in two variables.

Why Should I Practice NCERT Solutions Class 9 Maths Linear Equations in Two Variables Chapter 4?

Questions asked in the CBSE maths exams are based on NCERT textbook. Practicing with NCERT Solutions Class 9 Maths Chapter 4 Linear Equations in Two Variables ensures that you have covered each chapter’s topic in detail. It also helps to develop the problem-solving skills in students that are most important for maths studies.

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

• NCERT Solutions
• Chapter 4 Linear Equations In Two Variables

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

Class 9 Mathematics Chapter 4 Linear Equations in Two Variables is an important subject to master in this curriculum. This chapter is intended to help students understand the ideas involved in producing graphs of linear equations on a Cartesian plane. Follow the expert-designed Class 9 Mathematics Chapter 4 Linear Equations in Two Variables NCERT Answers to grasp the context and how to solve the questions in this chapter. Discover the correct answers to all of the questions included to the chapter's exercise with ease.

The NCERT Solutions for Class 9 Maths Chapter 4 - Linear Equations in Two Variables provides you the richest range of questions along with the properly graded solutions for you to grasp the fundamentals and also acquire the problem solving and learning skills. To the point and straightforward approach is applied to make Linear Equations Class 9 easy and interesting. The experts of Vedantu have curated the solutions as per latest NCERT (CBSE) Book guidelines. NCERT Solutions for Class 9 Maths Chapter 4- Linear Equations in Two Variables always prove to be beneficial for your exam preparation and revision. 

Students can also download NCERT Solutions for Class 9 Science created by the best Teachers at Vedantu for Free. Every NCERT Solution is provided to make the study simple and interesting on Vedantu.

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

There are 4 exercises under NCERT Solutions for Class 9 Maths Chapter 4, which is Linear Equations in Two Variables. Students will get great practice by going through these solutions and will be able to master the concepts of Linear Equations in two Variables. Following are the details of question types and varieties that are included in each of the exercises:

Exercise 4.1: Exercise 4.1 majorly includes questions related to determining the values of different variables and constructing linear equations in two variables to represent given statements.

Exercise 4.2: Exercise 4.2 has problems comprising completing given statements and stating the reasons for choosing an answer, determining solutions for given equations, finding the actual solutions for given linear equations from provided solutions, and determining the values of constants.

Exercise 4.3: Exercise 4.3 mostly has questions based on graphs. For example, plotting graphs, answering questions based on given graphs, providing solutions for given graphs, formulating relations and plotting relevant graphs for the same, choosing correct equations for provided graphs, deriving linear equations to satisfy given data and plotting the graph, are some of the questions that are included in this exercise.

Exercise 4.4: Exercise 4.4 includes questions that are related to geometric representations, like giving or describing geometric representations as equations.

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

Exercise (4.1)

1: Construct a linear equation in two variables to express the following statement.

The cost of a textbook is twice the cost of an exercise book.

Ans. Let the cost of a textbook be $\text{x}$ rupees and the cost of an exercise book be $\text{y}$ rupees.

The given statement:  The cost of a textbook is twice the cost of an exercise book

So, in order to form a linear equation, 

the cost of the textbook $\text{=}\,\text{2 }\!\!\times\!\!\text{ }$ the cost of an exercise book.  

$\Rightarrow \text{x=2y}$

$\Rightarrow \text{x-2y=0}$.

2: Determine the values of $\text{a}$, $\text{b}$, $\text{c}$ from the following linear equations by expressing each of them in the standard form \[\text{ax+by+c=0}\].

(i) $\text{2x+3y=9}\text{.}\overline{\text{35}}$ 

Ans. The given linear equation is

$\text{2x+3y=9}\text{.}\overline{\text{35}}$

Subtracting $9.\overline{35}$ from both sides of the equation gives

$\text{2x+3y}-\text{9}\text{.}\overline{\text{35}}\text{=0}$ 

Now, by comparing the above equation with the standard form of the linear equation, $\text{ax+by+c=0}$, the values of $\text{a,}\,\text{b,}$ and $\text{c}$ are obtained as  

$\text{a=2}$, 

$\text{b=3}$, and

$\text{c}=-\text{9}\text{.}\overline{\text{35}}$

(ii) $\text{x-}\frac{\text{y}}{\text{5}}\text{-10=0}$

$\text{x-}\frac{\text{y}}{\text{5}}\text{-10}=\text{0}$ 

Now, by comparing the above equation with the standard form of the linear equation, $\text{ax+by+c=0}$, the values of $\text{a,}\,\text{b,}$ and $\text{c}$ are obtained as

$\text{a}=\text{1}$,

$\text{b}=-\frac{\text{1}}{\text{5}}$, and

$\text{c}=-\text{10}$.

(iii) $\text{-2x+3y=6}$

$\text{-2x+3y=6}$

Subtracting $6$ from both sides of the equation gives 

$-\text{2x+3y}-\text{6}=\text{0}$

Now, by comparing the above equation with the standard form of the linear equation, $\text{ax+by+c=0}$, the values of $\text{a,}\,\text{b,}$ and $\text{c}$ are obtained as 

$\text{a}=-\text{2}$,

$\text{b}=\text{3}$, and

$\text{c}=-\text{6}$.

(iv) $\text{x=3y}$

Ans. The given linear equation can be written as

$\text{1x}=\text{3y}$

Subtracting $3y$ from both sides of the equation gives 

$\text{1x-3y+0=0}$

Now, by comparing the above equation with the standard form of the linear equation $\text{ax+by+c=0}$, the values of $\text{a,}\,\text{b,}$ and $\text{c}$ are obtained as 

$\text{b}=-\text{3}$, and

$\text{c}=\text{0}$.

(v) \[\text{2x}\mathbf{=-}\,\text{5y}\]

\[\text{2x}=-\text{5y}\].

Adding $5y$ on both sides of the equation gives

\[\text{2x+5y+0=0}\].

$\text{a}=\text{2}$,

$\text{b}=\text{5}$, and

(vi) $\text{3x+2=0}$

$\text{3x+2=0}$. 

Rewriting the equation gives

$\text{3x+0y+2=0}$

Now, by comparing the above equation with the standard form of linear equation $\text{ax+by+c=0}$, the values of $\text{a,}\,\text{b,}$ and $\text{c}$ are obtained as

$\text{a}=\text{3}$,

$\text{b}=\text{0}$, and

$\text{c}=\text{2}$.

(vii) $\text{y-2=0}$

$\text{y-2=0}$ 

The equation can be expressed as

$\text{0x+1y-2}=\text{0}$

$\text{a}=\text{0}$,

$\text{b}=\text{1}$, and

$\text{c}=-\text{2}$.

(viii) $\text{5=2x}$

Ans: The given linear equation is

$\text{5=2x}$.

The equation can be written as 

$\text{-2x+0y+5=0}$.

Now, by comparing the above equation with the standard form of the linear equation $\text{ax+by+c=0}$, the values of $\text{a,}\,\text{b,}$ and $\text{c}$ are obtained as

$\text{c}=\text{5}$.

Exercise (4.2)  

1: Complete the following statement by choosing the appropriate answer and explain why it should be chosen?

$\text{y=3x+5}$ has ___________. 

(a) A unique solution, 

(b) Only two solutions, 

(c) Infinitely many solutions. 

Ans: Observe that, $\text{y}=\text{3x+5}$ is a linear equation. 

Now, note that, for $\text{x}=\text{0}$, $\text{y}=\text{0+5=5}$.

So, $\left( \text{0,5} \right)$ is a solution of the given equation.

If $\text{x=1}$, then $\text{y}=\text{3 }\!\!\times\!\!\text{ 1+5}=\text{8}$.

That is, $\left( \text{1,8} \right)$ is another solution of the equation. 

Again, when $\text{y}=\text{0}$, $\text{x}=-\frac{5}{3}$ .

Therefore, $\left( -\frac{5}{3},0 \right)$ is another solution of the equation.

Thus, it is noticed that for different values of $\text{x}$ and $\text{y}$, different solutions are obtained for the given equation.

So, there are countless different solutions exist for the given linear equation in two variables. Therefore, a linear equation in two variables has infinitely many solutions. 

Hence, option (c) is the correct answer.

2: Determine any four solutions for each of equations given below. 

(i) $\mathbf{2x}+\mathbf{y}=\mathbf{7}$. 

Ans: The given equation 

$\text{2x+y}=\text{7}$ is a linear equation.

Solving the equation for $y$ gives

$\text{y=7-2x}$. 

Now substitute $\text{x=0,1,2,3}$ in succession into the above equation.

For  $\text{x=0}$, 

$\text{2}\left( \text{0} \right)\text{+y=7}$

$\Rightarrow \text{y=7}$

So, one of the solutions obtained is $\left( \text{x,y} \right)\text{=}\left( \text{0,7} \right)$. 

For  $\text{x=1}$, 

$\text{2}\left( \text{1} \right)\text{+y=7}$

$\Rightarrow \text{y=5}$

Therefore, another solution obtained is $\left( \text{x,y} \right)\text{=}\left( \text{1,5} \right)$.

For $\text{x=2}$, 

$\text{2}\left( \text{2} \right)\text{+y=7}$

$\Rightarrow \text{y=3}$

That is, a solution obtained is $\left( \text{x,y} \right)\text{=}\left( \text{3,1} \right)$.

Also, for $\text{x=3}$, 

\[\text{2}\left( \text{3} \right)\text{+y=7}\]

$\Rightarrow \text{y=1}$ 

So, another one solution is $\left( \text{x,y} \right)\text{=}\left( \text{3,1} \right)$.

Thus, four solutions obtained for the given equations are $\left( \text{0,7} \right)$ , $\left( 1,5 \right)$, $\left( 2,3 \right)$, $\left( 3,1 \right)$.

(ii) $\mathbf{\pi x}+\mathbf{y}=\mathbf{9}$.

Ans. The given equation 

$\pi x+y=9$                                                                          …… (a)

is a linear equation in two variables.

By transposing, the above equation (a) can be written as

$\text{y=9- }\!\!\pi\!\!\text{ x}$. 

For $\text{x=0}$, 

$\text{y=9- }\!\!\pi\!\!\text{ }\left( \text{0} \right)$

$\Rightarrow \text{y=9}$

Therefore, one of the solutions obtained is $\left( \text{x,y} \right)\text{=}\left( \text{0,9} \right)$.

For $\text{x=1}$,

$\text{y}=\text{9}-\text{ }\!\!\pi\!\!\text{ }\left( \text{1} \right)$

$\Rightarrow \text{y}=\text{9}-\text{ }\!\!\pi\!\!\text{ }$.

So, another solution obtained is $\left( \text{x,y} \right)\text{=}\left( \text{1,9- }\!\!\pi\!\!\text{ } \right)$.

For  $\text{x=2}$, 

$\text{y}=\text{9}-\text{ }\!\!\pi\!\!\text{ }\left( \text{2} \right)$

$\Rightarrow \text{y}=\text{9}-\text{2 }\!\!\pi\!\!\text{ }$

That is, another solution obtained is $\left( \text{x,y} \right)\text{=}\left( \text{2,}\,\text{9-2 }\!\!\pi\!\!\text{ } \right)$.

$\text{y}=\text{9}-\text{ }\!\!\pi\!\!\text{ }\left( \text{3} \right)$

$\Rightarrow \text{y}=9-\text{3 }\!\!\pi\!\!\text{ }$.

Therefore, another one solution is $\left( \text{x,y} \right)\text{=}\left( \text{3,}\,\text{9-3 }\!\!\pi\!\!\text{ } \right)$.

Thus, four solutions obtained for the given equations are $\left( 0,9 \right)$, $\left( \text{1,9,- }\!\!\pi\!\!\text{ } \right)$, $\left( \text{2,9-2 }\!\!\pi\!\!\text{ } \right)$, $\left( \text{3,9-3 }\!\!\pi\!\!\text{ } \right)$.

(iii) $\mathbf{x}=\mathbf{4y}$.

$\text{x=4y}$ is a linear equation in two variables.

By transposing, the above equation can be written as

$\text{y=}\frac{\text{x}}{4}$ .

$\text{y}=\frac{0}{4}=0$.

Therefore, one of the solutions is $\left( \text{x,y} \right)\text{=}\left( \text{0,0} \right)$.

$\text{y}=\frac{1}{4}$.

So, another solution of the given equation is $\left( \text{x,y} \right)\text{=}\left( \text{1,}\frac{1}{4} \right)$.

For $\text{x=2}$,

$\text{y}=\frac{2}{4}=\frac{1}{2}$.

That is, another solution obtained is $\left( \text{x,y} \right)\text{=}\left( \text{2,}\frac{1}{2} \right)$.

Also, for $\text{x=3}$ . ,

$\text{y}=\frac{3}{4}$.

Therefore, another one solution is $\left( \text{x,y} \right)=\left( 3,\frac{3}{4} \right)$.

Thus, four solutions obtained for the given equations are $\left( 0,0 \right)$, $\left( \text{1,}\frac{1}{4} \right)$, $\left( \text{2,}\frac{1}{2} \right)$, $\left( 3,\frac{3}{4} \right)$.

3: Identify the actual solutions of the linear equation\[\text{ }\!\!~\!\!\text{ x-2y=4}\] from each of the following solutions.

(i)  $\left( \mathbf{0},\mathbf{2} \right)$

Ans: Substituting $\text{x=0}$ and $\text{y=2}$ in the Left-hand-side of the equation \[\text{ }\!\!~\!\!\text{ x-2y=4}\] gives

$\begin{align} & \text{x-2y}=0-2\left( 2 \right) \\ & =-4 \\ & \ne 4. \end{align}$

Therefore, Left-hand-side is not equal Right-hand-side of the given equation for$\left( \text{x,y} \right)=\left( 0,2 \right)$.

Hence, $\left( 0,2 \right)$ is not a solution of the equation \[\text{x-2y=4}\]. 

(ii) $\left( \mathbf{2},\mathbf{0} \right)$

Ans. Substituting $\text{x=2}$ and $\text{y=0}$ in the Left-hand-side of the equation \[\text{ }\!\!~\!\!\text{ x-2y=4}\] gives

$\begin{align} & \text{x-2y}=2-2\left( 0 \right) \\ & =2 \\ & \ne 4. \end{align}$

Therefore, Left-hand-side is not equal Right-hand-side of the given equation for$\left( \text{x,y} \right)=\left( 2,0 \right)$.

Hence, $\left( 2,0 \right)$ is not a solution of the equation \[\text{x-2y=4}\]. 

(iii) $\left( \mathbf{4},\mathbf{0} \right)$ 

Ans. Substituting $\text{x=4}$ and $\text{y=0}$ in the Left-hand-side of the equation\[\text{ }\!\!~\!\!\text{ x-2y=4}\] gives

$\begin{align} & \text{x-2y}=4-2\left( 0 \right) \\ & =4. \end{align}$

Therefore, Left-hand-side is equal Right-hand-side of the given equation for$\left( \text{x,y} \right)=\left( 4,0 \right)$.

Hence, $\left( 4,0 \right)$ is a solution of the equation \[\text{x-2y=4}\]. 

(iv) $\left( \sqrt{\mathbf{2}}\mathbf{,4}\sqrt{\mathbf{2}} \right)$

Ans. Substituting $\text{x=}\sqrt{2}$ and $\text{y=4}\sqrt{2}$ in the Left-hand-side of the equation\[\text{ }\!\!~\!\!\text{ x-2y=4}\] gives

$\begin{align} & \text{x-2y}=\sqrt{2}-2\left( 4\sqrt{2} \right) \\ & =\sqrt{2}-8\sqrt{2} \\ & =-7\sqrt{2} \\ & \ne 4. \end{align}$

Therefore, Left-hand-side is not equal Right-hand-side of the given equation for$\left( \text{x,y} \right)=\left( \sqrt{2},4\sqrt{2} \right)$.

Hence, $\left( \sqrt{2},4\sqrt{2} \right)$ is not a solution of the equation \[\text{x-2y=4}\]. 

(v) $\left( \mathbf{1},\mathbf{1} \right)$

Ans. Substituting $\text{x}=1$ and $\text{y}=1$ in the Left-hand-side of the equation\[\text{ }\!\!~\!\!\text{ x-2y=4}\] gives

$\begin{align} & \text{x-2y}=1-2\left( 1 \right) \\ & =1-2 \\ & =-1 \\ & \ne 4. \end{align}$

Therefore, Left-hand-side is not equal Right-hand-side of the given equation for$\left( \text{x,y} \right)=\left( 1,1 \right)$.

Hence, $\left( 1,1 \right)$ is not a solution of the equation \[\text{x-2y=4}\]. 

4: If $\left( \mathbf{x},\mathbf{y} \right)=\left( \mathbf{2},\mathbf{1} \right)$ is a solution of the equation \[\text{2x+3y=k}\], then what is the value of $\mathbf{k}$?

Ans: By substituting $\text{x}=2$, $\text{y}=1$ and into the equation

\[\text{2x+3y=k}\] gives

$\text{2}\left( \text{2} \right)\text{+3}\left( \text{1} \right)\text{=k}$

$\Rightarrow \text{4+3=k}$

$\Rightarrow \text{k=7}$.

Hence, the value of $\text{k}$ is $7$.

Exercise (4.3)

1: Graph each of the linear equations given below.

(i) \[\text{ }\!\!~\!\!\text{ x+y=4}\]

\[\text{x+y=4}\]

\[\Rightarrow \text{y=4--x}\]                                                                                           …… (a)

Substitute $\text{x}=0$ into the equation (a) gives

$\text{y}=4-0=4.$

Similarly, substituting $\text{x}=2,4$ in succession into the equation (a), the following table of $\text{y}$ -values are obtained:

Now, Plot the points $\left( 0,4 \right)$, $\left( 2,2 \right)$ and $\left( 4,0 \right)$ on a graph paper and connect the points by a straight line. 

Thus, the following graph of the straight line represents the required graph of the linear equation $\text{x+y}=4$. 

(ii) \[\text{x--y=2}\]

\[\text{x-y}=2\]

\[\Rightarrow \text{y}=\text{x}-2\]                                                                                           …… (a)

$\text{y}=0-2=-2.$

Now, Plot the points $\left( 0,-2 \right)$, $\left( 2,0 \right)$ and $\left( 4,2 \right)$ on a graph paper and connect the points by a straight line. 

Thus, the following graph of the straight line represents the required graph of the linear equation $\text{x}-\text{y}=2$. 

(iii) \[\text{y=3x}\]

\[\text{y}=3\text{x}\]                                                                                                  …… (a)

$\text{y}=3\left( 0 \right)=0.$

Similarly, substituting $\text{x}=2,-2$ in succession into the equation (a), the following table of $\text{y}$ -values are obtained:

Now, Plot the points $\left( 0,0 \right)$, $\left( 2,6 \right)$ and $\left( -2,-6 \right)$ on a graph paper and connect the points by a straight line. 

Thus, the following graph of the straight line represents the required graph of the linear equation $\text{y}=3\text{x}$.

(iv) \[\text{3=2x+y}\]

Ans.   The given linear equation is

\[3=2\text{x+y}\]

$\Rightarrow \text{y}=3-2\text{x}$                                                                                      …… (a)

$\text{y}=3-2\left( 0 \right)=3$.

Similarly, substituting $\text{x}=1,\,3$ in succession into the equation (a), the following table of $\text{y}$ -values are obtained:

Now, Plot the points $\left( 0,3 \right)$, $\left( 1,1 \right)$ and $\left( 3,-3 \right)$ on a graph paper and connect the points by a straight line. 

Thus, the following graph of the straight line represents the required graph of the linear equation $3=\text{2x}+\text{y}$.

2: Provided that the equations of two lines passing through the point \[\left( \mathbf{2,14} \right)\]. Can there exist more than two equations of such type? If it is, then state the reason. 

Ans. Provided that equations of two lines passing through \[\left( \text{2,14} \right)\]. 

It can be noted that the point \[\left( \text{2,14} \right)\] satisfies the equation \[\text{7x-y=0}\] and\[\text{x-y+12=0}\]. 

So, the equations \[\text{7x-y=0}\] and \[\text{x-y+12=0}\] represent two lines passing through a point \[\left( \text{2,14} \right)\]. 

Now, since we know that through infinite number of lines can pass through any one point, so, there are an infinite number such types of lines exist that passes through the point $\left( 2,14 \right)$. 

Hence, there exist more than two equations whose graph passes through the point $\left( 2,14 \right)$.

3: Determine the value of $\mathbf{a}$ in the linear equation \[\text{3y=ax+7}\] if the point \[\left( \mathbf{3,4} \right)\] lies on the graph of the equation.

Ans. Given that \[\text{3y=ax+7}\] is a linear equation and the point \[\left( 3,4 \right)\] lies on the equation.

Substituting $\text{x=3}$, \[\text{y=4}\] in the equation gives 

\[\text{3y=ax+7}\]

$\Rightarrow \text{3}\left( 4 \right)\text{=a}\left( 3 \right)\text{+7}$

$\Rightarrow \text{3a}=5$

$\Rightarrow \text{a}=\frac{5}{3}$.

Hence, the value of $\text{a}$ is $\frac{5}{3}$.

4: Derive a linear equation for the following situation: 

For the first kilometre, a cab take rent $\mathbf{8}$ rupees and for the subsequent distances it becomes $\mathbf{5}$ rupees per kilometres. Assume the distance covered is $\mathbf{x}$ km and total rent is $\mathbf{y}$ rupees. Hence, draw the graph of the linear equation.

Ans. Let the total distance covered $=$ $\text{x}$ km 

and the total cost for the distance travelled $=$$\text{y}$ rupees.

It is given that the rent for the 1st kilometre is $8$ rupees and for the subsequent km, it is $5$ rupees per kilometres.

Therefore, rent for the rest of the distance $=$ \[\left( \text{x-1} \right)\text{5}\] rupees.

Total cost for travelling $\text{x}$ km is given by

\[\text{y=}\left[ \text{8+}\left( \text{x-1} \right)\text{5} \right]\]

 \[\Rightarrow \text{y=8+5x-5}\]

 \[\Rightarrow \text{y=5x+3}\]                                                                                      …… (1)

\[\Rightarrow \text{5x-y+3=0}\],

which is the required linear equation.

Now, substituting $\text{x}=0$ into the equation (1) gives

$\text{y}=5\left( 0 \right)+3=3$.

Similarly, substituting $\text{x}=1,\,-1$ in succession into the equation (1), the following table of $\text{y}$ -values are obtained:

Now, Plot the points $\left( 0,3 \right)$, $\left( 1,8 \right)$ and $\left( -1,-2 \right)$ on a graph paper and connect the points by a straight line. 

Thus, the following graph of the straight line represents the required graph of the linear equation \[\text{5x-y+3=0}\].

It is concluded by observing the graph of the linear equations that the variable $\text{x}$ and $\text{y}$ represent the distance travelled by the car and the total cost of rent for the distance respectively. Therefore, $\text{x}$ and $\text{y}$ are non-negative quantities.

Thus, only the first quadrant of the graph of the linear equation \[\text{5x-y+3=0}\] is only valid. 

5: Choose the correct linear equation for the given graphs in (a) and (b). 

(a) (i) $\mathbf{y}=\mathbf{x}$ 

     (ii) $\mathbf{x}+\mathbf{y}=\mathbf{0}$ 

     (iii) $\mathbf{y}=\mathbf{2x}$

     (iv) $\mathbf{2}+\mathbf{3y}=\mathbf{7x}$

Ans. It is observed in the given graph that the points $\left( -1,1 \right)$, $\left( 0,0 \right)$, and $\left( 1,-1 \right)$ lie on the straight line. Also, the coordinates of the points satisfy the equation \[\text{x+y=0}\].

So, \[\text{x+y=0}\] is the required linear equation corresponding to the given graph.

Hence, option (ii) is the correct answer. 

(b) (i) $\mathbf{y}=\mathbf{x}+\mathbf{2}$

     (ii) $\mathbf{y}=\mathbf{x}-\mathbf{2}$ 

     (iii) $\mathbf{y}=-\mathbf{x}+\mathbf{2}$

     (iv) $\mathbf{x}+\mathbf{2y}=\mathbf{6}$

Ans. It is observed in the given graph that the points $\left( -1,3 \right)$, $\left( 0,2 \right)$, and $\left( 2,0 \right)$ lie on the straight line. Also, the coordinates of the points satisfy the equation \[\text{y}=-\text{x+2}\].

So, \[\text{y}=-\text{x+2}\] is the required linear equation corresponding to the given graph.

Hence, option (iii) is the correct answer.

6: The work done by a body on the application of a constant force is proportional to the distance moved by the body. Formulate this relation by a linear equation and graph the same by using a constant force of five units. Hence from the graph, determine the work done when the distance moved by the body is 

(i) $\mathbf{2}$ units 

(ii) $\mathbf{0}$ unit. 

Ans: Let the distance moved by the body be $\text{x}$ units and the work done be $\text{y}$ units. 

Now, given that, work done is proportional to the distance. 

Therefore, \[\text{y}\propto \text{x}\].

\[\Rightarrow \text{y}=\text{kx}\],                                                                                        …… (a) 

where, $\text{k}$ is a constant.

By considering constant force of five units, the equation (a) becomes

$\text{y}=\text{5x}$.                                                                                             …… (b)

Now, substituting $\text{x}=0$ into the equation (b) gives

$\text{y}=5\left( 0 \right)=0$.

Similarly, substituting $\text{x}=1,-1$ in succession into the equation (b), gives the following table of $\text{y}$-values. 

Now, Plot the points $\left( 0,0 \right)$, $\left( 1,5 \right)$ and $\left( -1,-5 \right)$ on a graph paper and connect the points by a straight line. 

Thus, the following graph of the straight line represents the required graph of the linear equation $\text{y}=\text{5x}$.

It can be concluded by observing the graph of the linear equation that the value of $\text{y}$ corresponding to \[\text{x=2}\] is $10$. Thus, when the distance moved by the body is $2$ units, then the work done by it is $10$ units. 

Also, the value of $\text{y}$ corresponding to \[\text{x=0}\] is $0$. So, when the distance travelled by the body is $0$ unit, then the work done by it is $0$ unit.

7: Derive a linear equation that satisfies the following data and graph it.

Sujata and Suhana, two students of Class X of a school, together donated $\mathbf{100}$ rupees to the Prime Minister’s Relief Fund for supporting the flood victims.

Ans: Let Sujata and Suhana donated $\text{x}$ rupees and $\text{y}$ rupees respectively to the Prime Minister’s Relief fund.  

Given that, the amount donated by Sujata and Suhana together is $100$ rupees.

Therefore, $\text{x+y}=100$. 

$\Rightarrow \text{y}=\text{100}-\text{x}$.                                                            …… (a)

Now, substituting $\text{x}=0$ into the equation (a) gives

$\text{y}=100-0=100$.

Similarly, substituting $\text{x}=50,100$ in succession into the equation (a), gives the following table of $\text{y}$-values. 

Now, Plot the points $\left( 0,100 \right)$, $\left( 50,50 \right)$ and $\left( 100,0 \right)$ on a graph paper and connect the points by a straight line. 

Thus, the following graph of the straight line represents the required graph of the linear equation $\text{x+y}=100$.

It is concluded by observing the graph of the linear equation that the variable $\text{x}$ and $\text{y}$ are showing the amount donated by Sujata and Suhana respectively and so, $\text{x}$ and $\text{y}$ are nonnegative quantities. 

Hence, the values of $\text{x}$ and $\text{y}$ lying in the first quadrant will only be considered.

8: The following linear equation converts Fahrenheit to Celsius: 

$\mathbf{F=}\left( \frac{\mathbf{9}}{\mathbf{5}} \right)\mathbf{C+32}$,

where $\mathbf{F}$ denotes the measurement of temperature in Fahrenheit and $\mathbf{C}$ in Celsius unit.

Then do as directed in the following questions.

(i) Graph the linear equation given above by taking $\mathbf{x}$-axis as Celsius and $\mathbf{y}$-axis as Fahrenheit. 

$\text{F=}\left( \frac{\text{9}}{\text{5}} \right)\text{C+32}$                                                                                 …… (a) 

Now, substituting $C=0$ into the equation (a) gives

$\text{F}=\left( \frac{9}{5} \right)\left( 0 \right)+32=32$.

Similarly, substituting $C=-40,10$ in succession into the equation (a) gives the following table of $\text{F}$-values.

Now, Plot the points $\left( 0,32 \right)$, $\left( -40,-40 \right)$ and $\left( 10,50 \right)$ on a graph paper and connect the points by a straight line. 

Thus, the following graph of the straight line represents the required graph of the linear equation $\text{F=}\left( \frac{\text{9}}{\text{5}} \right)\text{C+32}$.

(ii) Determine the temperature in Fahrenheit if it is $\mathbf{3}{{\mathbf{0}}^{\mathbf{o}}}\mathbf{C}$ in Celsius.

Ans. Given that the temperature $={{30}^{\circ }}\text{C}$.

Now, it is also provided that, $\text{F=}\left( \frac{\text{9}}{\text{5}} \right)\text{C+32}$. 

Substitute $\text{C}=\text{32}$, in the above linear equation.

$\text{F=}\left( \frac{\text{9}}{\text{5}} \right)\text{30+32=54+32=86}$.

Hence, the temperature in Fahrenheit obtained is \[\text{86  }\!\!{}^\circ\!\!\text{ F}\]. 

(iii) Determine the temperature in Celsius if it is \[\mathbf{9}{{\mathbf{5}}^{\mathbf{o}}}\mathbf{F}\] in Fahrenheit. 

Ans. The given temperature $=\text{9}{{\text{5}}^{\circ }}\text{F}$.

\[\text{F=?}\]

It is provided that, $\text{F=}\left( \frac{\text{9}}{\text{5}} \right)\text{C+32}$  

Now, substitute $\text{F}=9\text{5}$, into the above linear equation.

Then it gives

$\text{95=}\left( \frac{\text{9}}{\text{5}} \right)\text{C+32}$

$\Rightarrow \text{63=}\left( \frac{\text{9}}{\text{5}} \right)\text{C}$

$\Rightarrow \text{C}=\text{35}$.

Hence, the temperature in Celsius obtained is \[\text{3}{{\text{5}}^{\circ }}\text{C}\]. 

(iv) Calculate the temperature in Fahrenheit when it is \[{{\mathbf{0}}^{\mathbf{o}}}\mathbf{C}\] in Celsius. Also, determine the temperature in Celsius when it is \[{{\mathbf{0}}^{\mathbf{o}}}\mathbf{F}\] in Fahrenheit.

Ans. It is known that, 

$\text{F=}\left( \frac{\text{9}}{\text{5}} \right)\text{c+32}\text{.}$                                                                                  …… (a)

Now, substituting \[\text{C}=0\] in the above linear equation gives, 

$\text{F=}\left( \frac{\text{9}}{\text{5}} \right)\left( 0 \right)\text{+32=32}$.

So, if \[\text{C}={{\text{0}}^{\text{o}}}\text{C}\], then \[\text{F}=\text{3}{{\text{2}}^{\circ }}\text{F}\].

Again, substituting  \[\text{F}=\text{0}\] into the equation (a) gives 

$\text{0=}\left( \frac{\text{9}}{\text{5}} \right)\text{C+32}$

$\Rightarrow \left( \frac{\text{9}}{\text{5}} \right)\text{C=}-\text{32}$

$\Rightarrow \text{C=}\frac{-\text{160}}{9}=-\text{17}\text{.77}$

Hence, if \[\text{F}={{\text{0}}^{\circ }}\text{F}\], then \[\text{C}=-\text{17}\text{.}{{\text{8}}^{\circ }}\text{C}\].

(v) Does there exist a temperature that numerically gives the same value in both Fahrenheit and Celsius? If it is, then show it. 

Ans:  It is provided that,

 $\text{F=}\left( \frac{\text{9}}{\text{5}} \right)\text{c+32}$.

Let assume that \[\text{F=C}\].

Then, 

$\text{F=}\left( \frac{\text{9}}{\text{5}} \right)\text{F+32}$

$\Rightarrow \left( \frac{\text{9}}{\text{5}}-\text{1} \right)\text{F+32=0}$

$\Rightarrow \left( \frac{\text{4}}{\text{5}} \right)\text{F}=-\text{32}$

 $ \Rightarrow \text{F}=-\text{40}$.

Yes, there exists a temperature \[-40{}^\circ \] that gives numerically the same value in both Fahrenheit and Celsius.

Exercise (4.4)

1: Describe the geometric representation of \[\text{y=3}\] as an equation 

(i) in one variable 

Ans. The given equation is \[\text{y=3}\].

Note that, when \[\text{y=3}\] is considered as an equation in one variable, then actually it represents a number in the one-dimensional number line as shown in following figure.

(ii) in two variables.

Ans: The given equation is \[\text{y=3}\].

The above equation can be written as \[\text{0}\text{.x+y=0}\].

Note that when $\text{y=3}$ is considered in two variables, then it represents a straight line passing through point \[\left( 0,3 \right)\] and parallel to the $\text{x}$-axis. Therefore, all the points in the graph having the $\text{y}$-coordinate as $3$, contained in the collection. 

Hence, at \[\text{x=0}\], \[\text{y=3}\]; 

at \[\text{x=2}\], \[\text{y=3}\];  and

at \[\text{x}=-2\], \[\text{y=3}\] are the solutions for the given equation.

Now, Plot the points $\left( 0,3 \right)$, $\left( 2,3 \right)$ and $\left( -2,3 \right)$ on a graph paper and connect the points by a straight line. The graphical representation is shown below:

2: Give the geometric representations of \[\text{2x+9=0}\] as an equation 

Ans. The given equation is \[\text{2x+9=0}\].

Now, the equation can be written as

\[\text{2x+9=0}\]

\[\Rightarrow \text{2x=(-9)}\]

$\Rightarrow \text{x=}\frac{\text{-9}}{2}=\text{-4}\text{.5}$

Hence, when \[\text{2x+9=0}\] is considered as an equation in one variable, then actually it represents a number $\text{x}=-4.5$ in the one-dimensional number line as shown in following figure 

(ii) in two variables 

Ans:   The given equation is \[\text{2x+9}=0\].

The above equation can be written as \[\text{2x+0y=}-\text{9}\].

Note that when \[\text{2x+9}=0\] is considered in two variables, then it represents a straight line passing through point \[\left( -4.5,0 \right)\] and parallel to the $\text{y}$-axis. Therefore, all the points in the graph having the $\text{x}$-coordinate as $-4.5$, contained in the collection. 

Hence, at \[\text{y=3}\], $\text{x}=-4.5$; 

at \[\text{y}=-1\], $\text{x}=-4.5$;  and

at \[\text{y}=1\], $\text{x}=-4.5$ are the solutions for the given equation.

Now, Plot the points $\left( -4.5,3 \right)$, $\left( -4.5,-1 \right)$ and $\left( -4.5,1 \right)$ on a graph paper and connect the points by a straight line. The graphical representation is shown below:

You can opt for Chapter 4 -  Linear Equations in Two Variables NCERT Solutions for Class 9 Maths PDF for Upcoming Exams and also You can Find the Solutions of All the Maths Chapters below.

NCERT Solutions for Class 9 Maths

Chapter 1 - Number System

Chapter 2 - Polynomials  

Chapter 3 - Coordinate Geometry

Chapter 4 - Linear Equations in Two Variables

Chapter 5 - Introductions to Euclids Geometry

Chapter 6 - Lines and Angles

Chapter 7 - Triangles

Chapter 8 - Quadrilaterals

Chapter 9 - Areas of Parallelogram and Triangles

Chapter 10 - Circles

Chapter 11 - Constructions

Chapter 12 - Herons formula

Chapter 13 - Surface area and Volumes

Chapter 14 - Statistics

Chapter 15 - Probability

Linear Equation in Two Variables

The Chapter 4 Linear Equation in Two Variables Class 9 is divided into five sections and four exercises. The first section is the introduction with no exercise. The Second and Third section discusses Linear Equation and it’s solution whereas the Fourth and Fifth sections are advanced topics where we learn about the graph of linear equations in two variables and the equations of lines parallel to x-axis and y-axis.

List of Exercises and topics covered in Linear Equation In Two Variable Class 9:

Exercise 4.1 - Linear Equations

Exercise 4.2 - Solution of a Linear Equation

Exercise 4.3 - Graph of a Linear Equation in Two Variables

Exercise 4.4 - Equations of Lines Parallel to the x-axis and y-axis

Along with this, students can also download additional study materials provided by Vedantu, for Chapter 4 of CBSE Class 9 Maths Solutions –

Chapter 4: Important Questions

Chapter 4: Important Formulas

Chapter 4: Revision Notes

Chapter 4: NCERT Exemplar Solutions

Chapter 4: RD Sharma Solutions

Equations Can Be Linear - Linear Equations

An equation includes equal sign (=) which indicates that the terms on the left-hand side are equal to the terms on the right-hand side. A Linear equation is an equation for a straight line containing variables and constants in the form given below:

\[a_{1}x_{1} + a_{2}x_{2} + a_{3}x_{3}\]..... + b = 0

Where \[a_{1}, a_{2}, a_{3}\]... are coefficients, b is a constant and \[x_{1}, x_{2}, x_{3}\].... are the variables. If the value of any coefficient or variable is zero then the term containing that coefficient or variable becomes zero. This is because anything multiplied to zero is equal to zero.

A linear equation is a simple equation containing coefficients, constants and one or more variables, but a linear equation can never have exponents and roots.

One, Two, Three What Variables Would Be?- Types of Linear equations

One variable linear equation: A linear equation which has only one variable (unknown term) represented by alphabets or symbols is known as one variable linear equation. It is represented as ax+b = 0, where a is a coefficient of variable x and b is a constant. The coefficient can never be zero. 

Examples: 7x + 6 = 13

Two variable linear equation: A linear equation is an equation which has two variables (unknown terms) represented by alphabets or symbols known as a two-variable linear equation. It is represented as ax+by+c = 0, where a and b are coefficients,  x and y are variables and c is a constant. If any coefficient becomes zero then the two-variable linear equation changes to one variable linear equation.

Examples: 2x +3y = 24

Three or more variable linear equations: A linear equation containing more than two variables is called a linear equation of three or more variables. It can be represented as: 

\[a_{1}x_{1} + a_{2}x_{2} + a_{3}x_{3}\]..... + b = 0.

Examples: 5x+ 21y - 3z = -2

Linear Equation with Two Variables

Till now in the journey of algebraic equations, we have learned solving single equations with only one variable (unknown). For example something like 9x + 4 = 22. Simple, Isn’t it?

But what happens if there is more than one unknown in an equation something like 5x + 3y = 15. We solve it differently. So before looking into the solution of a linear equation with two variables let us understand the Linear equation with two variables mathematically.

An equation of the type ax+by+c = 0, where a,b,c are real numbers such that a and b are non-zero, is called a linear equation in two variables x and y.

Example: x+y-5 = 0 is a linear equation in the two variables(unknowns) x and y. Note that x=2 and y=3 satisfy this linear equation.

A Single Linear Equation with Two Variables Cannot be Solved.

There’s no way anyone could legitimately ask you to solve a single equation with two variables because that would give you infinite solutions. But two equations having two variables each can be solved to find the value of x and y simultaneously. A group of two or more equations is called a system of equations.

Each equation represents a straight line. If two lines are taken then there are high chances that those two lines intersect at a unique point which satisfies both the equations. In order to find the intersecting point, pick two random lines and solve.

Solution of a Linear Equation

We know that every linear equation in one variable has a unique solution. What about the solution of a linear equation with  two variables? There will always be a pair of values one for x and the other for y which satisfy the given equation. Also, note that there is no end to different solutions of a linear equation in two variables. That is, a linear equation in two variables has infinitely many solutions.

There are many ways to solve a system of linear equations with two variables. Given below are the two basic methods to solve a linear equation with two variables.

1. Graphical Method of Solving Linear Equation

Instead of finding the solution of two the linear equations separately we find the solution of the system instead. If we graph both the lines in the same coordinate system then the point of intersection of two lines will be the solution of the system.

For Example: To solve the system of equations having two equations -

2x+2 = y and x-1 = y, we need to consider a value of x and find its corresponding value of y for each equation. For equations 2x+2 = y and x-1 = y a random value of x is taken and its corresponding value of y is to be calculated. The points will be (1,4), (2,6), (3,8) for equation 2x+2 = y. And the points (1,0), (2,1), (3,2) for equation x-1 = y. The points are to be plotted on a graph. The point of intersection of these two lines will be the solution of the system.

2. Substitution Method of Solving Linear Equation

The other way of solving a system of linear equations is by substitution method. This system shows how to solve linear equations easily by finding the value of one variable in terms of another variable by using one equation and then replacing this value in another equation. 

Let's find the solution of the same system of linear equations.

From equation (2) we can say that y = x-1.

Substituting the value of y in equation (1).

2x-x = -2-1

Thus, we can find the actual value of y by substituting the value of x as -3 in equation (1).

2(-3) + 2 = y

Therefore, the solution of the system of linear equations is (-3,-4).

NCERT Solutions Chapter- 4 Class 9 Maths By Vedantu

Vedantu provides learning in an entertaining and interesting manner to assist you in developing a firm conceptual foundation on each topic. Subject matter specialists give Class 9 Mathematics Chapter 4 Answers by tirelessly curating correct, simple, and step-by-step solutions for every question in NCERT textbooks . The numerical challenges are presented to assist you in approaching the chapter correctly and improving your comprehension of the key ideas. Mathematics Chapter 4 Solutions are created with the goal of covering the full curriculum in the form of NCERT answers. It has been shown to be a vital resource for students seeking effective learning in order to pass Board examinations and difficult competitive exams like as JEE (Mains and Advanced), AIMs, and others.

Vedantu being the best online tutoring company in India tries its best to render you real help by providing the NCERT Solutions for class 9th Maths Chapter 4Linear Equations and aim to deliver sufficient problems and solutions to practice and build a strong foundation on the chapter. The subject matter experts provide the NCERT Solutions for Class 9 Maths in a simple and precise manner with detailed summary provided at the end of the chapter. 

Deeper Into the Exercise - Types Of Questions In NCERT Class 9 Chapter 4

NCERT Grade 9 CBSE Chapter 4 Linear Equations in Two Variables belongs to Algebra. The Introduction describes solving a linear equation in two variables and how does the solution look like on the Cartesian plane. The topic Linear Equations explains about the points that should be kept in mind while solving a linear problem.In this chapter, you will get hold of the Solution of a Linear Equation. This topic explains a solution of a linear equation with two variables with a pair of values, one for x and one for y which satisfies the given equation.

All of these topics were presented through guided examples, making the learning process more participatory. The solutions to the issues between the chapters help students grasp their level of learning even more. This chapter will teach you interesting subjects like Graph of a Linear Equation in Two Variables and Equations of Lines Parallel to x-axis and y-axis by displaying the two variables of a linear equation on a graph sheet. The answers for all of these subjects are supplied step by step so that the student may internalise the notion gradually. Each lesson is followed by a short set of activities.

The exercises aim to test your knowledge and depth of understanding of the different theorems and concepts that are introduced in this chapter. Regardless, it must be noted that the numerical problems of this chapter are mostly based on specific theorems and other associated concepts. 

All subjects are discussed through step-by-step solved examples in exercises. The solutions to the questions in this chapter will help you understand the fundamental notion of linear equations. A variety of solved examples of numerical problems are also provided to assist you enhance your comprehension of these topics and associated ideas. Furthermore, for each solved case, a detailed step-by-step explanation is provided. It can assist in understanding which strategies should be employed to approach various sorts of issues in order to solve them correctly. The Vedantu team validated the number of exercises and types of problems in class 9th mathematics chapter 4.

Section 1.2 - Exercise 1.1

The first exercise of this chapter consists of 2 questions with question number two having eight sub questions in exercise 1.2 of NCERT Solutions for Maths Class 9 Chapter 4. Most of the questions of this exercise are based on the standard form of linear equation with two variables which is a potent technique to compute the value of a,b and c of any given equation. There are basically three types of questions found from this section:

Type 1: Representing a statement in a linear equation with two variables.

Type 2: Expressing the linear equation in its standard form.

Type 3: Identification of a, b and c in a linear equation with two variables. 

These types of questions involve a lot of steps to reach the solution and hence comes with a risk of making a lot of silly mistakes. Make sure that you have a clear understanding of linear equations and variables. Also, a better understanding of the steps involved would help them clear their lingering doubts easily. Get all your doubts clear and strengthen your knowledge of the different concepts covered in the chapter by referring to our NCERT Solutions for Class 9th Maths Chapter 4. Each numerical problem has been explained step by step to make it easy for you to understand them and grasp the logic behind the same. Additionally, you will also find many helpful tips and alternative techniques to solve similar problems accurately and with more confidence.

Section 1.3 - Exercise 1.2

The second exercise in chapter 4 class 9 maths consists of 4 questions and is mostly based on the solution of linear equations with two variables. Once you grasp the concept of finding the solution, you will be able to identify equations having unique solutions, two solutions or many solutions.Maths class 9 chapter 4 has found wide application both in the field of mathematics and beyond. Given below are the types questions found related to the topic:

Type 1: Identification of the number of solutions of the given equations.

Type 2: Finding the solution of the given equation.

Type 3: Cross checking the solutions of the equation.

Type 4: Finding the value of an unknown term if the solution of the equation is given.

This exercise increases your knowledge in finding the solution of the equation. Doing so, you will gain more confidence as to how to find the unknown terms or how to identify the number of solutions of an equation. It will also prove useful in helping you solve similar types of numerical problems efficiently and in less time. Study the shortcut techniques from up close by taking a quick look at the NCERT Solutions for Class 9 Maths Chapter 4 pdf offered online in its PDF format.You can ace your upcoming board examination quite easily and with many conveniences by incorporating NCERT Solutions for Class 9 Maths Chapter 4 pdf into your revision plan. Download Vedantu’s study solutions from it’s learning portal with just a click and improve your learning experience without much ado.

Section 1.4 - Exercise 1.3

Third exercise of NCERT Solutions for Ch 4 Maths Class 9  consists of the maximum questions. The entire exercise is divided into eight questions, all of which are broadly based on the graph of linear equations with two variables.

Type 1: Drawing of the graph of a linear equation with two variables.

Type 2: Equation of a line passing through a point (x,y).

Type 3: Finding the value of the unknown term if the solution is given.

Type 4: Problem sums.

Type 5: Identification of equation from the graph.

Usually numerical problems involve lengthy steps and complex approaches, which is why it is vital to be well-versed with the fundamentals of the concepts they are based on.This exercise consists of important and complex of questions in Ch 4 Class 9 Maths Solutions so you grasp the fundamental concept of this section and start solving the questions effectively. Vedantu’s study guides like NCERT Solutions for Class 9 Maths Chapter 4 pdf have been engineered by the experts keeping in mind the needs and requirements of both the CBSE board examination and students.

Section 1.5 - Exercise 1.4

The last exercise of Chapter 4 Class 9 Maths consists of 2 questions with sub divisions and is mostly based on the concepts of geometrical representation of the equation of a line. The solution for Maths Class 9 Chapter 4 covers each of these concepts in depth to help students strengthen their grasp on the same. If you have a sound understanding of the topic then you will be able to apply the concept to solve different numerical problems. On the basis of theorems, the exercise can be segregated into the following question types:

Type 1: Geometrical representation of an equation in one variable and two-variable. 

It is advisable to solve the exercise and match your answer with our chapter-based solutions online, to gauge your understanding of the topics more effectively. Revising NCERT Solutions for Class 9th Maths Chapter 4 persistently will go a long way to help you ace your preparation for the upcoming board examination and will prove useful in scoring well in them. It will help you effectively solve this type of numerical problem accurately and in less time.

Download NCERT Solutions for Class 9 Maths from Vedantu, which are curated by master teachers. Also, you can revise and solve the important questions for class 9 Maths Exam 2023-24, using the updated NCERT Book Solutions provided by us. Subjects like Science, Maths, English will become easy to study if you have access to NCERT Solution for Class 9 Science, Maths solutions, and solutions of other subjects that are available on Vedantu only.

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

A linear equation of two variables is a function that two variables are linked. The power of the variables is one. This is why it is called a linear equation. Class 9 Maths Chapter 4 Linear Equations in Two Variables will explain how a linear equation can be plotted on a Cartesian coordinate plane for solutions. To understand the context of this chapter and the mathematical principles, refer to the NCERT solutions designed by the experts.

All the exercise questions are solved in a precise manner by following the CBSE Class 9 standards. These answers will enable students to learn and find out the context of the topics of this chapter. The solutions will also describe the ideal methods of decoding the requirement of a question and guide students to solve them in a stepwise manner.

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

Get the answers to all the exercise questions in one place. Find out how the experts have solved them in a precise way. Download the solution file or access it online to make your exercise solving sessions more productive.

Use the solutions as the perfect material to develop your answering skills. Grab hold of the mathematical principles used in this chapter to solve problems related to linear equations with two variables.

Resolve doubts related to the exercise questions on your own. Use the solutions to leave no queries unattended and take your preparation to the next level. Check how the experts have developed exclusive solving methods for particular questions and proceed accordingly.

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. 

A linear equation in two variables has infinitely many solutions. 

The graph of every linear equation in two variables is a straight line. 

x = 0 is the equation of the y-axis and y = 0 is the equation of the x-axis. 

The graph of x = a is a straight line parallel to the y-axis. 

The graph of y = a is a straight line parallel to the x-axis. 

An equation of the type y = mx represents a line passing through the origin. 

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.

Conclusion 

The solutions are available in PDF format and may be added to the study material. When you finish the chapter, you can use the solutions to solve the exercise problems. Examine where you need to learn and practise more in this chapter to improve. Before a test, go over the chapter exercises again and practise by following the solutions. Remember what you practised and score higher.

FAQs on NCERT Solutions for Class 9 Maths Chapter 4 - Linear Equations In Two Variables

1. How do Vedantu’s NCERT Solutions for Maths Class 9 Chapter 4 help all the students to score good marks?

The solutions are written in simple and easy-to-understand language keeping in mind every kind of the students.

If there is any complicated solution, that is broken down into simple steps to make it understandable for every student to grasp the concept in less time.

The solutions are updated as per the latest NCERT curriculum and guidelines.

It covers the entire syllabus and concept in the form of solutions.

The answers are treated systematically and in an interesting manner.

The content is designed in a concise manner, which is brief and self-explanatory.

Some answers include necessary infographics and images to facilitate the understanding of the concept.

It is handy and serves as a note during exam revision.

The solutions are kept easy for you to solve maximum questions and get a command of the chapter. Also, it Improves the problem solving speed

2. What can I learn from Maths Class 9 Chapter 4?

In this chapter, the knowledge of linear equations in one variable is recalled and extended to that of two variables. Any equation which can be put in the form ax + by + c = 0, where a, b and c are real numbers, and a and b are not both zero, is defined as a linear equation in two variables. The NCERT solutions for Class 9 Maths provide chapter-wise answers of the questions asked in the exercises in the textbook and students can get the command over this Linear Equation concept of Algebra with the help of easy examples provided.

3. Give me an overview of the chapter.

4.1 – Introduction

4.2 – Linear Equations

4.3 – Solution of a Linear Equation

4.4 – Graph of a Linear Equation In Two Variables

4.5 – Equations of Lines Parallel to the x-axis and y-axis

4. How many exercises are there in this chapter?

The Chapter 4 named Linear Equation in Two Variables Class 9 is divided into five sections and four exercises. Below is the list of exercises along with the number and types of questions.

Exercise 4.1: 2 Questions (1 Short Answer, 1 Main Question with 8 short answer questions)

Exercise 4.2: 4 Questions (2 MCQs, 1 Main Questions with 3 equations, 1 Short Answer Questions)

Exercise 4.3: 8 Questions (4 Long Answer Questions, 2 Short Answer Questions, 1 MCQ, 1 Main Question with 5 Part Questions)

Exercise 4.4: 2 Questions (1 main question with 2 sub-section, 1 Main question with 2 sub-section)

5. What are all the important topics that need to be covered to score in Class 9 Maths Chapter 4?

The topics that you must not miss in order to fully complete the Class 9 Maths Chapter 4 Linear Equations in Two Variables are - expressing linear equations in Ax + By + C = 0; different solutions of a linear equation with two variables; graph of a linear equation in two variables; and equations of the line parallel to x-axis and y-axis. To get clarity of these concepts, you need to solve all examples and exercise questions. 

6. Where can I get the best solutions for Class 9 Maths Chapter 4? 

Vedantu is the best place to get free solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables. Vedantu’s website and Vedantu app offers a comprehensive and detailed NCERT Solutions Class 9 Maths Chapter 4 Linear Equations in Two Variables that has explanations to all the different sub-topics. These solutions have complete step by step solutions to the examples and exercises that will give clarity to all the concepts. These are prepared by experts who have decades of experience, so these solutions are credible and up to date.

7. Is Chapter 4 of Linear Equation in Two Variables of Class 9 Maths tough?

No, the NCERT Class 9 Mathematics Linear Equation in Two Variables is not difficult for those who thoroughly practise the chapter. Your main aim should be to finish the chapter thoroughly, including all examples, exercise questions, and other questions. NCERT Solutions Class 9 Maths Chapter 4 Linear Equations in Two Variables by Vedantu is the best resource for completing Class 9 Maths Chapter 4 without problem.

8. How can I solve Class 9 Maths Chapter 4?

You can easily solve Class 9 Maths Chapter 4 Linear Equations in Two Variables with the help of Vedantu’s NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables. These solutions have full solved numerical problems step by step that will help you in clearing your concepts. You can also get access to many miscellaneous questions that will enhance your base even further.  

9. Can the graph of a linear equation with two variables be a curve?

No, the graph of a two-variable linear equation cannot be a curve. The path is always straight. Get the greatest companion available only at Vedantu for additional NCERT Class 9 Mathematics Chapter 3 questions that will assist you in completing comprehensive exercises along with examples and random questions. You may obtain entire solutions to the exercises by downloading the NCERT Solutions for Class 9 Maths Chapter 3 Coordinate Geometry.

NCERT Solutions for Class 9

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Unit 3: Linear equations in one variable

Solving linear equations.

• Equation with variables on both sides: fractions (Opens a modal)
• Two-step equation word problem: oranges (Opens a modal)
• Word problems linear equations (basic) 4 questions Practice
• Linear equations word problems (advanced) 4 questions Practice

• Math Article
• Linear Equations In Two Variables Class 9

Linear Equations in Two Variables Class 9 Notes

Cbse class 9 maths linear equations in two variables notes:- download pdf here.

Get the complete notes on linear equations in two variables Class 9 here. In this article, you are going to study the basics of linear equations involving one variable, two variables, and so on. Also, learn how to graph linear equations and how to find the solutions for linear equations in detail.

Linear equation in one variable

When an equation has only one variable of degree one, that equation is known as linear equation in one variable.

• Standard form: ax+b=0, where a and b ϵ R & a ≠ 0
• Examples of linear equations in one variable are :

– 3x-9 = 0 – 2t = 5

To know more about Linear equations in one variable, visit here .

Linear equation in 2 variables

When an equation has two variables both of degree one, that equation is known as linear equation in two variables.

Standard form: ax+by+c=0, where a, b, c ϵ R & a, b ≠ 0 Examples of linear equations in two variables are: – 7x+y=8 – 6p-4q+12=0

To know more about Linear equations in 2 variables, visit here .

Examples of Linear Equations

The solution of linear equation in 2 variables.

A linear equation in two variables has a pair of numbers that can satisfy the equation. This pair of numbers is called as the solution of the linear equation in two variables.

• The solution can be found by assuming the value of one of the variables and proceeding to find the other solution.
• There are infinitely many solutions for a single linear equation in two variables.

Graph of a Linear Equation

Graphical representation of a linear equation in 2 variables.

• Any linear equation in the standard form ax+by+c=0 has a pair of solutions in the form (x,y), that can be represented in the coordinate plane.
• When an equation is represented graphically, it is a straight line that may or may not cut the coordinate axes.

To know more about the Graphical representation of a linear equation, visit here .

Solutions of linear equation in 2 variables on a graph

• A linear equation ax+by+c=0 is represented graphically as a straight line.
• Every point on the line is a solution for the linear equation.
• Every solution of the linear equation is a point on the line.

Lines passing through the origin

• Certain linear equations exist such that their solution is (0, 0). Such equations, when represented graphically, pass through the origin.
• The coordinate axes, namely the x-axis and the y-axis, can be represented as y=0 and x=0, respectively.

Lines parallel to coordinate axes

• Linear equations of the form y=a, when represented graphically, are lines parallel to the x-axis and a is the y-coordinate of the points in that line.
• Linear equations of the form x=a, when represented graphically, are lines parallel to the y-axis and a is the x-coordinate of the points in that line.

Put your understanding of this concept to test by answering a few MCQs. Click ‘Start Quiz’ to begin!

Select the correct answer and click on the “Finish” button Check your score and answers at the end of the quiz

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thanks for the help

The linear equation in two variable please notes send me

Go through the article for notes: Linear equation in two variables

4xy-3y-8=0, is this linear equation?

4xy-3y-8=0 is not a linear equation. Reason: The term 4xy has two variables “x” and “y” multiplied together. Hence, the degree will be 1 + 1 = 2.

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