Frank Solutions for Class 10 Maths Chapter 5 Linear Inequations

Frank Solutions for Class 10 Maths Chapter 5 Linear Inequations, are available here. A thorough understanding and study of the Frank Solutions will allow students to prepare comprehensively for their examinations and in turn, ace exams effectively. Our topmost experts have answered these questions in such a way that students can understand clearly without any doubts. We suggest students download Frank Solutions for Class 10 Maths Chapter 5 offline as well.

Chapter 5 – Linear Inequations, linear inequality is an inequality which involves a linear function. A linear inequality contains one of the inequalities. This chapter deals with different types of equations and problems based on linear inequalities.

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Access answers to Frank Solutions for Class 10 Maths Chapter 5 Linear Inequations

1. Solve for x in the following in equations, if the replacement set is N<10:

(i) x + 5 > 11

Solution:-

x + 5 > 11

By transposing we get,

x > 11 – 5

x > 6

As per the condition given in the question, {x : x ∈ N;N < 10}

Therefore, solution set x = {7, 8, 9}

(ii) 2x + 1 < 17

Solution:-

2x + 1 < 17

By transposing we get,

2x < 17 – 1

x < 16/2

x < 8

As per the condition given in the question, {x : x ∈ N;N < 10}

Therefore, solution set x = {1, 2, 3, 4, 5, 6, 7}

(iii) 3x – 5 ≤ 7

Solution:-

3x – 5 ≤ 7

By transposing we get,

3x ≤ 7 + 5

x ≤ 12/3

x ≤ 4

As per the condition given in the question, {x : x ∈ N;N < 10}

Therefore, solution set x = {1, 2, 3, 4}

(iv) 8 – 3x ≥ 2

Solution:-

8 – 3x ≥ 2

By transposing we get,

3x ≥ 8 – 2

3x ≥ 6

x ≥ 6/3

x ≥ 2

As per the condition given in the question, {x : x ∈ N;N < 10}

Therefore, solution set x = {1, 2}

(v) 5 – 2x < 11

Solution:-

5 – 2x < 11

By transposing we get,

2x > 5 – 11

2x > -6

x > -6/2

x > -3

As per the condition given in the question, {x : x ∈ N;N < 10}

Therefore, solution set x = {1, 2, 3, 4, 5, 6, 7, 8, 9}

2. Solve for x in the following in-equations, if the replacement set is R;

(i) 3x > 12

Solution:-

3x > 12

By cross multiplication we get,

x > 12/3

x > 4

As per the condition given in the question, the replacement set is R.

Therefore, solution set x = {x : x ∈ R; x > 4}

(ii) 2x – 3 > 7

Solution:-

2x – 3 > 7

By transposing we get,

2x > 7 + 3

2x > 10

x > 10/2

x > 5

As per the condition given in the question, the replacement set is R.

Therefore, solution set x = {x : x ∈ R; x > 5}

(iii) 3x + 2 ≤ 11

Solution:-

3x + 2 ≤ 11

By transposing we get,

3x ≤ 11 – 2

3x ≤ 9

x ≤ 9/3

x ≤ 3

As per the condition given in the question, the replacement set is R.

Therefore, solution set x = {x : x ∈ R; x ≤ 3}

(iv) 14 – 3x ≥ 5

Solution:-

14 – 3x ≥ 5

By transposing we get,

3x ≤ 14 – 5

3x ≤ 9

x ≤ 9/3

x ≤ 3

As per the condition given in the question, the replacement set is R.

Therefore, solution set x = {x : x ∈ R; x ≤ 3}

(v) 7x + 11 > 16 – 3x

Solution:-

7x + 11 > 16 – 3x

By transposing we get,

7x + 3x > 16 – 11

10x > 5

x > 5/10

x > ½

x > 0.5

As per the condition given in the question, the replacement set is R.

Therefore, solution set x = {x : x ∈ R; x > 0.5}

(vi) 3x + 25 > 8x – 10

Solution:-

3x + 25 > 8x – 10

By transposing we get,

8x – 3x < 25 + 10

5x < 35

x < 35/5

x < 7

As per the condition given in the question, the replacement set is R.

Therefore, solution set x = {x : x ∈ R; x < 7}

(vii) 2(3x – 5) ≤ 8

Solution:-

2(3x – 5) ≤ 8

6x – 10 ≤ 8

By transposing we get,

6x ≤ 8 + 10

6x ≤ 18

x ≤ 18/6

x ≤ 3

As per the condition given in the question, the replacement set is R.

Therefore, solution set x = {x : x ∈ R; x ≤ 3}

(viii) x + 7 ≥ 15 + 3x

Solution:-

x + 7 ≥ 15 + 3x

By transposing we get,

3x – x ≤ 7 – 15

2x ≤ -8

x ≤ -8/2

x ≤ -4

As per the condition given in the question, the replacement set is R.

Therefore, solution set x = {x : x ∈ R; x ≤ -4}

(ix) 2x – 7 ≥ 5x + 8

Solution:-

2x – 7 ≥ 5x + 8

By transposing we get,

5x – 2x ≤ – 8 – 7

3x ≤ – 15

x ≤ – 15/3

x ≤ – 5

As per the condition given in the question, the replacement set is R.

Therefore, solution set x = {x : x ∈ R; x ≤ – 5}

(x) 9 – 4x ≤ 15 – 7x

Solution:-

9 – 4x ≤ 15 – 7x

By transposing we get,

7x – 4x ≤ 15 – 9

3x ≤ 6

x ≤ 6/3

x ≤ 2

As per the condition given in the question, the replacement set is R.

Therefore, solution set x = {x : x ∈ R; x ≤ 2}

3. Solve for x: 6 – 10x < 36, x ∈ {-3, -2, -1, 0, 1, 2}

Solution:-

From the question it is given that,

6 – 10x < 36

So, by transposing we get,

– 10x < 36 – 6

– 10x < 30

10x > -30

x > – 30/10

x > – 3

As per the condition given in the question, x ∈ {-3, -2, -1, 0, 1, 2}.

Therefore, solution set x = {-2, -1, 0, 1, 2}

4. Solve for x: 3 – 2x ≥ x – 12, x ∈ N

Solution:-

From the question it is given that,

3 – 2x ≥ x – 12

So, by transposing we get,

2x + x ≤ 12 + 3

3x ≤ 15

3x ≤ 15

x ≤ 15/3

x ≤ 5

As per the condition given in the question, x ∈ N.

Therefore, solution set x = {1, 2, 3, 4, 5}

5. Solve for x : 5x – 9 ≤ 15 – 7x, x ∈ W.

Solution:-

From the question it is given that,

5x – 9 ≤ 15 – 7x

So, by transposing we get,

5x + 7x ≤ 15 + 9

12x ≤ 24

x ≤ 24/12

x ≤ 2

As per the condition given in the question, x ∈ W.

Therefore, solution set x = {0, 1, 2}

6. Solve for x : 7 + 5x > x – 13, where x is a negative integer.

Solution:-

From the question it is given that,

7 + 5x > x – 13

So, by transposing we get,

5x – x > -13 – 7

4x > – 20

x > -20/4

x > -5

As per the condition given in the question, x is a negative integer.

Therefore, solution set x = {-4, -3, -2, -1}

7. Solve for x : 5x – 14 < 18 – 3x, x ∈ W.

Solution:-

From the question it is given that,

5x – 14 < 18 – 3x

So, by transposing we get,

5x + 3x < 18 +14

8x < 32

x < 32/8

x < 4

As per the condition given in the question, x is x ∈ W.

Therefore, solution set x = {0, 1, 2, 3}

8. Solve for x : 2x + 7 ≥ 5x – 14, where x is a positive prime number.

Solution:-

From the question it is given that,

2x + 7 ≥ 5x – 14

So, by transposing we get,

5x – 2x ≤ 14 + 7

3x ≤ 21

3x ≤ 21

x ≤ 21/3

x ≤ 7

As per the condition given in the question, x is a positive prime number.

Therefore, solution set x = {2, 3, 5, 7}

9. Solve for x : x/4 + 3 ≤ x/3 + 4, where x is a negative odd number.

Solution:-

From the question it is given that,

x/4 + 3 ≤ x/3 + 4

So, by transposing we get,

x/4 – x/3 ≤ 4 – 3

(3x – 4x)/12 ≤ 1

-x ≤ 12

x ≥ -12

As per the condition given in the question, x is a negative odd number.

Therefore, solution set x = {-11, -9, -7, -5, -3, -1}

10. Solve for x : (x + 3)/3 ≤ (x + 8)/ 4, where x is a positive even number.

Solution:-

From the question it is given that,

(x + 3)/3 ≤ (x + 8)/ 4

So, by cross multiplication we get,

4(x + 3) ≤ 3(x + 8)

4x + 12 ≤ 3x + 24

Now, transposing we get

4x – 3x ≤ 24 – 12

x ≤ 12

As per the condition given in the question, x is a positive even number.

Therefore, solution set x = {2, 4, 6, 8, 10, 12}

11. If x + 17 ≤ 4x + 9, find the smallest value of x, when:

(i) x ∈ Z

Solution:-

From the question,

x + 17 ≤ 4x + 9

So, by transposing we get,

4x – x ≥ 17 – 9

3x ≥ 8

x ≥ 8/3

As per the condition given in the question, x ∈ Z.

Therefore, smallest value of x = {3}

(ii) x ∈ R

Solution:-

From the question,

x + 17 ≤ 4x + 9

So, by transposing we get,

4x – x ≥ 17 – 9

3x ≥ 8

x ≥ 8/3

As per the condition given in the question, x ∈ R.

Therefore, smallest value of x = {
Frank Solutions for Class 10 Maths Chapter 4 Image 1}

12. If (2x + 7)/3 ≤ (5x + 1)/4, find the smallest value of x, when:

(i) x ∈ R

Solution:-

From the question,

(2x + 7)/3 ≤ (5x + 1)/4

So, by cross multiplication we get,

4(2x + 7) ≤ 3(5x + 1)

8x + 28 ≤ 15x + 3

Now transposing we get,

15x – 8x ≥ 28 – 3

7x ≥ 25

x ≥ 25/7

As per the condition given in the question, x ∈ R.

Therefore, smallest value of x = {
Frank Solutions for Class 10 Maths Chapter 4 Image 2}

(ii) x ∈ Z

Solution:-

From the question,

(2x + 7)/3 ≤ (5x + 1)/4

So, by cross multiplication we get,

4(2x + 7) ≤ 3(5x + 1)

8x + 28 ≤ 15x + 3

Now transposing we get,

15x – 8x ≥ 28 – 3

7x ≥ 25

x ≥ 25/7

As per the condition given in the question, x ∈ Z.

Therefore, smallest value of x = {7}

13. Solve the following linear in-equations and graph the solution set on a real number line.

(i) 2x – 11 ≤ 7 – 3x, x ∈ N.

Solution:-

2x – 11 ≤ 7 – 3x

By transposing we get,

2x + 3x ≤ 7 + 11

5x ≤ 18

x ≤ 18/5

x ≤ 3.6

As per the condition given in the question, x ∈ N.

Therefore, solution set x = {1, 2, 3}

Set can be represented in number line as,

Frank Solutions for Class 10 Maths Chapter 4 Image 3

(ii) 3(5x + 3) ≥ 2(9x – 17), x ∈ W.

Solution:-

From the question it is given that,

3(5x + 3) ≥ 2(9x – 17)

15x + 9 ≥ 18x – 34

So, by transposing we get,

18x – 15x ≤ 34 + 9

3x ≤ 43

x ≤ 43/3

As per the condition given in the question, x ∈ W.

Therefore, solution set x ≤ 43/3

Set can be represented in number line as,

Frank Solutions for Class 10 Maths Chapter 4 Image 4

(iii) 2(3x – 5) > 5(13 – 2x), x ∈ W.

Solution:-

From the question it is given that,

2(3x – 5) > 5(13 – 2x)

6x – 10 > 65 – 10x

So, by transposing we get,

6x + 10x > 65 + 10

16x > 75

x > 75/16

x >
Frank Solutions for Class 10 Maths Chapter 4 Image 5

As per the condition given in the question, x ∈ W.

Therefore, solution set x >
Frank Solutions for Class 10 Maths Chapter 4 Image 6

Set can be represented in number line as,

Frank Solutions for Class 10 Maths Chapter 4 Image 7

(iv) 3x – 9 ≤ 4x – 7 < 2x + 5, x ∈ R.

Solution:-

From the question,

Consider 3x – 9 ≤ 4x – 7

So, by transposing we get,

4x – 3x ≥ -9 + 7

x ≥ -2

Now, consider 4x – 7 < 2x + 5

By transposing we get,

4x – 2x < 5 + 7

2x < 12

x < 12/2

x < 6

As per the condition given in the question, x ∈ R.

Therefore, solution set = [-2 ≤ x < 6]

Set can be represented in number line as,

Frank Solutions for Class 10 Maths Chapter 4 Image 8

(v) 2x – 7 < 5x + 2 ≤ 3x + 14, x ∈ R.

Solution:-

From the question,

Consider 2x – 7 < 5x + 2

By transposing we get,

5x – 2x > – 7 – 2

3x < – 9

x < -9/3

x < -3

Now, consider 5x + 2 ≤ 3x + 14

So, by transposing we get,

5x – 3x ≤ 14 – 2

2x ≤ 12

x ≤ 12/2

x ≤ 6

As per the condition given in the question, x ∈ R.

Therefore, solution set = [-3 < x ≤ 6]

Set can be represented in number line as,

Frank Solutions for Class 10 Maths Chapter 4 Image 9

(vi) – 3 ≤ ½ – (2x/3) ≤
Frank Solutions for Class 10 Maths Chapter 4 Image 10 x ∈ N.

Solution:-

From the question,

Consider – 3 ≤ ½ – (2x/3)

-3 ≤ (3 – 4x)/6

– 18 ≤ (3 – 4x)

So, by transposing we get,

– 18 – 3 ≤ – 4x

– 21 ≤ – 4x

x ≤ 21/4

x ≤ 5¼

Now, consider ½ – (2x/3) ≤
Frank Solutions for Class 10 Maths Chapter 4 Image 11

(3 – 4x)/6 ≤ 8/3

By cross multiplication we get,

3 (3 – 4x) ≤ 48

9 – 12x ≤ 48

By transposing we get,

– 12x ≤ 48 – 9

– 12x ≤ 39

12x ≥ – 39

x ≥ – 39/12

x ≥ -3¼

As per the condition given in the question, x ∈ N.

Therefore, solution set = [-3¼ ≤ x ≤ 5¼]

Set can be represented in number line as

Frank Solutions for Class 10 Maths Chapter 4 Image 12

(vii) 4¾ ≥ x + 5/6 > 1/3, x ∈ R

Solution:-

From the question,

Consider, 4¾ ≥ x + 5/6

19/4 ≥ (6x + 5)/6

114 ≥ 24x + 20

By transposing we get,

114 – 20 ≥ 24x

94 ≥ 24x

x ≤ 94/24

x ≤
Frank Solutions for Class 10 Maths Chapter 4 Image 13

Now, consider x + 5/6 > 1/3

(6x + 5)/6 > 1/3

18x + 15 > 6

By transposing we get,

18x > 6 – 15

18x > – 9

x > – 9/18

x > -½

As per the condition given in the question, x ∈ R.

Therefore, solution set = [- ½ < x ≤
Frank Solutions for Class 10 Maths Chapter 4 Image 14]

Set can be represented in number line as

Frank Solutions for Class 10 Maths Chapter 4 Image 15

(viii) 1/3 (2x – 1) < ¼ (x + 5) < 1/6 (3x + 4), x ∈ R.

Solution:-

From the question it is given that,

Consider 1/3 (2x – 1) < ¼ (x + 5)

By cross multiplication we get,

4(2x – 1) < 3(x + 5)

8x – 4 < 3x + 15

By transposing we get,

8x – 3x < 15 + 4

5x < 19

x < 19/5

x <
Frank Solutions for Class 10 Maths Chapter 4 Image 16

Then, consider ¼ (x + 5) < 1/6 (3x + 4)

6(x + 5) < 4(3x + 4)

6x + 30 < 12x + 16

By transposing we get,

6x – 12x < 16 – 30

– 6x < – 14

x >
Frank Solutions for Class 10 Maths Chapter 4 Image 17

As per the condition given in the question, x ∈ R.

Therefore, solution set = [
Frank Solutions for Class 10 Maths Chapter 4 Image 18< x <
Frank Solutions for Class 10 Maths Chapter 4 Image 19]

Set can be represented in number line as

Frank Solutions for Class 10 Maths Chapter 4 Image 20

(ix) 1/3(5x – 8) ≥ ½ (4x – 7), x ∈ R.

Solution:-

From the question it is given that,

1/3(5x – 8) ≥ ½ (4x – 7)

By cross multiplication we get,

2(5x – 8) ≥ 3(4x – 7)

10x – 16 ≥ 12x – 21

Transposing we get,

12x – 10x ≤ 21 – 16

2x ≤ 5

x ≤ 5/2

x ≤ 2½

As per the condition given in the question, x ∈ R.

Therefore, solution set = {x < – 8}

Set can be represented in number line as

Frank Solutions for Class 10 Maths Chapter 4 Image 21

(x) 5/4x > 1 + 1/3 (4x – 1), x ∈ R.

Solution:-

From the question,

Consider, (5/4)x > 1 + 1/3 (4x – 1)

(5/4)x > (3 +(4x – 1)/3)

15x > 12 + 16x – 4

By transposing we get,

15x – 16x > 8

– x > 8

x < – 8

As per the condition given in the question, x ∈ R.

Therefore, solution set = {x < – 8}

Set can be represented in number line as,

Frank Solutions for Class 10 Maths Chapter 4 Image 22

14. If P = {x : -3 < x ≤ 7, x ∈ R} and Q = {x : – 7 ≤ x < 3, x ∈ R}, represent the following solution set on the different number lines:

(i) P ⋂ Q

(ii) Q’ ⋂ P

(iii) P – Q

Solution:-

As per the condition given in the question,

P = {x : -3 < x ≤ 7, x ∈ R}

So, P = {-2, – 1, 0, 1, 2, 3, 4, 5, 6, 7}

Then, Q = {x : – 7 ≤ x < 3, x ∈ R}

Q = {-7, -6, -5, -4, -3, -2, -1, 0, 1, 2}

(i) P ⋂ Q = {-2, – 1, 0, 1, 2, 3, 4, 5, 6, 7} ⋂ {-7, -6, -5, -4, -3, -2, -1, 0, 1, 2}

= {-2, -1, 0, 1, 2}

Frank Solutions for Class 10 Maths Chapter 4 Image 23

(ii) Q’ ⋂ P

Q’ = {3, 4, 5, 6, 7}

Q’ ⋂ P = {3, 4, 5, 6, 7} ⋂ {-2, – 1, 0, 1, 2, 3, 4, 5, 6, 7}

= {3, 4, 5, 6, 7}

Frank Solutions for Class 10 Maths Chapter 4 Image 24

(iii) P – Q

P – Q = {-2, – 1, 0, 1, 2, 3, 4, 5, 6, 7} – {-7, -6, -5, -4, -3, -2, -1, 0, 1, 2}

= {3, 4, 5, 6, 7}

Frank Solutions for Class 10 Maths Chapter 4 Image 25

15. If P = {x : 7x – 4 > 5x + 2, x ∈ R} and Q = {x : x – 19 ≥ 1 – 3x, x ∈ R}, represent the following solution set on the different number lines:

(i) P ⋂ Q

(ii) P’ ⋂ Q

Solution:-

As per the condition given in the question,

P = { x : 7x – 4 > 5x + 2, x ∈ R}

7x – 4 > 5x + 2

By transposing we get,

7x – 5x > 4 + 2

2x > 6

x > 6/2

x > 3

Therefore, P = {4, 5, 6, 7, ….}

Q = { x : x – 19 ≥ 1 – 3x, x ∈ R}

x – 19 ≥ 1 – 3x

By transposing we get,

x + 3x ≥ 1 + 19

4x ≥ 20

x ≥ 20/4

x ≥ 5

Q = {5, 6, 7, 8, …}

Then,

(i) P ⋂ Q = {2, 3, 4, 5, ….} ⋂ {5, 6, 7, 8, …}

= {5, 6, 7, 8, …}

Frank Solutions for Class 10 Maths Chapter 4 Image 26

(ii) P’ ⋂ Q = {∅}

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