NCERT Solutions for Class 9 Maths Exercise 4.2 Chapter 4 Linear Equation in Two Variables

The concept that is being discussed in detail under section 4.2 of the chapter “Linear Equations in Two Variables” is that a linear equation in two variables has infinitely many solutions. Exercise 4.2 provided below, contains questions that explain this concept a bit more clearly. NCERT Solutions have been designed by subject experts at BYJU’S. This NCERT Solutions for Class 9 Maths Chapter 4 – Linear Equations in Two Variables Exercise 4.2 was developed with an aim to help the students to learn the subject more thoroughly.

In order to score good marks in the final exams, students are advised to make use of the solutions provided at BYJU’S. Designed as per the Class 9 syllabus and NCERT guidelines, the NCERT Solutions for Class 9 Maths Chapter 4 – Linear Equations are accurate and are prepared after proper research. It helps students to comprehend the concepts well. It uses clear and lucid language to elucidate the topic.

NCERT Solutions for Class 9 Maths Chapter 4 – Linear Equations in Two Variables Exercise 4.2

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Exercise 4.1 Solutions

Exercise 4.3 Solutions

Exercise 4.4 Solutions

NCERT Solutions for Class 9 Maths Chapter 4 – Linear Equations in Two Variables Exercise 4.2

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

y = 3x+5 has

1. A unique solution
2. Only two solutions
3. Infinitely many solutions

Solution:

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

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, option (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 equation 2x+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.

Now, students are well aware from earlier lessons that a linear equation in one variable has a unique solution. However, what about the solution of a linear equation involving two variables? Exercise 4.2 of Chapter 4 from the NCERT textbook tries to explain this concept to you with the help of 4 questions and some solved examples. To make it easier for the students to follow the concept and learn it thoroughly, we have given step-by-step explanations in the NCERT Solutions for Class 9 Maths.

Question number 1 has a main question followed by 3 sub-questions, while question number 2 is also of the same format. Meanwhile, question number 3 is of MCQ format, and question number 4 is a short answer question.

Solving these questions helps the students to be more confident while writing exams. Other benefits are as follows:

• Help students to get practice in solving a variety of Maths problems
• Gain efficiency and accuracy in solving questions
• Get good scores in Mathematics as they understand the subject better
• Understand the concepts more clearly