NCERT Solutions for Class 11 Maths Chapter 6 - Linear Inequalities Exercise 6.2

*According to the CBSE Syllabus 2023-24, this chapter has been renumbered as Chapter 5.

Chapter 6 Linear Inequalities of Class 11 Maths is categorised under the CBSE Syllabus for 2023-24. The second exercise of this chapter is based on the topic Graphical Solution of Linear Inequalities in Two Variables. The graph of an inequality in two variables is the set of points that represents all solutions to the inequality. A linear inequality divides the coordinate plane into two halves by a boundary line, where one half represents the solutions of the inequality. Learn the graphical method of problem-solving by practising the questions given in Exercise 6.2 of NCERT Solutions for Class 11 Maths Chapter 6 – Linear Inequalities.

The NCERT textbook provides plenty of questions for the students to solve and practise. Solving the NCERT Solutions for Class 11 Maths and practising is more than enough to score high in the Class 11 board examinations. However, the students should make sure that they practise every problem given in the textbook repeatedly till the concept gets clear.

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Solutions for Class 11 Maths Chapter 6 – Exercise 6.2

Solve the following inequalities graphically in a two-dimensional plane.

1. x + y < 5

Solution:

Given, x + y < 5

Consider,

Now, draw a dotted line x + y = 5 in the graph (∵ x + y = 5 is excluded in the given question.)

Now, consider x + y < 5

Select a point (0, 0).

⇒ 0 + 0 < 5

⇒ 0 < 5 (this is true)

∴ The solution region of the given inequality is below the line x + y = 5. (That is, the origin is included in the region.)

The graph is as follows:

2. 2x + y ≥ 6

Solution:

Given 2x + y ≥ 6

Now, draw a solid line 2x + y = 6 in the graph (∵ 2x + y = 6 is included in the given question.)

Now, consider 2x + y ≥6

Select a point (0, 0).

⇒ 2 × (0) + 0 ≥ 6

⇒ 0 ≥ 6 (This is false.)

∴ The solution region of the given inequality is above the line 2x + y = 6. (Away from the origin.)

The graph is as follows:

3. 3x + 4y ≤ 12

Solution:

Given, 3x + 4y ≤ 12

Now, draw a solid line 3x + 4y = 12 in the graph (∵ 3x + 4y = 12 is included in the given question.)

Now, consider 3x + 4y ≤ 12

Select a point (0, 0).

⇒ 3 × (0) + 4 × (0) ≤ 12

⇒ 0 ≤ 12 (This is true.)

∴ The solution region of the given inequality is below the line 3x + 4y = 12. (That is, the origin is included in the region.)

The graph is as follows:

4. y + 8 ≥ 2x

Solution:

Given, y + 8 ≥ 2x

Now, draw a solid line y + 8 = 2x in the graph (∵ y + 8 = 2x is included in the given question.)

Now, consider y + 8 ≥ 2x

Select a point (0, 0).

⇒ (0) + 8 ≥ 2 × (0)

⇒ 0≤ 8 (This is true.)

∴ The solution region of the given inequality is above the line y + 8 = 2x. (That is, the origin is included in the region.)

The graph is as follows:

5. x – y ≤ 2

Solution:

Given, x – y ≤ 2

Now, draw a solid line x – y = 2 in the graph (∵ x – y = 2 is included in the given question.)

Now, consider x – y ≤ 2

Select a point (0, 0).

⇒ (0) – (0) ≤ 2

⇒ 0 ≤ 2 (This is true.)

∴ The solution region of the given inequality is above the line x – y = 2. (That is, the origin is included in the region.)

The graph is as follows:

6. 2x – 3y > 6

Solution:

Given, 2x – 3y > 6

Now, draw a dotted line 2x – 3y = 6 in the graph (∵ 2x – 3y = 6 is excluded in the given question.)

Now, consider 2x – 3y > 6

Select a point (0, 0).

⇒ 2 × (0) – 3 × (0) > 6

⇒ 0 > 6 (This is false.)

∴ The solution region of the given inequality is below the line 2x – 3y > 6. (Away from the origin.)

The graph is as follows:

7. – 3x + 2y ≥ – 6

Solution:

Given, – 3x + 2y ≥ – 6

Now, draw a solid line – 3x + 2y = – 6 in the graph (∵– 3x + 2y = – 6 is included in the given question.)

Now, consider – 3x + 2y ≥ – 6

Select a point (0, 0).

⇒ – 3 × (0) + 2 × (0) ≥ – 6

⇒ 0 ≥ – 6 (This is true.)

∴ The solution region of the given inequality is above the line – 3x + 2y ≥ – 6. (That is, the origin

is included in the region)

The graph is as follows:

8. y – 5x < 30

Solution:

Given, y – 5x < 30

Now, draw a dotted line 3y – 5x = 30 in the graph (∵3y – 5x = 30 is excluded in the given question.)

Now, consider 3y – 5x < 30

Select a point (0, 0).

⇒ 3 × (0) – 5 × (0) < 30

⇒ 0 < 30 (This is true.)

∴ The solution region of the given inequality is below the line 3y – 5x < 30. (That is, the origin is included in the region.)

The graph is as follows:

9. y < – 2

Solution:

Given, y < – 2

Now, draw a dotted line y = – 2 in the graph (∵ y = – 2 is excluded in the given question.)

Now, consider y < – 2

Select a point (0, 0).

⇒ 0 < – 2 (This is false)

∴ The solution region of the given inequality is below the line y < – 2. (That is, away from the origin.)

The graph is as follows:

10. x > – 3

Solution:

Given, x > – 3

Now, draw a dotted line x = – 3 in the graph (∵x = – 3 is excluded in the given question.)

Now, consider x > – 3

Select a point (0, 0).

⇒ 0 > – 3

⇒ 0 > – 3 (This is true.)

∴ The solution region of the given inequality is right to the line x > – 3. (That is, the origin is included in the region.)

The graph is as follows:

Access other exercise solutions of Class 11 Maths Chapter 6 – Linear Inequalities

NCERT Class 11 Solutions for Maths Chapter 6 exercise wise problems can be referred by clicking on the links below.

Exercise 6.1 Solutions 26 Questions

Exercise 6.3 Solutions 15 Questions

Miscellaneous Exercise on Chapter 6 Solutions 14 Questions