NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables
The solutions have important questions and exercises intrinsic to the chapter that include definitions, operations, and theorems. Each solution is well-explained and easy to comprehend; thus, it will help students to understand concepts given in the chapter.
Access Answers to NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables
Students can access the NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables. Curated by experts according to the CBSE syllabus for 2023–2024, these step-by-step solutions make Maths much easier to understand and learn for the students. These solutions can be used in practice by students to attain skills in solving problems, reinforce important learning objectives, and be well-prepared for tests.
Exercise 4.1
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).
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.
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 =
(ii)
(iii) – 2x + 3y = 6
(iv) x = 3y
(v) 2x = -5y
(vi) 3x + 2 = 0
(vii) y – 2 = 0
(viii) 5 = 2x
(i) We have 2x + 3y =
or (2)x + (3)y + (
Comparing it with ax + by +c= 0, we geta = 2,
b = 3 and c= –
(ii) We have
or x + (- 
Comparing it with ax + by + c = 0, we get
a =1, b =- 
(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.
Exercise 4.2
Question 1. Which one of the following options is true, and why?
(i) a unique solution,
(ii) only two solutions,
(iii) infinitely many solutions
We need to the number of solutions of the linear equation
We know that any linear equation has infinitely many solutions.
Justification:
If 
If 
If 
Similarly, we can find infinite many solutions by putting the values of
Write four solutions for each of the following equations:
(i)
(ii)
(iii)
We know that any linear equation has infinitely many solutions.
Let us putin the linear equation
Thus, we get first pair of solution as
Let us put
Thus, we get second pair of solution as
Let us put
Thus, we get third pair of solution as
Let us put
Thus, we get fourth pair of solution as
Therefore, we can conclude that four solutions for the linear equation
(ii)
We know that any linear equation has infinitely many solutions.
Let us put
Thus, we get first pair of solution as
Let us put
Thus, we get second pair of solution as
Let us put
Thus, we get third pair of solution as
Let us put
Thus, we get fourth pair of solution as
Therefore, we can conclude that four solutions for the linear equation
(iii)
We know that any linear equation has infinitely many solutions.
Let us put
Thus, we get first pair of solution as
Let us put
Thus, we get second pair of solution as
Let us put
Thus, we get third pair of solution as
Let us put
Thus, we get fourth pair of solution as
Therefore, we can conclude that four solutions for the linear equation
Check which of the following are solutions of the equation
(i)
(ii)
(iii)
(iv)
(v)
(i)
We need to put
Therefore, we can conclude that
(ii)
We need to put
Therefore, we can conclude that
(iii)
We need to put
Therefore, we can conclude that
(iv)
We need to put
Therefore, we can conclude that
(v)
We need to put
Therefore, we can conclude that
Find the value ofk, if x = 2, y = 1 is a solution of the equation
We know that, if 
Therefore, we can conclude that the value of k, for which the linear equation 
Exercise 4.3
Question 1. 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
(i) x + y = 4 ⇒ y = 4 – x
If we have x = 0, then y = 4 – 0 = 4
x = 1, then y =4 – 1 = 3
x = 2, then y = 4 – 2 = 2
∴ We get the following table:
Plot the ordered pairs (0, 4), (1,3) and (2,2) on the graph paper. Joining these points, we get a straight line AB as shown.
Thus, the line AB is the required graph of x + y = 4
(ii) x – y = 2 ⇒ y = x – 2
If we have x = 0, then y = 0 – 2 = -2
x = 1, then y = 1 – 2 = -1
x = 2, then y = 2 – 2 = 0
∴ We get the following table:
Plot the ordered pairs (0, -2), (1, -1) and (2, 0) on the graph paper. Joining these points, we get a straight line PQ as shown.
Thus, the ime is the required graph of x – y = 2
(iii) y = 3x
If we have x = 0,
then y = 3(0) ⇒ y = 0
x = 1, then y = 3(1) = 3
x= -1, then y = 3(-1) = -3
∴ We get the following table:
Plot the ordered pairs (0, 0), (1, 3) and (-1, -3) on the graph paper. Joining these points, we get a straight line LM as shown.
Thus, the line LM is the required graph of y = 3x.
(iv) 3 = 2x + y ⇒ y = 3 – 2x
If we have x = 0, then y = 3 – 2(0) = 3
x = 1,then y = 3 – 2(1) = 3 – 2 = 1
x = 2,then y = 3 – 2(2) = 3 – 4 = -1
∴ We get the following table:
Plot the ordered pairs (0, 3), (1, 1) and (2, – 1) on the graph paper. Joining these points, we get a straight line CD as shown.
Thus, the line CD is the required graph of 3 = 2x + y.
Give the equations of two lines passing through (2, 14). How many more such lines are there, and why?
(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.
If the point (3, 4) lies on the graph of the equation 3y = ax + 7, find the value of a.
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 =
Thus, the required value of a is 5/3.
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.
Here, total distance covered = x km and total taxi fare = Rs. y
Fare for 1km = Rs. 8
Remaining distance = (x – 1) km
∴ Fare for (x – 1)km = Rs.5 x(x – 1)
Total taxi fare = Rs. 8 + Rs. 5(x – 1)
According to question,
y = 8 + 5(x – 1) = y = 8 + 5x – 5
⇒ y = 5x + 3,
which is the required linear equation representing the given information.
Graph: We have y = 5x + 3
Wben x = 0, then y = 5(0) + 3 ⇒ y = 3
x = -1, then y = 5(-1) + 3 ⇒ y = -2
x = -2, then y = 5(-2) + 3 ⇒ y = -7
∴ We get the following table:
Now, plotting the ordered pairs (0, 3), (-1, -2) and (-2, -7) on a graph paper and joining them, we get a straight line PQ as shown.
Thus, the line PQ is the required graph of the linear equation y = 5x + 3.
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
For Fig. (2)
(i) y = x + 2
(ii) y = x – 2
(iii) y = -x + 2
(iv) x + 2y = 6
For Fig. (1), the correct linear equation is x + y = 0
[As (-1, 1) = -1 + 1 = 0 and (1,-1) = 1 + (-1) = 0]
For Fig.(2), the correct linear equation is y = -x + 2
[As(-1,3) 3 = -1(-1) + 2 = 3 = 3 and (0,2)
⇒ 2 = -(0) + 2 ⇒ 2 = 2]
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
Constant force is 5 units.
Let the distance travelled = x units and work done = y units.
Work done = Force x Distance
⇒ y = 5 x x ⇒ y = 5x
For drawing the graph, we have y = 5x
When x = 0, then y = 5(0) = 0
x = 1, then y = 5(1) = 5
x = -1, then y = 5(-1) = -5
∴ We get the following table:
Ploffing the ordered pairs (0, 0), (1, 5) and (-1, -5) on the graph paper and joining the points, we get a straight line AB as shown.
From the graph, we get
(i) Distance travelled =2 units i.e., x = 2
∴ If x = 2, then y = 5(2) = 10
⇒ Work done = 10 units.
(ii) Distance travelled = 0 unit i.e., x = 0
∴ If x = 0 ⇒ y = 5(0) – 0
⇒ Work done = 0 unit.
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.xand Rs.y.) Draw the graph of the same.
Let the contribution of Yamini = Rs. x
and the contribution of Fatima Rs. y
∴ We have x + y = 100 ⇒ y = 100 – x
Now, when x = 0, y = 100 – 0 = 100
x = 50, y = 100 – 50 = 50
x = 100, y = 100 – 100 = 0
∴ We get the following table:
Plotting the ordered pairs (0,100), (50,50) and (100, 0) on a graph paper using proper scale and joining these points, we get a straight line PQ as shown.
Thus, the line PQ is the required graph of the linear equation x + y = 100.
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 = (
(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.
(i) We have
F = (
When C = 0 , F = (
When C = 15, F = (
When C = -10, F = 
We have the following table:
Plotting the ordered pairs (0, 32), (-15, 5) and (-10,14) on a graph paper. Joining these points, we get a straight line AB.
(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 = (
Also, F = 0
From (1), we get
0 = (
(V) When F = C (numerically)
From (1), we get
F = 
⇒ 
∴ Temperature is – 40° both in F and C.
Exercise 4.4
Give the geometric representation of
(i) In one variable
(ii) In two variables
(i) We need to represent the linear equation
We can conclude that in one variable, the geometric representation of the linear equation
Given below is the representation of number 3 on the number line.
(ii) We need to represent the linear equation
We know that the linear equation
We can conclude that in two variables, the geometric representation of the linear equation
Given below is the representation of the linear equation
We can optionally consider the given below table for plotting the linear equation
|
X |
1 |
0 |
|
y |
3 |
3 |
Give the geometric representations of 
(i) In one variable
(ii) In two variables
(i) We need to represent the linear equation
We know that the linear equation
We can conclude that in one variable, the geometric representation of the linear equation
Given below is the representation of number 3 on the number line.
(ii) We need to represent the linear equation
We know that the linear equation
We can conclude that in two variables, the geometric representation of the linear equation
Given below is the representation of the linear equation
We can optionally consider the given below table for plotting the linear equation
|
X |
1 |
0 |
|
y |
4.5 |
4.5 |
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Frequently Asked Questions
The NCERT solution for Class 9 Chapter 4: Linear Equations in Two Variables is important as it provides a structured approach to learning, ensuring that students develop a strong understanding of foundational concepts early in their academic journey. By mastering these basics, students can build confidence and readiness for tackling more difficult concepts in their further education.
Yes, the NCERT solution for Class 9 Chapter 4: Linear Equations in Two Variables is quite useful for students in preparing for their exams. The solutions are simple, clear, and concise allowing students to understand them better. They can solve the practice questions and exercises that allow them to get exam-ready in no time.
You can get all the NCERT solutions for Class 9 Maths Chapter 4 from the official website of the Orchids International School. These solutions are tailored by subject matter experts and are very easy to understand.
Yes, students must practice all the questions provided in the NCERT solution for Class 9 Maths Chapter 4: Linear Equations in Two Variables as it will help them gain a comprehensive understanding of the concept, identify their weak areas, and strengthen their preparation.
Students can utilize the NCERT solution for Class 9 Maths Chapter 4 effectively by practicing the solutions regularly. Solve the exercises and practice questions given in the solution.