# 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 Frank Solutions will help students to prepare comprehensively for the board examinations and, in turn, score high marks. Our topmost subject experts have designed these questions in such a way that students can understand clearly without having any doubts. We suggest that students download Frank Solutions for Class 10 Maths Chapter 5 in PDF so that they can practise 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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1. Solve for x in the following inequations, if the replacement set is N<10:

(i) x + 5 > 11

Solution:-

x + 5 > 11

By transposing, we get,

x > 11 â€“ 5

x > 6

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

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

(ii) 2x + 1 < 17

Solution:-

2x + 1 < 17

By transposing, we get,

2x < 17 â€“ 1

x < 16/2

x < 8

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

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

(iii) 3x – 5 â‰¤ 7

Solution:-

3x – 5 â‰¤ 7

By transposing, we get,

3x â‰¤ 7 + 5

x â‰¤ 12/3

x â‰¤ 4

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

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

(iv) 8 â€“ 3x â‰¥ 2

Solution:-

8 â€“ 3x â‰¥ 2

By transposing, we get,

3x â‰¥ 8 – 2

3x â‰¥ 6

x â‰¥ 6/3

x â‰¥ 2

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

Therefore, solution set x = {1, 2}

(v) 5 â€“ 2x < 11

Solution:-

5 â€“ 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 âˆˆ N;N < 10}

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

2. Solve for x in the following inequations, 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 âˆˆ 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 âˆˆ R; x > 5}

(iii) 3x + 2 â‰¤ 11

Solution:-

3x + 2 â‰¤ 11

By transposing, we get,

3x â‰¤ 11 – 2

3x â‰¤ 9

x â‰¤ 9/3

x â‰¤ 3

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

Therefore, solution set x = {x : x âˆˆ R; x â‰¤ 3}

(iv) 14 â€“ 3x â‰¥ 5

Solution:-

14 â€“ 3x â‰¥ 5

By transposing, we get,

3x â‰¤ 14 – 5

3x â‰¤ 9

x â‰¤ 9/3

x â‰¤ 3

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

Therefore, solution set x = {x : x âˆˆ R; x â‰¤ 3}

(v) 7x + 11 > 16 â€“ 3x

Solution:-

7x + 11 > 16 â€“ 3x

By transposing, we get,

7x + 3x > 16 – 11

10x > 5

x > 5/10

x > Â½

x > 0.5

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

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

(vi) 3x + 25 > 8x â€“ 10

Solution:-

3x + 25 > 8x â€“ 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 âˆˆ R; x < 7}

(vii) 2(3x – 5) â‰¤ 8

Solution:-

2(3x – 5) â‰¤ 8

6x â€“ 10 â‰¤ 8

By transposing, we get,

6x â‰¤ 8 + 10

6x â‰¤ 18

x â‰¤ 18/6

x â‰¤ 3

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

Therefore, solution set x = {x : x âˆˆ R; x â‰¤ 3}

(viii) x + 7 â‰¥ 15 + 3x

Solution:-

x + 7 â‰¥ 15 + 3x

By transposing, we get,

3x â€“ x â‰¤ 7 – 15

2x â‰¤ -8

x â‰¤ -8/2

x â‰¤ -4

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

Therefore, solution set x = {x : x âˆˆ R; x â‰¤ -4}

(ix) 2x – 7 â‰¥ 5x + 8

Solution:-

2x – 7 â‰¥ 5x + 8

By transposing, we get,

5x â€“ 2x â‰¤ – 8 – 7

3x â‰¤ – 15

x â‰¤ – 15/3

x â‰¤ – 5

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

Therefore, solution set x = {x : x âˆˆ R; x â‰¤ – 5}

(x) 9 – 4x â‰¤ 15 â€“ 7x

Solution:-

9 – 4x â‰¤ 15 â€“ 7x

By transposing, we get,

7x â€“ 4x â‰¤ 15 – 9

3x â‰¤ 6

x â‰¤ 6/3

x â‰¤ 2

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

Therefore, solution set x = {x : x âˆˆ R; x â‰¤ 2}

3. Solve for x: 6 â€“ 10x < 36, x âˆˆ {-3, -2, -1, 0, 1, 2}

Solution:-

From the question, it is given that,

6 â€“ 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 âˆˆ {-3, -2, -1, 0, 1, 2}.

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

4. Solve for x: 3 â€“ 2x â‰¥ x – 12, x âˆˆ N

Solution:-

From the question, it is given that,

3 â€“ 2x â‰¥ x – 12

So, by transposing, we get,

2x + x â‰¤ 12 + 3

3x â‰¤ 15

3x â‰¤ 15

x â‰¤ 15/3

x â‰¤ 5

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

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

5. Solve for x : 5x â€“ 9 â‰¤ 15 â€“ 7x, x âˆˆ W.

Solution:-

From the question, it is given that,

5x â€“ 9 â‰¤ 15 â€“ 7x

So, by transposing, we get,

5x + 7x â‰¤ 15 + 9

12x â‰¤ 24

x â‰¤ 24/12

x â‰¤ 2

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

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

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

Solution:-

From the question, it is given that,

7 + 5x > x â€“ 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 â€“ 14 < 18 â€“ 3x, x âˆˆ W.

Solution:-

From the question, it is given that,

5x – 14 < 18 â€“ 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 âˆˆ W.

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

8. Solve for x : 2x + 7 â‰¥ 5x â€“ 14, where x is a positive prime number.

Solution:-

From the question, it is given that,

2x + 7 â‰¥ 5x â€“ 14

So, by transposing, we get,

5x – 2x â‰¤ 14 + 7

3x â‰¤ 21

3x â‰¤ 21

x â‰¤ 21/3

x â‰¤ 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 â‰¤ x/3 + 4, where x is a negative odd number.

Solution:-

From the question, it is given that,

x/4 + 3 â‰¤ x/3 + 4

So, by transposing, we get,

x/4 â€“ x/3 â‰¤ 4 – 3

(3x â€“ 4x)/12 â‰¤ 1

-x â‰¤ 12

x â‰¥ -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 â‰¤ (x + 8)/ 4, where x is a positive even number.

Solution:-

From the question, it is given that,

(x + 3)/3 â‰¤ (x + 8)/ 4

So, by cross multiplication, we get,

4(x + 3) â‰¤ 3(x + 8)

4x + 12 â‰¤ 3x + 24

Now, by transposing, we get

4x â€“ 3x â‰¤ 24 – 12

x â‰¤ 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 â‰¤ 4x + 9, find the smallest value of x, when:

(i) x âˆˆ Z

Solution:-

From the question,

x + 17 â‰¤ 4x + 9

So, by transposing, we get,

4x – x â‰¥ 17 – 9

3x â‰¥ 8

x â‰¥ 8/3

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

Therefore, smallest value of x = {3}

(ii) x âˆˆ R

Solution:-

From the question,

x + 17 â‰¤ 4x + 9

So, by transposing, we get,

4x – x â‰¥ 17 – 9

3x â‰¥ 8

x â‰¥ 8/3

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

Therefore, the smallest value of x = {
}

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

(i) x âˆˆ R

Solution:-

From the question,

(2x + 7)/3 â‰¤ (5x + 1)/4

So, by cross multiplication, we get,

4(2x + 7) â‰¤ 3(5x + 1)

8x + 28 â‰¤ 15x + 3

Now transposing, we get,

15x â€“ 8x â‰¥ 28 – 3

7x â‰¥ 25

x â‰¥ 25/7

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

Therefore, the smallest value of x = {
}

(ii) x âˆˆ Z

Solution:-

From the question,

(2x + 7)/3 â‰¤ (5x + 1)/4

So, by cross multiplication, we get,

4(2x + 7) â‰¤ 3(5x + 1)

8x + 28 â‰¤ 15x + 3

Now, by transposing, we get,

15x â€“ 8x â‰¥ 28 – 3

7x â‰¥ 25

x â‰¥ 25/7

As per the condition given in the question, x âˆˆ 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 â€“ 11 â‰¤ 7 â€“ 3x, x âˆˆ N.

Solution:-

2x â€“ 11 â‰¤ 7 â€“ 3x

By transposing, we get,

2x + 3x â‰¤ 7 + 11

5x â‰¤ 18

x â‰¤ 18/5

x â‰¤ 3.6

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

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

The set can be represented in a number line as,

(ii) 3(5x + 3) â‰¥ 2(9x – 17), x âˆˆ W.

Solution:-

From the question, it is given that,

3(5x + 3) â‰¥ 2(9x – 17)

15x + 9 â‰¥ 18x – 34

So, by transposing, we get,

18x â€“ 15x â‰¤ 34 + 9

3x â‰¤ 43

x â‰¤ 43/3

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

Therefore, solution set x â‰¤ 43/3

The set can be represented in a number line as,

(iii) 2(3x – 5) > 5(13 â€“ 2x), x âˆˆ W.

Solution:-

From the question, it is given that,

2(3x – 5) > 5(13 â€“ 2x)

6x – 10 > 65 â€“ 10x

So, by transposing, we get,

6x + 10x > 65 + 10

16x > 75

x > 75/16

x >

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

Therefore, solution set x >

The set can be represented in a number line as,

(iv) 3x â€“ 9 â‰¤ 4x â€“ 7 < 2x + 5, x âˆˆ R.

Solution:-

From the question,

Consider 3x â€“ 9 â‰¤ 4x â€“ 7

So, by transposing, we get,

4x – 3x â‰¥ -9 + 7

x â‰¥ -2

Now, consider 4x â€“ 7 < 2x + 5

By transposing, we get,

4x â€“ 2x < 5 + 7

2x < 12

x < 12/2

x < 6

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

Therefore, solution set = [-2 â‰¤ x < 6]

The set can be represented in a number line as,

(v) 2x â€“ 7 < 5x + 2 â‰¤ 3x + 14, x âˆˆ R.

Solution:-

From the question,

Consider 2x â€“ 7 < 5x + 2

By transposing, we get,

5x â€“ 2x > – 7 – 2

3x < – 9

x < -9/3

x < -3

Now, consider 5x + 2 â‰¤ 3x + 14

So, by transposing, we get,

5x – 3x â‰¤ 14 – 2

2x â‰¤ 12

x â‰¤ 12/2

x â‰¤ 6

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

Therefore, solution set = [-3 < x â‰¤ 6]

The set can be represented in a number line as,

(vi) â€“ 3 â‰¤ Â½ – (2x/3) â‰¤
x âˆˆ N.

Solution:-

From the question,

Consider â€“ 3 â‰¤ Â½ – (2x/3)

-3 â‰¤ (3 â€“ 4x)/6

– 18 â‰¤ (3 â€“ 4x)

So, by transposing, we get,

– 18 â€“ 3 â‰¤ – 4x

– 21 â‰¤ – 4x

x â‰¤ 21/4

x â‰¤ 5Â¼

Now, consider Â½ – (2x/3) â‰¤

(3 â€“ 4x)/6 â‰¤ 8/3

By cross multiplication, we get,

3 (3 â€“ 4x) â‰¤ 48

9 â€“ 12x â‰¤ 48

By transposing, we get,

– 12x â‰¤ 48 â€“ 9

– 12x â‰¤ 39

12x â‰¥ – 39

x â‰¥ – 39/12

x â‰¥ -3Â¼

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

Therefore, solution set = [-3Â¼ â‰¤ x â‰¤ 5Â¼]

Set can be represented in a number line as

(vii) 4Â¾ â‰¥ x + 5/6 > 1/3, x âˆˆ R

Solution:-

From the question,

Consider, 4Â¾ â‰¥ x + 5/6

19/4 â‰¥ (6x + 5)/6

114 â‰¥ 24x + 20

By transposing, we get,

114 â€“ 20 â‰¥ 24x

94 â‰¥ 24x

x â‰¤ 94/24

x â‰¤

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 > -Â½

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

Therefore, solution set = [- Â½ < x â‰¤
]

Set can be represented in a number line as

(viii) 1/3 (2x – 1) < Â¼ (x + 5) < 1/6 (3x + 4), x âˆˆ R.

Solution:-

From the question, it is given that,

Consider 1/3 (2x – 1) < Â¼ (x + 5)

By cross multiplication, we get,

4(2x – 1) < 3(x + 5)

8x â€“ 4 < 3x + 15

By transposing, we get,

8x â€“ 3x < 15 + 4

5x < 19

x < 19/5

x <

Then, consider Â¼ (x + 5) < 1/6 (3x + 4)

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

6x + 30 < 12x + 16

By transposing, we get,

6x â€“ 12x < 16 â€“ 30

– 6x < – 14

x >

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

Therefore, solution set = [
< x <
]

Set can be represented in a number line as

(ix) 1/3(5x – 8) â‰¥ Â½ (4x – 7), x âˆˆ R.

Solution:-

From the question, it is given that,

1/3(5x – 8) â‰¥ Â½ (4x – 7)

By cross multiplication, we get,

2(5x – 8) â‰¥ 3(4x – 7)

10x â€“ 16 â‰¥ 12x â€“ 21

Transposing we get,

12x â€“ 10x â‰¤ 21 â€“ 16

2x â‰¤ 5

x â‰¤ 5/2

x â‰¤ 2Â½

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

Therefore, solution set = {x < – 8}

Set can be represented in a number line as

(x) 5/4x > 1 + 1/3 (4x – 1), x âˆˆ R.

Solution:-

From the question,

Consider, (5/4)x > 1 + 1/3 (4x – 1)

(5/4)x > (3 +(4x – 1)/3)

15x > 12 + 16x â€“ 4

By transposing, we get,

15x â€“ 16x > 8

– x > 8

x < – 8

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

Therefore, solution set = {x < – 8}

Set can be represented in a number line as,

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

(i) P â‹‚ Q

(ii) Qâ€™ â‹‚ P

(iii) P â€“ Q

Solution:-

As per the condition given in the question,

P = {x : -3 < x â‰¤ 7, x âˆˆ R}

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

Then, Q = {x : – 7 â‰¤ x < 3, x âˆˆ R}

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

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

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

(ii) Qâ€™ â‹‚ P

Qâ€™ = {3, 4, 5, 6, 7}

Qâ€™ â‹‚ P = {3, 4, 5, 6, 7} â‹‚ {-2, – 1, 0, 1, 2, 3, 4, 5, 6, 7}

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

(iii) P â€“ Q

P â€“ 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}

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

(i) P â‹‚ Q

(ii) Pâ€™ â‹‚ Q

Solution:-

As per the condition given in the question,

P = { x : 7x – 4 > 5x + 2, x âˆˆ R}

7x â€“ 4 > 5x + 2

By transposing, we get,

7x â€“ 5x > 4 + 2

2x > 6

x > 6/2

x > 3

Therefore, P = {4, 5, 6, 7, â€¦.}

Q = { x : x – 19 â‰¥ 1 â€“ 3x, x âˆˆ R}

x â€“ 19 â‰¥ 1 â€“ 3x

By transposing, we get,

x + 3x â‰¥ 1 + 19

4x â‰¥ 20

x â‰¥ 20/4

x â‰¥ 5

Q = {5, 6, 7, 8, â€¦}

Then,

(i) P â‹‚ Q = {2, 3, 4, 5, â€¦.} â‹‚ {5, 6, 7, 8, â€¦}

= {5, 6, 7, 8, â€¦}

(ii) Pâ€™ â‹‚ Q = {âˆ…}