case study questions linear equations 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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Case Study Questions for Class 9 Maths Chapter 4 Linear Equations in Two Variables

• Last modified on: 11 months ago
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Case Study Questions

Question 1:

Anil went to buy some vegetables, he bought ‘x’ kgs. of tomato and ‘y’ kgs. of potato. The total cost of vegetables comes out to be of Rs. 200. Now if the cost of 1 kg of tomato is Rs. 50 and 1 kg of potato is Rs. 20, then answer the following questions.

(i) Which of the following equations represent the total cost. (a) 5x – 2y = 20 (b) 5y + 2x = 20 (c) 5x + 2y = 20 (d) 2x + 5y = 20

(ii) If Anil bought ‘x’ kgs of tomato and 2.5 kgs. of potato, then find the value of ‘x’. (a) 5 (b) 2 (c) 3 (d) 4

(iii) If Anil bought ‘2’ kgs of tomato and ‘y’ kgs of potato, then find the value of ‘y’. (a) 5 (b) 2 (c) 3 (d) 4

(iv) The graph of 5x + 2y = 20 cuts x-axis at the point. (a) (10, 0) (b) (4, 0) (c) (0, 0) (d) it is parallel to x-axis

(v) The graph of 5x + 2y = 20 cuts y-axis at the point. (a) (0, 10) (b) (0, 4) (c) (0, 0) (d) it is parallel to y-axis

(i) C (ii) C (iii) A (iv) B (v) A

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4 thoughts on “ Case Study Questions for Class 9 Maths Chapter 4 Linear Equations in Two Variables ”

Where are the answers?

Answer key (i)(c) 5x + 2y = 20 (ii)(c) 3 (iii)(a) 5 (iv)(b) (4, 0) (v)(a) (0, 10)

Hlo sir I have doubt

kindly solve ii and iii part

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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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CBSE Class 9 Mathematics Case Study Questions

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If you’re looking for a comprehensive and reliable study resource and case study questions for class 9 CBSE, myCBSEguide is the perfect door to enter. With over 10,000 study notes, solved sample papers and practice questions, it’s got everything you need to ace your exams. Plus, it’s updated regularly to keep you aligned with the latest CBSE syllabus . So why wait? Start your journey to success with myCBSEguide today!

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)

I	NUMBER SYSTEMS	10
II	ALGEBRA	20
III	COORDINATE GEOMETRY	04
IV	GEOMETRY	27
V	MENSURATION	13
VI	STATISTICS & PROBABILITY	06

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)

1.	Exhibit memory of previously learned material by recalling facts, terms, basic concepts, and answers.  Demonstrate understanding of facts and ideas by organizing, comparing, translating, interpreting, giving descriptions, and stating main ideas	43	54
2.	Solve problems to new situations by applying acquired knowledge, facts, techniques and rules in a different way.	19	24
3.	Examine and break information into parts by identifying motives or causes. Make inferences and find evidence to support generalizations Present and defend opinions by making judgments about information, validity of ideas, or quality of work based on a set of criteria. Compile information together in a different way by combining elements in a new pattern or proposing alternative solutions	18	22
		80	100

myCBSEguide: Blessing in disguise

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

All questions was easy but search ? hard questions. These questions was not comparable with cbse. It was totally wastage of time.

Where is search ? bar

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Can I have more questions without downloading the app.

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CBSE Class 9th Maths 2023 : 30 Most Important Case Study Questions with Answers; Download PDF

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CBSE Class 9 Maths exam 2022-23 will have a set of questions based on case studies in the form of MCQs. CBSE Class 9 Maths Question Bank on Case Studies given in this article can be very helpful in understanding the new format of questions.

Each question has five sub-questions, each followed by four options and one correct answer. Students can easily download these questions in PDF format and refer to them for exam preparation.

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Important Questions Class 9 Maths Chapter 4

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Important Questions Class 9 Mathematics Chapter 4 – Linear Equations in Two Variables

Mathematics Chapter 4 of Class 9 introduces students to Linear Equations In Two Variables. A required linear equation in two variables has two numbers that can satisfy the given equation. These two numbers are called the solution of the required linear equation in two variables.

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 Following are the key topics covered in Chapter 4 of CBSE Class 9 Mathematics syllabus:

Graphical representation of a linear equation in 2 variables

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

Solutions of Linear equation in 2 variables on a graph

• A given linear equation ax+by+c=0 is illustrated graphically as a straight line.
• Each point on the given line is a solution for the linear equation.
• Each solution of the given linear equation is a given point on the required line.

Lines passing through the origin

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

Lines parallel to coordinate axes

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

Extramarks’ credibility lies in providing reliable and trusted study material related to NCERT is the most useful study companion for students and enables students with their comprehensive online study solutions from Class 1 to Class 12. Our qualified Mathematics subject experts have prepared various NCERT explanations to help students in their studies and exam preparation. Students can direct to our Important Questions Class 9 Mathematics Chapter 4 to practise exam-oriented questions. We have collected questions from diverse sources such as NCERT textbooks and exemplars, CBSE sample papers, CBSE past year question papers, etc. Students can prepare excellently for their exams and quizzes by solving various chapter questions from our Important Questions Class 9 Mathematics Chapter 4.

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Important Questions Class 9 Mathematics Chapter 4 – With Solutions

Our in-house Mathematics faculty experts have collected an entire list of Important Questions Class 9 Mathematics Chapter 4 by referring to various sources. For every question, the team has prepared a step-by-step explanation that will aid students in understanding the concepts used in each question. Also, the given questions are prepared in a way that will cover the entire chapter.. By practising from our question bank, students should be able to revise the chapter and self-assess their strong and weak points. And improvise by further focusing on weak areas of the chapter and practising harder to maximise their potential.

 Following are some of the questions and explanations from our question bank of Mathematics Class 9 Chapter 4 Important Questions:

Question 1: Define the following linear equations in the form ax + by + c = 0 and show the values of a, b and c in every individual case:

(i) x – y/5 – 10 = 0

(ii) -2x+3y = 6

(iii) y – 2 = 0

(i) The equation x-y/5-10 = 0 

(1)x + (-1/5) y + (-10) = 0

Directly compare the above equation with ax + by + c = 0

Therefore, we get;

(ii) –2x + 3y = 6

Re-arranging the provided equation, we obtain,

–2x + 3y – 6 = 0

The required equation –2x + 3y – 6 = 0 can be written as,

(–2)x + 3y +(– 6) = 0

Directly comparing (–2)x + 3y +(– 6) = 0 with ax + by + c = 0

We obtain a = –2

The required equation y – 2 = 0 can be written as,

0x + 1y + (–2) = 0

Directly comparing 0x + 1y + (–2) = 0 with ax + by + c = 0

We obtain a = 0

Question 2: The price of a notebook is twice the cost of a pen. Note a linear equation in two variables to illustrate this statement.

(Taking the price of a notebook to be ₹ x and that of a pen to be ₹ y)

Answer 2: Let the price of one notebook be = ₹ x

Let the price of one pen be = ₹ y

As per the question,

The price of one notebook is twice the cost of one pen.

i.e., the price of one notebook = 2×price of a pen

x-2y = 0 is the required linear equation in two variables to illustrate the statement, ‘The price of one given notebook is twice the cost of a pen.

Question 3: Give the geometric representations of 2x + 9 = 0 as an equation

(i) in one variable

(ii) in two variables

Answer 3:  

(i) 2x + 9 = 0

We have, 2x + 9 = 0 

                 2x = – 9 

                   x = -9/2

which is the required linear equation in one variable, that is, x only.

Therefore, x= -9/2 is a unique solution on the number line as shown below:

(ii) 2x +9=0

We can write 2x + 9 = 0 in the two variables as 2x + 0, y + 9 = 0

or x = −9−0.y/2

∴ When y = 1, x =  −9−0.(1)/2 = -9/2

               y=2 , x = −9−0.(2)/2 =  -9/2

                y = 3, x = −9−0.(3)/2= -9/2

Therefore, we obtain the following table:

X	-9/2	-9/2	-9/2
Y	1	2	3

Now, plotting the ordered pairs (−9/2,3), (−9/2,3) and (−9/2,3) on graph paper and connecting them, we get a line PQ as the solution of 2x + 9 = 0.

Image source: https://www.learncbse.in/wp-content/uploads/2020/10/NCERT-Solutions-for-Class-9-Maths-Chapter-4-Linear-Equations-in-Two-Variables-Ex-4.4-Q2.3.png

Question 4: Note four solutions individually for the following equations:

(i) 2x + y = 7

Answer 4: For the four answers of 2x + y = 7, we replace different values for x and y

(2×0)+y = 7

(2×1)+y = 7

The answers are (0, 7), (1,5), (3,1), (2,3)

Question 5: The linear equation 2x -5y = 7 has

(A) A unique solution

(B) Two solutions

(C) Infinitely many solutions

Answer 5: (C) Infinitely many solutions

Linear equation: The equation of two variables which gives a straight line graph is called a linear equation.

Here the linear equation is 2x – 5y = 7

Let y = 0, then the value of x is:

2x – 5(0)=7

Now, let y = 1, then the value of x is:

2x – 5 (1) =7

Here for different values of y, we are getting different values of x

Therefore the equation has infinitely many solutions

Question 6: Represent the following linear equations in the form ax + by + c = 0 and show the required values of a, b and c in every case:

Answer 6: (i) x –(y/5)–10 = 0

The required equation x –(y/5)-10 = 0 can be written as,

1x+(-1/5)y +(–10) = 0

Comparing the given equation x+(-1/5)y+(–10) = 0 with ax+by+c = 0

(ii) –2x+3y = 6

Rearranging the equation, we obtain,

–2x+3y–6 = 0

The required equation –2x+3y–6 = 0 can be written as,

(–2)x+3y+(– 6) = 0

Comparing the given equation (–2)x+3y+(–6) = 0 with ax+by+c = 0

(iii) x = 3y

The required equation x-3y=0 can be written as,

1x+(-3)y+(0)c = 0

Comparing the given equation 1x+(-3)y+(0)c = 0 with ax+by+c = 0

We obtain a = 1

(iv) 2x = –5y

The required equation 2x+5y = 0 can be written as,

2x+5y+0 = 0

Comparing the given equation 2x+5y+0= 0 with ax+by+c = 0

We obtain a = 2

(v) 3x+2 = 0

The required equation 3x+2 = 0 can be written as,

3x+0y+2 = 0

Comparing the given equation 3x+0+2= 0 with ax+by+c = 0

We obtain a = 3

(vi) y–2 = 0

The required equation y–2 = 0 can be written as,

0x+1y+(–2) = 0

Comparing the given equation 0x+1y+(–2) = 0 with ax+by+c = 0

(vii) 5 = 2x

i.e., 2x–5 = 0

The required equation 2x–5 = 0 can be written as,

2x+0y–5 = 0

Comparing the given equation 2x+0y–5 = 0 with ax+by+c = 0

Question 7: Note four solutions individually for the following equations:

 πx + y = 9

Answer 7: For the four answers of πx + y = 9, we replace other values for x and y

(π × 0)+y = 9

(π×1)+y = 9

Put x=2, we have 

π(2) + y = 9

y = 9 – 2π

The answers are (0,9), (1,9-π),(9/π,0),(2,9 – 2π)

Question 8:  Find out the value of k, if x = 2, y = 1 is a given solution of the equation 2x + 3y = k.

Answer 8: The provided equation is

2x + 3y = k

As per the given question, x = 2 and y = 1.

Then, Replacing the values of x and y in the equation 2x + 3y = k,

⇒(2 x 2)+ (3 × 1) = k

The required value of k, if x = 2, y = 1 is a given solution of the equation 2x + 3y = k, is 7.

Question 9:  Establish that the required points A (1, 2), B ( – 1, – 16) and C (0, – 7) lie on the graph of the required linear equation y = 9x – 7.

Answer 9: We include the equation,

For A (1, 2),

Replacing (x,y) = (1, 2),

2 = 9(1) – 7

For B (–1, –16),

Replacing (x,y) = (–1, –16),

–16 = 9(–1) – 7

-16 = – 9 – 7

For C (0, –7),

Replacing (x,y) = (0, –7),

– 7 = 9(0) – 7

Therefore, the points A (1, 2), B (–1, –16) and C (0, –7) satisfy the line y = 9x – 7.

Therefore, A (1, 2), B (–1, –16) and C (0, –7) are answers to the linear equation y = 9x – 7

Thus, points A (1, 2), B (–1, –16), and C (0, –7) lie on the graph of the linear equation y = 9x – 7.

Question 10: Note the linear equation such that every point on its graph has a coordinate 3 times its abscissa.

A given linear equation such that every point on its graph has a coordinate(y) which is 3 times its

abscissa(x).

So we obtain

Therefore, y = 3x is the required linear equation.

Question 11: Illustrate the graph of the given linear equation 3x + 4y = 6. At what points does the graph cut the X and Y-axis?

Answer 11:  Given the equation,

3x + 4y = 6.

We need at least 2 points on the graph to illustrate the graph of this equation,

Therefore, the points the graph cuts

The given point is on the x-axis. We have y = 0.

Replacing y = 0 in the equation, 3x + 4y = 6,

3x + 4×0 = 6

Therefore, the point at which the graph cuts the x-axis = (2, 0).

(ii) y-axis

Since the point is on the y-axis, we have x = 0.

Replacing x = 0 in the equation, 3x + 4y = 6,

3×0 + 4y = 6

Thus, the point at which the graph cuts the x-axis = (0, 1.5).

By plotting the points (0, 1.5) and (2, 0) on the graph.

image source: Online 

Question 12: Show that the required points A (1, 2), B ( – 1, – 16) and C (0, – 7) lie on the given graph of the linear equation y = 9x – 7.

Answer 12: We have the given equation,

Substitute the values of (x,y) = (1, 2),

2 = 9 (1) – 7 = 9 – 7 = 2

Substitute the values of (x,y) = (–1, –16),

–16 = 9(–1) – 7 = – 9 – 7 = – 16

Substitute the values of (x,y) = (0, –7),

– 7 = 9(0) – 7 = 0 – 7 = – 7

Thus, we locate that points A (1, 2), B (–1, –16) and C (0, –7) satisfy the line y = 9x – 7.

Thus, A (1, 2), B (–1, –16), and C (0, –7) are required solutions of the linear equation y = 9x – 7

Hence, the given points A (1, 2), B (–1, –16) and C (0, –7) lie on the graph of the required linear equation y = 9x – 7.

Question 13: Show the required geometric representations of y = 3 as an equation

Answer 13: (i) In one variable

∵ y = 3 is the required equation in one variable, that is, y only.

∴ y = 3 is the required unique solution on the number line as displayed below:

Image source: https://www.learncbse.in/wp-content/uploads/2020/10/NCERT-Solutions-for-Class-9-Maths-Chapter-4-Linear-Equations-in-Two-Variables-Ex-4.4-Q1.png

(ii) In two variables 

When an equation in two variables, it can be expressed as,

0.x + y – 3 = 0

which is a linear equation in the variables x and y. 

When x = 0, y = 3

When x = 1, y = 3

Image source: https://cdn1.byjus.com/wp-content/uploads/2020/07/ncert-solutions-class-9-chapter-4-13.png

Question 14: In countries like the USA and Canada, the temperature is measured In the required Fahrenheit, whereas in countries like India, it is calculated in Celsius. Given here is a linear equation that converts Fahrenheit to Celsius:

F = (95 )C + 32

(i) Draw the linear equation graph above using Celsius for the x-axis and Fahrenheit for the y-axis.

(ii) If the required temperature Is 30°C, what is the temperature in Fahrenheit?

(iii) If the required temperature is 95°F, what is the temperature in Celsius?

(iv) If the required temperature is 0°C, what is Fahrenheit? If the required temperature is 0°F, what Is the temperature In Celsius?

(v) Is a required temperature numerically the same in Fahrenheit and Celsius? If yes, find It.

Answer 14: (i)F=9/5C+32

When C=0 then F=32

also, when C=5 then F=41

C	0	5
F	32	41

Image source: https://s3-us-west-2.amazonaws.com/infinitestudent-images/ckeditor_assets/pictures/64720/content_degree.png

(ii) Placing the value of C=30 in F=9/5C+32, we obtain

F=9/5×30+32

(iii) Putting the value of F=95 in F=9/5C+32, we get

Question 15:  If the work accomplished by a body on application of a steady force is directly proportional to the required distance traversed by the body, convey this in the required form of an equation in two variables and sketch the exact graph by taking the steady force as 5 units. Likewise, read from the graph the work done when the distance traversed by the body is

(i) 2 units

(ii) 0 unit

Answer 15: Steady force is 5 units.

Let the distance traversed = x units and work done = y units.

Work done = Force x Distance

 y = 5 x x 

For sketching the graph, we are having y = 5x

When x = 0, 

 y = 5(0) = 0

x = 1, then y = 5(1) = 5

x = -1, then y = 5(-1) = -5

∴ We obtain the following given table:

x	0	1	-1
y	0	5	-5

Plotting the required ordered pairs (0, 0), (1, 5) and (-1, -5) on the graph paper and joining the points, we acquire a straight line AB as shown.

Image source: https://www.learncbse.in/wp-content/uploads/2020/10/NCERT-Solutions-for-Class-9-Maths-Chapter-4-Linear-Equations-in-Two-Variables-Ex-4.3-Q6.1.png

From the required graph, we obtain

(i) The required distance traversed =2 units, that is, x = 2

∴ If x = 2, 

then y = 5(2) = 10

 The required work done = 10 units when the distance travelled by it is 2 units.

(ii) The required distance traversed = 0 unit ,that is , x = 0

∴ If x = 0 

          y = 5(0) – 0

 The required work done = 0 unit when the distance travelled by it is 0 units.

Benefits Of Solving Important Questions Class 9 Mathematics Chapter 4 

Practice is the key to scoring good marks in Mathematics.  Mathematics taught in Classes 8, 9, and 10 prepares the foundation for higher classes and also many other educational streams. We recommend students register on our website to get access to our Important Questions Class 9 Mathematics Chapter 4. By rigorously solving questions and going through the required solutions, students will get enough  confidence by clarifying their doubts to solve any other complicated questions in the given chapter, Linear Equations In Two Variables.

Given below are a few benefits of frequently solving questions from our question bank of Important Questions Class 9 Mathematics Chapter 4:

• Our expert Mathematics teacher faculty has carefully compiled the most important questions in our questionnaire Important Questions Class 9 Mathematics Chapter 4. These questions are  picked after referring to many past years’ question  papers, NCERT textbooks, and other Mathematics reference books.
• The given questions and solutions provided are completely based on the NCERT book and in accordance with the latest CBSE syllabus and guidelines. So the students can confidently bank  on our study resources. 
• The given questions covered in our set of important Questions Class 9 Mathematics Chapter 4 are entirely based on several topics covered in the chapter Linear Equations in Two Variables. It is suggested that students revise and clear all their doubts before solving all these important questions. 
• By going through all the questions given in our Chapter 4 Class 9 Mathematics Important Questions, students will get an idea of the question paper pattern. . Practising questions comparable to the exam questions would aid students in gaining confidence, scoring better marks, and eventually acing their exams

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Q.1 If the point (4, 3) lies on the graph of the equation 3x ay = 6, find whether (2, 6) also lies on the same graph.

Marks: 3 Ans

Since 4, 3 liesonline3x ay=6 So,puttingx=4andy=3,weget 3 4 a 3 =6 12 3a=6 12 6=3a a= 6 3 =2 So,equationbecomes 3x-2y=6 Puttingx= 2andy= 6inLHS, LHS=3x y =3 2 2 6 = 6+12 =6=RHS Thus, 2, 6 alsolieson the graph of the sameline.

Q.2 For what value of k, the point (k, 5) lies on the line 4x 5y = 10

Marks: 2 Ans

Point (k, 5) lies on the line 4x 5y = 10, Since the point satisfies the line, so, we have 4(k) 5(5) =10 4k 25 = 10 4k = 35 k =

Q.3 Find four different solutions of the equation x+2y = 6.

Let us consider x = 0 Then the given equation reduces to 2y = 6 y = 3 So, (0,3) is a solution of x+2y = 6. At x = 2, the given equation reduces to 2+2y = 6 y = 2 So, (0,2) is a solution of given equation. At y = 0, the given eqution reduces to x = 6. So (6,0) is a solution of given equation. At y = 1 given equation reduces to x+2 = 6 x = 4. So (4,1) is a solution of given equation.’

Thus (2,2), (0,3), (6,0) and (4,1) are the solutions of given equation.

Q.4 Draw the graph of x+y = 7.

Marks: 5 Ans

Todrawthegraph,weneedatleasttwosolutions. x+y=7 y=7 x x=0 y=7 0 y=7 x=7 y=7 7 y=0 0,7 and 7,0 aresolutionsofx+y=7 We​canusethefollowingtabletodrawthegraph: x 0 7 y 7 0

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Faqs (frequently asked questions), 1. what is the standard equation in linear equations in two variables.

  The standard equation for Equations in two variables is ax + by +c =0

2. How can I get good grades in Class 9 Mathematics examinations?

Mathematics is such a subject that requires plenty of practice. To achieve well in Mathematics, one must have a potent conceptual knowledge of the chapter, be good enough at calculations, practice questions regularly, give mock tests from time to time, get feedback and avoid silly mistakes. The more you practice, the better you will get. Everyday practice with discipline, working diligently and conscientiously towards your ambition, will ensure 100% in your exams.  

3. What can I get from the Extramarks website?

Extramarks is one of the best educational platforms and it has its archive of academic resources, which also assists students in accomplishing their exams. You can acquire all the NCERT-related material like NCERT solutions, solved exemplar solutions, NCERT-based mock tests, CBSE revision notes, and  Class 9 Mathematics Chapter 4 important questions on the Extramarks website. Besides this, you can get great suggestions from our subject matter experts during doubt-clearing sessions when you sign up on Extramarks official website. .        

4. How many total chapters will students study in Class 9 Mathematics?

There are 15 chapters in Class 9 Mathematics. The list is given below:

• Chapter 1- Number System
• Chapter 2 –Polynomials
• Chapter 3 – Coordinate Geometry
• Chapter 4 –Linear Equations In Two Variables
• Chapter 5 – Introduction To Euclid’s Geometry
• Chapter 6 – Lines And Angles
• Chapter 7 –Triangles
• Chapter 8 – Quadrilaterals
• Chapter 9 –Areas Of Parallelograms And Triangles
• Chapter 10 – Circles
• Chapter 11- Constructions
• Chapter 12- Heron’s Formula
• Chapter 13-Surface Area And Volumes
• Chapter 14- Statistics
• Chapter 15- Probability

5. What are the important chapters in Class 9 Mathematics Chapter 4?

The NCERT Mathematics book has 15 chapters. When it comes to grasping the fundamentals and taking the test, each and every chapter is equally important. Additionally, because CBSE does not specify marks distribution.  Students need to study all the chapters in order to receive a good grade. 

6. Where can I get important questions for Class 9 Mathematics Chapter 4 online?

On the Extramarks website, you can find all the important questions for Class 9 Mathematics Chapter 4, along with their answers. You can even find many other NCERT-based study solutions on the website for Classes 1 to 12.

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

By QB365 on 08 Sep, 2022

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.

QB365 - Question Bank Software

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.

	+ x = 1

(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.

*****************************************

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

(i) (b) 10x, 5y Area of one bedroom = 5x sq.m Area of two bedrooms = 10x sq.m Area of kitchen = 5y sq. m (ii) (d) 54 m Length of outer boundary = 12 + 15 + 12 + 15 = 54 m (iii) (d) None of the above Area of two bedrooms = 10x sq.m Area of kitchen = 5y sq. m So, 10x + 5y = 95  2x + y = 19 Also, x + 2 + y = 15 x + y = 13 (iv) (c) x = 6, y = 7 x + y = 6 + 7 = 13 2x + y = 2(6) + 7 = 19 x = 6, y = 7 x + y = 6 + 7 = 13 2x + y = 2(6) + 7 = 19  x = 6, y = 7 (v) (a) 30 sq. m Area of living room = (15 x 7) – 30 = 105 – 30 =75 sq. m

(i) (a) 3x + 2y = 80 and 4x + 3y = 110 Here, the cost of one notebook be Rs. x and that of pen be Rs. y. According to the statement, we have 3x + 2y = 80 and 4x + 3y = 110 (ii) (b) x = 20, y = 10 3x + 2y = 3(20) + 2(10) = 60 + 20 = 80 4x + 3y = 4(20) + 3(10) = 80 + 30 = 110 (b) x = 20, y = 10 (iii) (b) Rs. 10 Cost of 1 pen = Rs. 10 (b) Rs. 10 (iv) (d) Rs. 420 Total cost = Rs. 15 x 20 + Rs. 12 x 10 = 300 + 120 = Rs. 420 (v) (a) Deepak Ram said that price of each notebook could be Rs. 25. Ajay felt that Rs. 2.50 for one pen was too little. It should be at least Rs. 16 Deepak guess the cost of one pen is Rs. 10 and Lohith guess the cost of one notebook is Rs. 30 Therefore, estimation of Deepak is correct

(a) (iii) x2 + x = 1 (b) (ii) x + y = 200 Here, x represents Sita's contribution and y represents Gita's contribution. (c) (iii) Rs. 124 If x = 76 then 76 + y = 200 y = 200 - 76 y = 124 (d) (ii) Rs. 100, Rs. 100 If x = y then x + x = 200 2x = 200 x = 200/2 = 100 (e) (iii) 1.x + 0.y + 5 = 0 Since, x = -5 ⇒ x + 5 = 0 Thus, standard form of x = -5 is 1.x + 0.y + 5 = 0.

(i) (d) 5x + 2y = 120 Here, the cost of one notebook be Rs. x and that of pen be Rs. y. According to the statement, we have 5x + 2y = 120 (ii) (b) x = 20, y = 10 5x + 2y = 5(10) + 2(20) = 50 + 40 = 90 ≠ 120 5x + 2y = 5(20) + 2(10) = 100 + 20 = 120 5x + 2y = 5(15) + 2(15) = 75 + 30 = 105 ≠ 120 (c) (ii) Rs. 95 5x + 2y = 5(15) + 2(10) = 75 + 20 = 95 (d) (b) Rs. 10 Here, x = 2y 5(2y) + 2y = 10y + 2y = 12y = 120 ⇒ y = 10 (e) (iv) 0.x + 1.y – 4 = 0 Since, y = 4 ⇒ y – 4 = 0 Thus, standard form of y = 4 is 0.x + 1.y – 4 = 0

(a) (iii) 2x + 3y = 60 Let the number of children be x and the number of adults be y then the linear equation in two variable for the given situation is  2x + 3y = 60. (b) (i) 10 2x + 3y =60 ⇒ 2(15) + 3y = 60 ⇒ 3y = 60 - 30 = 30  ⇒ y = 10 (c) (i) 12 2x + 3y = 60 ⇒ 2x + 3(12) = 60 ⇒ 2x 60 - 36 = 24 ⇒ x = 12 (d) (iii) 15 On putting x = 5 and y = 0 in the equation 3x + 5y = b, we have  3 x 5 + 5 x 0 = b ⇒ 15 + 0 = b ⇒ b = 15 (e) (ii) 1.x - 1.y + 5 = 0 y - x = 5 ⇒ y = x + 5 ⇒ x - y + 5 = 0 ⇒ 1.x - 1.y + 5 = 0

(a) (iii) x + y = 7 (b) (iv) 2 x + y = 7 ⇒ 5 + y = 7 ⇒ y = 7 - 5 = 2 (c) (i) (3, 4) (d) (iv) (0, 7) (e) (iii) -1 On putting x = 2 and y = -1 in the equation x + 3y - k = 0, we have 2 + 3(-1) - k = 0 ⇒ 2 - 3 = k ⇒ k = -1

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Linear Equations for Two Variables Class 9 Extra Questions Maths Chapter 4 with Solutions Answers

September 7, 2020 by Prasanna

Here we are providing Linear Equations for Two Variables Class 9 Extra Questions Maths Chapter 4 with Answers Solutions, Extra Questions for Class 9 Maths  was designed by subject expert teachers.

Extra Questions for Class 9 Maths Linear Equations for Two Variables with Answers Solutions

Extra Questions for Class 9 Maths Chapter 4 Linear Equations for Two Variables with Solutions Answers

Linear Equations for Two Variables Class 9 Extra Questions Very Short Answer Type

Question 1. Linear equation x – 2 = 0 is parallel to which axis ? Solution: Here, linear equation is x – 2 ⇒ 0 x = 2 Thus, it is parallel to the y-axis.

Question 2. Express x in term of y: \(\frac{x}{7}\) + 2y = 6 Solution: Given equation is \(\frac{x}{7}\) + 2y = 6 ⇒ \(\frac{x}{7}\) = 6 – 2y Thus, x = 7(6 – 2y).

Question 3. If we multiply or divide both sides of a linear equation with a non-zero number, then what will happen to the solution of the linear equation ? Solution: Solution remains the same.

Question 4. Find the value of k for which x = 0, y = 8 is a solution of 3x – 6y = k. Solution: Since x = 0 and y = 8 is a solution of given equation 3x – 6y = k 3(0) – 6(8) = k ⇒ k = – 48

Question 5. Write the equation of a line which is parallel to x-axis and is at a distance of 2 units from the origin. Solution: Here, required line is parallel to x-axis and at a distance of 2 units from the origin. ∴ Its equation is y + 2 = 0 or y – 2 = 0

Question 6. Find ‘a’, if linear equation 3x – ay = 6 has one solution as (4, 3). Solution: Since (4, 3) is a solution of given equation. ∴ 3(4) – a(3) = 6 ⇒ 12 – 3a = 6 ⇒ a = \(\frac{-6}{-3}\) Hence, a = 2

Question 7. Cost of a pen is two and half times the cost of a pencil. Express this situation as a linear equation in two variables. Solution: Let cost of a pen be ₹ x and cost of a pencil be ₹ y. According to statement of the question, we have x = 2\(\frac{1}{2}\) y ⇒ 2x = 5y or 2x – 5y = 0

Question 8. In an one day international cricket match, Raina and Dhoni together scored 198 runs. Express the statement as a linear equation in two variables. Solution: Let runs scored by Raina be x and runs scored by Dhoni be y. According to statement of the question, we have x + y = 198 x + y – 198 = 0

Question 9. The cost of a table is 100 more than half the cost of a chair. Write this statement as a linear equation in two variables. Solution: Let the cost price of a table be ₹ x and that of a chair be ₹ y. Since the cost price of a table is 100 more than half the cost price of a chair. ∴x = \(\frac{1}{2}\)y + 100 ⇒ 2x = y + 200 or 2x – y – 200 = 0.

Linear Equations for Two Variables Class 9 Extra Questions Short Answer Type 1

Question 1. Write linear equation representing a line which is parallel to y-axis and is at a distance of 2 units on the left side of y-axis. Solution: Here, required equation is parallel to y-axis at a distance of 2 units on the left side of y-axis. x = -2 or x + 2 = 0

Question 2. In some countries temperature is measured in Fahrenheit, whereas in countries like India it is measured in Celsius. Here is a linear equation that converts Fahrenheit to Celsius : F = \(\left(\frac{9}{5}\right)\)C + 32° If the temperature is – 40°C, then what is the temperature in Fahrenheit? Solution: Given linear equation is F = \(\left(\frac{9}{5}\right)\)C + 32° Put C = -40°, we have F = \(\frac{9}{5}\)(-40°) + 32° F = – 72° + 32° F= -40°

Question 3. Give equation of two lines on same plane which are intersecting at the point (2, 3). Solution: Since there are infinite lines passing through the point (2, 3). Let, first equation is x + y = 5 and second equation is 2x + 3y = 13. Clearly, the lines represented by both equations intersect at the point (2, 3).

Question 6. Let y varies directly as x. If y = 12 when x = 4, then write a linear equation. What is the value of y, when x = 5 ? Solution: Given y varies directly as x implies y = kx But y = 12 for x = 4 ⇒ 4k = 12 = k = 3 Put k = 3 in y = kx, we have y = 3x Now, when x = 5, y = 3 x 5 = y = 15 …(i)

Linear Equations for Two Variables Class 9 Extra Questions Short Answer Type 2

Question 1. A fraction becomes , when 2 is subtracted from the numerator and 3 is added to the denominator. Represent this situation as a linear equation in two variables. Also, find two solutions for this. Solution: Let numerator and denominator of the given fraction be respectively x and y. According to the statement, we obtain \(\frac{x-2}{y+3}\) = \(\frac{1}{4}\) ⇒ 4x – 8 = y + 3 ⇒ 4x – y – 11 = 0 Which is the required linear equation. When y = 1, then x = 3. When y = 5, then x = 4. Hence, the two solutions are (3, 1) and (4, 5).

Question 2. The path of an aeroplane is given by the equation 3x – 4y = 1 2: Represent the path graphically. Also, show that the point (-4,-6) lies on the graph. Solution: Given equation is 3x – 4y = 12 ∴ y = \(\frac{3 x-12}{4}\).

Linear Equations for Two Variables Class 9 Extra Questions Long Answer Type

Linear Equations for Two Variables Class 9 Extra Questions HOTS

Question 2. A and B are friends. A is elder to B by 5 years. B’s sister C is half the age of B while A’s father D is 8 years older than twice the age of B. If the present age of D is 48 years, find the present ages of A, B and C. Solution: Let the age of B’s sister i.e., C be x years. ∴Age of B be 2x years, age of A be (2x + 5) years. And age of A’s father i.e., D be 2(2x) + 8 = 4x + 8 years According to the statement of the question, we have 4x + 8 = 48 ⇒ 4x = 48 – 8 = 40 ⇒ x = \(\frac{40}{4}\) = 10 Age of A = (2x + 5) years i.e., (2 × 10 + 5) years = 25 years Age of B = 2x years = 2 × 10 years = 20 years and Age of C = \(\frac{1}{2}\) of B’s age = \(\frac{1}{2}\) × 20 years = 10 years

Linear Equations for Two Variables Class 9 Extra Questions Value Based (VBQs)

Question 2. In an election, a good candidate may lose because 40% of voters do not cast their votes due to various reasons. Form an equation and draw the graph with data. From the graph, find : (i) The total number of voters, if 720 voters cast their votes. (ii) The number of votes cast, if the total number of voters are 1000. (iii) What message did you get from above information ? Solution: (i) We have, total number of voters who do not cast their votes = 40% ⇒ Total number of voters who cast their votes = 60% Let the total number of voters be x and number of voters cast their votes be y

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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.

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Chapter Name:	Chapter 4 - Linear Equations in Two Variables
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Academic Year:	2024-25
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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:

$\text{x}$	$0$	$2$	$4$
$\text{y}$	$4$	$2$	$0$

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)

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

$\text{x}$	$0$	$2$	$4$
$\text{y}$	$-2$	$0$	$2$

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:

$\text{x}$	$0$	$2$	$-2$
$\text{y}$	$0$	$6$	$-6$

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:

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

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:

$\text{x}$	$0$	$1$	$-1$
$\text{y}$	$3$	$8$	$-2$

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. 

$\text{x}$	$0$	$1$	$-1$
$\text{y}$	$0$	$5$	$-5$

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. 

$\text{x}$	$0$	$50$	$100$
$\text{y}$	$100$	$50$	$0$

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.

$\text{C}$	$0$	$-40$	$10$
$\text{F}$	$32$	$-40$	$50$

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

NCERT Solutions for Class 6, 7, 8, 9, 10, 11 and 12

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

NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.1 are part of NCERT Solutions for Class 9 Maths . Here we have given NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.1.

NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.1

Ex 4.1 Class 9 Maths   Question 1. The cost of a notebook is twice the cost of a pen. Write a linear equation in two variables to represent this statement. (Take the cost of a notebook to be Rs. x and that of a pen to be Rs.y). Solution: Let the cost of a notebook = Rs. x and the cost of a pen = Rs. y According to the condition, we have [Cost of a notebook] =2 x [Cost of a pen] i. e„ (x) = 2 x (y) or, x = 2y or, x – 2y = 0 Thus, the required linear equation is x – 2y = 0.

Ex 4.1 Class 9 Maths   Question 2 Express the following linear equations in the form ax + by + c = 0 and indicate the values of a, b and c in each case: (i) 2x + 3y = \(9.3\overline { 5 }\) (ii) \(x-\frac { y }{ 5 } -10\quad =\quad 0\) (iii) – 2x + 3y = 6 (iv) x = 3y (v) 2x = -5y (vi) 3x + 2 = 0 (vii) y – 2 = 0 (viii) 5 = 2x Solution: (i) We have 2x + 3y = \(9.3\overline { 5 }\) or (2)x + (3)y + (\(-9.3\overline { 5 }\) ) = 0 Comparing it with ax + by +c= 0, we geta = 2, b = 3 and c= –\(9.3\overline { 5 }\) .

(ii) We have \(x-\frac { y }{ 5 } -10\quad =\quad 0\) or x + (- \(\frac { 1 }{ 5 }\)) y + (10) = 0 Comparing it with ax + by + c = 0, we get a =1, b =- \(\frac { 1 }{ 5 }\) and c= -10

(iii) Wehave -2x + 3y = 6 or (-2)x + (3)y + (-6) = 0 Comparing it with ax – 4 – by + c = 0,we get a = -2, b = 3 and c = -6.

(iv) We have x = 3y or (1)x + (-3)y + (0) = 0 Comparing it with ax + by + c = 0, we get a = 1, b = -3 and c = 0. (v) We have 2x = -5y or (2)x + (5)y + (0) = 0 Comparing it with ax + by + c = 0, we get a = 2, b = 5 and c = 0. (vi) We have 3x + 2 = 0 or (3)x + (0)y + (2) = 0 Comparing it with ax + by + c = 0, we get a = 3, b = 0 and c = 2. (vii) We have y – 2 = 0 or (0)x + (1)y + (-2) = 0 Comparing it with ax + by + c = 0, we get a = 0, b = 1 and c = -2. (viii) We have 5 = 2x ⇒ 5 – 2x = 0 or -2x + 0y + 5 = 0 or (-2)x + (0)y + (5) = 0 Comparing it with ax + by + c = 0, we get a = -2, b = 0 and c = 5.

NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables (दो चरों में रैखिक समीकरण) (Hindi Medium) Ex 4.1

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

Question 1 Which one of the following options is true, and why? y = 3x + 5 has (i) a unique solution, (ii) only two solutions, (iii) infinitely many solutions Solution: Option (iii) is true because for every value of x, we get a corresponding value of y and vice-versa in the given equation. Hence, given linear equation has an infinitely many solutions.

Question 2 Write four solutions for each of the following equations: (i) 2x + y = 7 (ii) πx + y = 9 (iii) x = 4y Solution: (i) 2x + y = 7 When x = 0, 2(0) + y = 7 ⇒ y = 7 ∴ Solution is (0, 7) When x =1, 2(1) + y = 7 ⇒ y = 7 – 2 ⇒ y = 5 ∴ Solution is (1, 5) When x = 2, 2(2) + y =7y = 7 – 4 ⇒ y = 3 ∴ Solution is (2, 3) When x = 3, 2(3) + y = 7y = 7 – 6 ⇒ y = 1 ∴ Solution is (3, 1).

(ii) πx + y = 9 When x = 0, π(0) + y = 9 ⇒ y = 9 – 0 ⇒ y = 9 ∴ Solution is (0, 9) When x = 1, π(1) + y = 9 ⇒ y = 9 – π ∴ Solution is (1, (9 – π)) When x = 2, π(2) + y = 9 ⇒ y = 9 – 2π ∴ Solution is (2, (9 – 2π)) When x = -1,π(-1) + y = 9 ⇒ y = 9 + π ∴ Solution is (-1, (9 + π))

(iii) x = 4y When x = 0, 4y = 1 ⇒ y = 0 ∴ Solution is (0, 0) When x = 1, 4y = 1 ⇒ y = \(\frac { 1 }{ 4 }\) ∴ Solution is (1,\(\frac { 1 }{ 4 }\) ) When x = 4, 4y = 4 ⇒ y = 1 ∴ Solution is (4, 1) When x = 4, 4y = 4 ⇒ y = -1 ∴ Solution is (-4, -1)

Question 3 Check which of the following are solutions of the equation x – 2y = 4 and which are not: (i) (0,2) (ii) (2,0) (iii) (4, 0) (iv) (√2, 4√2) (v) (1, 1) Solution: (i) (0,2) means x = 0 and y = 2 Puffing x = 0 and y = 2 in x – 2y = 4, we get L.H.S. = 0 – 2(2) = -4. But R.H.S. = 4 ∴ L.H.S. ≠ R.H.S. ∴ x =0, y =2 is not a solution.

(ii) (2, 0) means x = 2 and y = 0 Putting x = 2 and y = 0 in x – 2y = 4, we get L.H:S. 2 – 2(0) = 2 – 0 = 2. But R.H.S. = 4 ∴ L.H.S. ≠ R.H.S. ∴ (2,0) is not a solution.

(iii) (4, 0) means x = 4 and y = 0 Putting x = 4 and y = o in x – 2y = 4, we get L.H.S. = 4 – 2(0) = 4 – 0 = 4 =R.H.S. ∴ L.H.S. = R.H.S. ∴ (4, 0) is a solution.

(iv) (√2, 4√2) means x = √2 and y = 4√2 Putting x = √2 and y = 4√2 in x – 2y = 4, we get L.H.S. = √2 – 2(4√2) = √2 – 8√2 = -7√2 But R.H.S. = 4 ∴ L.H.S. ≠ R.H.S. ∴ (√2 , 4√2) is not a solution.

(v) (1, 1)means x =1 and y = 1 Putting x = 1 and y = 1 in x – 2y = 4, we get LH.S. = 1 – 2(1) = 1 – 2 = -1. But R.H.S = 4 ∴ LH.S. ≠ R.H.S. ∴ (1, 1) is not a solution.

Question 4 Find the value of k, if x = 2, y = 1 ¡s a solution of the equation 2x + 3y = k. Solution: We have 2x + 3y = k putting x = 2 and y = 1 in 2x+3y = k,we get 2(2) + 3(1) ⇒ k = 4 + 3 – k ⇒ 7 = k Thus, the required value of k is 7.

NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.3

Question 2 Give the equations of two lines passing through (2, 14). How many more such lines are there, and why? Solution: (2, 14) means x = 2 and y = 14 Equations which have (2,14) as the solution are (i) x + y = 16, (ii) 7x – y = 0 There are infinite number of lines which passes through the point (2, 14), because infinite number of lines can be drawn through a point.

Question 3 If the point (3, 4) lies on the graph of the equation 3y = ax + 7, find the value of a. Solution: The equation of the given line is 3y = ax + 7 ∵ (3, 4) lies on the given line. ∴ It must satisfy the equation 3y = ax + 7 We have, (3, 4) ⇒ x = 3 and y = 4. Putting these values in given equation, we get 3 x 4 = a x 3 + 7 ⇒ 12 = 3a + 7 ⇒ 3a = 12 – 7 = 5 ⇒ a = \(\frac { 5 }{ 3 }\) Thus, the required value of a is \(\frac { 5 }{ 3 }\)

Question 5 From the choices given below, choose the equation whose graphs are given ¡n Fig. (1) and Fig. (2). For Fig. (1) (i) y = x (ii) x + y = 0 (iii) y = 2x (iv) 2 + 3y = 7x

(ii) Distance travelled = 0 unit i.e., x = 0 ∴ If x = 0 ⇒ y = 5(0) – 0 ⇒ Work done = 0 unit.

(ii) From the graph, we have 86°F corresponds to 30°C. (iii) From the graph, we have 95°F corresponds 35°C. (iv) We have, C = 0 From (1), we get F = (\(\frac { 9 }{ 5 }\))0 + 32 = 32 Also, F = 0 From (1), we get 0 = (\(\frac { 9 }{ 5 }\))C + 32 ⇒ \(\frac { -32\times 5 }{ 9 } \) = C ⇒ C = -17.8 (V) When F = C (numerically) From (1), we get F = \(\frac { 9 }{ 5 }\)F + 32 ⇒ F – \(\frac { 9 }{ 5 }\)F = 32 ⇒ \(\frac { -4 }{ 5 }\)F = 32 ⇒ F = -40 ∴ Temperature is – 40° both in F and C.

NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.4

NCERT Solutions for Class 9 Maths

• Chapter 1 Number systems
• Chapter 2 Polynomials
• Chapter 3 Coordinate Geometry
• Chapter 4 Linear Equations in Two Variables
• Chapter 5 Introduction to Euclid Geometry
• Chapter 6 Lines and Angles
• Chapter 7 Triangles
• Chapter 8 Quadrilaterals
• Chapter 9 Areas of Parallelograms and Triangles
• Chapter 10 Circles
• Chapter 11 Constructions
• Chapter 12 Heron’s Formula
• Chapter 13 Surface Areas and Volumes
• Chapter 14 Statistics
• Chapter 15 Probability
• Class 9 Maths (Download PDF)

We hope the NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.1, help you. If you have any query regarding NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.1, drop a comment below and we will get back to you at the earliest.

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Test: Linear Equations in Two Variables- Case Based Type Questions- 1 - EmSAT Achieve MCQ

10 questions mcq test - test: linear equations in two variables- case based type questions- 1.

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Prime Minister ’s National Relief Fund (also called PMNRF in short) is the fund raised to provide support for people affected by natural and manmade disasters. Natural disasters that are covered under this include flood, cyclone, earthquake etc. Man-made disasters that are included are major accidents, acid attacks, riots, etc. Two friends Sita and Gita, together contributed ₹ 200 towards Prime Minister’s Relief Fund. Answer the following : Q. How to represent the above situation in linear equations in two variables?

• A. 2x + y = 200
• B. x + y = 200
• C. 200x = y
• D. 200 + x = y

Here, x represents Sita’s contribution and y represents Gita’s contribution.

Prime Minister ’s National Relief Fund (also called PMNRF in short) is the fund raised to provide support for people affected by natural and manmade disasters. Natural disasters that are covered under this include flood, cyclone, earthquake etc. Man-made disasters that are included are major accidents, acid attacks, riots, etc. Two friends Sita and Gita, together contributed ₹ 200 towards Prime Minister’s Relief Fund. Answer the following : Q. If both contributed equally, then how much is contributed by each?

• A. ₹ 50, ₹ 150
• B. ₹ 100, ₹ 100
• C. ₹ 50, ₹ 50
• D. ₹ 120, ₹ 120

Thus, each contributed is ₹ 100.

Prime Minister ’s National Relief Fund (also called PMNRF in short) is the fund raised to provide support for people affected by natural and manmade disasters. Natural disasters that are covered under this include flood, cyclone, earthquake etc. Man-made disasters that are included are major accidents, acid attacks, riots, etc. Two friends Sita and Gita, together contributed ₹ 200 towards Prime Minister’s Relief Fund. Answer the following : Q. Which out of the following is not the linear equation in two variables ?

4 = 5x – 4y

x 2 + x = 1

x − √2y = 3

x 2 + x = 1 is not linear as highest power is 2. Also, it is an equation in one variable.

Thus, it is not a linear equation in two variables.

Prime Minister ’s National Relief Fund (also called PMNRF in short) is the fund raised to provide support for people affected by natural and manmade disasters. Natural disasters that are covered under this include flood, cyclone, earthquake etc. Man-made disasters that are included are major accidents, acid attacks, riots, etc.

Two friends Sita and Gita, together contributed ₹ 200 towards Prime Minister’s Relief Fund. Answer the following :

Q. If Sita contributed ₹ 76, then how much was contributed by Gita ?

y = 200 – 76

Q. Which is the standard form of linear equation x = – 5 ?

• A. x + 5 = 0
• B. 1.x – 5 = 0
• C. x + 0.y + 5 = 0
• D. 1.x + 0.y = 5

Since, x = 5

⇒ x + 5 = 0

Thus, standard form of x = – 5 is

1.x + 0.y + 5 = 0.

Rainwater harvesting system is a technology that collects and stores rainwater for human use.

Anup decided to do rainwater harvesting. He collected rainwater in the underground tank at the rate of 30 cm 3 /sec.

Q. What is the type of solution of the equation formed?

• A. a unique solution
• B. only two solutions
• C. no solution
• D. infinitely many solutions

Q. How much water will be collected in 60 sec ?

• A. 1500 cm 3
• C. 1800 cm 3

If x = 60, then, y = 30 × 60

Required volume is 1800 cm 3

Q. What will be the equation formed if volume of water collected in x seconds is taken as y cm 3 ?

• C. 30 – x = y
• D. 30 + y = x

Amount of stored water = y cubic cm

Water stored per second = 30 cubic cm

Water stored in ‘x’ second = 30x

Amount of stored water = water in 'x' second

Q. Write the equation in standard form.

30x – y + 0 = 0

30x + y + 0 = 0

30x – y – 0 = 0

30x – y = 0

Rainwater is collected in the  underground tank at the rate of 30 cm³ / sec

Water collected in 1 sec  = 30 cm³

Water collected in x secs = 30x  cm³

Water collected in x secs =  y cm³

=> y = 30x

a linear equation y = 30x

30x - y = 0 ,30x -y +0=0

is standard form

Q. How much time will it take to collect water in 900 cm 3 ?

Since y = 30x

If y = 900, then, 900 = 30x

X = 900 / 30 = 30

Required time is 30 sec.

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Important Questions for Linear Equations in Two Variables- Case Based Type Questions- 1

Linear equations in two variables- case based type questions- 1 mcqs with answers, online tests for linear equations in two variables- case based type questions- 1.

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Unit 1: Algebra foundations

Unit 2: solving equations & inequalities, unit 3: working with units, unit 4: linear equations & graphs, unit 5: forms of linear equations, unit 6: systems of equations, unit 7: inequalities (systems & graphs), unit 8: functions, unit 9: sequences, unit 10: absolute value & piecewise functions, unit 11: exponents & radicals, unit 12: exponential growth & decay, unit 13: quadratics: multiplying & factoring, unit 14: quadratic functions & equations, unit 15: irrational numbers, unit 16: creativity in algebra.

• 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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