Learn how Forensic Science & Reasoning helps solves crimes - Learn how Forensic Science & Reasoning helps solves crimes -

# NCERT Solutions for Class 11 Maths Chapter 6 Linear Inequalities

## NCERT Solutions Class 11 Maths Linear Inequalities – Free PDF Download

NCERT Solutions for Class 11 Maths Chapter 6 Linear Inequalities are curated by the highly qualified expert teachers at BYJU’S. These NCERT Solutions of Maths help the students in solving the problems efficiently. The solutions for NCERT textbook questions contain steps with appropriate formulas and examples. All these solutions are written as per the latest update of the CBSE Syllabus 2022-23 to help students in scoring full marks.

The PDF of Class 11 Maths NCERT Solutions for Chapter 6 Linear Inequalities serves as a detailed study material for students. BYJU’S designs all the concepts, and solutions in simple words so that students can easily learn the chapter in less time. Also, the solutions of BYJU’S are accurate and students need not worry about the quality and correctness of the solutions. The knowledge of fundamental concepts helps in gaining knowledge and remembering more effectively for their upcoming exams.

Linear Inequalities are used in many real-life applications such as income and expenditure problems to find the proportion of the amount spent on various things. Two types of linear inequalities are explained in NCERT Solutions for Class 11 Maths of Chapter 6, i.e., linear inequalities in one variable and linear inequalities in two variables.

### Access answers of NCERT Solutions for Class 11 Maths Chapter 6 Linear Inequalities

Exercise 6.1 Page No: 122

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 on 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 on both sides, we get,

5x – 3 + 3 < 7 + 3

The 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 on both sides, 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 numbers greater than -2 are -1, 0, 1, 2,…

Thus, solution of 3x + 8 > 2 is -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 means 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, ∞)

6. 3x – 7 > 5x – 1

Solution:

Given that,

3x – 7 > 5x – 1

Now, by adding 7 to both the sides, we get

3x – 7 +7 > 5x – 1 + 7

3x > 5x + 6

Again, by subtracting 5x from both the sides,

3x – 5x > 5x + 6 – 5x

-2x > 6

Dividing both sides by -2 to simplify, we get

-2x/-2 < 6/-2

x < -3

∴ The solutions of the given inequality are defined by all the real numbers less than -3.

Hence the required solution set is (-∞, -3)

7. 3(x – 1) ≤ 2 (x – 3)

Solution:

Given that, 3(x – 1) ≤ 2 (x – 3)

By multiplying, the above inequality can be written as

3x – 3 ≤ 2x – 6

Now, by adding 3 to both the sides, we get

3x – 3+ 3 ≤ 2x – 6+ 3

3x ≤ 2x – 3

Again, by subtracting 2x from both the sides,

3x – 2x ≤ 2x – 3 – 2x

x ≤ -3

Therefore, the solutions of the given inequality are defined by all the real numbers less than or equal to -3.

Hence, the required solution set is (-∞, -3]

8. 3 (2 – x) ≥ 2 (1 – x)

Solution:

Given that, 3 (2 – x) ≥ 2 (1 – x)

By multiplying, we get

6 – 3x ≥ 2 – 2x

Now, by adding 2x to both the sides,

6 – 3x + 2x ≥ 2 – 2x + 2x

6 – x ≥ 2

Again, by subtracting 6 from both the sides, we get

6 – x – 6 ≥ 2 – 6

– x ≥ – 4

Multiplying throughout inequality by negative sign, we get

x ≤ 4

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

Hence the required solution set is (- ∞, 4]

9. x + x/2 + x/3 < 11

Solution:

x < 6

The solutions of the given inequality are defined by all the real numbers less than 6.

Hence the solution set is (-∞, 6)

10. x/3 > x/2 + 1

Solution:

– x/6 > 1

– x > 6

x < – 6

∴ The solutions of the given inequality are defined by all the real numbers less than – 6.

Hence, the required solution set is (-∞, -6)

11. 3(x – 2)/5 ≤ 5 (2 – x)/3

Solution:

Given that,

Now by cross-multiplying the denominators, we get

9(x- 2) ≤ 25 (2 – x)

9x – 18 ≤ 50 – 25x

Now adding 25x both the sides,

9x – 18 + 25x ≤ 50 – 25x + 25x

34x – 18 ≤ 50

34x – 18 + 18 ≤ 50 + 18

34x ≤ 68

Dividing both sides by 34,

34x/34 ≤ 68/34

x ≤ 2

The solutions of the given inequality are defined by all the real numbers less than or equal to 2.

Required solution set is (-∞, 2]

Solution:

120 ≥ x

∴ The solutions of the given inequality are defined by all the real numbers less than or equal to 120.

Thus, (-∞, 120] is the required solution set.

13. 2 (2x + 3) – 10 < 6 (x – 2)

Solution:

Given that,

2 (2x + 3) – 10 < 6 (x – 2)

By multiplying, we get

4x + 6 – 10 < 6x – 12

On simplifying, we get

4x – 4 < 6x – 12

– 4 + 12 < 6x – 4x

8 < 2x

4 < x

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

Hence the required solution set is (4, -∞)

14. 37 – (3x + 5) ≥ 9x – 8 (x – 3)

Solution:

Given that, 37 – (3x + 5) ≥ 9x – 8 (x – 3)

On simplifying, we get

= 37 – 3x – 5 ≥ 9x – 8x + 24

= 32 – 3x ≥ x + 24

On rearranging,

= 32 – 24 ≥ x + 3x

= 8 ≥ 4x

= 2 ≥ x

All the real numbers of x which are less than or equal to 2 are the solutions of the given inequality

Hence, (-∞, 2] will be the solution for the given inequality

Solution:

= 15x < 4 (4x – 1)

= 15x < 16x – 4

= 4 < x

All the real numbers of x which are greater than 4 are the solutions of the given inequality

Hence, (4, ∞) will be the solution for the given inequality

Solution:

= 20 (2x – 1) ≥ 3 (19x – 18)

= 40x – 20 ≥ 57x – 54

= – 20 + 54 ≥ 57x – 40x

= 34 ≥ 17x

= 2 ≥ x

∴ All the real numbers of x which are less than or equal to 2 are the solutions of the given inequality

Hence, (-∞, 2] will be the solution for the given inequality

Solve the inequalities in Exercises 17 to 20 and show the graph of the solution in each case on number line.

17. 3x – 2 < 2x + 1

Solution:

Given,

3x – 2 < 2x + 1

Solving the given inequality, we get

3x – 2 < 2x + 1

= 3x – 2x < 1 + 2

= x < 3

Now, the graphical representation of the solution is as follows:

18. 5x – 3 ≥ 3x – 5

Solution:

We have,

5x – 3 ≥ 3x – 5

Solving the given inequality, we get

5x – 3 ≥ 3x – 5

On rearranging, we get

= 5x – 3x ≥ -5 + 3

On simplifying,

= 2x ≥ -2

Now, dividing by 2 on both sides, we get

= x ≥ -1

The graphical representation of the solution is as follows:

19. 3 (1 – x) < 2 (x + 4)

Solution:

Given,

3 (1 – x) < 2 (x + 4)

Solving the given inequality, we get

3 (1 – x) < 2 (x + 4)

Multiplying, we get

= 3 – 3x < 2x + 8

On rearranging, we get

= 3 – 8 < 2x + 3x

= – 5 < 5x

Now by dividing 5 on both sides, we get

-5/5 < 5x/5

= – 1 < x

Now, the graphical representation of the solution is as follows:

Solution:

On computing we get

= 15x ≥ 2 (4x – 1)

= 15x ≥ 8x -2

= 15x -8x ≥ 8x -2 -8x

= 7x ≥ -2

= x ≥ -2/7

Now, the graphical representation of the solution is as follows:

21. Ravi obtained 70 and 75 marks in the first two unit tests. Find the minimum marks he should get in the third test to have an average of at least 60 marks.

Solution:

Let us assume that x is the marks obtained by Ravi in his third unit test.

According to the question, all the students should have an average of at least 60 marks

(70 + 75 + x)/3 ≥ 60

= 145 + x ≥ 180

= x ≥ 180 – 145

= x ≥ 35

Hence, all the students must obtain 35 marks in order to have an average of at least 60 marks

22. To receive Grade ‘A’ in a course, one must obtain an average of 90 marks or more in five examinations (each of 100 marks). If Sunita’s marks in the first four examinations are 87, 92, 94 and 95, find the minimum marks that Sunita must obtain in the fifth examination to get Grade ‘A’ in the course.

Solution:

Let us assume Sunita scored x marks in her fifth examination

Now, according to the question, in order to receive A grade in the course, she must obtain an average of 90 marks or more in her five examinations

(87 + 92 + 94 + 95 + x)/5 ≥ 90

= (368 + x)/5 ≥ 90

= 368 + x ≥ 450

= x ≥ 450 – 368

= x ≥ 82

Hence, she must obtain 82 or more marks in her fifth examination

23. Find all pairs of consecutive odd positive integers both of which are smaller than 10 such that their sum is more than 11.

Solution:

Let us assume x to be the smaller of the two consecutive odd positive integers.

∴ The other integer is = x + 2

It is also given in the question that both the integers are smaller than 10.

∴ x + 2 < 10

x < 8 … (i)

Also, it is given in the question that the sum of two integers is more than 11.

∴ x + (x + 2) > 11

2x + 2 > 11

x > 9/2

x > 4.5 … (ii)

Thus, from (i) and (ii), we have,

x is an odd integer and it can take values 5 and 7.

Hence, possible pairs are (5, 7) and (7, 9)

24. Find all pairs of consecutive even positive integers, both of which are larger than 5 such that their sum is less than 23.

Solution:

Let us assume x is the smaller of the two consecutive even positive integers.

∴ The other integer = x + 2

It is also given in the question that both the integers are larger than 5.

∴ x > 5   ….(i)

Also, it is given in the question that the sum of two integers is less than 23.

∴ x + (x + 2) < 23

2x + 2 < 23

x < 21/2

x < 10.5   … (ii)

Thus, from (i) and (ii) we have x is an even number and it can take values 6, 8 and 10.

Hence, possible pairs are (6, 8), (8, 10) and (10, 12).

25. The longest side of a triangle is 3 times the shortest side and the third side is 2 cm shorter than the longest side. If the perimeter of the triangle is at least 61 cm, find the minimum length of the shortest side.

Solution:

Let us assume the length of the shortest side of the triangle to be x cm.

∴ According to the question, the length of the longest side = 3x cm

And, length of third side = (3x – 2) cm

As, the least perimeter of the triangle = 61 cm

Thus, x + 3x + (3x – 2) cm ≥ 61 cm

= 7x – 2 ≥ 61

= 7x ≥ 63

Now dividing by 7, we get

= 7x/7 ≥ 63/7

= x ≥ 9

Hence, the minimum length of the shortest side will be 9 cm.

26. A man wants to cut three lengths from a single piece of board of length 91cm. The second length is to be 3cm longer than the shortest and the third length is to be twice as long as the shortest. What are the possible lengths of the shortest board if the third piece is to be at least 5cm longer than the second?

Solution:

Let us assume the length of the shortest piece to be x cm

∴ According to the question, length of the second piece = (x + 3) cm

And, length of third piece = 2x cm

As all the three lengths are to be cut from a single piece of board having a length of 91 cm

∴ x + (x + 3) + 2x ≤ 91 cm

= 4x + 3 ≤ 91

= 4x ≤ 88

= 4x/4 ≤ 88/4

= x ≤ 22 … (i)

Also, it is given in the question that, the third piece is at least 5 cm longer than the second piece.

∴ 2x ≥ (x+3) + 5

2x ≥ x + 8

x ≥ 8 … (ii)

Thus, from equation (i) and (ii), we have:

8 ≤ x ≤ 22

Hence, it is clear that the length of the shortest board is greater than or equal to 8 cm and less than or equal to 22 cm.

Exercise 6.2 Page No: 127

Solve the following inequalities graphically in two-dimensional plane:

1. x + y < 5

Solution:

Given x + y < 5

Consider

 X 0 5 y 5 0

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)

∴ Solution region of the given inequality is below the line x + y = 5. (i.e., 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)

∴ 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)

∴ Solution region of the given inequality is below the line 3x + 4y = 12. (i.e., 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)

∴ Solution region of the given inequality is above the line y + 8 = 2x. (i.e., 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)

∴ Solution region of the given inequality is above the line x – y = 2. (i.e., 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)

∴ 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)

∴ Solution region of the given inequality is above the line – 3x + 2y ≥ – 6. (i.e., 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)

∴ Solution region of the given inequality is below the line 3y – 5x < 30. (i.e., 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)

∴ Solution region of the given inequality is below the line y < – 2. (i.e., 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)

∴ Solution region of the given inequality is right to the line x > – 3. (i.e., origin is included in the region)

The graph is as follows:

Exercise 6.3 Page No: 129

Solve the following system of inequalities graphically:

1. x ≥ 3, y ≥ 2

Solution:

Given x ≥ 3……… (i)

y ≥ 2…………… (ii)

Since x ≥ 3 means for any value of y the equation will be unaffected, so similarly for y ≥ 2, for any value of x the equation will be unaffected.

Now putting x = 0 in (i)

0 ≥ 3 which is not true

Putting y = 0 in (ii)

0 ≥ 2 which is not true again

This implies the origin doesn’t satisfy in the given inequalities. The region to be included will be on the right side of the two equalities drawn on the graphs.

The shaded region is the desired region.

2. 3x + 2y ≤ 12, x ≥ 1, y ≥ 2

Solution:

Given 3x+ 2y ≤ 12

Solving for the value of x and y by putting x = 0 and y = 0 one by one, we get

y = 6 and x = 4

So the points are (0, 6) and (4, 0)

Now checking for (0, 0)

0 ≤ 12 which is also true.

Hence, the origin lies in the plane and the required area is toward the left of the equation.

Now checking for x ≥ 1, the value of x would be unaffected by any value of y.

The origin would not lie on the plane.

⇒ 0 ≥ 1 which is not true

The required area to be included would be on the left of the graph x ≥1

Similarly, for y ≥ 2

Value of y will be unaffected by any value of x in the given equality. Also, the origin doesn’t satisfy the given inequality.

⇒ 0 ≥ 2 which is not true. Hence origin is not included in the solution of the inequality.

The region to be included in the solution would be towards the left of the equality y≥ 2

The shaded region in the graph will give the answer to the required inequalities as it is the region which is covered by all the given three inequalities at the same time satisfying all the given conditions.

3. 2x + y ≥ 6, 3x + 4y ≤ 12

Solution:

Given 2x + y ≥ 6…………… (i)

3x + 4y ≤ 12 ……………. (ii)

2x + y ≥ 6

Putting value of x = 0 and y = 0 in equation one by one, we get value of

y = 6 and x = 3

So the point for the (0, 6) and (3, 0)

Now checking for (0, 0)

0 ≥ 6 which is not true, hence the origin does not lie in the solution of the equality. The required region is on the right side of the graph.

Checking for 3x + 4y ≤ 12,

Putting value of x= 0 and y = 0 one by one in equation,

We get y = 3, x = 4

The points are (0, 3), (4, 0)

Now, checking for origin (0, 0)

0 ≤ 12 which is true,

So the origin lies in solution of the equation.

The region on the right of the equation is the region required.

The solution is the region which is common to the graphs of both the inequalities.

The shaded region is the required region.

4. x + y ≥ 4, 2x – y < 0

Solution:

Given x + y ≥ 4

Putting value of x = 0 and y = 0 in equation one by one, we get value of

y = 4 and x = 4

The points for the line are (0, 4) and (4, 0)

Checking for the origin (0, 0)

0 ≥ 4

This is not true,

So the origin would not lie in the solution area. The required region would be on the right of line’s graph.

2x – y < 0

Putting value of x = 0 and y = 0 in equation one by one, we get value of

y= 0 and x = 0

Putting x = 1 we get y = 2

So, the points for the given inequality are (0, 0) and (1, 2)

Now that the origin lies on the given equation, we will check for (4, 0) point to check which side of the line’s graph will be included in the solution.

⇒ 8 < 0 which is not true, hence the required region would be on the left side of the line 2x-y < 0

The shaded region is the required solution of the inequalities.

5. 2x – y >1, x – 2y < – 1

Solution:

Given 2x – y >1……………… (i)

Putting value of x = 0 and y = 0 in equation one by one, we get value of

y = -1 and x = 1/2 = 0.5

The points are (0,-1) and (0.5, 0)

Checking for the origin, putting (0, 0)

0 >1, which is false

Hence the origin does not lie in the solution region. The required region would be on the right of the lines graph.

x – 2y < – 1………… (ii)

Putting value of x = 0 and y = 0 in equation one by one, we get value of

y = ½ = 0.5 and x = -1

The required points are (0, 0.5) and (-1, 0)

Now checking for the origin, (0, 0)

0 < -1 which is false.

Hence, the origin does not lie in the solution area; the required area would be on the left side of the line’s graph.

∴ The shaded area is the required solution of the given inequalities.

6. x + y ≤ 6, x + y ≥ 4

Solution:

Given x + y ≤ 6,

Putting value of x = 0 and y = 0 in equation one by one, we get value of

y = 6 and x = 6

The required points are (0, 6) and (6, 0)

Checking further for origin (0, 0)

We get 0 ≤ 6, this is true.

Hence the origin would be included in the area of the line’s graph. So, the required solution of the equation would be on the left side of the line graph which will be including origin.

x + y ≥ 4

Putting value of x = 0 and y = 0 in equation one by one, we get value of

y = 4 and x = 4

The required points are (0, 4) and (4, 0)

Checking for the origin (0, 0)

0 ≥ 4 which is false

So, the origin would not be included in the required area. The solution area will be above the line graph or the area on the right of line graph.

Hence, the shaded area in the graph is required graph area.

7. 2x + y ≥ 8, x + 2y ≥ 10

Solution:

Given 2x + y ≥ 8

Putting value of x = 0 and y = 0 in equation one by one, we get value of

y = 8 and x = 4

The required points are (0, 8) and (4, 0)

Checking if the origin is included in the lines graph (0, 0)

0 ≥ 8, which is false.

Hence, the origin is not included in the solution area and the required area would be the area to the right of the line’s graph.

x + 2y ≥ 10

Putting value of x = 0 and y = 0 in equation one by one, we get value of

y = 5 and x = 10

The required points are (0, 5) and (10, 0)

Checking for the origin (0, 0)

0 ≥ 10 which is false,

Hence the origin would not lie in the required solution area. The required area would be to the left of the line graph.

The shaded area in the graph is the required solution of the given inequalities.

8. x + y ≤ 9, y > x, x ≥ 0

Solution:

Given x + y ≤ 9,

Putting value of x = 0 and y = 0 in equation one by one, we get value of

y = 9 and x = 9

The required points are (0, 9) and (9, 0)

Checking if the origin is included in the line’s graph (0, 0)

0 ≤ 9

Which is true. So, the required area would be including the origin and hence, will lie on the left side of the line`s graph.

y > x,

Solving for y = x

We get x= 0, y = 0, so the origin lies on the line’s graph.

The other points would be (0, 0) and (2, 2)

Checking for (9, 0) in y > x,

We get 0 > 9 which is false, since the area would not include the area below the line’s graph and hence, would be on the left side of the line.

We have x ≥ 0

The area of the required line’s graph would be on the right side of the line’s graph.

Therefore, the shaded area is the required solution of the given inequalities.

9. 5x + 4y ≤ 20, x ≥ 1, y ≥ 2

Solution:

Given 5x + 4y ≤ 20,

Putting value of x = 0 and y = 0 in equation one by one, we get value of

y = 5 and x= 4

The required points are (0, 5) and (4, 0)

Checking if the origin lies in the solution area (0, 0)

0 ≤ 20

Which is true, hence the origin would lie in the solution area. The required area of the line’s graph is on the left side of the graph.

We have x ≥ 1,

For all the values of y, x would be 1,

The required points would be (1, 0), (1, 2) and so on.

Checking for origin (0, 0)

0 ≥ 1, which is not true.

So, the origin would not lie in the required area. The required area on the graph will be on the right side of the line’s graph.

Consider y ≥ 2

Similarly for all the values of x, y would be 2.

The required points would be (0, 2), (1, 2) and so on.

Checking for origin (0, 0)

0 ≥ 2, this is not true.

Hence, the required area would be on the right side of the line’s graph.

The shaded area on the graph shows the required solution of the given inequalities.

10. 3x + 4y ≤ 60, x + 3y ≤ 30, x ≥ 0, y ≥ 0

Solution:

Given 3x + 4y ≤ 60,

Putting value of x = 0 and y = 0 in equation one by one, we get value of

y = 15 and x = 20

The required points are (0, 15) and (20, 0)

Checking if the origin lies in the required solution area (0, 0)

0 ≤ 60, this is true.

Hence the origin would lie in the solution area of the line’s graph.

The required solution area would be on the left of the line’s graph.

We have x + 3y ≤ 30,

Putting value of x = 0 and y = 0 in equation one by one, we get value of

y = 10 and x = 30

The required points are (0, 10) and (30, 0).

Checking for the origin (0, 0)

0 ≤ 30, this is true.

Hence the origin lies in the solution area which is given by the left side of the line’s graph.

Consider x ≥ 0,

y ≥ 0,

The given inequalities imply the solution lies in the first quadrant only.

Hence the solution of the inequalities is given by the shaded region in the graph.

11. 2x + y ≥ 4, x + y ≤ 3, 2x – 3y ≤ 6

Solution:

Given 2x + y ≥ 4,

Putting value of x = 0 and y = 0 in equation one by one, we get value of

y = 4 and x =2

The required points are (0, 4) and (2, 0)

Checking for origin (0, 0)

0 ≥ 4, this is not true

Hence the origin doesn’t lie in the solution area of the line’s graph. The solution area would be given by the right side of the line’s graph.

x + y ≤ 3,

Putting value of x = 0 and y = 0 in equation one by one, we get value of

y = 3 and x = 3

The required points are (0, 3) and (3, 0)

Checking for the origin (0, 0)

0 ≤ 3, this is true.

Hence the solution area would include the origin and hence, would be on the left side of the line’s graph.

2x – 3y ≤ 6

Putting value of x = 0 and y = 0 in equation one by one, we get value of

y = -2 and x = 3

The required points are (0, -2), (3, 0).

Checking for the origin (0, 0)

0 ≤ 6 this is true

So the origin lies in the solution area and the area would be on the left of the line’s graph.

Hence, the shaded area in the graph is the required solution area for the given inequalities.

12. x – 2y ≤ 3, 3x + 4y ≥ 12, x ≥ 0 , y ≥ 1

Solution:

Given, x – 2y ≤ 3

Putting value of x = 0 and y = 0 in the equation one by one, we get value of

y = -3/2 = -1.5 and x = 3

The required points are (0, -1.5) and (3, 0)

Checking for the origin (0, 0)

0 ≤ 3, this is true.

Hence, the solution area would be on the left of the line’s graph

3x + 4y ≥ 12,

Putting value of x = 0 and y = 0 in equation one by one, we get value of

y = 3 and x = 4

The required points are (0, 3) and (4, 0)

Checking for the origin (0, 0)

0 ≥ 12, this is not true.

So, the solution area would include the origin and the required solution area would be on the right side of the line’s graph.

We have x ≥ 0,

For all the values of y, the value of x would be same in the given inequality, which would be the region above the x axis on the graph.

Consider, y ≥ 1

For all the values of x, the value of y would be same in the given inequality.

The solution area of the line would not include origin as 0 ≥ 1 is not true.

The solution area would be on the left side of the line’s graph.

The shaded area in the graph is the required solution area which satisfies all the given inequalities at the same time.

13. 4x + 3y ≤ 60, y ≥ 2x, x ≥ 3, x, y ≥ 0

Solution:

Given, 4x + 3y ≤ 60,

Putting value of x = 0 and y = 0 in equation one by one, we get value of

y = 20 and x = 15

The required points are (0, 20) and (15, 0).

Checking for the origin (0, 0)

0 ≤ 60, this is true.

Hence the origin would lie in the solution area. The required area would be on the left of the line’s graph.

We have y ≥ 2x,

Putting value of x = 0 and y = 0 in equation one by one, we get value of

y = 0 and x = 0

Hence the line would pass through origin.

To check which side would be included in the line’s graph solution area, we would check for point (15, 0)

⇒ 0 ≥ 15, this is not true. So the required solution area would be to the left of the line’s graph.

Consider, x ≥ 3,

For any value of y, the value of x would be same.

Also the origin (0, 0) doesn’t satisfiy the inequality as 0 ≥ 3.

So, the origin doesn’t lie in the solution area. Hence, the required solution area would be on the right of the line’s graph.

We have x, y ≥ 0

Since it is given both x and y are greater than 0

∴ the solution area would be in the first Ist quadrant only.

The shaded area in the graph shows the solution area for the given inequalities.

14. 3x + 2y ≤ 150, x + 4y ≤ 80, x ≤ 15, y ≥ 0, x ≥ 0

Solution:

Given, 3x + 2y ≤ 150

Putting value of x = 0 and y = 0 in the equation one by one, we get value of

y = 75 and x = 50

The required points are (0, 75) and (50, 0).

Checking for the origin (0, 0)

0 ≤ 150, this is true.

Hence, the solution area for the line would be on the left side of the line’s graph, which would be including the origin too.

We have x + 4y ≤ 80,

Putting value of x = 0 and y = 0 in equation one by one, we get value of

y = 20 and x = 80

The required points are (0, 20) and (80, 0).

Checking for the origin (0, 0)

0 ≤ 80, this is also true. So, the origin lies in the solution area.

The required solution area would be toward the left of the line’s graph.

Given x ≤ 15,

For all the values of y, x would be same.

Checking for the origin (0, 0)

0 ≤ 15, this is true. So, the origin would be included in the solution area. The required solution area would be towards the left of the line’s graph.

Consider y ≥ 0, x ≥ 0

Since x and y are greater than 0, the solution would lie in the 1st quadrant.

The shaded area in the graph satisfies all the given inequalities, and hence is the solution area for given inequalities.

15. x + 2y ≤ 10, x + y ≥ 1, x – y ≤ 0, x ≥ 0, y ≥ 0

Solution:

Given, x + 2y ≤ 10,

Putting value of x = 0 and y = 0 in equation one by one, we get value of

y = 5 and x = 10

The required points are (0, 5) and (10, 0).

Checking for the origin (0, 0)

0 ≤ 10, this is true.

Hence, the solution area would be toward origin including the same. The solution area would be toward the left of the line’s graph.

We have x + y ≥ 1,

Putting value of x = 0 and y = 0 in equation one by one, we get value of

y = 1 and x = 1

The required points are (0, 1) and (1, 0)

Checking for the origin (0, 0)

0 ≥ 1, this is not true.

Hence, the origin would not be included in the solution area. The required solution area would be toward the right of the line’s graph.

Consider x – y ≤ 0,

Putting value of x = 0 and y = 0 in equation one by one, we get value of

y = 0 and x = 0

Hence, the origin would lie on the line.

To check which side of the line graph would be included in the solution area, we would check for the (10, 0)

10 ≤ 0, which is not true. Hence, the solution area would be on the left side of the line’s graph.

Again, we have x ≥ 0, y ≥ 0

Since both x and y are greater than 0, the solution area would be in the 1st quadrant.

Hence, the solution area for the given inequalities would be the shaded area of the graph satisfying all the given inequalities.

Miscellaneous Exercise Page No: 129

Solve the inequalities in Exercises 1 to 6

1. 2 ≤ 3x – 4 ≤ 5

Solution:

According to the question,

The inequality given is,

2 ≤ 3x – 4 ≤ 5

⇒ 2 ≤ 3x – 4 ≤ 5

⇒ 2 + 4 ≤ 3x – 4 + 4 ≤ 5 + 4

⇒ 6 ≤ 3x ≤ 9

⇒ 6/3 ≤ 3x/3 ≤ 9/3

⇒ 2 ≤ x ≤ 3

Hence, all real numbers x greater than or equal to 2, but less than or equal to 3 are solutions of given equality.

x ∈ [2, 3]

2. 6 ≤ –3 (2x – 4) < 12

Solution:

According to the question,

The inequality given is,

6 ≤ –3 (2x – 4) < 12

⇒ 6 ≤ -3 (2x – 4) < 12

Dividing the inequality by 3, we get.

⇒ 2 ≤ – (2x – 4) < 4

Multiplying the inequality by -1,

⇒ -2 ≥ 2x – 4 > -4 [multiplying the inequality with -1 changes the inequality sign.]

⇒ -2 + 4 ≥ 2x – 4 + 4 > -4 + 4

⇒ 2 ≥ 2x > 0

Dividing the inequality by 2,

⇒ 0 < x ≤ 1

Hence, all real numbers x greater than 0, but less than or equal to 1 are solutions of given equality.

x ∈ (0, 1]

3. – 3 ≤ 4 – 7x/2 ≤ 18

Solution:

According to the question,

The inequality given is,

– 3 ≤ 4 – 7x/2 ≤ 18

⇒ – 3 – 4 ≤ 4 – 7x/2 – 4 ≤ 18 – 4

⇒ – 7 ≤ – 7x/2 ≤ 18 – 14

Multiplying the inequality by -2,

⇒ 14 ≥ 7x ≥ -28

⇒ -28 ≤ 7x ≤ 14

Dividing the inequality by 7,

⇒ -4 ≤ x ≤ 2

Hence, all real numbers x greater than or equal to -4, but less than or equal to 2 are solutions of given equality.

x ∈ [-4, 2]

4. – 15 ≤ 3(x – 2)/5 ≤ 0

Solution:

According to the question,

The inequality given is,

– 15 ≤ 3(x – 2)/5 ≤ 0

⇒ – 15 < 3(x – 2)/5 ≤ 0

Multiplying the inequality by 5,

⇒ -75 < 3(x – 2) ≤ 0

Dividing the inequality by 3, we get,

⇒ -25 < x – 2 ≤ 0

⇒ – 25 + 2 < x – 2 + 2 ≤ 0 + 2

⇒ – 23 < x ≤ 2

Hence, all real numbers x greater than -23, but less than or equal to 2 are solutions of given equality.

x ∈ (-23, 2]

5. – 12 < 4 – 3x/ (-5) ≤ 2

Solution:

According to the question,

The inequality given is,

Hence, all real numbers x greater than -80/3, but less than or equal to -10/3 are solutions of given equality.

x ∈ (-80/3, -10/3]

6. 7 ≤ (3x + 11)/2 ≤ 11

Solution:

According to the question,

The inequality given is,

⇒ 14 ≤ 3x + 11 ≤ 22

⇒ 14 – 11 ≤ 3x + 11 – 11 ≤ 22 – 11

⇒ 3 ≤ 3x ≤ 11

⇒ 1 ≤ x ≤ 11/3

Hence, all real numbers x greater than or equal to -4, but less than or equal to 2 are solutions of given equality.

x ∈ [1, 11/3]

Solve the inequalities in Exercises 7 to 11 and represent the solution graphically on number line.

7. 5x + 1 > – 24, 5x – 1 < 24

Solution:

According to the question,

The inequalities given are,

5x + 1 > -24 and 5x – 1 < 24

5x + 1 > -24

⇒ 5x > -24 – 1

⇒ 5x > -25

⇒ x > -5 ……… (i)

5x – 1 < 24

⇒ 5x < 24 + 1

⇒ 5x < 25

⇒ x < 5 ……….(ii)

From equations (i) and (ii),

We can infer that the solution of given inequalities is (-5, 5).

8. 2 (x – 1) < x + 5, 3 (x + 2) > 2 – x

Solution:

According to the question,

The inequalities given are,

2 (x – 1) < x + 5 and 3 (x + 2) > 2 – x

2 (x – 1) < x + 5

⇒ 2x – 2 < x + 5

⇒ 2x – x < 5 + 2

⇒ x < 7 ……… (i)

3 (x + 2) > 2 – x

⇒ 3x + 6 > 2 – x

⇒ 3x + x > 2 – 6

⇒ 4x > -4

⇒ x > -1 ………. (ii)

From equations (i) and (ii),

We can infer that the solution of given inequalities is (-1, 7).

9. 3x – 7 > 2(x – 6), 6 – x > 11 – 2x

Solution:

According to the question,

The inequalities given are,

3x – 7 > 2(x – 6) and 6 – x > 11 – 2x

3x – 7 > 2(x – 6)

⇒ 3x – 7 > 2x – 12

⇒ 3x – 2x > 7 – 12

⇒ x > -5 ………… (i)

6 – x > 11 – 2x

⇒ 2x – x > 11 – 6

⇒ x > 5 ……….(ii)

From equations (i) and (ii),

We can infer that the solution of given inequalities is (5, ∞).

10. 5(2x – 7) – 3(2x + 3) ≤ 0, 2x + 19 ≤ 6x + 47

Solution:

According to the question,

The inequalities given are,

5(2x – 7) – 3(2x + 3) ≤ 0 and 2x + 19 ≤ 6x + 47

5(2x – 7) – 3(2x + 3) ≤ 0

⇒ 10x – 35 – 6x – 9 ≤ 0

⇒ 4x – 44 ≤ 0

⇒ 4x ≤ 44

⇒ x ≤ 11 ……(i)

2x + 19 ≤ 6x +47

⇒ 6x – 2x ≥ 19 – 47

⇒ 4x ≥ -28

⇒ x ≥ -7 ……….(ii)

From equations (i) and (ii),

We can infer that the solution of given inequalities is (-7, 11).

11. A solution is to be kept between 68° F and 77° F. What is the range in temperature in degree Celsius (C) if the Celsius / Fahrenheit (F) conversion formula is given by F = (9/5) C + 32?

Solution:

According to the question,

The solution has to be kept between 68° F and 77° F

So, we get, 68° < F < 77°

Substituting,

⇒ 20 < C < 25

Hence, we get,

The range of temperature in degree Celsius is between 20° C to 25° C.

12. A solution of 8% boric acid is to be diluted by adding a 2% boric acid solution to it. The resulting mixture is to be more than 4%, but less than 6% boric acid. If we have 640 litres of the 8% solution, how many litres of the 2% solution will have to be added?

Solution:

According to the question,

8% of solution of boric acid = 640 litres

Let the amount of 2% boric acid solution added = x litres

Then we have,

Total mixture = x + 640 litres

We know that,

The resulting mixture has to be more than 4% but less than 6% boric acid.

∴ 2% of x + 8% of 640 > 4% of (x + 640) and

2% of x + 8% of 640 < 6% of (x + 640)

2% of x + 8% of 640 > 4% of (x + 640)

⇒ (2/100) × x + (8/100) × 640 > (4/100) × (x + 640)

⇒ 2x + 5120 > 4x + 2560

⇒ 5120 – 2560 > 4x – 2x

⇒ 2560 > 2x

⇒ x < 1280 ….(i)

2% of x + 8% of 640 < 6% of (x + 640)

⇒ (2/100) × x + (8/100) × 640 < (6/100) × (x + 640)

⇒ 2x + 5120 < 6x + 3840

⇒ 6x – 2x > 5120 – 3840

⇒ 4x > 1280

⇒ x > 320 ……….(i)

From (i) and (ii)

320 < x < 1280

Therefore, the number of litres of 2% of boric acid solution that has to be added will be more than 320 litres but less than 1280 litres.

13. How many litres of water will have to be added to 1125 litres of the 45% solution of acid so that the resulting mixture will contain more than 25% but less than 30% acid content?

Solution:

According to the question,

45% of solution of acid = 1125 litres

Let the amount of water added = x litres

Resulting mixture = x + 1125 litres

We know that,

The resulting mixture has to be more than 25% but less than 30% acid content.

Amount of acid in resulting mixture = 45% of 1125 litres.

∴ 45% of 1125 < 30% of (x + 1125) and 45% of 1125 > 25% of (x + 1125)

45% of 1125 < 30% of (x + 1125)

⇒ 45 × 1125 < 30x + 30 × 1125

⇒ (45 – 30) × 1125 < 30x

⇒ 15 × 1125 < 30x

⇒ x > 562.5 ………..(i)

45% of 1125 > 25% of (x + 1125)

⇒ 45 × 1125 > 25x + 25 × 1125

⇒ (45 – 25) × 1125 > 25x

⇒ 25x < 20 × 1125

⇒ x < 900 …..(ii)

∴ 562.5 < x < 900

Therefore, the number of litres of water that has to be added will have to be more than 562.5 litres but less than 900 litres.

14. IQ of a person is given by the formula
, Where MA is mental age and CA is chronological age. If 80 ≤ IQ ≤ 140 for a group of 12-year-old children, find the range of their mental age.

Solution:

According to the question,

Chronological age = CA = 12 years

IQ for age group of 12 is 80 ≤ IQ ≤ 140.

We get that,

80 ≤ IQ ≤ 140

Substituting,

⇒ 9.6 ≤ MA ≤ 16.8

∴ Range of mental age of the group of 12 year-old-children is 9.6 ≤ MA ≤ 16.8

 Also Access NCERT Exemplar for Class 11 Maths Chapter 6 CBSE Notes for Class 11 Maths Chapter 6

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

The Chapter Linear Inequalities belongs to the unit Algebra of Class 11 Maths CBSE Syllabus for 2022-23, and there exist 3 exercises and a miscellaneous exercise in this chapter, which help the students understand the concepts related to the chapter in detail. The topics covered in Chapter 6 Linear Inequalities of NCERT Solutions for Class 11 include:

6.1 Introduction

Two algebraic expressions or real numbers related by any of the symbols ≤, ≥, <, and > form an inequality. For example: px + qy > 0, 3a –19b < 0. Here, students will be able to know how to solve word problems by converting them into inequalities.

6.2 Inequalities

This topic is well explained with real-life scenarios which can be transformed to linear inequalities. Also, enough practice problems are provided along with solved examples.

6.3 Algebraic Solutions of Linear Inequalities in 1 Variable and their Graphical Representation

In this exercise, students can learn the meaning of a solution of linear inequalities and the graphical representation of these solutions. Besides, the methods of finding the solutions for linear inequalities have been explained with examples.

6.4 Graphical Solution of Linear Inequalities in Two Variables

After this section, students can understand the representation of the solution of linear inequalities in two variables on the Cartesian plane. Also, they can identify the solution region for the given inequalities.

6.5 Solution of System of Linear Inequalities in Two Variables

The solution of the system of linear inequalities in two variables using graphical methods is explained with many examples to help the students to understand the concept in a better way.

Exercise 6.1 Solutions 26 Questions

Exercise 6.2 Solutions 10 Questions

Exercise 6.3 Solutions 15 Questions

Miscellaneous Exercise On Chapter 6 Solutions 14 Questions

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

The summary of the concepts involved in the NCERT Solutions for Class 11 Maths Chapter 6 of BYJU’S is given below:

1. Two real numbers or two algebraic expressions related by the symbols <, >, ≤ or ≥ form an inequality.
2. Equal numbers may be added to (or subtracted from) both sides of an inequality.
3. Both sides of an inequality can be multiplied (or divided) by the same positive number. But, when both sides are multiplied (or divided) by a negative number, then the inequality is reversed.
4. The values of x, which make an inequality a true statement, are called solutions of the inequality.
5. To represent x < a (or x > a) on a number line, put a circle on the number a and dark line to the left (or right) of the number a.
6. To represent x ≤ a (or x ≥ a) on a number line, put a dark circle on the number a and darken the line to the left (or right) of the number x.
7. If inequality is having ≤ or ≥ symbol, then the points on the line are also included in the solutions of the inequality, and the graph of the inequality lies to the left (below) or right (above) of the graph of the equality represented by a dark line that satisfies an arbitrary point in that part.
8. If an inequality is having < or > symbol, then the points on the line are not included in the solutions of the inequality, and the graph of the inequality lies to the left (below) or right (above) of the graph of the corresponding equality represented by a dotted line that satisfies an arbitrary point in that part.
9. The solution region of a system of inequalities is the region which satisfies all the given inequalities in the system simultaneously.

### Key Features of NCERT Solutions for Class 11 Maths Chapter 6 – Linear Inequalities

The solutions for NCERT questions provided by BYJU’S for Class 11 Maths Chapter 6 have covered the below concepts. These solutions are prepared meticulously to avoid mistakes so that students are assured of getting full marks after practising them.

• Meaning of Linear inequalities.
• Algebraic solutions of linear inequalities in one variable and their representation on the number line.
• Graphical solution of linear inequalities in two variables.
• Graphical method of finding a solution of systems of linear inequalities in two variables.

Disclaimer –

Dropped Topics –

6.4 Graphical Solution of Linear Inequalities in Two Variables
6.5 Solution of System of Linear Inequalities in Two Variables
Last three points in the Summary

## Frequently Asked Questions on NCERT Solutions for Class 11 Maths Chapter 6

### List out the number of exercises present in NCERT Solutions for Class 11 Maths Chapter 6 Linear Inequalities.

The chapter consists of 3 exercises and one miscellaneous exercise which covers all the topics. The number of questions present in each exercise is mentioned below.
Exercise 6.1 – 26 Questions
Exercise 6.2 – 10 Questions
Exercise 6.3 – 15 Questions
Miscellaneous Exercise – 14 Questions

### Mention the important topics covered in the NCERT Solutions for Class 11 Maths Chapter 6.

The important topics covered in the NCERT Solutions for Class 11 Maths Chapter 6 are as follows:
1. Introduction
2. Inequalities
3. Algebraic Solutions of Linear Inequalities in One Variable and Their Graphical Representation
4. Graphical Solution of Linear Inequalities in Two Variables
5. Solution of System of Linear Inequalities in Two Variables

### Explain the steps to plot a graph of linear inequality covered in NCERT Solutions for Class 11 Maths Chapter 6.

The equation should be arranged in such a way that ‘y’ is on LHS and the other terms are on the RHS. Next, plot the ‘y’ line (continuous line when y <= or y >= ) and dotted line (when y < or y >). Shade the area above the ‘y’ line when y > or y >= or down the line when y < or y <=. Students can download the solutions PDF in order to get a clear idea about the steps to be followed in plotting the graph of a linear inequality.