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
Answer
(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
Answer
(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}



Answer

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



Answer

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.



Answer

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.



Answer

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.



Answer

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.



Answer

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.



Answer

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.



Answer

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



Answer

(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



Answer

(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}
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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.



Answer

(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



Answer

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



Answer

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 = {∅}