Frank Solutions for Class 10 Maths Chapter 5 Linear Inequations

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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 – 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 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 ∈ 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, by 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, the smallest value of x = {
}

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, the smallest value of x = {
}

(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, by 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}

The set can be represented in a number line as,

(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

The set can be represented in a number line as,

(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 >

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

Therefore, solution set x >

The set can be represented in a number line as,

(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]

The set can be represented in a number line as,

(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]

The set can be represented in a number line as,

(vi) – 3 ≤ ½ – (2x/3) ≤
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) ≤

(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 a number line as

(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 ≤

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 ≤
]

Set can be represented in a number line as

(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 <

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 >

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

Therefore, solution set = [
< x <
]

Set can be represented in a number line as

(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 a number line as

(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 a number line as,

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}

(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}

(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}

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, …}

(ii) P’ ⋂ Q = {∅}