# 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.

### 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 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!