LINEAR INEQUALITY

Class 11 - Mathematics
Time Allowed: 3 hours Maximum Marks: 164

1. If x is a real number and | x | < 3 , then [1]

a) x ≥ 3 b) − 3 ≤ x ≤ 3

c) - 3 < x < 3 d) x ≥ − 3
2. If a, b, c are real numbers such that a > b, c < 0 [1]

a) ac > bc b) ac ≥ bc

c) ac < bc d) ac ≠ bc
3. The solution set of the inequation: is: [1]
2x−1 3x
− + 1 < 0, x ∈ W
3 5

a) null set b) x ∈ S
le
c) x ∈ W d) x ∈ N
dh
4. If x and a are real numbers such that a > 0 and |x| > a, then [1]

a) x ∈ [−∞, a] b) x ∈ (-a, a)
Pa

c) x ∈ (−a, ∞) d) x ∈ (−∞, −a) ∪ (a, ∞)

5. Solve the system of inequalities x - 2 > 0, 3x < 18 [1]

a) -6 < x < -2 b) 2 < x < 6

c) 1 < x < 3 d) 3 < x < 18

6. If -3x + 17 < -13, then [1]

a) x ∈ (∞, −10] b) x ∈ (−∞, 10]

c) x ∈ (10, ∞) d) x ∈ [10, ∞)

7. Solutions of the inequalities comprising a system in variable x are represented on number lines as given below, [1]
then

a) x ∈ (– ∞, – 4] ∪ [3, ∞) b) x ∈ (–∞, – 4) ∪ [3, ∞)

c) x ∈ [– 3, 1] d) x ∈ [– 4, 3]

8. The solution set for:

|x|−1
≥ 0, x ≠ ±2 [1]
|x|−2

a) (-2, 2) b) (−∞, −2) ∪(−1, 1) ∪(2, ∞)

c) (−∞, −2) ∪(2, ∞) d) (−1, 2) ∪(3, ∞)

9. If x is a real number and |x| < 5, then [1]

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 a) x ≥ 5 b) -5 ≤ x ≤ 5

c) x ≤ -5 d) - 5 < x < 5
10. If x and b are real numbers . If b > 0 and |x| > b , then [1]

a) x ∈ (−∞, −b) ∪ (
b, ∞) b) x ∈ [−∞, b)

c) x ∈ (−b, b) d) x ∈ (−b, ∞)

11. State whether the given statement is True or False: [2]

(a) If a > b and c > 0, then a

c
> b

c
and ac > bc. [1]
(b) The half plane of y ≥ 1 does not contain the origin. [1]
12. Solve: [1]
3x−2 4x−3
≤
5 2

13. Solve: 12x < 50, when x ∈ R. [1]

14. Solve: . [1]
4+2x x
≥ − 3
3 2

15. Solve inequation and represent the solution set on the number line: 5x + 2 < 17 where x ∈ Z [1]
16. Solve: 4x - 2 < 8, when x ∈ R. [1]
17. Solve: 12x < 50, when x ∈ Z. [1]
18. Solve: - 4x > 30, when x ∈ Z. [1]
19. Solve inequation and represent the solution set on the number line: 3x + 8 > 2, x ∈ R [1]
20. Solve: 12 + 1 5

6
x ≤ 5 + 3x when x ∈ R [1]
21. Solve inequation and represent the solution set on the number line: 5x + 2 < 17 where x ∈ R [1]
le
22. Solve: 3x - 7 > x + 1 [1]
23. Solve: - 4x > 30, when x ∈ N. [1]
dh
24. Solve the inequalities: 2 ⩽ 3x − 4 ⩽ 5 [1]
25. Solve: -4x > 30, when x ∈ R [1]
Pa

26. Solve: 4x - 2 < 8, when x ∈ N. [1]

27. Solve each of the following system of equations in R. x + 3 > 0, 2x < 14. [2]
28. The water acidity in a pool is considerd normal when the average pH reading of three daily measurements is [2]
between 8.2 and 8.5. If the first two pH readings are 8.48 and 8.35, find the range of pH value for the third
reading that will result in the acidity level being normal.
29. Solve the inequality and show the graph for the solution on number line: 5x – 3 ≥ 3x – 5 [2]
30. Solve the given system of equations in R. 4x - 1 < 0, 3 - 4x < 0. [2]
31. Solve the given system of equations in R. x - 2 > 0, 3x < 18. [2]
32. Solve the given inequation:
3(x−2)
≥
5(2−x)
. [2]
5 3

33. Solve the given system of equations in R. 10 < - 5 (x - 2 ) < 20. [2]
34. Find all pairs of consecutive odd natural numbers, both of which are larger than 10, such that their sum is less [2]
than 40.
35. Solve the linear inequality 4x + 2 ≥ 14. [2]
36. Solve the inequality 1
(
3x
+ 4) ≥
1
(x − 6) for real x. [2]
2 5 3

37. Solve
|x|−1
≥ 0, x ∈ R − {−2, 2} [2]
|x|−2

38. Solve [2]

2
> 5, x ∈ R
|x−3|

39. Solve the linear inequality: [2]

3
< 1
x−2

40. Solve the inequality 3(x – 1) ≤ 2 (x – 3) for real x. [2]

41. Solve: −1

|x|−2
≥ 1 where x ∈ R, x ≠ 2. [2]

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42. Solve the system of inequations 2x - 1 > x + Represent the solution set on the number line [2]
7−x
> 2, x ∈ R
3

43. Solve 24 x < 100 when [2]

(i) x is a natural number
(ii) x is an integer
44. Find all pairs of consecutive odd positive integers, both of which are smaller than 18, such that their sum is more [2]
than 20.
45. Solve the given system of equations in R:
|x+2|−x
< 2 . [2]
x

46. Solve system of linear inequation: [2]

7x−1 3x+8
< −3, + 11 < 0
2 5

47. The longest side of a triangle is 3 times the shortest side and the third side is 2 cm shorter than the longest side. [3]
If the perimeter of the triangle is at least 61 cm. Find the minimum length of the shortest side.
48. Solve the inequality and show the graph for the solution on number line: x
≥
(5x−2)
−
(7x−3)
[3]
2 3 5

49. Solve 5x - 3 < 3x + 1 when [3]

i. x is a real number
ii. x is an integer
iii. x is a natural number
50. Solve the inequality x

3
>
x

2
+ 1 for real x. [3]
51. Solve the following linear inequations: [3]
i. 2x − 4 ⩽ 0

ii. -5x + 15 < 0

le
52. If a, b, c be non-negative numbers, show that: a2(a - b)(a - c) + b2(b - c)(b - a) + c2(c - a)(c - b) ≥ 0. [3]
dh
53. Solve 3x + 8 > 2 when [3]
(i) x is integer
(ii) x is a real number
Pa

54. To receive grade A in a course one must obtain an average of 90 marks or more in five papers, each of 100 [3]
marks. If Tanvy scored 89, 93, 95 and 91 marks in first four papers, find the minimum marks that she must score
in the last paper to get grade A in the course.
2

55. solve:
x −2x+5

2
>
1
[3]
3x −2x−5 2

56. Solve the following system of linear inequations: [3]

3x - 6 ≥ 0
4x - 10 ≤ 6
[3]
(2x−1) (3x−2) (2−x)
57. Solve the inequality 3
≥
4
−
5
for real x.
2

58. Solve
8x +16x−51

2
> 3 [3]
2x +5x−12

59. Solve the inequality x + for real x. [3]

x x
+ < 11
2 3

60. Solve 1
≥ 1, x ∈ R − [−2, 2] . [3]
2−|x|

61. Find all pairs of consecutive even positive integers, both of which are larger than 5 such that their sum is less [3]
than 23.
[3]
|x|−1
62. Slove: ≥ 0 x ∈ R, x ≠ ±2
|x|−2

63. Solve the system of inequation: , [3]

6x 1 x 1
< ≥
4x−1 2 2x+1 4

64. Solve the following system of inequations: [3]

5x 3x 39
+ >
4 8 8
2x−1 x−1 3x+1
− <
12 3 4

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65. Solve
|x+2|−x
< 2, x ∈ R . [3]
x

66. Find all pairs of consecutive odd positive integers both of which are smaller than 10 such that their sum is more [3]
than 11.
67. Solve the following system of linear inequalities [5]
-2- and 3 - x < 4(x-3)
x 1+x
≥
4 3

[5]
|x+3|+x
68. Solve for x, x+2
>1
69. Solve the following system of linear inequalities. [5]
2(2x + 3) - 10 < 6(x - 2)
and
2x−3 4x
+ 6 ≥ 2+
4 3

70. Solve for x, |x + 1| + |x| > 3 [5]

71. Solve the following system of linear inequalities [5]
<x+ and - > x.
4x 9 3 7x−1 7x+2
−
3 4 4 3 6

72. Read the following text carefully and answer the questions that follow: [4]
In drilling world’s deepest hole, the Kola Superdeep Borehole, the deepest manmade hole on Earth and deepest
artificial point on Earth, as a result of a scientific drilling project, it was found that the temperature T in degree
Celsius, x km below the surface of Earth, was given by:
T = 30 + 25 (x – 3), 3 < x < 15.

If the required temperature lies between 200o C and 300o C, then

le
dh
Pa

i. Find the depth, x will lie between 200o C and 300o C. (1)
ii. Solve forx.-9x + 2 > 18 (1)
iii. If |x| < 5 then find the value of x lies in the interval. (2)
OR
Graph the inequality on the number line: x > −32 (2)
73. Read the following text carefully and answer the questions that follow: [4]
A company produces certain items. The manager in the company used to make a data record on daily basis about
the cost and revenue of these items separately. The cost and revenue functions of a product are given by C(x) =
20x + 4000 and R(x) = 60x + 2000, respectively, where x is the number of items produced and sold.

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 The company manager wants to know:

i. How many items must be sold to realize some profit? (1)

ii. Solve for x: 12x + 7 <-11 (1)
iii. Solve the x:5x - 8 > 40 (2)
OR
Graph the x > 27 inequality on the number line.(2)
74. Read the following text carefully and answer the questions that follow: [4]
Plumbers are responsible for installing and maintaining water systems within buildings, including drinking
water, drainage, heating, sanitation, and sewage systems. Plumbers are not only involved with the installation
and development of new houses and plumbing systems, but also with assessing and fixing problems in existing
and older systems.
A plumber can be paid under two schemes as given below:
le
I: ₹ 600 and ₹ 50 per hour.
II: ₹ 170 per hour.
dh
Pa

i. If the job takes n hours, for how many values of n for which the scheme I will give the plumber the better
wages? (1)
ii. Solve for x: 3x - 91 > -87 and 17x - 16 > 18. (1)
iii. Show the graph represents the system of inequalities. y ≥ x + 2 and y ≥ 3x (2)
−1

OR
Solve: f(x) = {(x - 1) × (2 - x)}/(x - 3) ≥ 0. (2)

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 Solution

LINEAR INEQUALITY

Class 11 - Mathematics

1.
(c) - 3 < x < 3
Explanation:
We have |x| < a ⇔ −a < x < a

2.
(c) ac < bc
Explanation:
The sign of the inequality is to be reversed (< to > or > to <) if both sides of an inequality are multiplied by the same negative
real number.

3. (a) null set

Explanation:
2x−1 3x
− + 1 < 0
3 5
2x−1 3x
⇒ 15 ⋅
3
− 15 ⋅
5
+ 15 < 0 [Multiply the inequality throughout by the L.C.M]
⇒ 5(2x - 1) -3(3x) + 15 < 0
⇒ 10x - 5 - 9x + 15 < 0

⇒ x + 10 < 0
le
⇒ x < -10, but given x ∈ W

Hence the solution set will be null set.

dh
4.
(d) x ∈ (−∞, −a) ∪ (a, ∞)
Explanation:
Pa

|x| > a
⇒ x < -a or x > a

⇒ x ∈ (−∞ , -a) ∪ (a, ∞ )

5.
(b) 2 < x < 6
Explanation:
x-2>0
⇒ x>2
⇒ x ∈ (2, ∞)

Now 3x < 18
⇒ x < 6

⇒ x ∈ (−∞, 6)

So solution set is (2, ∞) ∩ (−∞, 6) = (2, 6)

⇒ 2 < x < 6

6.
(c) x ∈ (10, ∞)
Explanation:
- 3x + 17 < - 13
⇒ - 3x + 17 - 17 < - 13 - 17

⇒ - 3x < - 30
−3x −30
⇒ >
−3 −3

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 ⇒ x > 10
⇒ x ∈ (10, ∞)

7. (a) x ∈ (– ∞ , – 4] ∪ [3, ∞ )
Explanation:
Common solution of the inequalities is from – ∞ to – 4 and 3 to ∞
{(– ∞ , – 4] ∪ [3, ∞ )} ∩ {(– ∞ , – 3] ∪ [1, ∞ )} = (– ∞ , – 4] ∪ [3, ∞ )
8.
(b) (−∞, −2) ∪(−1, 1) ∪(2, ∞)
Explanation:
|x|−1
Given |x|−2
≥ 0, x ≠ ±2

|x|−1
≥ 0
|x|−2

a
⇒ |x| - 1 ≥ 0 and |x| - 2 ≥ 0 or |x| - 1 ≤ 0 and |x| - 2 ≤ 0 [∵ b
≥ 0 ⇒ (a ≥ 0 and b ≥ 0) or (a ≤ 0 and b ≤ 0)]
⇒ |x| ≥ 1 and |x| ≥ 2 or |x| ≤ 1 and |x| ≤ 2
⇒ |x| ≥ 2 or |x| ≤ 1 [∵ |x| ≥ a ⇒ x ≥ a or x ≤ -a and |x| ≤ a ⇒ −a ≤ x ≤ a ]
⇒ x ≥ 2 or x ≤ -2 or −1 ≤ x ≤ 1

⇒ x ∈ (2, ∞) or x ∈ (−∞, −2) or xϵ(−1, 1)

⇒ x ∈ (2, ∞) ∪ (−∞, −2) ∪ (−1, 1)

9.
(d) - 5 < x < 5
Explanation:
le
|x| < 5
⇒ -5 < x < 5
dh

10. (a) x ∈ (−∞, −b) ∪ b, ∞) (

Explanation:
Pa

We have |x| > a ⇔ x < −a or x > a

So |x| > b ⇒ x < −b or x > b
(
⇒ x ∈ (−∞, −b) ∪ b, ∞)

11. State whether the given statement is True or False:

(i) (a) True
Explanation: {
True
(ii) (a) True
Explanation: {
True
12. ⇒ 3x

5
−
2

5
≤
4x

2
−
3

2
3x 4x −3 2
⇒ − ≤ +
5 2 2 5
6x−20x −15+4
⇒ ≤
10 10

⇒ -14x ≤ -11
⇒ 14x ≥ 11
11
⇒ x≥ 14

Therefore, [ 11

14
, ∞) is the solution set.
13. Given 12x < 50
⇒
12x
<
12
[divide both sides by 12]
50

12
25
∴ x <
6

x∈R
When x is a real number, the soluṁṁtion of the given inequation is(−∞, 25

6
) .

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 4+2x x
14. 3
≥
2
− 3 ,
8 + 4x ≥ 3x -18
8 + x ≥ -18
x ≥ -26
Thus, the solution set of the given incquation is [−26, ∞) .
15. Given, 5x + 2 < 17
Subtracting 2 from both the sides in the above equation,
⇒ 5x + 2 – 2 < 17 – 2
⇒ 5x < 15

Dividing both the sides by 5 in the above equation,

5x 15
⇒ <
5 5

⇒ x<3
Since x is an integer.
Therefore, possible values of x can be
x = {…, -2, -1, 0, 1, 2}

16. Given 4x – 2 < 8

⇒ 4x – 2 + 2 < 8 + 2

⇒ 4x < 10

⇒ <
4x

4
[divide both sides by 4]
10

4
5
∴ x <
2

x∈R
le
When x is a real number, the solution of the given inequation is(−∞, 5

2
) .
17. Given 12x < 50
dh
⇒ <
12x

12
[divide both sides by 12]
50

12
25
∴ x <
6

x∈Z
Pa

As 4 < 25

6
< 5 , when x is an integer, the maximum possible value of x is 4.
Thus, the solution of the given inequation is {…, -2, -1, 0, 1, 2, 3, 4}.
18. Given, -4x > 30
−30
⇒ x <
4
[divide both sides by -4]
−15
⇒ x <
2

If x ∈ Z, ⇒ x ∈ {… , −10, −9 − 8}
19. Given, 3x + 8 > 2, x ∈ R
Subtracting 8 from both the sides in above equation
⇒ 3x + 8 – 8 > 2 – 8

⇒ 3x > - 6
Dividing both the sides by 3 in above equation
3x −6
⇒ >
3 3

Thus, x > -2
x ∈ (-2, ∞ )

5
20. 12 + 1 6
x ≤ 5 + 3x when x ∈ R
11x
⇒ 12 +
6
≤ 5 + 3x
⇒ 72 + 11x ≤ 30 + 18x
⇒ 7x ≥ 42

⇒ x ≥ 6

Solution set is {x ∈ R : x ≥ 6} = [6, ∞)

The graph of this set is shown below

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21. Given, 5x + 2 < 17
Subtracting 2 from both the sides in above equation
⇒ 5x + 2 – 2 < 17 – 2

⇒ 5x < 15
Dividing both the sides by 5 in above equation
5x 15
⇒ <
5 5

⇒ x<3
Therefore, x ∈ (-∞ , 3)

22. ⇒ 3x - x > 1 + 7
⇒ 2x > 8
8
⇒ x> 2

⇒ x>4
∴ (4, ∞ ) is the solution set.

23. Given –4x > 30

⇒ − >
4x

4
[divide both sides by 4]
30

4
15
⇒ −x >
2
15
∴ x < −
2
[multiply both sides by -1]
x∈N
As natural numbers start from 1 and can never be negative, when x is a natural number, the solution of the given inequation is ϕ.
24. We have 2 ⩽ 3x − 4 ⩽ 5
⇒ 2 + 4 ⩽ 3x ⩽ 5 + 4 ⇒ 6 ⩽ 3x ⩽ 9

⇒ 2 ⩽ x ⩽ 3
le
25. Now, -4x > 30
−30 −15
⇒ x < =
4 2
dh
15
If x ∈ R, ⇒ x ∈ (−∞, − 2
)

26. Given, 4x - 2 < 8, x ∈ N

4x < 10
Pa

10
x< 4
[divide both sides by 4]
5
x< 2

As 2 < 5

2
< 3 , when x is a natural number, the maximum possible value of x is 2 and we know the natural numbers start from 1.
[divide both sides by 4]
Thus, the solution of the given inequation is {1, 2}.
27. x + 3 > 0
x > -3
x ∈ (−3, ∞)

Also. 2x < 14
x<7
x ∈ (−∞, 7)

Thus, the solution of the given set of inequalities is the intersection (-3, ∞ ) ∩ (−∞ , 7) = (-3, 7}
Therefore, x ∈ (-3, 7) Thus, the solution of the given set of inequalities is (-3, 7)
28. Let the third pH value be x
Given that first pH value = 8.48
and second pH value = 8.35
8⋅48+8⋅35+x
∴ The average value of pH = 3

But the average value of pH lies between 8.2 and 8.5

8⋅48+8⋅35+x
∴ 8 ⋅ 2 < < 8 ⋅ 5
3

⇒ 24.6 < 16.83 + x < 25.5 [multiply inequality by 3]

⇒ 24.6 - 16.83 < x < 25.5 - 16.83

⇒ 7.77 < x < 8.67

Hence, the third pH value lies between 7.77 and 8.67

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29. Here 5x − 3 ⩾ 3x − 5
⇒ 5x − 3x ⩾ −5 + 3

⇒ 2x ⩾ −2

Dividing both sides by 2, we have

x ⩾ −1

The solution set is [−1, ∞)

The representation of the solution set on the number line is

30. We have, 4x - 1 ≤ 0
⇒ 4x ≤ 1
⇒x ≤ ⇒ x ∈ (-∞ , ] ...... (i)
1 1

4 4

Also, 3 - 4x < 0
⇒ 0 > 3 - 4x

⇒ 4x > 3

⇒x >
3

⇒ x ∈ ( , ∞ ) ......(ii)
3

4
3
Hence, the solution set of inequalities is the intersection of (i) and (ii). But, (- ∞ , 1

4
) ∩ ( , ∞) = ϕ
4

Thus, the given set of inequations has no solution.

31. We have, x - 2 > 0
⇒ x > 2

⇒ x ∈ (2, ∞ )

Also. 3x < 18
⇒ x<6
le
⇒ x ∈ (-∞ , 6)

Solution of the given set of the inequations is intersection of (2, ∞ ) ∩ ( - ∞ , 6) = (2, 6)

dh
Thus, (2, 6) is the solution of the given set of inequalities.
32. We have,
3(x−2) 5(2−x)
≥
5 3
Pa

3x−6 10−5x
⇒ ≥
5 3

⇒ 3(3x - 6) ≥ 5 (10 - 5x ) [Multiplying both sides by 15]

⇒ 9x - 18≥ 50 - 25x

⇒ 9x + 25x ≥ 50 +18

⇒ 34x ≥ 68
34x 68
⇒ ≥
34 34

⇒ x≥2
⇒ x ∈ [2, ∞ )

Hence, [2, ∞ ) is the solution set of the given inequation. This solution set can be graphed on real line as shown in Figure.

33. Consider the first inequation,

⇒ 10 ≤ -5(x - 2)

⇒ 2 ≤ -(x - 2)

⇒ 2 ≤ -x + 2
⇒ 2 - 2 ≤ -x

⇒ 0 ≤ -x .

x ≥ 0 ......(i)
Consider the second inequation,
- 5(x - 2) < 20
⇒ -5x + 10 < 20

⇒ -5x < 20 - 10

⇒ -5x < 10

⇒ -x < 2

x > - 2 .......(ii)

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 From (i) and (ii),
(- 2, 0) is the solution set of the simultaneous equations.
34. Let x be the smaller of the two consecutive odd natural numbers. Then the other odd integer is x+2.
It is given that both the natural number are greater than 10 and their sum is less than 40.
∴ x > 10 and, x + x + 2 < 40

⇒ x > 10 and 2x < 38

⇒ x > 10 and x <19

⇒ 10 < x < 19

⇒ x = 11, 13, 15, 17 [∵ x is an odd number]

Hence, the required pairs of odd natural number are (11, 13), (13, 15), (15, 17) and (17, 19).
35. We have, 4x + 2 ≥ 14
⇒ 4x + 2 − 2 ≥ 14 − 2 [subtracting 2 from both sides]

⇒ 4x ≥ 12

⇒
4x

4
≥
12

4
[dividing both sides by 4]
∴ x ≥ 3

Thus, any value of x greater than or equal to 3 satisfies the inequality.

So, the solution set is x ∈ [3, ∞).
This solution set can be represented on number line as given below

Hence, the dark portion on the number line represents the solution of given inequality.
36. Here 1

2
(
3x

5
+ 4) ⩾
1

3
(x − 6)
le
3x x
⇒ + 2 ⩾ − 2
10 3
3x x
⇒ − ⩾ −2 − 2
10 3
dh
9x−10x
⇒ ⩾ −4
30
−x
⇒ ⩾ −4
30

Multiplying both sides by 30, we have

Pa

−x ⩾ −120

Dividing both sides by -1, we have

x ⩽ 120

Thus the solution set is [−∞, 120]

|x|−1
37. We have, ≥ 0
|x|−2

y−1
⇒
y−2
≥ 0
⇒ y ≤ 1 or y > 2
⇒ |x| ≤ 1 or |x| > 2
⇒ (-1 ≤ x ≤ 1) or (x < -2 or x > 2)
⇒ x ∈ [-1, 1] or x ∈ (−∞ , -2) ∪ (2, ∞ )

⇒ x ∈ [-1, 1] ∪ (−∞ , -2) ∪ (2, ∞ )

Hence, the solution set is [-1, 1] ∪ (−∞ , -2) ∪ (2, ∞ )

2
38. We have, |x−3|
> 5

Since x - 3 is positive, we have,

2 > 5|x − 3|
2
⇔ > |x − 3|
5
2
⇔ |x − 3| <
5
−2 2
⇔ < x − 3 <
5 5
−2 2
⇔
5
< x − 3 and x − 3 < 5
−2
⇔
5
+ 3 < x and x < 2

5
+ 3

⇔
13

5
< x and x < 17

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Tapasya Batch
 13 17
⇔ < x <
5 5

13 17
= {x ∈ R : < x < }
5 5

39. Now,
3

x−2
<1
3

x−2
-1<0
3−(x−2)

x−2
<0
3−x+2

x−2
<0
5−x

x−2
<0
x−5

x−2
>0
Case 1: x - 5 > 0 and x - 2 > 0
⇒ x > 5 and x > 2
⇒ x>5
Case 2: x - 5 < 0 and x - 2 < 0
⇒ x < 5 and x < 2

⇒ x < 2

∴ Solution set is (-∞ , 2) ∪ (5, ∞ )

40. It is given in the question that,

3(x – 1) ≤ 2 (x – 3)
⇒ 3x – 3 ≤ 2x – 6

⇒ 3x – 3+ 3 ≤ 2x – 6+ 3

⇒ 3x ≤ 2x – 3

⇒ 3x – 2x ≤ 2x – 3 – 2x
le
⇒ x ≤ -3

∴ The solutions of the given inequality are defined by all the real numbers less than or equal to -3.

Thus, the solutions set to the given inequality are (-∞ , - 3]

dh
−1
41. We have, |x|−2
≥ 1

−1
⇒ − 1 ≥ 0
|x|−2
Pa

−1−(|x|−2)
⇒ ≥ 0
|x|−2

1−|x|
⇒ ≥ 0
|x|−2

|x|−1
⇒ ≤ 0
|x|−2

y−1
⇒
y−2
≤ 0 where y = |x|
⇒ 1 ≤ y < 2 {see figure}
⇒ 1 ≤ |x| < 2 [∵ y = |x|]
⇒ |x| < 2 ⇒ -2 < x < 2
and ⇒ |x| ≥ 1
x ≤ -1 or x ≥ 1
⇒ x ∈ (-2 , -1] ∪ [1 , 2)

Hence, the solution set of the given inequation is (-2, -1] ∪ [1, 2)
42. we have
7−x
2x − 1 > x +
3
> 2
7−x 7x
⇒ 2x − 1 > x +
3
and x + 3
> 2

7−x
Now, 2x − 1 > x + 3
= 6x - 3> 3x + 7 - x
⇒ 6x - 3 > 2x + 7
⇒ 4x - 3 > 7

⇒ 4x > 10
5
⇒ x >
2

5
x ∈ ( , ∞)
2

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Tapasya Batch
 7−x
and x + 3
> 2 = 3x + 7- x > 6
⇒ 2x + 7 > 6
⇒ 2x > -1
−1
⇒ x >
2

−1
⇒ x ∈ ( , ∞)
2

5 −1 5
= ( , ∞) ∩ ( , ∞) = ( , ∞)
2 2 2

The solution set on the line may be represented as shown below.

43. Here 24x < 100

Dividing both sides by 24, we have
100 25
x < ⇒ x <
24 6

(i) when x is a natural number then values of x that make the statement true are 1, 2, 3, 4. The solution set of inequality is {1, 2, 3,
4}
(ii) When x is an integer then values of x that make the statement true are . . . , -2, -1, 0, 1, 2, 3, 4.
The solution set of inequality is {..., -2, -1, 0, 1, 2, 3, 4}
44. Let x be the smaller of the two consecutive odd positive integers. Then, the other odd integers is x + 2.
It is given that both the integers are smaller than 18 and their sum is more than 20. Therefore,
x + 2 < 18 and, x + (x + 2) > 20
⇒ x < 16 and 2x + 2 > 20

⇒ x < 16 and x > 9 ⇒ 9 < x < 16 ⇒ x = 11, 13, 15 [∵ x is an odd integers]

Hence, the requiered pairs odd odd integers are (11, 13), (13, 15) and (15, 17).
le
45. When x ≥ -2 then |x + 2| = x + 2
x+2−x
< 0
x
2
dh
< 0
x

x < 0 and x ≥ -2
x ∈ [-2,0)
when x < -2 then |x + 2| = -(x + 2)
Pa

−x−2−x
< 0
x
−2x−2
< 0
x
x+1
> 0
x

x ∈ (−∞ , 0) ∪(1, ∞ )
Hence, the solution is (−∞ , 0) ∪(1, ∞ ).
46. When,
7x−1
< −3
2

7x−1
⇒ ( ) (2) < −3(2)
2

⇒ 7x -1 < -6
⇒ 7x – 1 + 6 < -6 + 6
⇒ 7x + 5 < 0

⇒ 7x + 5 – 5 < 0 – 5

⇒ 7x < - 5
7x −5
⇒ <
7 7
−9
Therefore, x < 7

Now when,
3x+8

5
+ 11 < 0
3x+8
⇒
5
+ 11 -11 < 0 – 11
3x+8
⇒
5
< -11
3x+8
⇒ (
5
) (5) < -11(5)
⇒ 3x + 8 < -55
⇒ 3x + 8 – 8 < -55 – 8

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Tapasya Batch
 ⇒ 3x < -63
3x −63
⇒
3
<
3
therefore,
⇒ x < -21
−9
Combining both conditions x < -21 and x < 7

⇒ x < -21
⇒ x∈ (-∞ , -21)

47. Let the length of the shortest side be x cm.

Then length of longest side = 3x cm
length of third side = (3x - 2)cm
Perimeter of triangle = x + 3x + 3x - 2
= (7x - 2)cm
Now 7x − 2 ⩾ 61
⇒ 7x ⩾ 61 + 2 ⇒ 7x ⩾ 63 ⇒ x ⩾ 9

Thus the minimum length of shortest side = 9 cm

(5x−2) (7x−3)
48. Here x

2
⩾
3
−
5
x 5x 2 7x 3
⇒ ⩾ − − +
2 3 3 5 5
15x−50x+42x −10+9
⇒ ⩾
30 15
7x −1
⇒ ⩾
30 15

Multiplying both sides by 30, we have

7x ⩾ −2

Dividing both sides by 7, we have

−2
The solution set is [ 7
, ∞)

The representation of the solution set on the number line is

le
dh
49. Here 5x - 3 < 3x + 1
⇒ 5x - 3x < 1 + 3
⇒ 2x < 4
Pa

Dividing both sides by 2

2x 4
∴ <
2 2

⇒ x<2
i. If x is a real number,
Thus solution set of the given inequation is (−∞, 2)
ii. If x is an integer,
Thus solution set of the given inequation is (−∞, 1]
iii. If x is a natural number,
Thus solution set of the given inequation is [1].
50. Here x

3
>
x

2
+ 1

x x
⇒ − > 1
3 2
2x−3x
⇒ > 1
6
−x
⇒ > 1
6

Multiplying both sides by 6, we have

-x > 6
Dividing both sides by -1, we have x < -6
Thus the solution set is (−∞, −6)
51. i. Here 2x − 4 ⩽ 0
⇒ 2x ⩽ 4

Dividing both sides by 2

2x 4
∴ ⩽
2 2

⇒ x ⩽ 2

Thus solution set of the given inequation is (−∞, 2]

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Tapasya Batch
 ii. Here -5x + 15 < 0
⇒ −5x < −15

Dividing both sides by -5

−5x −15
∴ >
5 −5

⇒ x>3
Thus solution set of the given inequation is (3, ∞)
52. Since the give inequality is symmetrical in a, b, c there is no loss of generality in assuming that a ≥ b ≥ c.
∴ a2(a - b)(a - c) + b2(b - c)(b - a) + c2(c - a)(c - b)
= (a - b) {(a2(a - c) - b2(b - c)} + c2(c - a) (c - b)
= (a - b) {(a3 - b3) - c(a2 - b2)} + c2(a - c) (b- c)
= (a - b) {(a - b)(a2 + ab + b2) - c(a - b)(a + b)} + c2(a - c) (b - c)
= (a - b)2 {(a2 + ab + b2 - c(a + b)} + c2(a - c) (b - c)
= (a - b)2 {a2 + ab + b2 - ca - bc} + c2 (a - c) (b - c)
= (a - b)2 {a(a - c) + b (b - c) + ab} + c2(a - c) (b - c)
which is non-negative because each term of RHS is nonnegative since a ≥ b ≥ c and a, b, c be non-negative
Thus, a2(a - b)(a - c) + b2(b - c)(b - a) + c2(c - a)(c - b) ≥ 0.
53. Here 3x + 8 > 2
⇒ 3x > 2 − 8 ⇒ 3x > −6

Dividing both sides by 3, we have

x > -2
(i) When x is an integer then values of x that make the statement true are -1, 0, 1, 2, 3, . . . The solution set of inequality is {-1, 0,
1, 2, 3, . . . }
le
(ii) When x is a real number. The solution set of inequality is x ∈ (−2, ∞)
54. Let x marks be scored by Tanvy in her last paper.
dh
It is given that Tanvy scored 89, 93, 95 and 91 marks in the first 4 papers.
To receive grade A, she must obtain an average of 90 marks or more.
Therefore, the average of there marks must more than equal to 90
Pa

89+93+95+91+x

5
≥ 90
Multiplying both the sides by 5 in the above equation
89+93+95+91+x
⇒(
5
) (5) ≥ 90(5)
⇒ 368 + x ≥ 450
Subtracting 368 from both the sides in the above equation
⇒ 368 + x – 368 ≥ 450 – 368

⇒ x ≥ 82

Therefore, Tanvy should score a minimum of 82 marks in her last paper to get grade A in the course.
2
x −2x+5 1
55. Solve 2
>
2
3x −2x−5
2
x −2x+5 1
>
2 2
3x −2x−5
2
x −2x+5 1
⇒ − > 0
2 2
3x −2x−5
2 2
2(x −2x+5)−(3x −2x−5)
⇒ > 0
2
2(3x −2x−5)

2
− x −2x+15
⇒ > 0
2(3x2 −2x−5)

2
−( x +2x−15)
⇒ > 0
2
2(3x −2x−5)

Multiplying inequality by -ve sign

2
x +2x−15
⇒ < 0
2
2(3x −2x−5)

2
x +2x−15
⇒ < 0
3x2 −2x−5

(x+5)(x−3)
⇒ < 0
(x+1)(3x−5)

On equating all factors to zero, we get x = - 5, -1 ,5 / 3, 3. Plotting these points on number line

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Tapasya Batch
 5
x ∈ (−5, −1) ∪ ( , 3)
3

56. The given system of inequations is 3x - 6 ≥ 0 ...(i)

4x - 10 ≤ 6 .....(ii)
Now 3x - 6 ≥ 0 ⇒ 3x ≥ 6 ⇒ ≥ ⇒ x ≥ 2
3x

3
6

Solution set of inequation (i) is [2, ∞ )

and, 4x -10 < 6 ⇒ 4x ≤ 16 ⇒ x < 4
∴ The solution set of inequation (ii) is (∞ , 4]

The solution sets of inequations (i) and (ii) are represented graphically on the real line in the above figure.
Clearly, the intersection of these solution sets is the set [2,4].
Hence, the solution set of the given system of inequations is the interval [2,4].
(2x−1) (3x−2) (2−x)
57. Here 3
⩾
4
−
5
2x 1 3x 2 2 x
⇒ − ⩾ − − +
3 3 4 4 5 5
2x 3x x −2 2 1
⇒ − − ⩾ − +
3 4 5 4 5 3
40x−45x−12x −30−24+20
⇒ ⩾
60 60
−17x −34
⇒ ⩾
60 60

Multiplying both sides by 60, we have

−17x ⩾ −34

Dividing both sides by -17, we have

−17x −34
le
⩽
−17 −17

⇒ x ⩽ 2
dh
Thus the solution set is (−∞, 2]
58. Solution We have,
2
8x +16x−51
> 3
2
2x +5x−12
Pa

2
8x +16x−51
⇔ − 3 > 0
2
2x +5x−12
2 2
8x +16x−51−6x −15x+36
⇔ > 0
2
2x +5x−12
2
2x +x−15
⇔ > 0
2
2x +5x−12
2
2x +6x−5x−15
⇔ > 0
2
2x +8x−3x−12

(x+3)(2x−5)
⇔
(x+4)(2x−3)
> 0 ...(i)
Sign of expression on number line

Hence
x ∈ (−∞, −4) ∪ (−3, 3/2) ∪ (5/2, ∞)

59. Here x + x

2
+
x

3
< 11

6x+3x+2x
⇒ < 11
6
11x
⇒ < 11
6

Multiplying both sides by 6, we have

11x < 66
Dividing both sides by 11, we have
x<6
Thus the solution set is (−∞, 6)
60. Case I: When x ≥ 0
Then, |x| = x and so 2−|x|
1
≥ 1
x−1
⇒
1

2−x
-1≥0⇒ 2−x
≥ 0
∴ (x - 1 ≥ 0 and 2 - x > 0) or (x - 1 ≤ 0 and 2 - x < 0)

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Tapasya Batch
 ⇒ (x ≥ 1 and x < 2) or (x ≤ 1 and x > 2)
⇒ (1 ≤ x < 2)
⇒ x ∈ (1, 2)

Case II: When x < 0

Then, |x| = -x
so, 1
≥ 1 ⇒
2+x
1
-1≥0
2−|x|

−1−x 1+x
⇒
2+x
≥ 0⇒ 2+x
≤ 0
⇒ (1 + x ≤ 0 and 2 + x > 0) or (1 + x ≥ 0 and 2 + x > 0)
⇒ (-2 < x ≤ -1) ⇒ x ∈ (-2, -1)

∴ solution set = (-2, -1) ∪ (1, 2)

61. Let x and x + 2 be two consecutive even positive integers

Then x > 5 and x + x + 2 < 23
⇒ x > 5 and 2x + 2 < 23

⇒ x > 5 and 2 x < 23 - 2

⇒ x > 5 and 2 x < 21

⇒ x > 5 and x <

21

2
21
⇒ 5 < x <
2
21
⇒ 5 < x <
2

⇒ x = 6, 8, 10
Thus required pairs of even positive integers are (6, 8), (8, 10), and (10, 12).
|x|−1
62. We have, |x|−2
≥ 0

y−1
⇒
y−2
≥ 0 where y = |x|
or y > 2
le
⇒ y ≤ 1

or |x| > 2
dh
⇒ |x| ≤ 1

⇒ (−1 ≤ x ≤ 1) or (x < -2 or x > 2)

⇒ x ∈ [−1, 1] or x ∈ (−∞, −2) ∪ (2, ∞)
Pa

⇒ x ∈ [−1, 1] ∪ (−∞, −2) ∪ (2, ∞)

Hence, solution set of the given inequation is

[−1, 1] ∪ (−∞, −2) ∪ (2, ∞)

63. The first inequation of the given system is

6x 1
⇒ <
4x−1 2

6x 1
⇒ − < 0
4x−1 2

12x−4x+1
⇒ < 0
2(4x−1)

8x+1
⇒ < 0
2(4x−1)

8x+1
⇒ < 0
4x−1

⇒ (8x +1 > 0 and 4x - 1 < 0) or (8x +1 < 0 and 4x - 1 > 0)

−1
(x <
8
and x > 1

4
)

−1 1
⇒ < x <
8 4

−1
⇒ x ∈ (
8
,
1

4
) ........(1)
The second inequation of the given system is
x 1
≥
2x+1 4

x 1
⇒ − ≥ 0
2x+1 4
4x−2x−1 2x−1
⇒ ≥ 0 ⇒ ≥ 0
4(2x+1) 2x+1

⇒ ( 2 x - 1 < 0 and 2x + 1 < 0) or ( 2 x - 1 > 0 and 2 x + 1 > 0)

⇒ (2x < 1 and 2x < -1) or (2x > 1 and 2x > -10)
−1 −1
⇒ (x ≤
1

2
and x < 2
) or (x ≥
1

2
and x > 2
)

−1
⇒ (x <
2
) or (x ≥
1

2
)

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Tapasya Batch
 −1
⇒ x ∈ (−∞,
2
) or x ∈ [ 1

2
, ∞) ......(2)
From (1) and (2)
−1 1 −1 1
x ∈ ( , ) ∩ {(−∞, ) ∪ [ , ∞)} = ϕ
8 4 2 2

Hence, the given system of inequations has no solution

64. The given system of inequation is
5x

4
+ >
8
...(i)
3x 39

8
2x−1 x−1 3x+1

12
−
3
<
4
...(ii)
Now, 5x

4
+
3x

8
>
39

8
10x+3x 39
⇒ >
8 8

⇒ 13x > 39
⇒ x > 3

⇒ x ∈ (3, ∞ )

So, the solution set of inequation (i) is the interval (3, ∞ ).

2x−1 x−1 3x+1
and, 12
−
3
<
4
(2x−1)−4(x−1) 3x+1
⇒ <
12 4
−2x+3 3x+1
⇒ <
12 4

⇒ -2x + 3 < 3(3x + 1) [Multiplying both sides by 12]

⇒ -2x + 3 < 9x + 3

⇒ -2x - 9x < 3 - 3

⇒ -11x < 0
⇒ x > 0

⇒ x ∈ (0, ∞ )
le
So, the solution set of inequation (ii) is the interval (0, ∞ ).
These solution sets are graphed on the real line in Figure (i) and (ii) respectively.
dh
From Fig. (i) and (ii), we observe that the intersection of the solution sets of inequations (i) and (ii) is the interval (3, ∞ )
represented by the common thick line.
Hence, the solution set of the given system of inequations is the interval (3, ∞ ).
Pa

i.
ii.
65. Given:
|x−3|−x

x
< 2, x ∈ R.
Intervals of |x - 3|
|x - 3| = -(x - 3) or (x - 3)
When |x - 3| = x - 3
x-3≥0
Therefore, x ≥ 3
When |x - 3| = -(x - 3)
(x - 3) < 0
Therefore, x < 3
Intervals: x ≥ 3 or x < 3
|x−3|−x
Domain of x
< 2:
|x−3|−x

x
is not defined for x = 0
Therefore, x > 0 or x < 0
Now, combining intervals and domain:
x < 0 or 0 < x < 3 or x ≥ 3
For x = 0
|x−3|−x −(x−3)−x
< 2 < 2
x x

Now, subtracting 2 from both the sides

−(x−3)−x
− 2 < 2 − 2
x
−x+3−x−2x
< 2 − 2
x

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3−4x
< 0
x

Signs of 3 - 4x:
3
3 - 4x = 0 → x = 4

(Subtracting 3 from both the sides and then dividing both sides by -1)
3
3 - 4x > 0 → x < 4

(Subtracting 3 from both the sides and then multiplying both sides by -1)
3
3 - 4x < 0 → x > 4

(Subtracting 3 from both the sides and then multiplying both sides by -1)
Signs of x:
x=0
x<0
x>0
Intervals satisfying the required condition: < 0
3
x < 0 or x > 4

Combining the intervals:

3
x < 0 or x > 4
and x < 0
Merging the overlapping intervals:
x<0
Similarly, for 0 < x < 3
x < 0 or X > and 0 < x < 3 3

Merging the overlapping intervals:

3

4
<x<3
For, x ≥ 3
le
|x−3|−x (x−3)−x
< 2 → < 2
x x

Now, subtracting 2 from both the sides

dh
(x−3)−x
− 2 < 2 − 2
x
x−3−x−2x
< 2 − 2
x
−3−2x
< 0
Pa

Signs of -3 - 2x:
−3
-3 - 2x = 0 → x = 2

(Adding 3 to both the sides and then dividing both sides by -2)
−3
-3 - 2x > 0 → x < 2

(Adding 3 to both the sides and then multiplying both sides by -1)
−3
-3 - 2x < 0 → x > 2

(Adding 3 to both the sides and then multiplying both sides by -1)
Signs of x:
x=0
x<0
x>0
Intervals satisfying the required condition: < 0
−3
x <
2
or x > 0
Combining the intervals:
−3
x < or x > 0 and x ≥ 3
2

Merging the overlapping intervals:

x≥3
Combining all the intervals:
3
x < 0 or 4
< x < 3 or x ≥ 3
Merging overlapping intervals:
3
x < 0 and x > 4

Therefore,
3
x ∈ (−∞, 0) ∪ ( , ∞)
4

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Tapasya Batch
66. Let x and x + 2 be two consecutive odd positive integers
Then x + 2 < 10 and x + x + 2 > 11.
⇒ x < 8 and 2x + 2 > 11

⇒ x < 8 and 2x > 9

⇒ x <8 and 2x > 9
⇒ x < 8 and x >
9

2
9
⇒ < x < 8
2

⇒ x = 5 and 7
Thus required pairs of odd positive integers are 5, and 7.
67. The given system of linear inequalities is
1+x
-2- ≥ x

4
... (i)
3

and 3 - x < 4 (x - 3) ... (ii)

From inequality (i), we get
1+x
-2- ≥ x

4 3

⇒ - 24 - 3x ≥ 4 + 4x [multiplying both sides by 12]

⇒ - 24 - 3x - 4 ≥ 4 + 4x - 4 [subtracting 4 from both sides]

⇒ - 28 - 3x ≥ 4x

⇒ - 28 - 3x + 3x ≥ 4x + 3x [adding 3x on both sides]

⇒ - 28 ≥ 7x

⇒ - [dividing both sides by 7]

28 7x
≥
7 7

⇒ - 4 ≥ x or x ≤ - 4 ... (iii)
Thus, any value of x less than or equal to - 4 satisfied the inequality.
So, solution set is x ∈ (−∞, −4]
le
dh
From inequality (ii), we get
3 - x < 4 (x - 3)
⇒ 3 - x < 4x - 12
Pa

⇒ 3 - x + 12 < 4x - 12 + 12 [adding 12 on both sides]

⇒ 15 - x < 4x

⇒ 15 - x + x < 4x + x [adding x on both sides]

⇒ 15 < 5x

⇒ 3 < x [dividing both sides by 3]

or x > 3 ... (iv)

Thus, any value of x greater than 3 satisfies the inequality.
So, the solution set is x ∈ (3, ∞)

The solution set of inequalities (i) and (ii) are represented graphically on number line as given below:

As no region is common, hence the given system has no solution.

|x+3|+x
68. We have, x+2
>1
|x+3|+x
⇒
x+2
-1>0
|x+3|+x−x−2
⇒
x+2
>0
|x+3|−2
⇒
x+2
>0
Let x + 3 = 0
⇒ x = - 3

∴ x = - 3 is a critical point.

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Tapasya Batch
 So, here we have two intervals (−∞, −3) and [−3, ∞)
Case I: When - 3 ≤ x < ∞ , then |x + 3| = (x + 3)
|x+3|−2
∴
x+2
>0
x+3−2
⇒
x+2
>0
x+1
⇒
x+2
>0
2

> 0 × (x + 2)2
(x+1)(x+2)
⇒
(x+2)

⇒ (x + 1) (x + 2) > 0
Product of (x + 1) and (x + 2) will be positive, if both are of same sign.
∴ (x + 1) > 0 and (x + 2) > 0

or (x + 1) < 0 and (x + 2) < 0

⇒ x > - 1 and x > - 2
or x < - 1 and x < - 2
On number line, these inequalities can be represented as,

Thus, - 1 < x < ∞ or - ∞ < x < - 2

But, here - 3 ≤ x < ∞
- 1 < x < ∞ or - 3 ≤ x < - 2
le
∴

Then, solution set in this case is

x ∈ [- 3, - 2) ∪ (- 1, ∞ )
dh
Case II: When x < - 3, then |x + 3| = - (x + 3)
|x+3|−2
∴
x+2
>0
−x−3−2
>0
Pa

⇒
x+2

−(x+5)
⇒
x+2
>0
x+5
⇒
x+2
<0
2

< 0 × (x + 2)2
(x+5)(x+2)
⇒
x+2

⇒ (x + 5) (x + 2) < 0
Product of (x + 5) and (x + 2) will be negative, if both are of opposite sign.
∴ (x + 5) > 0 and (x + 2) < 0

or (x + 5) < 0 and (x + 2) > 0

⇒ x > - 5 and x < - 2
or x < - 5 and x > - 2
On number line, these inequalities can be represented as,

Thus, - 5 < x < - 2 i.e., solution set in the case is x ∈ (- 5, - 2).

On combining cases I and II, we get the required solution set of given inequality, which is
x ∈ (- 5, - 2) ∪ ( - 1, ∞ )

69. The given system of linear inequalities is

2(2x + 3) -10 < 6(x - 2) ....(i)
2x−3
and 4
+ 6 ≥ 2 + ...(ii)
4x

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 From inequality (i), we get
2(2x + 3) - 10 < 6(x - 2)
⇒ 4x + 6 - 10 < 6x - 12

⇒ 4x - 4 < 6x - 12
⇒ 4x - 4 + 4 < 6x - 12 + 4 [adding 4 on both sides]
⇒ 4x < 6x - 8

⇒ 4x - 6x < 6x - 8 - 6x [subtracting 6x from both sides]

⇒ -2x < -8

⇒ 2x > 8 [dividing both sides by - 1 and then inequality sign will change]
2x 8
⇒
2
> [dividing both sides by 2]
2

∴ x > 4 ...(iii)
Thus, any value of x greater than 4 satisfies the inequality.
∴ Solution set is x ∈ (4, ∞)

The representation of solution of inequality (i) is

From inequality (ii), we get

2x−3 4x 2x−3+24 6+4x
+ 6 ≥ 2 + ⇒ ≥
4 3 4 3
2x+21 6+4x
⇒ ≥ ⇒ 3(2x + 21) ≥ 4(6 + 4x)
4 3

⇒ 6x + 63 ≥ 24 + 16x

⇒ −10x ≥ −39 ⇒ 10x ≤ 39

10x 39
⇒ ≤
10 10
le
⇒ ...(iv)
x ≤ 3.9

Thus, any value of x less than or equal to 3.9 satisfies the inequality.
dh
∴ Solution set is x ∈ (−∞, 3.9] .

Its representation on number line is

Pa

From Eqs. (iii) and (iv), it is clear, that there is no common value of x, which satisfies both inequalities (iii) and (iv).
Hence, the given system of inequalities has no solution.
70. We have, |x + 1| + |x| > 3
Put x + 1 = 0 and x = 0 ⇒ x = - 1 and x = 0
∴ x = - 1, 0 are critical point.

So, we will consider three intervals (−∞ , - 1), [- 1, 0), [0, ∞ )

Case I: When - ∞ < x < - 1, then |x + 1| = - (x + 1) and |x| = - x
∴ |x + 1| + |x| > 3

⇒ - x - 1 - x > 3

⇒ - 2x - 1 > 3

⇒ - 2x > 4

⇒ x < - 2

Case II: When - 1 ≤ x < 0, then |x + 1| = x + 1 and |x| = - x

∴ |x + 1| + |x| > 3

⇒ x + 1 - x > 3 ⇒ 1 > 3 [not possible]

Case III: When 0 ≤ x < ∞ , then |x + 1| = x + 1 and |x| = x

∴ |x + 1| + |x| > 3

⇒ x+1+x>3
⇒ 2x + 1 > 3 ⇒ 2x > 2
∴ x > 1

On combining the results of cases I, II and III, we get

x < - 2 and x > 1
∴ x ∈ (−∞ , - 2) ∪ (1, ∞ )

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71. We have, 4x

3
−
9

4
<x+ 3

4
... (i)
and 7x−1

3
−
7x+2

6
> x ... (ii)
From inequality (i), we get
16x−27 4x+3
4x

3
−
9

4
<x+ 3

4
⇒
12
< 4

⇒ 16x - 27 < 12x + 9 [multiplying both sides by 12]

⇒ 16x - 27 + 27 < 12x + 9 + 27 [adding 27 on both sides]

⇒ 16x < 12x +36

⇒ 16x - 12x < 12x + 36 - 12x [ subtracting 12x from bot sides]
⇒ 4x < 36 ⇒ x < 9 [dividing both sides by 4]

Thus, any value of x less than 9 satisfies the inequality. So, the solution of inequality (i) is given by x ∈ (−∞, 9)

From inequality (ii) we get,

7x−1 7x+2 14x−2−7x−2

3
- 6
>x⇒ 6
>x
⇒ 7x - 4 > 6x [multiplying by 6 on both sides]
⇒ 7x - 4 + 4 > 6x + 4 [adding 4 on both sides]

⇒ 7x > 6x + 4

⇒ 7x - 6x > 6x + 4 - 6x [subtracting 6x from both sides]

∴ x > 4

Thus, any value of x greater than 4 satisfies the inequality.

So, the solution set is x ∈ (4, ∞)
le
The solution set of inequalities (i) and (ii) are represented graphically on number line as given below:
dh

Clearly, the common value of x lie between 4 and 9.

Hence, the solution of the given system is, 4 < x < 9 i.e., x ∈ (4, 9)
Pa

72. i. Here T = 30 + 25 (x - 3), 3 < x < 15

Now 200 < 30 + 25 (x - 3) < 300
⇒ 170 < 25(x − 3) < 270
170 270
⇒ < (x − 3) <
25 25

⇒ 6.8 < (x - 3) < 10.8

⇒ 6.8 + 3 < x < 10.8 + 3

⇒ 9.8 < x < 13.8

Thus required depth will be between 9.8 km and 13.8 km.

ii. -9x + 2 > 18
-9x > 18 - 2
-9x > 16
−16
x> 9

iii. If x is real number and |x| < 5

Then |x| < 5 will give us
x < 5 and -x < 5 for x ≥ and x < 0
0 ≤ x ≤ 5 and 0 > x > -5
So, combine and get -5 < x < 5

OR
The inequality on the number line: x > −32

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73. i. We have, profit = Revenue - Cost
= (60x + 2000) - (20x + 4000)
= 40x - 2000
To earn some profit, 40x - 2000 > 0.
⇒ x > 50

Hence, the manufacturer must sell more than 50 items to realise some profit.
ii. 12x + 7 < -11
12x < -11 - 7
12x < -18
−3
x< 2

iii. 5x - 8 > 40
5x > 40 + 8
5x > 48
x> 48

OR

74. i. Determine the equations:

The total number of hours a job takes is n hours.
Therefore, the wage earned according to scheme I = 600 + (50 × n)
The wage earned according to scheme II = 170 × n
Find the values of n:
For scheme I to be better for wages:
Wages earned according > Wages earned according to the scheme II to the scheme I
le
⇒ 600 + (50 × n) > 170 × n

⇒ 600 + 50n > 170n

dh
⇒ 600 > 170n - 51 [Subtracting 50n on both sides]
⇒ 600 > 120n

⇒ > n [Dividing both sides by 120]

600
Pa

120

⇒ 5>n
∴ n < 5

Therefore, for n < 5 hours, scheme I gives the plumber better wages.
ii. Given: 3x - 91 > -87 and 17x - 16 > 18
3x - 91 > -87...(i)
17x - 16 > 18 ...(ii)
Consider (i),
3x - 91 > -87
Adding 91 to both sides,
⇒ 3x - 91 + 91 > -87 + 91

⇒ 3x > 4

Dividing by 3 on both sides,

3x 4
⇒ >
3 3

⇒ x> 4

⇒ x > 1.33333
Consider (ii),
17x - 16 > 18
Adding 16 to both sides,
⇒ 17x - 16 + 16 > 18 + 16

⇒ 17x > 34

Dividing by 17 on both sides,

17x 34
⇒ >
17 17

⇒ x>2

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 We got x > 1.33333 and x > 2
Hence, x > 2
−1
iii. y ≥ 3
x + 2 and y ≥ 3x

OR
(-∞ , 1] ∪ (2, 3)

le
dh
Pa

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