# Linear Inequalities Chapter 6 Class 11 Maths NCERT Textbook With Solutions PDF

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Table of Contents

## NCERT Class 11 Maths Textbook Chapter 6 With Answer Book PDF Free Download

### Chapter 6: Linear Inequalities

#### 6.1 Introduction

In earlier classes, we have studied equations in one variable and two variables and also solved some statement problems by translating them into the form of equations.

Now a natural question arises: ‘Is it always possible to translate a statement problem into the form of an equation?

For example, the height of all the students in your class is less than 160 cm. Your classroom can occupy at most 60 tables or chairs or both.

Here we get certain statements involving a sign ‘<’ (less than), ‘>’ (greater than), ‘≤’ (less than or equal), and ≥ (greater than or equal) which are known as inequalities. In this chapter, we will study linear inequalities in one and two variables.

The study of inequalities is very useful in solving problems in the field of science, mathematics, statistics, optimization problems, economics, psychology, etc.

6.4 Graphical Solution of Linear Inequalities in Two Variables

In the earlier section, we have seen that a graph of an inequality in one variable is a visual representation and is a convenient way to represent the solutions to inequality.

Now, we will discuss a graph of a linear inequality in two variables.

e know that a line divides the Cartesian plane into two parts. Each part is known as a half-plane. A vertical line will divide the plane into left and right half-planes and a non-vertical line will divide the plane into lower and upper half-planes.

### NCERT Solutions Class 11 Maths Chapter 6 Linear Inequalities

1. Solve 24x < 100, when

(i) x is a natural number.

(ii) x is an integer.

Solution:

(i) Given that 24x < 100

Now we have to divide the inequality by 24 then we get x < 25/6

Now when x is a natural integer then

It is clear that the only natural number less than 25/6 are 1, 2, 3, 4.

Thus, 1, 2, 3, 4 will be the solution of the given inequality when x is a natural number.

Hence {1, 2, 3, 4} is the solution set.

(ii) Given that 24x < 100

Now we have to divide the inequality by 24 then we get x < 25/6

now when x is an integer then

It is clear that the integer number less than 25/6 are…-1, 0, 1, 2, 3, 4.

Thus, solution of 24 x < 100 are…,-1, 0, 1, 2, 3, 4, when x is an integer.

Hence {…, -1, 0, 1, 2, 3, 4} is the solution set.

2. Solve – 12x > 30, when

(i) x is a natural number.

(ii) x is an integer.

Solution:

(i) Given that, – 12x > 30

Now by dividing the inequality by -12 into both sides we get, x < -5/2

When x is a natural integer then

It is clear that there is no natural number less than -2/5 because -5/2 is a negative number and natural numbers are positive numbers.

Therefore there would be no solution of the given inequality when x is a natural number.

(ii) Given that, – 12x > 30

Now by dividing the inequality by -12 on both sides we get, x < -5/2

When x is an integer then

It is clear that the integer number less than -5/2 are…, -5, -4, – 3

Thus, solution of – 12x > 30 is …,-5, -4, -3, when x is an integer.

Therefore the solution set is {…, -5, -4, -3}

3. Solve 5x – 3 < 7, when

(i) x is an integer

(ii) x is a real number

Solution:

(i) Given that, 5x – 3 < 7

Now by adding 3 both side we get,

5x – 3 + 3 < 7 + 3

Above inequality becomes

5x < 10

Again by dividing both sides by 5 we get,

5x/5 < 10/5

x < 2

When x is an integer then

It is clear that that the integer number less than 2 are…, -2, -1, 0, 1.

Thus, solution of 5x – 3 < 7 is …,-2, -1, 0, 1, when x is an integer.

Therefore the solution set is {…, -2, -1, 0, 1}

(ii) Given that, 5x – 3 < 7

Now by adding 3 both side we get,

5x – 3 + 3 < 7 + 3

Above inequality becomes

5x < 10

Again by dividing both sides by 5 we get,

5x/5 < 10/5

x < 2

When x is a real number then

It is clear that the solutions of 5x – 3 < 7 will be given by x < 2 which states that all the real numbers that are less than 2.

Hence the solution set is x ∈ (-∞, 2)

4. Solve 3x + 8 >2, when

(i) x is an integer.

(ii) x is a real number.

Solution:

(i) Given that, 3x + 8 > 2

Now by subtracting 8 from both sides we get,

3x + 8 – 8 > 2 – 8

The above inequality becomes,

3x > – 6

Again by dividing both sides by 3 we get,

3x/3 > -6/3

Hence x > -2

When x is an integer then

It is clear that the integer number greater than -2 are -1, 0, 1, 2,…

Thus, solution of 3x + 8 > 2is -1, 0, 1, 2,… when x is an integer.

Hence the solution set is {-1, 0, 1, 2,…}

(ii) Given that, 3x + 8 > 2

Now by subtracting 8 from both sides we get,

3x + 8 – 8 > 2 – 8

The above inequality becomes,

3x > – 6

Again by dividing both sides by 3 we get,

3x/3 > -6/3

Hence x > -2

When x is a real number.

It is clear that the solutions of 3x + 8 >2 will be given by x > -2 which states that all the real numbers that are greater than -2.

Therefore the solution set is x ∈ (-2, ∞)

Solve the inequalities in Exercises 5 to 16 for real x.

5. 4x + 3 < 5x + 7

Solution:

Given that, 4x + 3 < 5x + 7

Now by subtracting 7 from both the sides, we get

4x + 3 – 7 < 5x + 7 – 7

The above inequality becomes,

4x – 4 < 5x

Again by subtracting 4x from both the sides,

4x – 4 – 4x < 5x – 4x

x > – 4

∴The solutions of the given inequality are defined by all the real numbers greater than -4.

The required solution set is (-4, ∞)

NCERT Class 11 Maths Textbook Chapter 6 With Answer Book PDF Free Download