The National Council of Educational Research and Training (NCERT) is an autonomous body of the Indian government that formulates the curricula for schools in India that are governed by the Central Board of Secondary Education (CBSE) and certain state boards. Therefore, students who will be taking the Class 10 tests administered by various boards should consult this NCERT Syllabus in order to prepare for those examinations, which in turn will assist those students get a passing score.
When working through the exercises in the NCERT textbook, if you run into any type of difficulty or uncertainty, you may use the swc NCERT Solutions for class 9 as a point of reference. While you are reading the theory form textbook, it is imperative that you always have notes prepared. You should make an effort to understand things from the very beginning so that you may create a solid foundation in the topic. Use the NCERT as your parent book to ensure that you have a strong foundation. After you have finished reading the theoretical section of the textbook, you should go to additional reference books.
NCERT Solutions for Class 9 Maths Exercise 4.1
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.
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 = 
(ii) 
(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 =
or (2)x + (3)y + (
) = 0
Comparing it with ax + by +c= 0, we geta = 2,
b = 3 and c= – .
(ii) We have
or x + (- 
) y + (10) = 0
Comparing it with ax + by + c = 0, we get
a =1, b =- 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 Exercise 4.2
Question 1. Which one of the following options is true, and why?
has
(i) a unique solution,
(ii) only two solutions,
(iii) infinitely many solutions
Solution:
We need to the number of solutions of the linear equation.
We know that any linear equation has infinitely many solutions.
Justification:
If 
then 
If 
then 
If 
then 
= -1
Similarly, we can find infinite many solutions by putting the values of 
Question 2. Write four solutions for each of the following equations:
(i) 
(ii) 
(iii) 
Solution:
We know that any linear equation has infinitely many solutions.
Let us putin the linear equation, to get
Thus, we get first pair of solution as
.
Let us put
in the linear equation, to get
Thus, we get second pair of solution as.
Let us putin the linear equation, to get
Thus, we get third pair of solution as
.
Let us put
in the linear equation, to get
Thus, we get fourth pair of solution as.
Therefore, we can conclude that four solutions for the linear equationare
.
(ii) 
We know that any linear equation has infinitely many solutions.
Let us put.png "NCERT SOLUTIONS FOR CLASS 9 MATHS CHAPTER 4 LINEAR EQUATIONS IN TWO VARIABLES 43")
in the linear equation, to get
Thus, we get first pair of solution as
.
Let us put
in the linear equation, to get
Thus, we get second pair of solution as
.
Let us put
in the linear equation, to get
Thus, we get third pair of solution as
.
Let us put
in the linear equation, to get
Thus, we get fourth pair of solution as
.
Therefore, we can conclude that four solutions for the linear equationare
.
(iii) 
We know that any linear equation has infinitely many solutions.
Let us put
in the linear equation, to get
Thus, we get first pair of solution as
.
Let us putin the linear equation, to get
Thus, we get second pair of solution as
.
Let us put
in the linear equation, to get
Thus, we get third pair of solution as
.
Let us put
in the linear equation, to get
Thus, we get fourth pair of solution as
.
Therefore, we can conclude that four solutions for the linear equationare
.
Question 3. Check which of the following are solutions of the equationand which are not :
(i) 
(ii) 
(iii) 
(iv) 
(v) 
Solution:
(i)
We need to put
in the L.H.S. of linear equation
, to get
L.H.S. 
R.H.S.
Therefore, we can conclude thatis not a solution of the linear equation .
(ii)
We need to put
in the L.H.S. of linear equation, to get
L.H.S. R.H.S.
Therefore, we can conclude thatis not a solution of the linear equation .
(iii)
We need to put
in the linear equation, to get
L.H.S. 
R.H.S.
Therefore, we can conclude thatis a solution of the linear equation .
(iv)
We need to put
in the linear equation, to get
L.H.S. R.H.S.
Therefore, we can conclude thatis not a solution of the linear equation .
(v)
We need to put
in the linear equation, to get
L.H.S. R.H.S.
Therefore, we can conclude thatis not a solution of the linear equation .
Question 4. Find the value ofk, if x = 2, y = 1 is a solution of the equation.
Solution:
We know that, if is a solution of the linear equation, then on substituting the respective values of x and y in the linear equation , the LHS and RHS of the given linear equation will not be effected.
Therefore, we can conclude that the value of k, for which the linear equation has as one of its solutions is 7.
NCERT Solutions for Class 9 Maths 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
Solution:
(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.
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 =
Thus, the required value of a is
Question 4. 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.
Solution:
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.
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
For Fig. (2)
(i) y = x + 2
(ii) y = x – 2
(iii) y = -x + 2
(iv) x + 2y = 6
Solution:
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]
Question 6. 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
Solution:
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.
Question 7. 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.
Solution:
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.
Question 8. 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 = ( )C + 32
(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.
Solution:
(i) We have
F = ( )C + 32
When C = 0 , F = ( ) x 0 + 32 = 32
When C = 15, F = ( )(-15) + 32= -27 + 32 = 5
When C = -10, F = (-10)+32 = -18 + 32 = 14
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 = ()0 + 32 = 32
Also, F = 0
From (1), we get
0 = ()C + 32 ⇒ = C ⇒ C = -17.8
(V) When F = C (numerically)
From (1), we get
F = F + 32 ⇒ F – F = 32
⇒ F = 32 ⇒ F = -40
∴ Temperature is – 40° both in F and C.
NCERT Solutions for Class 9 Maths Exercise 4.4
Question 1. Give the geometric representation ofas an equation
(i) In one variable
(ii) In two variables
Solution:
(i) We need to represent the linear equationgeometrically in one variable.
We can conclude that in one variable, the geometric representation of the linear equationwill be same as representing the number 3 on a number line.
Given below is the representation of number 3 on the number line.
(ii) We need to represent the linear equationgeometrically in two variables.
We know that the linear equationcan also be written as.
We can conclude that in two variables, the geometric representation of the linear equationwill be same as representing the graph of linear equation.
Given below is the representation of the linear equationon a graph.
We can optionally consider the given below table for plotting the linear equationon the graph.
| X | 1 | 0 |
| y | 3 | 3 |
Question 2. Give the geometric representations of as an equation
(i) In one variable
(ii) In two variables
Solution:
(i) We need to represent the linear equationgeometrically in one variable.
We know that the linear equationcan also be written as.
We can conclude that in one variable, the geometric representation of the linear equationwill be same as representing the numberon a number line.
Given below is the representation of number 3 on the number line.
(ii) We need to represent the linear equationgeometrically in two variables.
We know that the linear equationcan also be written as.
We can conclude that in two variables, the geometric representation of the linear equationwill be same as representing the graph of linear equation.
Given below is the representation of the linear equationon a graph.
We can optionally consider the given below table for plotting the linear equationon the graph.
| X | 1 | 0 |
| y | 4.5 | 4.5 |
Conclusions for NCERT SOLUTIONS FOR CLASS 9 MATHS CHAPTER 4 LINEAR EQUATIONS IN TWO VARIABLES
SWC academic staff has developed NCERT answers for this chapter of the ninth grade mathematics curriculum. We have solutions prepared for all of the exercises of this chapter. The answers, broken down into steps, to all of the questions included in the NCERT textbook’s chapter are provided here. Read this chapter on theory. Be certain that you have read the theory section of this chapter of the NCERT textbook and that you have learnt the formulas for the chapter that you are studying.
