# Frank Solutions for Chapter 5 Linear Inequations Class 10 ICSE Mathematics

### Exercise 5

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

(i) x + 5 > 11

(ii) 2x + 1 < 17

(iii) 3x – 5 ≤ 7

(iv) 8 – 3x ≥ 2

(v) 5 – 2x < 11

(i) 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

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

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

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

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

(ii) 2x – 3 > 7

(iii) 3x + 2 ≤ 11

(iv) 14 – 3x ≥ 5

(v) 7x + 11 > 16 – 3x

(vii) 2(3x – 5) ≤ 8

(viii) x + 7 ≥ 15 + 3x

(ix) 2x – 7 ≥ 5x + 8

(x) 9 – 4x ≤ 15 – 7x

(i) 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

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

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

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

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

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

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

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

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

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}

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

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.

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.

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.

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.

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.

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.

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

(ii) x R

(i) x ∈ Z

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

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 = {8/3}

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

(i) x R

(ii) x Z

(i) x ∈ R

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 = {25/7 }

(ii) x ∈ Z

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 = {4}

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

(i) 2x – 11 ≤ 7 – 3x, x N.

(ii) 3(5x + 3) ≥ 2(9x – 17), x W.

(iii) 2(3x – 5) > 5(13 – 2x), x W.

(iv) 3x – 9 ≤ 4x – 7 < 2x + 5, x R.

(v) 2x – 7 < 5x + 2 ≤ 3x + 14, x R.

(vi) – 3 ≤ ½ – (2x/3) ≤ 2.2/3, x N.

(vii) 4¾ ≥ x + 5/6 > 1/3, x R

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

(ix) 1/3(5x – 8) ≥ ½ (4x – 7), x R.

(x) 5/4x > 1 + 1/3 (4x – 1), x R.

(i) 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,

(ii) 3(5x + 3) ≥ 2(9x – 17), x ∈ W.

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,

(iii) 2(3x – 5) > 5(13 – 2x), x ∈ W.

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 > 4.6875

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

Therefore, solution set x > 4.6875

Set can be represented in number line as,

(iv) 3x – 9 ≤ 4x – 7 < 2x + 5, x ∈ R.

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,

(v) 2x – 7 < 5x + 2 ≤ 3x + 14, x ∈ R.

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,

(vi) – 3 ≤ ½ – (2x/3) ≤ 2.2/3 x ∈ N.

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) ≤ 2.2/3

(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 number line as

(vii) 4¾ ≥ x + 5/6 > 1/3, x ∈ R

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 ≤ 3.92

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 ≤ 3.92]

Set can be represented in number line as

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

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 < 3.8

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 > 2.1/3

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

Therefore, solution set = [ 2.1/3 < x < 3.4/5]

Set can be represented in number line as

(ix) 1/3(5x – 8) ≥ ½ (4x – 7), x ∈ R.

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

(x) 5/4x > 1 + 1/3 (4x – 1), x ∈ R.

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,

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

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

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 = {∅}