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).

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, $x=2y$

$\Rightarrow x-2y=0$

Given : $2x + 3y = 9.3\bar{5}$

$\Rightarrow 2x + 3y - 9.3\bar{5}=0$

Here , a=2, b=3  and c =$- 9.3\bar{5}$

Given:

$x - \frac{y}{5} - 10 = 0$

$\Rightarrow x - \frac{y}{5} - 10 = 0$

Here ,

a=1,

$b=\frac{-1}{5}$

c = -10

Given :

$-2x + 3y = 6$

$\Rightarrow -2x + 3y - 6=0$

Here , a= -2, b=3  and c = -6

Given : $x =3y$

$\Rightarrow x -3y=0$

Here , a= 1, b= -3  and c =0

Given : $2x = - 5y$

$\Rightarrow 2x + 5y=0$

Here , a=2, b= 5  and c =0

Given : $3x + 2 = 0$

$\Rightarrow 3x + 2 = 0$

Here , a= 3, b=0  and c =2

Given : $y -2 = 0$

$\Rightarrow 0.x+y -2 = 0$

Here , a=0, b= 1  and c = -2

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

(iii) infinitely many solutions

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.

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

(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 $(3,\frac{3}{4})$.

(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 $x - 2y = 4$.

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 $x - 2y = 4$.

Given : $x - 2y = 4$

Putting  (4,0),

we have ,   $x - 2y = 4-2(0)=4=4$

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

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 $x - 2y = 4$.

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 $x - 2y = 4$.

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: $x + y = 4$

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.

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.

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.

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.

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.

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

Put x=3 and y=4

$3y = ax+ 7$

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

$\Rightarrow 12 = 3a+ 7$

$\Rightarrow 12 -7 = 3a$

$\Rightarrow 5 = 3a$

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

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 :

(i) $y = x$

(ii) $x + y = 0$

(iii) $y = 2x$

(iv) $2 + 3 y = 7x$

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.

(i) $y = x + 2$

(ii) $y = x - 2$

(iii) $y = -x + 2$

(iv) $x + 2y = 6$

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.

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.

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.

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)

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

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).

If the temperature is 30°C, what is the temperature in Fahrenheit?

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

Put c=30,

$F = \left(\frac{9}{5} \right )30 + 32=54+32=86$

Thus, the temperature = 30°C, then the temperature is 86 in Fahrenheit.

If the temperature is 95°F, what is the temperature in Celsius?

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

Put F=95,

$95 = \left(\frac{9}{5} \right )C + 32$

$\Rightarrow \left(\frac{9}{5} \right )C=95 -32$

$\Rightarrow \left(\frac{9}{5} \right )C=63$

$\Rightarrow C=35$

If the temperature is 95°F, then 35 is the temperature in Celsius.

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?

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 .

(v) Is there a temperature which is numerically the same in both Fahrenheit and Celsius? If yes, find it.

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

Let temperature be x in both Fahrenheit and celsius.

$x= \left(\frac{9}{5} \right )x + 32$

$\Rightarrow 5x= 9x + 160$

$\Rightarrow 5x-9x = 160$

$\Rightarrow -4x = 160$

$\Rightarrow x = -40 \degree$

Thus, 40 is the temperature which is numerically the same in both Fahrenheit and celsius.

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

in one variable

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

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.

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.

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

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!