Linear Inequalities

Chapter 5.1: Introduction to Inequalities

• Inequality: A mathematical statement that compares the values of two expressions using symbols like <, >, ≤, or ≥.
• Solution Set: The set of all values that satisfy an inequality.
• Graphing Inequalities: Representing the solution set on a number line.

Chapter 5.2: Linear Inequalities in One Variable

• Linear Inequality: An inequality involving a single variable raised to the power of 1.
• Solving Linear Inequalities: Similar to solving linear equations, but with additional rules for handling inequalities.
• Interval Notation: A concise way to represent solution sets using intervals.

Chapter 5.3: Linear Inequalities in Two Variables

• Linear Inequality in Two Variables: An inequality involving two variables with a degree of 1.
• Graphing Linear Inequalities in Two Variables: Shading the region in the coordinate plane that satisfies the inequality.
• System of Linear Inequalities: A set of two or more linear inequalities.
• Solution Region: The region that satisfies all inequalities in a system.

Key Concepts:

• Inequalities and their solution sets
• Solving linear inequalities in one and two variables
• Graphing linear inequalities
• Systems of linear inequalities and their solution regions

Exercise 5.1

1. Solve 24x < 100, when 

(i) x is a natural number. (ii) x is an integer. 

Ans : 

(i) When x is a natural number

1. Divide both sides by 24:
   • x < 100/24
   • x < 25/6
2. Approximate the value:
   • 25/6 ≈ 4.16
3. Find the natural numbers less than 4.16:
   • x = 1, 2, 3, 4

Therefore, the solution set for x is {1, 2, 3, 4}.

(ii) When x is an integer

1. Use the same inequality as before:
   • x < 25/6
2. Find all integers less than 4.16:
   • x = … -2, -1, 0, 1, 2, 3, 4

Therefore, the solution set for x is {…, -2, -1, 0, 1, 2, 3, 4}.

2. Solve – 12x > 30, when 

(i) x is a natural number. (ii) x is an integer.

Ans : 

(i) When x is a natural number

1. Divide both sides by -12. Remember to reverse the inequality sign when dividing by a negative number:
   • x < -30/12
   • x < -5/2
2. Approximate the value:
   • -5/2 ≈ -2.5
3. Find the natural numbers less than -2.5:
   • There are no natural numbers less than -2.5.

Therefore, there is no solution for x when x is a natural number.

(ii) When x is an integer

1. Use the same inequality as before:
   • x < -5/2
2. Find all integers less than -2.5:
   • x = … -4, -3, -2

Therefore, the solution set for x when x is an integer is {…, -4, -3, -2}.

3. Solve 5x – 3 < 7, when 

(i) x is an integer. (ii) x is a real number.

Ans : 

(i) When x is an integer

1. Add 3 to both sides:
   • 5x < 10
2. Divide both sides by 5:
   • x < 2
3. Find all integers less than 2:
   • x = … -2, -1, 0, 1

Therefore, the solution set for x when x is an integer is {…, -2, -1, 0, 1}.

(ii) When x is a real number

1. Use the same inequality as before:
   • x < 2

Therefore, the solution set for x when x is a real number is (-∞, 2).

4. Solve 3x + 8 >2, when 

(i) x is an integer. (ii) x is a real number

Ans : 

(i) When x is an integer

1. Subtract 8 from both sides:
   • 3x > -6
2. Divide both sides by 3:
   • x > -2
3. Find all integers greater than -2:
   • x = -1, 0, 1, 2, 3, …

Therefore, the solution set for x when x is an integer is {-1, 0, 1, 2, 3, …}.

(ii) When x is a real number

1. Use the same inequality as before:
   • x > -2

Therefore, the solution set for x when x is a real number is (-2, ∞).

Solve the inequalities in Exercises 5 to 16 for real x. 

5. 4x + 3 < 5x + 7

Ans : 

Subtract 4x from both sides:

• 4x + 3 – 4x < 5x + 7 – 4x
• 3 < x + 7

Subtract 7 from both sides:

• 3 – 7 < x + 7 – 7
• -4 < x

6. 3x – 7 > 5x – 1

Ans :

Subtract 3x from both sides:

• 3x – 7 – 3x > 5x – 1 – 3x
• -7 > 2x – 1

Add 1 to both sides:

• -7 + 1 > 2x – 1 + 1
• -6 > 2x

Divide both sides by 2:

• -6/2 > 2x/2
• -3 > x

7. 3(x – 1) ≤ 2 (x – 3)

Ans :  

1. Distribute the terms on both sides:
   • 3x – 3 ≤ 2x – 6
2. Subtract 2x from both sides:
   • 3x – 3 – 2x ≤ 2x – 6 – 2x
   • x – 3 ≤ -6
3. Add 3 to both sides:
   • x – 3 + 3 ≤ -6 + 3
   • x ≤ -3

8. 3 (2 – x) ≥ 2 (1 – x)

Ans : 

Distribute the terms on both sides:

• 6 – 3x ≥ 2 – 2x

Add 2x to both sides:

• 6 – 3x + 2x ≥ 2 – 2x + 2x
• 6 – x ≥ 2

Subtract 6 from both sides:

• 6 – x – 6 ≥ 2 – 6
• -x ≥ -4

Multiply both sides by -1 (remember to reverse the inequality sign):

• (-1)(-x) ≤ (-1)(-4)
• x ≤ 4

9. 

Ans : 

10. 

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11. 

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12. 

Ans :

13. 2 (2x + 3) – 10 < 6 (x – 2)

Ans : 

Distribute the terms on both sides:

• 4x + 6 – 10 < 6x – 12

Combine like terms:

• 4x – 4 < 6x – 12

Subtract 4x from both sides:

• -4 < 2x – 12

Add 12 to both sides:

• 8 < 2x

Divide both sides by 2:

• 4 < x

14. 37 – (3x + 5) > 9x – 8 (x – 3)

Ans : 

Distribute the terms on both sides:

• 37 – 3x – 5 > 9x – 8x + 24

Combine like terms:

• 32 – 3x > x + 24

Add 3x to both sides:

• 32 – 3x + 3x > x + 24 + 3x
• 32 > 4x + 24

Subtract 24 from both sides:

• 32 – 24 > 4x + 24 – 24
• 8 > 4x

Divide both sides by 4:

• 8/4 > 4x/4
• 2 > x

15. 

Ans : 

16. 

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Solve the inequalities in Exercises 17 to 20 and show the graph of the solution in each case on number line

17. 3x – 2 < 2x + 1

Ans :

18. 5x – 3 > 3x – 5

Ans : 

19. 3 (1 – x) < 2 (x + 4)

Ans : 

20. 

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21. Ravi obtained 70 and 75 marks in first two unit test. Find the minimum marks he should get in the third test to have an average of at least 60 marks

Ans : 

Let x be the marks Ravi gets in the third test.

Average marks = (70 + 75 + x)/3

We need the average to be at least 60, so:

(70 + 75 + x)/3 ≥ 60

Multiply both sides by 3:

70 + 75 + x ≥ 180

Combine like terms:

145 + x ≥ 180

Subtract 145 from both sides:

x ≥ 35

Therefore, Ravi should get at least 35 marks in the third test to have an average of at least 60 marks.

22. To receive Grade ‘A’ in a course, one must obtain an average of 90 marks or more in five examinations (each of 100 marks). If Sunita’s marks in first four examinations are 87, 92, 94 and 95, find minimum marks that Sunita must obtain in fifth examination to get grade ‘A’ in the course.

Ans : 

1. Calculate the total marks obtained in the first four examinations:
   Total marks = 87 + 92 + 94 + 95 = 368
2. Calculate the minimum total marks required for an average of 90:
   Minimum total marks = 90 * 5 = 450
3. Calculate the minimum marks required in the fifth examination:
   Minimum marks in the fifth examination = Minimum total marks – Total marks obtained in the first four examinations
   Minimum marks in the fifth examination = 450 – 368 = 82

Therefore, Sunita needs to obtain at least 82 marks in the fifth examination to get grade ‘A’ in the course.

23. Find all pairs of consecutive odd positive integers both of which are smaller than 10 such that their sum is more than 11.

Ans : 

The other integer is x + 2.

Since both integers are smaller than 10:

• x < 10

Since the sum of the two integers is more than 11:

• x + (x + 2) > 11
• 2x + 2 > 11
• 2x > 9
• x > 4.5

Combining the inequalities:

• 4.5 < x < 10

Since x is an odd integer, x can take the values 5 and 7.

24. Find all pairs of consecutive even positive integers, both of which are larger than 5 such that their sum is less than 23.

Ans : 

The other integer is x + 2.

Since both integers are larger than 5:

• x > 5

Since the sum of the two integers is less than 23:

• x + (x + 2) < 23
• 2x + 2 < 23
• 2x < 21
• x < 10.5

Combining the inequalities:

• 5 < x < 10.5

Since x is an even integer, x can take the values 6, 8, and 10.

Therefore, the required possible pairs are (6, 8), (8, 10).

25. The longest side of a triangle is 3 times the shortest side and the third side is 2 cm shorter than the longest side. If the perimeter of the triangle is at least 61 cm, find the minimum length of the shortest side.

Ans : 

The longest side is 3x cm 

and the third side is (3x – 2) cm.

The perimeter is given by:

x + 3x + (3x – 2) = 7x – 2

We are given that the perimeter is at least 61 cm, so:

7x – 2 ≥ 61

Adding 2 to both sides:

7x ≥ 63

Dividing both sides by 7:

x ≥ 9

Therefore, the minimum length of the shortest side is 9 cm.

26. A man wants to cut three lengths from a single piece of board of length 91cm. The second length is to be 3cm longer than the shortest and the third length is to be twice as long as the shortest. What are the possible lengths of the shortest board if the third piece is to be at least 5cm longer than the second? [Hint: If x is the length of the shortest board, then x , (x + 3) and 2x are the lengths of the second and third piece, respectively. Thus, x + (x + 3) + 2x ≤ 91 and 2x ≥ (x + 3) + 5]

Ans : 

• The second piece is x + 3 cm long.
• The third piece is 2x cm long.

We are given two conditions:

1. The total length of the three pieces must be less than or equal to 91 cm.
2. The third piece must be at least 5 cm longer than the second.

Condition 1:

• x + (x + 3) + 2x ≤ 91

Condition 2:

• 2x ≥ (x + 3) + 5

Solving the Inequalities

1. Solving the first inequality:

• 4x + 3 ≤ 91
• 4x ≤ 88
• x ≤ 22

2. Solving the second inequality:

• 2x ≥ x + 8
• x ≥ 8

Combining the inequalities:

• 8 ≤ x ≤ 22

Therefore, the possible lengths of the shortest board are any real number between 8 cm and 22 cm, inclusive.