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- In previous classes, you have studied linear equations in one variable. In this particular chapter, you will study linear equations in two variables of the type ax+by+c=0 where a, b and c are the real numbers, and a and b both are not zero. Solutions of NCERT for class 9 maths chapter 4 Linear Equations in Two Variables are there to make your task easy while preparing for the exams. In this chapter, there are a total of 4 exercises which consist of 16 questions. CBSE NCERT solutions for class 9 maths chapter 4 Linear Equations in Two Variables is covering the detailed solutions to each and every questions present in the practice exercises. This is an important chapter as this created a foundation for the higher level of algebra. Algebra is a unit in class 9 maths which holds 20 marks in the final examination. This chapter always comes with a good number of questions in competitive exams like the Indian National Olympiad (INO), National Talent Search Examination (NTSE). NCERT solutions for class 9 maths chapter 4 Linear Equations in Two Variables are designed in such a manner that a student can get maximum marks assigned to that particular question. NCERT solutions are also available classwise and subject wise which can be downloaded by clicking on the given link.

NCERT solutions for class 9 maths chapter 4 Linear Equations in Two Variables Excercise: 4.1

Q1 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).

Answer:

Let the cost of a notebook be Rs x and that of a pen be Rs y.

According to the given condition: The cost of a notebook is twice the cost of a pen.

Thus, x=2y

   \Rightarrow x-2y=0

Q2 (iv) Express the following linear equations in the form ax + by + c = 0 and indicate the values of a, b and c in each case: x =3y

Answer:

Given : x =3y

       \Rightarrow x -3y=0

Here , a= 1, b= -3  and c =0

Q2 (vii) Express the following linear equations in the form ax + by + c = 0 and indicate the values of a, b and c in each case: y -2 = 0

Answer:

Given : y -2 = 0

       \Rightarrow 0.x+y -2 = 0

Here , a=0, b= 1  and c = -2

Q2 (viii) Express the following linear equations in the form ax + by + c = 0 and indicate the values of a, b and c in each case: 5 = 2x

Answer:

Given : 5 = 2x

       \Rightarrow 2x+0.y-5=0

Here , a=2, b= 0 and c = -5

NCERT solutions for class 9 maths chapter 4 Linear Equations in Two Variables Excercise: 4.2

Q1 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

Answer:

Given : y = 3x + 5

This equation is of a line and a line has infinite points on it and each point is a solution Thus, (iii) infinitely many solutions is the correct option.

Q2 Write four solutions for each of the following equations:

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

Answer:

(i) Given : 2x + y = 7

Putting x=0, we have ,y=7-2\times 0=7  means (0,7)  is a solution.

Putting x=1, we have ,y=7-2\times 1=5  means (1,5)  is a solution.

Putting x=2, we have ,y=7-2\times 2=3  means (2,3)  is a solution.

Putting x=3, we have ,y=7-2\times 3=1  means (3,1)  is a solution.

The four solutions are :(0,7),(1,5),(2,3),(3,1).

(ii) Given : \pi x +y = 9

Putting x=0, we have ,y=9-\pi \times 0=9  means (0,9)  is a solution.

Putting x=1, we have ,y=9-\pi \times 1=9-\pi  means (1,9-\pi )  is a solution.

Putting x=2, we have ,y=9-\pi \times 2=9-2\pi  means (2,9-2\pi ) is a solution.

Putting x=3, we have ,y=9-\pi \times 3=9-3\pi  means (3,9-3\pi ) is a solution.

The four solutions are :(0,9),(1,9-\pi ),(2,9-2\pi ),(3,9-3\pi ).

 

(iii) Given : x = 4y

Putting x=0, we have ,y=\frac{0}{4}=0  means (0,0)  is a solution.

Putting x=1, we have ,y=\frac{1}{4}  means (1,\frac{1}{4})  is a solution.

Putting x=2, we have ,y=\frac{2}{4}=\frac{1}{2}  means (2,\frac{1}{2})  is a solution.

Putting x=3, we have ,y=\frac{3}{4}  means (3,\frac{3}{4})  is a solution.

The four solutions are :(0,0),(1,\frac{1}{4}),(2,\frac{1}{2}) and (3,\frac{3}{4}).

Q3 (i) Check which of the following are solutions of the equation x - 2y = 4and which are not: ((0,2)

Answer:

(i) Given : x - 2y = 4

Putting (0,2),

we have ,   x - 2y = 0-2(2)=-4\neq 4

Therefore,(0,2) is not a solution of x - 2y = 4.

Q3 (ii) Check which of the following are solutions of the equation x - 2y = 4 and which are not: (2,0)

Answer:

 Given : x - 2y = 4

Putting  (2,0),

we have ,   x - 2y = 2-2(0)=2\neq 4

Therefore, (2,0) is not a solution of x - 2y = 4.

Q3 (iii) Check which of the following are solutions of the equation x - 2y = 4and which are not: (4,0)

Answer:

 Given : x - 2y = 4

Putting  (4,0),

we have ,   x - 2y = 4-2(0)=4=4

Therefore, (4,0) is a solution of x - 2y = 4.

Q3 (iv) Check which of the following are solutions of the equation x - 2y = 4and which are not: (\sqrt2 , 4\sqrt2)

Answer:

 Given : x - 2y = 4

Putting  (\sqrt2 , 4\sqrt2),

we have ,   x - 2y =\sqrt{2}-2(4\sqrt{2})=\sqrt{2}-8\sqrt{2}=-7\sqrt{2}\neq 4

Therefore, (\sqrt2 , 4\sqrt2) is not a solution of x - 2y = 4.

Q3 (v) Check which of the following are solutions of the equation x - 2y = 4and which are not: (1,1) 

Answer:

Given : x - 2y = 4

Putting  (1,1) ,

we have ,   x - 2y = 1-2(1)=-1\neq 4

Therefore,   (1,1)  is not a solution of x - 2y = 4.

Q4 Find the value of k, if x = 2, y = 1 is a solution of the equation 2x + 3y = k.

Answer:

Given : 2x + 3y = k

Putting  (2,1),

we have ,   k=2x + 3y = 2(2)+3(1)=4+3=7

Therefore, k=7 for 2x + 3y = k putting x=2 and y=1.

NCERT solutions for class 9 maths chapter 4 Linear Equations in Two Variables Excercise: 4.3

Q1 (i) Draw the graph of each of the following linear equations in two variables: x + y = 4

Answer:

Given :x + y = 4

Putting x=0,we have y=4-0=4

Putting x=1,we have y=4-1=3

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

Q1 (ii) Draw the graph of each of the following linear equations in two variables: x - y = 2

Answer:

Given :x - y = 2

Putting x=0,we have y=0-2=-2

Putting x=1,we have y=1-2=-1

Thus, (0,-2) and (1,-1) are solutions of given equation.

Q1 (iii) Draw the graph of each of the following linear equations in two variables: y = 3x

Answer:

Given :y = 3x

Putting x=0,we have y=3(0)=0

Putting x=1,we have y=3(1)=3

Thus, (0,0) and (1,3) are solutions of given equation.

Q1 (iv) Draw the graph of each of the following linear equations in two variables: 3 = 2x + y

Answer:

Given :3 = 2x + y

Putting x=0,we have y=3-2.(0)=3

Putting x=1,we have y=3-2.(1)=1

Thus, (0,3) and (1,1) are solutions of given equation.

Q2 Give the equations of two lines passing through (2, 14). How many more such lines are there, and why?

Answer:

The equations of two lines passing through (2, 14) are given by : x+y=16\, \, and\, \, x-y=-12

There are infinite lines passing through (2, 14) because infinite lines pass through a point.

Q3 If the point (3, 4) lies on the graph of the equation 3y = ax+ 7, find the value of a.

Answer:

Given : the point (3, 4) lies on the graph of the equation 3y = ax+ 7

Put x=3 and y=4

3y = ax+ 7

\Rightarrow 3(4) = a(3)+ 7

\Rightarrow 12 = 3a+ 7

\Rightarrow 12 -7 = 3a

\Rightarrow 5 = 3a

\Rightarrow a=\frac{5}{3}

Q4 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.

Answer:

Given: The distance covered as x km and total fare is Rs. y.

Total fare =the fare for first km + the fare of the remaining  distance 

\therefore y=8+5\times (x-1)

\Rightarrow y=8+5x-5

\Rightarrow y=3+5x

For graph,

Putting x=0, we have y=3+5(0)=3

Putting x=1, we have y=3+5(1)=8

Putting x=2, we have y=3+5(2)=13

Hence,(0,3),(1,8) and (2,13)  are solutions of equation.

The graph is as shown :

Q5 (A) From the choices given below, choose the equation whose graph is given in Fig. 4.6.

(i) y = x

(ii) x + y = 0

(iii) y = 2x

(iv) 2 + 3 y = 7x

  

Answer:

For the given figure :

Points on line are (-1,1) ,  (0,0 ) and (1,-1)

x + y = 0  satisfies all the above points.

Thus, x + y = 0 is the correct equation of the line.

Q5 (B) From the choices given below, choose the equation whose graph is given in Fig. 4.7

(i) y = x + 2

(ii) y = x - 2

(iii) y = -x + 2

(iv) x + 2y = 6

 

Answer:

For the given figure :

Points on line are (-1,3) ,  (0,2 ) and (2,0)

y = -x + 2  satisfies all the above points.

Thus, y = -x + 2 is the correct equation of the line.

Q6 (i) 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: 2 units

Answer:

Let work done be y and distance be x.

Given : Constant force = 5 units 

Work done by a body on application of a constant force is directly proportional to the distance travelled by the body.  

                                     i.e.  y\, \, \alpha \, \, x

                                     \Rightarrow y=kx

k=Constant force = 5

Then,   

  \Rightarrow y=5x

For graph,

Put x=0,we have y=5(0)=0

Put x=1,we have y=5(1)=5

Put x=2,we have y=5(2)=10

Points are (0,0)   ,  (0,5)   and  (2,10)

 If diatance travelled is 2 units than work done is 10  units.

Q6 (ii) 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 : 0 unit

Answer:

Let work done be y and distance be x.

Given : Constant force = 5 units 

Work done by a body on application of a constant force is directly proportional to the distance travelled by the body.  

                                     i.e.  y\, \, \alpha \, \, x

                                     \Rightarrow y=kx

k=Constant force = 5

Then,      \Rightarrow y=5x

For graph,

Put x=0,we have y=5(0)=0

Put x=1,we have y=5(1)=5

Put x=2,we have y=5(2)=10

Points are (0,0)   ,  (0,5)   and  (2,10)

 If diatance travelled is 0 units than work done is 0  units.

Q7 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.

Answer:

Let the contribution of Yamini be x.

      contribution of Yamini be y.

According to question,

                                  x+y=100

For x=0 , we have y=100-0=100

For x=10 , we have y=100-10=90

For x=20 , we have y=100-20=80

Hence, (0,100) , (10,90)  and  (20,80)

Q8 (i) In countries like the USA and Canada, the 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

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

Answer:

Let celsius be on x-axis and Fahrenheit be on y-axis.

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

For graph,

Putting  x=0, we get  y= \left(\frac{9}{5} \right )(0)+ 32=32

Putting  x=5, we get  y= \left(\frac{9}{5} \right )(5)+ 32=41

Putting  x=10, we get  y= \left(\frac{9}{5} \right )(10)+ 32=50

Hence, points are  (0,32) , (5,41)   and (10,50).

Q8 (iv) 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 = \left(\frac{9}{5} \right )C + 32

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?

Answer:

If the temperature is 0°C,

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

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

                      \Rightarrow F = 32

if the temperature is 0°F,

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

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

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

                        \Rightarrow C=- 17.8

Thus, if the temperature is 0°C ,then temperature in fahrenheit is 32 and  if the temperature is 0°F ,then temperature in celsius is -17.8 .

NCERT solutions for class 9 maths chapter 4 Linear Equations in Two Variables Excercise: 4.4

Q1 (i) Give the geometric representations of y = 3 as an equation

in one variable

Answer:

Equation y = 3 can be represented in one variable on number line.

Q1 (ii) Give the geometric representations of y = 3 as an equation: in two variables

Answer:

For 2 variable representation of y = 3.

  Equation :     0.x+y=3

   For graph, 

     x=0,we have y=3         

     x=1,we have y=3   

     x=2,we have y=3     

Hence, (0,3) ,(1,3) and (2,3)  are solutions of equation.

        

Q2 (i) Give the geometric representations of 2x + 9 = 0 as an equation: in one variable

Answer:

Equation 2x + 9 = 0 can be represented in one variable on the number line.

               \Rightarrow x=\frac{-9}{2}

The yellow mark represents x=  - 4.5.

Give the geometric representations

Q2 (ii) Give the geometric representations of 2x + 9 = 0 as an equation: in two variables

Answer:

For 2 variable representation of 2x + 9 = 0.

  Equation :     2x +0.y =-9

   For graph, 

     y=0,we have x=\frac{-9}{2}        

     y=1,we have  x=\frac{-9}{2}  

     y=2,we have x=\frac{-9}{2}    

Hence, (\frac{-9}{2},0) ,   (\frac{-9}{2},1)and    (\frac{-9}{2},2)  are solutions of equation.

NCERT solutions for class 9 maths chapter wise

Chapter No.

Chapter Name

Chapter 1

NCERT solutions for class 9 maths chapter 1 Number Systems

Chapter 2

CBSE NCERT solutions for class 9 maths chapter 2 Polynomials

Chapter 3

Solutions of NCERT class 9 maths chapter 3 Coordinate Geometry

Chapter 4

NCERT solutions for class 9 maths chapter 4 Linear Equations In Two Variables

Chapter 5

CBSE NCERT solutions for class 9 maths chapter 5 Introduction to Euclid's Geometry

Chapter 6

Solutions of NCERT class 9 maths chapter 6 Lines And Angles

Chapter 7

NCERT solutions for class 9 maths chapter 7 Triangles

Chapter 8

CBSE NCERT solutions for class 9 maths chapter 8 Quadrilaterals

Chapter 9

Solutions of NCERT class 9 maths chapter 9 Areas of Parallelograms and Triangles

Chapter 10

NCERT solutions for class 9 maths chapter 10 Circles

Chapter 11

CBSE NCERT solutions for class 9 maths chapter 11 Constructions

Chapter 12

Solutions of NCERT class 9 maths chapter 12 Heron’s Formula

Chapter 13

NCERT solutions for class 9 maths chapter 13 Surface Area and Volumes

Chapter 14

CBSE NCERT solutions for class 9 maths chapter 14 Statistics

Chapter 15

Solutions of NCERT class 9 maths chapter 15 Probability

NCERT solutions for class 9 subject wise

CBSE NCERT Solutions for Class 9 Maths

Solutions of NCERT Solutions for Class 9 Science

How to use NCERT solutions for class 9 maths chapter 4 Linear Equations in Two Variables

  • First of all, go through the conceptual text given in the NCERT textbook.

  • Then understand the application of concepts in the examples.

  • After this, you should come over practice exercises.

  • While practicing the exercises, you can use NCERT solutions for class 9 maths chapter 4 Linear Equations in Two Variables as a reference.

  • After the completion of all the above points, you can practice the questions of past year papers.

Keep Working Hard & Happy Learning!

 

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