## NCERT Solutions for Class 9 Maths Chapter 4 – CBSE Free PDF Download

**NCERT Solutions for Class 9 Maths Chapter 4 **Linear Equations in Two Variables are considered to be very useful when students are preparing for the CBSE Class 9 Maths exams. Here, we bring detailed answers to the exercises of NCERT Class 9 Maths Chapter 4. Subject matter experts who created these NCERT Solutions have collected these questions for students to revise from Chapter 4 of the NCERT Textbook. We provide accurate solutions to all the questions that are covered in the NCERT books. These NCERT Solutions for Class 9 Maths will rely on the latest update on the CBSE syllabus for 2023-24 and its guidelines. Students will get enough practice solving these exercises, and it will also help them to score high marks.

### Download Exclusively Curated Chapter Notes for Class 9 Maths Chapter – 4 Linear Equations In Two Variables

### Download Most Important Questions for Class 9 Maths Chapter – 4 Linear Equations In Two Variables

The **NCERT Solutions for Class 9 Maths** helps to give students proper knowledge about the subject and the topic “Linear equations”. Does a linear equation in two variables have a solution? If yes, is it unique? What does the solution look like on the Cartesian plane? Students shall also use the concepts they studied in Chapter 3, and the NCERT Solutions will also give them an idea about these concepts. These questions have been devised as per the updated CBSE syllabus.

## NCERT Solutions for Class 9 Maths Chapter 4- Linear Equations in Two Variables

**List of Exercises in Class 9 Maths Chapter 4**

Exercise 4.1 Solutions 2 Questions (1 Short Answer, 1 Main Question with 8 Short Answer questions under it)

Exercise 4.2 Solutions 4 Questions (2 MCQs, 1 Main Question with 3 equations to solve as part of it, 1 Short Answer Question)

Exercise 4.3 Solutions 8 Questions (4 Long Answer Questions, 2 Short Answer Questions, 1 MCQ, 1 Main question with 5 sub-questions under it)

Exercise 4.4 Solutions 2 Questions (1 main question with 2 sub-section under it, 1 Main question with 2 sub-sections under it)

### Access answers of Maths NCERT Class 9 Chapter 4 – Linear Equations in Two Variables

## Exercise 4.1 Page: 68

**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 ₹ x and that of a pen to be ₹ y)**

Solution:

Let the cost of a notebook be = ₹ x

Let the cost of a pen be = ₹ y

According to the question,

The cost of a notebook is twice the cost of a pen.

i.e., cost of a notebook = 2×cost of a pen

x = 2×y

x = 2y

x-2y = 0

x-2y = 0 is the linear equation in two variables to represent the statement, ‘The cost of a notebook is twice the cost of a pen.’

**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.**

**(ii) x –(y/5)–10 = 0**

Solution:

The equation x –(y/5)-10 = 0 can be written as,

1x+(-1/5)y +(–10) = 0

Now comparing x+(-1/5)y+(–10) = 0 with ax+by+c = 0

We get,

a = 1

b = -(1/5)

c = -10

**(iii) –2x+3y = 6**

Solution:

–2x+3y = 6

Re-arranging the equation, we get,

–2x+3y–6 = 0

The equation –2x+3y–6 = 0 can be written as,

(–2)x+3y+(– 6) = 0

Now, comparing (–2)x+3y+(–6) = 0 with ax+by+c = 0

We get, a = –2

b = 3

c =-6

**(iv) x = 3y**

Solution:

x = 3y

Re-arranging the equation, we get,

x-3y = 0

The equation x-3y=0 can be written as,

1x+(-3)y+(0)c = 0

Now comparing 1x+(-3)y+(0)c = 0 with ax+by+c = 0

We get a = 1

b = -3

c =0

**(v) 2x = –5y**

Solution:

2x = –5y

Re-arranging the equation, we get,

2x+5y = 0

The equation 2x+5y = 0 can be written as,

2x+5y+0 = 0

Now, comparing 2x+5y+0= 0 with ax+by+c = 0

We get a = 2

b = 5

c = 0

**(vi) 3x+2 = 0**

Solution:

3x+2 = 0

The equation 3x+2 = 0 can be written as,

3x+0y+2 = 0

Now comparing 3x+0+2= 0 with ax+by+c = 0

We get a = 3

b = 0

c = 2

** (vii) y–2 = 0**

Solution:

y–2 = 0

The equation y–2 = 0 can be written as,

0x+1y+(–2) = 0

Now comparing 0x+1y+(–2) = 0with ax+by+c = 0

We get a = 0

b = 1

c = –2

** (viii) 5 = 2x**

Solution:

5 = 2x

Re-arranging the equation, we get,

2x = 5

i.e., 2x–5 = 0

The equation 2x–5 = 0 can be written as,

2x+0y–5 = 0

Now comparing 2x+0y–5 = 0 with ax+by+c = 0

We get a = 2

b = 0

c = -5

## Exercise 4.2 Page: 70

** 1. Which one of the following options is true, and why?**

** y = 3x+5 has**

**A unique solution****Only two solutions****Infinitely many solutions**

Solution:

Let us substitute different values for x in the linear equation y = 3x+5

x | 0 | 1 | 2 | …. | 100 |

y, where y=3x+5 | 5 | 8 | 11 | …. | 305 |

From the table, it is clear that x can have infinite values, and for all the infinite values of x, there are infinite values of y as well.

Hence, (iii) infinitely many solutions is the only option true.

**2. Write four solutions for each of the following equations:**

**(i) 2x+y = 7**

Solution:

To find the four solutions of 2x+y =7, we substitute different values for x and y.

Let x = 0

Then,

2x+y = 7

(2×0)+y = 7

y = 7

(0,7)

Let x = 1

Then,

2x+y = 7

(2×1)+y = 7

2+y = 7

y = 7-2

y = 5

(1,5)

Let y = 1

Then,

2x+y = 7

(2x)+1 = 7

2x = 7-1

2x = 6

x = 6/2

x = 3

(3,1)

Let x = 2

Then,

2x+y = 7

(2×2)+y = 7

4+y = 7

y =7-4

y = 3

(2,3)

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

**(ii) πx+y = 9**

Solution:

To find the four solutions of πx+y = 9, we substitute different values for x and y.

Let x = 0

Then,

πx+y = 9

(π×0)+y = 9

y = 9

(0,9)

Let x = 1

Then,

πx +y = 9

(π×1)+y = 9

π+y = 9

y = 9-π

(1, 9-π)

Let y = 0

Then,

πx+y = 9

πx+0 = 9

πx = 9

x = 9/π

(9/π,0)

Let x = -1

Then,

πx + y = 9

(π×-1) + y = 9

-π+y = 9

y = 9+π

(-1,9+π)

The solutions are (0,9), (1,9-π), (9/π,0), (-1,9+π)

**(iii) x = 4y**

Solution:

To find the four solutions of x = 4y, we substitute different values for x and y.

Let x = 0

Then,

x = 4y

0 = 4y

4y= 0

y = 0/4

y = 0

(0,0)

Let x = 1

Then,

x = 4y

1 = 4y

4y = 1

y = 1/4

(1,1/4)

Let y = 4

Then,

x = 4y

x= 4×4

x = 16

(16,4)

Let y = 1

Then,

x = 4y

x = 4×1

x = 4

(4,1)

The solutions are (0,0), (1,1/4), (16,4), (4,1)

**3. Check which of the following are solutions of the equation x–2y = 4 and which are not:**

**(i) (0, 2)**

**(ii) (2, 0)**

**(iii) (4, 0)**

**(iv) (√2, 4√2)**

**(v) (1, 1)**

Solutions:

**(i) (0, 2)**

(x,y) = (0,2)

Here, x=0 and y=2

Substituting the values of x and y in the equation x–2y = 4, we get,

x–2y = 4

⟹ 0 – (2×2) = 4

But, -4 ≠ 4

(0, 2) is **not** a solution of the equation x–2y = 4

**(ii) (2, 0)**

(x,y) = (2, 0)

Here, x = 2 and y = 0

Substituting the values of x and y in the equation x -2y = 4, we get,

x -2y = 4

⟹ 2-(2×0) = 4

⟹ 2 -0 = 4

But, 2 ≠ 4

(2, 0) is **not** a solution of the equation x-2y = 4

**(iii) (4, 0)**

Solution:

(x,y) = (4, 0)

Here, x= 4 and y=0

Substituting the values of x and y in the equation x -2y = 4, we get,

x–2y = 4

⟹ 4 – 2×0 = 4

⟹ 4-0 = 4

⟹ 4 = 4

(4, 0) is a solution of the equation x–2y = 4

** (iv) (√2,4√2)**

Solution:

(x,y) = (√2,4√2)

Here, x = √2 and y = 4√2

Substituting the values of x and y in the equation x–2y = 4, we get,

x –2y = 4

⟹ √2-(2×4√2) = 4

√2-8√2 = 4

But, -7√2 ≠ 4

(√2,4√2) is **not** a solution of the equation x–2y = 4

**(v) (1, 1)**

Solution:

(x,y) = (1, 1)

Here, x= 1 and y= 1

Substituting the values of x and y in the equation x–2y = 4, we get,

x –2y = 4

⟹ 1 -(2×1) = 4

⟹ 1-2 = 4

But, -1 ≠ 4

(1, 1) is **not** a solution of the equation x–2y = 4

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

Solution:

The given equation is

2x+3y = k

According to the question, x = 2 and y = 1

Now, substituting the values of x and y in the equation2x+3y = k,

We get,

(2×2)+(3×1) = k

⟹ 4+3 = k

⟹ 7 = k

k = 7

The value of k, if x = 2, y = 1 is a solution of the equation 2x+3y = k, is 7.

## Exercise 4.3 Page: 74

**1. Draw the graph of each of the following linear equations in two variables:**

**(i) x+y = 4**

Solution:

To draw a graph of linear equations in two variables, let us find out the points to plot.

To find out the points, we have to find the values which x and y can have, satisfying the equation.

Here,

x+y = 4

Substituting the values for x,

When x = 0,

x+y = 4

0+y = 4

y = 4

When x = 4,

x+y = 4

4+y = 4

y = 4–4

y = 0

x | y |

0 | 4 |

4 | 0 |

The points to be plotted are (0, 4) and (4,0)

**(ii) x–y = 2**

Solution:

To draw a graph of linear equations in two variables, let us find out the points to plot.

To find out the points, we have to find the values which x and y can have, satisfying the equation.

Here,

x–y = 2

Substituting the values for x,

When x = 0,

x–y = 2

0 – y = 2

y = – 2

When x = 2,

x–y = 2

2–y = 2

– y = 2–2

–y = 0

y = 0

x | y |

0 | – 2 |

2 | 0 |

The points to be plotted are (0, – 2) and (2, 0)

**(iii) y=3x**

Solution:

To draw a graph of linear equations in two variables, let us find out the points to plot.

To find out the points, we have to find the values which x and y can have, satisfying the equation.

Here,

y = 3x

Substituting the values for x,

When x = 0,

y = 3x

y = 3×0

y = 0

When x = 1,

y = 3x

y = 3×1

y = 3

x | y |

0 | 0 |

1 | 3 |

The points to be plotted are (0, 0) and (1, 3)

**(iv) 3 = 2x+y**

Solution:

To draw a graph of linear equations in two variables, let us find out the points to plot.

To find out the points, we have to find the values which x and y can have, satisfying the equation.

Here,

3 = 2x+y

Substituting the values for x,

When x = 0,

3 = 2x+y

3 = 2×0+y

3 = 0+y

y = 3

When x = 1,

3= 2x+y

3 = 2×1+y

3 = 2+y

y = 3–2

y = 1

x | y |

0 | 3 |

1 | 1 |

The points to be plotted are (0, 3) and (1, 1)

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

Solution:

We know that an infinite number of lines pass through a point.

The equation of 2 lines passing through (2,14) should be in such a way that it satisfies the point.

Let the equation be 7x = y

7x–y = 0

When x = 2 and y = 14

(7×2)-14 = 0

14–14 = 0

0 = 0

L.H.S. = R.H.S.

Let another equation be 4x = y-6

4x-y+6 = 0

When x = 2 and y = 14

(4×2–14+6 = 0

8–14+6 = 0

0 = 0

L.H.S. = R.H.S.

Since both the equations satisfy the point (2,14), then we can say that the equations of two lines passing through (2, 14) are 7x = y and 4x = y-6

We know that an infinite number of line passes through one specific point. Since there is only one point (2,14) here, there can be infinite lines that pass through the point.

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

Solution:

The given equation is

3y = ax+7

According to the question, x = 3 and y = 4

Now, substituting the values of x and y in the equation 3y = ax+7,

We get,

(3×4) = (a×3)+7

⟹ 12 = 3a+7

⟹ 3a = 12–7

⟹ 3a = 5

⟹ a = 5/3

The value of a, if the point (3,4) lies on the graph of the equation 3y = ax+7 is 5/3.

**4. The taxi fare in a city is as follows: For the first kilometre, the fare is **₹**8, and for the subsequent distance, it is **₹**5 per km. Taking the distance covered as x km and total fare as **₹** y, write a linear equation for this information, and draw its graph.**

Solution:

Given,

Total distance covered = x

Total fare = y

Fare for the first kilometre = 8 per km

Fare after the first 1km = 5 per km

If x is the total distance, then the distance after one km = (x-1)km

i.e., fare after the first km = 5(x-1)

According to the question,

The total fare = Fare of first km+ fare after the first km

y = 8+5(x-1)

y = 8+5(x-1)

y = 8+5x – 5

y = 5x+3

Solving the equation,

When x = 0,

y = 5x+3

y = 5×0+3

y = 3

When y = 0,

y = 5x+3

o = 5x+3

5x = -3

x = -3/5

x | y |

0 | 3 |

-3/5 | 0 |

The points to be plotted are (0, 3) and (-3/5, 0)

**5. From the choices given below, choose the equation whose graphs are given in Fig. 4.6 and Fig. 4.7.**

For Fig. 4. 6

(i) y = x

(ii) x+y = 0

(iii) y = 2x

(iv) 2+3y = 7x

Solution:

The points given in figure 4.6 are (0,0), (-1,1), (1,-1)

Substituting the values for x and y from these points in the equations, we get,

(i) y = x

(0,0) ⟹ 0 = 0

(-1, 1) ⟹ -1 ≠ 1 ————————— equation not satisfied

(1, -1) ⟹ 1≠ -1 ————————— equation not satisfied

(ii) x+y = 0

(0,0) ⟹ 0+0 = 0

(-1, 1) ⟹ -1+1 = 0

(1, -1) ⟹ 1+(-1) =0

(iii) y = 2x

(0,0) ⟹ 0 = 2×0

0 = 0

(-1, 1) ⟹ 1 = 2×(-1)

1≠ -2 ————————— equation not satisfied

(1, -1) ⟹ -1 = 2×1

-1 ≠ 2 ————————— equation not satisfied

(iv) 2+3y = 7x

(0,0) ⟹ 2+(30) = 7×0

2 ≠ 0 ————————— equation not satisfied

(-1, 1) ⟹ 2+(3×1) = 7×-1

5 ≠ -7 ————————— equation not satisfied

(1, -1) ⟹ 2+(3×-1) = 7×1

-1 ≠ 7 ————————— equation not satisfied

Since only equation x+y = 0 satisfies all the points, the equation whose graphs are given in Fig. 4.6 is

x+y = 0

For Fig. 4. 7

(i) y = x+2

(ii) y = x–2

(iii) y = –x+2

(iv) x+2y = 6

Solution:

The points given in figure 4.7 are (0,2), (2,0), (-1,3)

Substituting the values for x and y from these points in the equations, we get,

(i) y = x+2

(0,2) ⟹2 = 0+2

2 = 2

(2, 0) ⟹ 0= 2+2

0 ≠ 4 ————————— equation not satisfied

(-1, 3) ⟹ 3 = -1+2

3 ≠ 1 ————————— equation not satisfied

(ii) y = x–2

(0,2) ⟹ 2 = 0–2

2 ≠ -2 ————————— equation not satisfied

(2, 0) ⟹ 0 = 2–2

0= 0

(-1, 3) ⟹ 3= –1–2

3 ≠ –3 ————————— equation not satisfied

(iii) y = –x+2

(0,2) ⟹ 2 = -0+2

2 = 2

(2, 0) ⟹ 0 = -2+2

0 = 0

(-1, 3) ⟹ 3= -(-1)+2

3 = 3

(iv) x+2y = 6

(0,2) ⟹ 0+(2×2) = 6

4 ≠ 6 ————————— equation not satisfied

(2, 0) ⟹ 2+(2×0) = 6

2 ≠ 6 ————————— equation not satisfied

(-1, 3) ⟹ -1+(2×3) = 6

5 ≠ 6 ————————— equation not satisfied

Since only equation y = –x+2 satisfies all the points, the equation whose graphs are given in Fig. 4.7 is

y = –x+2

**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:

Let the distance travelled by the body be x and the force applied on the body be y.

It is given that,

The work done by a body is directly proportional to the distance travelled by the body.

According to the question,

y ∝ x

y = 5x (5 is a constant of proportionality)

Solving the equation,

(i) when x = 2 units,

then y = 5×2 = 10 units

(2, 10)

(ii) when x = 0 units,

then y = 5×0 = 0 units.

(0, 0)

The points to be plotted are (2, 10) and (0, 0)

**7. Yamini and Fatima, two students of Class IX of a school, together contributed ₹ 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 ₹ x and ₹ y). Draw the graph of the same.**

Solution:

Let Yamini’s donation be ₹x and Fatima’s donation be ₹y

According to the question,

x+y = 100

We know that,

when x = 0 , y = 100

when x = 50, y = 50

when x = 100, y = 0

The points to be plotted are (0,100), (50,50), (100,0)

**8. In countries like 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:**

**(i) Draw the graph of the linear equation above using Celsius for the x-axis and Fahrenheit for the 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) According to the question,

F = (9/5)C + 32

Solving the equation,

We get,

When C = 0, F = 32

When C = -10 , F = 14

The points to be plotted are (0, 32), (-10, 14)

(ii) When C = 30,

F = (9/5)C +32

F = (9×30)/5+32

= (9×6)+32

= 54+32

= 86^{o}F

(iii) When F = 95,

95 = (9/5)C +32

(9/5)C = 95-32

(9/5)C =63

C = (63×5)/9

=35^{o}C

(iv) When C = 0,

F = (9/5)C +32

F = (9×0)/5 +32

=0+32

=32^{o}F

When F = 0,

0 = (9/5)C+32

(9/5)C = 0-32

(9/5)C = -32

C = (-32×5)/9

=-17.7777

=-17.8^{o}C

(v) When F = C,

C = (9/5)C+32

C – (9/5)C = 32

(5-9)C/5 =32

(-4/5)C = 32

(-4/5)C = (-32×5)/4

= – 40^{o}C

Hence, -40^{o} is the temperature which is numerically the same in both Fahrenheit and Celsius.

## Exercise 4.4 Page: 77

**1. Give the geometric representations of y = 3 as an equation**

**(i) in one variable**

**(ii) in two variables**

Solution:

- In one variable, y = 3

(ii) In two variables, 0x+y = 3

When x = 0, y = 3

When x = 1, y = 3

**2. Give the geometric representations of 2x+9 = 0 as an equation**

**(i) in one variable**

**(ii) in two variables**

Solution:

(i) In one variable,

2x+9 = 0

2x = -9

x = -9/2

x = -4.5

(ii) In two variables,

2x+9 = 0

2x+0y+9 = 0

When y = 0, x = -4.5

When y = 1, x = -4.5

### Summary of NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables

“Linear Equations in Two Variables” is the 4th chapter of the Class 9 Maths **NCERT Textbook**, and it falls under Unit 2 Algebra. For the board exams, from the unit Algebra, students can expect 7 questions that are 1 multiple choice question for 1 mark, 2 short answers with reasoning for a total of 4 marks, 3 short answer questions for a total of 9 marks and 1 long answer question for 6 marks. Thus, the total weightage for the unit is 20 marks. Topics covered in this chapter are listed below.

Chapter 4 |
Linear Equations in Two Variables |

4.1 | Introduction |

4.2 | Linear Equations |

4.3 | Solution of a Linear Equation |

4.4 | Graph of a Linear Equation in Two Variables |

4.5 | Equations of Lines Parallel to the x-axis and y-axis |

### NCERT Solutions for Class 9 Maths Chapter 4 – Linear Equations in Two Variables

In this chapter, the knowledge of linear equations in one variable is recalled and extended to that of two variables. Any equation which can be put in the form ax + by + c = 0, where a, b and c are real numbers, and a and b are not both zero, is called a linear equation in two variables. The Maths NCERT Solutions of Class 9 offers chapter-wise solutions with precise explanations of the exercises provided in the textbook. Students can easily understand the concept of linear equations of Algebra with the help of easy examples provided in these **NCERT Solutions**.

### Key Benefits of NCERT Solutions for Class 9 Maths Chapter 4 – Linear Equations in Two Variables

In this chapter, students study Linear Equations in Two Variables. And here, we will see how the **NCERT Solutions for Class 9 Maths Chapter 4** can benefit the students.

- They are created on the basis of the CBSE syllabus for 2023-24
- They feature all questions under each exercise section of the textbook
- Students get a clear idea about the concept and topic
- Solving the questions of the solutions will help them self-evaluate their performance
- Students can prepare for the exams on the basis of their knowledge gap

Students can also access the **NCERT Solutions for Class 9** of other subjects in a chapter-wise format as well. Referring to these solutions will boost board exam preparations for the students.

Practising more problems is very important when it comes to exam preparation. For this reason, students can also solve the questions from other textbooks prescribed by the CBSE Board.

**Disclaimer:**

**Dropped Topics – **4.4 Graph of linear equations in two variables and 4.5 Equations of lines parallel–x–axis and y–axis.

## Frequently Asked Questions on NCERT Solutions for Class 9 Maths Chapter 4

### Give me a summary of exercises in NCERT Solutions for Class 9 Maths Chapter 4.

4.1 – Introduction

4.2 – Linear Equations

4.3 – Solution of a Linear Equation

4.4 – Graph of a Linear Equation in Two Variables

4.5 – Equations of Lines Parallel to the x-axis and y-axis

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