# NCERT Solutions for Class 9 Mathematics Chapter 4 – Linear equations in two variables

Mathematics is an important subject for all competitive examinations. The subject cannot be mugged up and hence, needs a detailed approach to every problem.  Students must study this subject by understanding the concepts well. NCERT Solutions Class 9 Mathematics Chapter 4 provides accurate answers to the NCERT questions of the chapter which  gives you in-depth knowledge about the subject and concepts to be studied.

The Linear chapter equations in two variables cover the following topics; solutions to linear equations, graphs of linear equations in two variables, and equations of the line passing through the X and Y axis. It is directly connected with the Linear chapter equations in one variable covered in NCERT Class 8 Mathematics. Hence, students are advised to revise the concepts given in the NCERT Class 8 textbook before reading this chapter and brush up on their concepts.

You can find quick revision points along with the core concepts required from previous classes in our NCERT Solutions for Class 9 Mathematics Chapter 4. This would help you give a clear-cut idea of what terminologies you have to keep in mind while reading this chapter. Apart from this, it also includes highlight points of all the topics and sub-topics from NCERT Class 9 Mathematics textbook, which will help in building the strong concept of the chapter.

Extramarks’ website is a complete source of information for the students studying in Class 1st to Class 12th. It has NCERT study materials, NCERT revision notes, NCERT textbooks, NCERT Exemplar and Past years’ papers, making it the most useful platform online for the teachers and students.

## Key Topics Covered In NCERT Solutions Class 9 Mathematics Chapter 4

Students can understand Linear equations in two variables in a proper way only if they have a proper understanding of Linear equations in one variable already covered in NCERT Class 8 Mathematics textbook. Hence, students are advised to revisit  the important points given in NCERT Class 8 Mathematics to quickly recall the concepts. The chapter strongly requires students to use their Mathematical solving skills. It also requires one to be comfortable with solving the equations.

We cover everything required to be noted in this chapter in NCERT Solutions for Class 9 Mathematics Chapter 4, available on the Extramarks website. Extramarks website is the right source to rely upon for your preparation as it includes all the learning material required to score well and upgrade your performance in your academics.

We have included the following key topics in our NCERT Solutions for Class 9 Mathematics Chapter 4:

 Exercise Topic 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 X-axis and Y-axis

Once you complete Chapter 4 Mathematics Class 9, you will be capable of solving some of the important topics of NCERT Class 11 and NCERT Class 12 Mathematics. As a result, you will start developing an interest in Algebra and look for in-depth knowledge of the topics.

4.1 Introduction

The chapter begins with the general introduction of Linear equations in one variable covered in NCERT Class 8 Mathematics. Further, it states its interrelation with Linear equations in two variables. The chapter is included in NCERT Class 9 Mathematics.

You will find the definition of linear equations in one and two variables, along with examples in our NCERT Solutions for Class 9 Mathematics Chapter 4.

4.2 Linear Equations

An equation is a combination of a constant and variables under equality. When the equation includes its variables in their simplest form, like x,y,z etc., it is called a linear equation.

For example, 7x+5=6 is a linear equation.

When a linear equation has two variables instead of one, it is called linear equations in two variables.

For example, 3y-2z=9 is a linear equation in two variables

### Steps to note while solving an equation:

• If a number is added to or subtracted from LHS, the same number will be added to or subtracted from RHS
• Similarly, if a number is multiplied to or divided from LHS, the same number will be multiplied to or divided from RHS
• You can solve the linear equations in two variables only and only if you have two given equations

4.3 Solution of a Linear Equation

You have already studied that a linear equation in one variable has a unique solution. Does this hold true for linear equations in two variables too? Let us explore this in this section.

As we know, a linear equation in two variables has two variables. There are high chances of finding more than one solution.

For example: Consider the equation 2x+3y = 12.

The solution for this equation is x=3 and y=2 as it satisfies RHS. It is written in the ordered pair as (3,2). Similarly, the other solutions for this equation are (0,4) and (6,0).

Hence, it is concluded that a linear equation in two variables may or may not have a unique solution.

You can practice more such examples from our NCERT Solutions for Class 9 Mathematics Chapter 4.

4.4 Graph of linear equations in two variables

In the last section, you have already learnt how to find different sets of solutions for linear equations in two variables. In this section, we will learn how to plot the obtained solutions on a graph.

Just like Coordinate Geometry, you have to draw a Cartesian system on the graph, and then by locating the different values of solutions obtained, you have to plot points by drawing proper lines.

4.5 Equations of lines parallel to X-axis and Y-axis

When a linear equation in two variables is written in the form y=0, it is called an equation of a line parallel to the X-axis. Likewise, when a linear equation in two variables is written in the form x=0, it is called an equation of a line parallel to the Y-axis.

Get the complete chapter study material along with NCERT Solutions for Class 9 Mathematics Chapter 4 on the Extramarks website today and see your child clear every examination with flying colours.

### NCERT Solutions for Class 9 Mathematics Chapter 4 Exercise &  Solutions

Extramarks website has NCERT Solutions for Class 9 Mathematics Chapter 4, which includes a detailed analysis of the Chapter along with revision notes. This will help you to level up your performance in different stages of preparation and go beyond what is expected as a Class 9th student.

You can find for exercise specific questions and solutions for NCERT Solutions for Class 9 Mathematics Chapter 4 by referring to the following links:

•  Chapter 4: Exercise 4.1 Question and answers
•  Chapter 4: Exercise 4.2 Question and answers
• Chapter 4: Exercise 4.3 Question and answers
• Chapter 4: Exercise 4.4 Question and answers

Along with NCERT Solutions for Class 9 Mathematics Chapter 4, students can explore NCERT Solutions on our Extramarks website for all primary and secondary classes.

• NCERT Solutions Class 1
• NCERT Solutions Class 2
• NCERT Solutions Class 3
• NCERT Solutions Class 4
• NCERT Solutions Class 5
• NCERT Solutions Class 6
• NCERT Solutions Class 7
• NCERT Solutions Class 8
• NCERT Solutions Class 9
• NCERT solutions Class 10
• NCERT solutions Class 11
• NCERT solutions Class 12

#### NCERT Exemplar for Class 9 Mathematics

NCERT Exemplar Class 9 Mathematics book is the collection of various questions required to be good performers in the subject of Mathematics. The different types and ranges of questions covered throughout the book will prove to be a milestone in every aspect of your Mathematics studies.

These books will help to develop an interest in the subject at large. You will also learn to get insights and learn to use various concepts of Mathematics in your day to day life. The benefits provided in this book will make an overall difference in your score in competitive examinations. It covers all the Chapters in-depth, which makes it fruitful for all the students.

After referring to the NCERT Solutions and NCERT Exemplar, the students develop more practical oriented thinking, which is the key aspect of anything you study as it helps in better understanding of the concepts. As a result, students can jump to more advanced and higher-level conceptual questions. By studying from the Exemplar, you can constantly assure yourself to be the next topper.

#### Key Features of NCERT Solutions for Class 9 Mathematics Chapter 4

You can gain momentum in your preparation only and only if all your doubts are cleared. Hence, NCERT Solutions for Class 9 Mathematics Chapter 4 offers a complete solution for all problems. The key features are provided:

• You can find quick referral links to all the exercises questions in NCERT Solutions for Class 9 Mathematics Chapter 4.
• You can solve numerical type as well as Integer type questions with the help of Extramarks NCERT Solutions.
• After completing the NCERT Solutions for Class 9 Mathematics Chapter 4, you will enhance  your problem solving skills and will be able to retain formulas and concepts involved in it for a long time.

Q.1 The cost of a notebook is twice the cost of a pen. Write a linear equation in two variables to represent this statement.The cost of a notebook is twice the cost of a pen. Write a linear equation in two variables to represent this statement.

Let the cost of a notebook and a pen be x and y respectively.
Since, cost of a notebook = 2 x cost of a pen
x = 2y or x – 2y = 0
Which is required linear equation.

$\begin{array}{l}\text{Q.2 Express the following linear equations in the form}\\ \text{ax\hspace{0.17em}+\hspace{0.17em}by\hspace{0.17em}+\hspace{0.17em}c\hspace{0.17em}=\hspace{0.17em}0 and indicate the values of \hspace{0.17em}\hspace{0.17em}a,b\hspace{0.17em}\hspace{0.17em}and c in}\\ \text{each case:}\\ \left(\text{i}\right)\text{\hspace{0.17em}}2\mathrm{x}+3\mathrm{y}=9.3\overline{5}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(\text{ii}\right)\text{\hspace{0.17em}}\mathrm{x}-\frac{\mathrm{y}}{5}-10=0\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(\text{iii}\right)\text{\hspace{0.17em}}-2\mathrm{x}+3\mathrm{y}=6\\ \left(\text{iv}\right)\text{\hspace{0.17em}}\mathrm{x}=3\mathrm{y}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}\left(\text{v}\right)\text{\hspace{0.17em}}2\mathrm{x}=-5\mathrm{y}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(\text{vi}\right)\text{\hspace{0.17em}}3\mathrm{x}+2=0\\ \left(\text{vii}\right)\text{\hspace{0.17em}}\mathrm{y}-2=0\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(\text{viii}\right)\text{\hspace{0.17em}}5=2\mathrm{x}\end{array}$

Ans-

$\begin{array}{l}\left(\text{i}\right)\text{\hspace{0.17em}}2x+3y=9.3\overline{5}\\ \text{or}2x+3y-9.3\overline{5}=0\\ \text{Comparing with ax + by + c = 0, we get}\\ \text{a = 2, b = 3 and c =}-9.3\overline{5}\\ \left(\text{ii}\right)\text{\hspace{0.17em}}x-\frac{y}{5}-10=0\\ \text{Comparing with ax + by + c = 0, we get}\\ \text{a = 1, b =}-\frac{1}{5}\text{and c =}-10\\ \left(\text{iii}\right)\text{\hspace{0.17em}}-2x+3y=6\\ \text{or}\text{\hspace{0.17em}}\text{\hspace{0.17em}}-2x+3y-6=0\\ \text{Comparing with ax + by + c = 0, we get}\\ \text{a =}-2\text{, b =}3\text{and c =}-6.\\ \left(\text{iv}\right)x=3y\\ \text{or}\text{\hspace{0.17em}}\text{\hspace{0.17em}}x-3y=0\\ \text{Comparing with ax + by + c = 0, we get}\\ \text{a =}1\text{, b =}-3\text{and c =}\text{\hspace{0.17em}}0.\\ \left(\text{v}\right)\text{\hspace{0.17em}}2x=-5y\\ \text{or}2x+5y=0\\ \text{Comparing with ax + by + c = 0, we get}\\ \text{a}=2\text{, b}=5\text{and c}=\text{\hspace{0.17em}}0.\\ \left(\text{vi}\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}3x+2=0\text{}\\ \text{or 3x}+0.y+2=0\\ \text{Comparing with ax + by + c = 0, we get}\\ \text{a}=3\text{, b}=0\text{and c}=\text{\hspace{0.17em}}2.\\ \left(\text{vii}\right)y-2=0\\ \text{or 0}\text{.x}+y-2=0\\ \text{Comparing with ax + by + c = 0, we get}\\ \text{a}=0\text{, b}=1\text{and c}=\text{\hspace{0.17em}}-2.\\ \left(\text{viii}\right)5=2x\\ \text{or}\text{\hspace{0.17em}}\text{\hspace{0.17em}}2x+0.y-5=0\\ \text{Comparing with ax + by + c = 0, we get}\\ \text{a}=2\text{, b}=0\text{and c}=\text{\hspace{0.17em}}-5.\end{array}$

Q.3 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

Ans-

Since, y = 3x + 5 is a linear equation in two variables and it has infinite possible solutions. As for every value of x, there will be a value of y satisfying the above equation and vice-versa.

Therefore, the correct option is (iii).

$\begin{array}{l}\text{Check which of the following are solutions of the equation}\\ x-\text{2y = 4 and which are not:}\\ \left(\text{i}\right)\text{\hspace{0.17em}}\left(\text{0,2}\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(\text{ii}\right)\text{\hspace{0.17em}}\left(\text{2,0}\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(\text{iii}\right)\text{\hspace{0.17em}}\left(\text{4,0}\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(\text{iv}\right)\text{\hspace{0.17em}}\left(\sqrt{\text{2}}\text{,}\text{\hspace{0.17em}}\text{4}\sqrt{\text{2}}\right)\\ \left(\text{v}\right)\text{\hspace{0.17em}}\left(\text{1,1}\right)\end{array}$

Ans-

$\begin{array}{l}\left(i\right)\text{\hspace{0.17em}}\left(0,2\right)\\ \text{Putting x}=0\text{and y}=2\text{in L}\text{.H}\text{.S}\text{. of}x-2y=4,\text{​}\text{\hspace{0.17em}}\text{we get}\\ \text{L}\text{.H}\text{.S}\text{.}=x-2y=0-2\left(2\right)=-4\ne \text{R}\text{.H}\text{.S}\\ \text{Therefore,}\left(\text{0,2}\right)\text{is not a solution of the given equation}\text{.}\\ \left(ii\right)\text{\hspace{0.17em}}\left(2,0\right)\\ \text{Putting x}=2\text{and y}=0\text{in L}\text{.H}\text{.S}\text{. of}x-2y=4,\text{​}\text{\hspace{0.17em}}\text{we get}\\ \text{L}\text{.H}\text{.S}\text{.}=x-2y=2-2\left(0\right)=2\ne \text{R}\text{.H}\text{.S}\\ \text{Therefore,}\left(2,0\right)\text{is not a solution of the given equation}\text{.}\\ \left(iii\right)\text{\hspace{0.17em}}\left(4,0\right)\\ \text{Putting x}=4\text{and y}=0\text{in L}\text{.H}\text{.S}\text{. of}x-2y=4,\text{​}\text{\hspace{0.17em}}\text{we get}\\ \text{L}\text{.H}\text{.S}\text{.}=x-2y=4-2\left(0\right)=4=\text{R}\text{.H}\text{.S}\\ \text{Therefore,}\left(\text{4,0}\right)\text{is a solution of the given equation}\text{.}\\ \left(iv\right)\text{\hspace{0.17em}}\left(\sqrt{2},\text{\hspace{0.17em}}4\sqrt{2}\right)\\ \text{Putting x}=\sqrt{2}\text{and y}=4\sqrt{2}\text{in L}\text{.H}\text{.S}\text{. of}x-2y=4,\text{​}\text{\hspace{0.17em}}\text{we get}\\ \text{L}\text{.H}\text{.S}\text{.}=x-2y\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\sqrt{2}-2\left(4\sqrt{2}\right)\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=-7\sqrt{2}\ne R.H.S\\ \text{Therefore,}\left(\sqrt{2},\text{\hspace{0.17em}}4\sqrt{2}\right)\text{is not a solution of the given equation}\text{.}\\ \left(v\right)\text{\hspace{0.17em}}\left(1,1\right)\\ \text{Putting x}=1\text{and y}=1\text{in L}\text{.H}\text{.S}\text{. of}x-2y=4,\text{​}\text{\hspace{0.17em}}\text{we get}\\ \text{L}\text{.H}\text{.S}\text{.}=x-2y=1-2\left(1\right)=-1\ne \text{R}\text{.H}\text{.S}\\ \text{Therefore,}\left(\text{1,1}\right)\text{is not a solution of the given equation}\text{.}\end{array}$

Q.5 Find the value of k, if x = 2, y = 1 is a solution of the equation 2x + 3y = k.

Ans-

Putting x = 2 and y = 1 in 2x + 3y = k, we get
2(2) + 3(1) = k
4 + 3 = k
Or k = 7
Therefore the value of k is 7.

Q.6 Draw the graph of each of the following linear equations in two variables:
(i) x + y = 4
(ii) x – y = 2
(iii) y = 3x
(iv) 3 = 2x + y

Ans-

(i) Graph of x + y = 4 is given below:

(ii) Graph of x – y = 2 is given below:

(iii) Graph of y = 3x is given below:

(iv) Graph of 3 = 2x + y is given below:

Q.7 Give the equations of two lines passing through (2, 14). How many more such lines are there, and why?

Ans-

The point (2, 14) satisfies the equations x + y = 16 and y – x = 12.
Therefore, x + y = 16 and y – x = 12 are the two lines passing through the point (2, 14).
Since, infinite number of lines can pass through one point, therefore, there are infinite lines of such type passing through the given point (2, 14).

Q.8 If the point (3, 4) lies on the graph of the equation 3y = ax + 7, find the value of a.

Ans-

Since, the point (3, 4) lies on the graph of the equation 3y = ax + 7, so it will satisfy the equation.
Putting x = 3 and y = 4 in given equation, we get
3(4) = a(3) + 7
12 = 3a + 7
5 = 3a
a = (5/3)

Q.9 The taxi fare in a city is as follows: For the first kilometre, the fare is Rs 8 and for the subsequent distance it is Rs 5 per km. Taking the distance covered as x km and total fare as Rs y, write a linear equation for this information, and draw its graph.

Ans-

Total covered distance = x km
Fare of first km distance = Rs. 8
Fare of the fare rest distance = (x – 1)5
Total fare of x km distance = 8 + (x – 1)5
y = 3 + 5x
or 5x – y + 3 = 0
which is the linear equation of the given informations.

 x 0 –(3/5) y 3 0

The graph of 5x – y + 3 = 0 is given below:

Q.10 From the choices given below, choose the equation whose graphs are given in Fig. 4.6 and Fig. 4.7.
For Fig. 4. 6 For Fig. 4.7
(i) y = x (i) y = x + 2
(ii) x + y = 0 (ii) y = x – 2
(iii) y = 2x (iii) y = –x + 2
(iv) 2 + 3y = 7x (iv) x + 2y = 6

Ans-

(i) Points on the given line are (−1, 1), (0, 0), and (1, −1).
We see that these points satisfies equation
x + y = 0. So, x + y =0 is the required equation of the given graph.
Hence, the correct answer is (ii) x + y = 0.

(ii) Since, points on the given line are (−1, 3), (0, 2), and (2, 0). We see that these points satisfies equation y = – x + 2. So, y = – x + 2 is the required equation of the given graph.
Therefore, the correct answer (iii) y = – x + 2.

Q.11 Yamini and Fatima, two students of Class IX of a school, together contributed Rs 100 towards the Prime Minister’s Relief Fund to help the earthquake victims. Write a linear equation which satisfies this data. (You may take their contributions as Rs x and Rs y.). Draw the graph of the same.

Ans-

Let money donated by Yamini = Rs. x
And money donated by Fatima = Rs. y
Total contributed money = Rs. 100
Then, x + y = 100
The graph of linear equation x + y = 100 is:

Q.12 Give the geometric representations of y = 3 as an equation
(i) in one variable
(ii) in two variables

Ans-

(i) y = 3 in one variable can be represented as given below:

(ii) y = 3 in two variables can be represented as
0.x + y = 3
So, the solutions are:

 x 0 4 -4 y 3 3

The graphical representation of the line is as given below:

Q.13 Write four solutions for each of the following equations

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

Ans-

$\begin{array}{l}\left(\text{i}\right)\text{2x}+\text{y}=\text{7}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{or}\text{\hspace{0.17em}}\mathrm{y}=7-2\mathrm{x}\\ \mathrm{We}\text{can get 4 solutions of the equation by putting x = 0, 1, 2, 3}\\ \text{y}=7-2\left(0\right)=7\\ \text{y}=7-2\left(1\right)=5\\ \text{y}=7-2\left(2\right)=3\\ \text{y}=7-2\left(3\right)=1\\ \mathrm{Therefore},\text{the four solutions of given equation are:}\\ \left(0,7\right),\text{\hspace{0.17em}}\left(1,5\right),\text{\hspace{0.17em}}\left(2,3\right)\text{and}\left(3,1\right).\\ \left(\text{ii}\right)\mathrm{\pi }\text{x}+\text{y}=\text{9}\\ \text{\hspace{0.17em}y}=9-\mathrm{\pi }\text{x}\\ \mathrm{We}\text{can get 4 solutions of the equation by putting x = 0, 1, 2, 3}\\ \text{\hspace{0.17em}y}=9-\mathrm{\pi }\left(0\right)=9\\ \text{\hspace{0.17em}y}=9-\mathrm{\pi }\left(1\right)=9-\mathrm{\pi }\\ \text{\hspace{0.17em}y}=9-\mathrm{\pi }\left(2\right)=9-2\mathrm{\pi }\\ \text{\hspace{0.17em}y}=9-\mathrm{\pi }\left(3\right)=9-3\mathrm{\pi }\\ \mathrm{Therefore},\text{the four solutions of given equation are:}\\ \left(0,9\right),\text{\hspace{0.17em}}\left(1,9-\mathrm{\pi }\right),\text{\hspace{0.17em}}\left(2,9-2\mathrm{\pi }\right)\text{and}\left(3,9-3\mathrm{\pi }\right).\\ \left(\text{iii}\right)\text{x}=\text{4y}\\ \mathrm{We}\text{can get 4 solutions of the equation by putting y = 0, 1, 2, 3}\\ \text{x}=\text{4}\left(0\right)=0\\ \text{x}=\text{4}\left(1\right)=4\\ \text{x}=\text{4}\left(2\right)=8\\ \text{x}=\text{4}\left(3\right)=12\\ \mathrm{Therefore},\text{the four solutions of given equation are:}\\ \left(0,\text{\hspace{0.17em}\hspace{0.17em}}0\right),\text{\hspace{0.17em}}\left(1,\text{\hspace{0.17em}\hspace{0.17em}}4\right),\text{\hspace{0.17em}}\left(2,\text{\hspace{0.17em}\hspace{0.17em}}8\right)\text{and}\left(3,\text{\hspace{0.17em}\hspace{0.17em}}12\right).\end{array}$

Q.14 If the work done by a body on application of a constant force is directly proportional to the distance travelled by the body, express this in the form of an equation in two variables and draw the graph of the same by taking the constant force as 5 units. Also read from the graph the work done when the distance travelled by the body is
(i) 2 units (ii) 0 unit

Ans-

$\begin{array}{l}\text{Let the distance travelled and the work done by the}\mathrm{}\text{body be x}\\ \text{and y respectively}.\\ \text{Work done}\mathrm{}\text{\hspace{0.17em}}\mathrm{\alpha }\text{\hspace{0.17em}\hspace{0.17em}distance travelled}\\ ⇒\text{y}\mathrm{\alpha }\text{\hspace{0.17em}x}\\ ⇒\text{y}=\text{kx}\mathrm{},\text{where k is constant.}\\ \mathrm{If}\text{contant force is 5 units, then work done}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}y}=5\mathrm{x}\\ \begin{array}{}\end{array}\end{array}$

 x 0 2 –2 y 0 10 –10

(i) From the graphs, it can be observed that the value of y corresponding to x = 2 is 10. This implies that the work done by the body is 10 units when the distance travelled by it is 2 units.

(ii) From the graphs, it can be observed that the value of y corresponding to x = 0 is 0. This implies that the work done by the body is 0 units when the distance travelled by it is 0 unit.

In countries like USA and Canada, 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 = (9/5) C + 32
Q.15 (i) Draw the graph of the linear equation above using Celsius for x-axis and Fahrenheit for y-axis.
(ii) If the temperature is 30°C, what is the temperature in Fahrenheit?
(iii) If the temperature is 95°F, what is the temperature in Celsius?
(iv) If the temperature is 0°C, what is the temperature in Fahrenheit and if the temperature is 0°F, what is the temperature in Celsius?
(v) Is there a temperature which is numerically the same in both Fahrenheit and Celsius? If yes, find it.

Ans-

(i) The given equation is: F = (9/5) C + 32
Let degree Celsius are taken on x-axis and Fahrenheit is taken on y- axis.
Then equation becomes:
y = (9/5) x + 32

 x 0 10 –10 y 32 50 14

The graph of the above equation is constructed as given below:

(ii) Temperature at 30°C
F = (9/5)30°C + 32
= 86
Therefore, temperature in Fahrenheit is 86°F.

(iii) Temperature = 95°F
95 = (9/5) C + 32
63 = (9/5) C
C = 63 × (5/9) = 35°
(iv) F = (9/5) C + 32
If C = 0°C, then
F = (9/5)(0) + 32
= 32
If F = 0, then
0 = (9/5) C + 32
–32 = (9/5) C
C = –32 x (5/9)
=–17.77
Therefore, if F = 0°F, then C = –17.77° C.
(v) F = (9/5) C + 32
If F = C, then
F = (9/5)F + 32
(1 – 9/5)F = 32
–(4/5)F = 32
F = 32 × (–5/4)
= – 40
Yes, there is a temperature, −40°, which is numerically the same in both Fahrenheit and Celsius.

Q.16 Give the geometric representations of 2x + 9 = 0 as an equation
(i) in one variable
(ii) in two variables

Ans-

(i) 2x + 9 = 0
⇒ x = – (9/2)
= – 4.5
2x + 9 = 0 in one variable can be represented as given below:

(ii) 2x + 9 = 0 in two variables can be represented as
2x + 0.y + 9 = 0
So, the solutions are:

 x –4.5 –4.5 –4.5 y 0 4 –4

The graphical representation of the line is as given below: