Linear Inequalities

Linear Inequalities: Comprehensive CBSE Class 11 Notes

Introduction and Basic Concepts

**Definition**: Two real numbers or two algebraic expressions related by the symbols '<' (less than), '>' (greater than), '≤' (less than or equal to), or '≥' (greater than or equal to) form an **inequality**.

**Key Distinction from Equations**:

Equations use the equality sign (=)

Inequalities use relationship symbols and represent a range of values, not a single value

**Examples of Inequalities**:

Numerical inequalities: 3 < 5, 7 > 5

Literal inequalities: x < 5, y > 2, x ≥ 3

Double inequalities: 3 < x < 5 (means x is simultaneously greater than 3 AND less than 5)

Compound form: 3 ≤ x < 5

**Real-Life Application**: If a student's marks in two exams are 62 and 48, and the student needs an average of at least 60 marks, we write: (62 + 48 + x)/3 ≥ 60, which is an inequality used to find minimum marks needed in the third exam.

Types of Inequalities

**Classification by Structure**:

**Strict inequalities**: Use < or > symbols (allow no equality)

Example: ax + b < 0, ax + by > c

**Slack (non-strict) inequalities**: Use ≤ or ≥ symbols (allow equality)

Example: ax + b ≤ 0, ax + by ≥ c

**Classification by Variables**:

**Linear inequalities in one variable**: ax + b < 0 (a ≠ 0)

Forms: ax + b > 0, ax + b ≤ 0, ax + b ≥ 0

**Linear inequalities in two variables**: ax + by < c (a ≠ 0, b ≠ 0)

Forms: ax + by > c, ax + by ≤ c, ax + by ≥ c

**Non-linear inequalities**: ax² + bx + c ≤ 0 (quadratic or higher degree)

**Note**: This chapter covers only linear inequalities

**Exam-Important**: CBSE typically asks questions on linear inequalities in one and two variables only. Quadratic inequalities are not part of this chapter's scope.

Rules for Solving Inequalities

**Rule 1 — Addition/Subtraction**:

Equal numbers may be added to or subtracted from both sides of an inequality **without changing the direction of the inequality sign**.

If a < b, then a + c < b + c

If a > b, then a – c > b – c

**Proof Concept**: This preserves the relative order of numbers.

**Rule 2 — Multiplication/Division by Positive Numbers**:

Both sides of an inequality can be multiplied or divided by the **same positive number without reversing the inequality sign**.

If a < b and c > 0, then ac < bc

If a < b and c > 0, then a/c < b/c

**Rule 2 (Critical) — Multiplication/Division by Negative Numbers**:

When both sides are multiplied or divided by the **same negative number, the inequality sign MUST be REVERSED**.

If a < b and c < 0, then ac > bc

If a < b and c < 0, then a/c > b/c

**Conceptual Basis**: Compare 3 > 2 with –3 < –2. When multiplying both sides by –1, the relationship reverses.

**Common Mistake to Avoid**: Students frequently forget to reverse the inequality when dividing by negative numbers. Always check: "Am I multiplying/dividing by a negative?"

Algebraic Solution of Linear Inequalities in One Variable

**General Method**:

1. Simplify both sides using Rule 1 (addition/subtraction)

2. Collect variable terms on one side and constants on the other

3. Use Rule 2 to isolate the variable

4. Remember to reverse the inequality if multiplying/dividing by a negative number

**Worked Example 1**: Solve 5x – 3 < 3x + 1

**Solution**:

5x – 3 < 3x + 1 (given)

5x – 3 + 3 < 3x + 1 + 3 (add 3 to both sides, Rule 1)

5x < 3x + 4

5x – 3x < 3x + 4 – 3x (subtract 3x from both sides, Rule 1)

2x < 4

x < 2 (divide by positive 2, Rule 2)

**Solution Set**: x ∈ (–∞, 2)

**Worked Example 2**: Solve 4x + 3 < 6x + 7

**Solution**:

4x + 3 < 6x + 7

4x – 6x < 7 – 3 (collect variables on left, constants on right)

–2x < 4

x > –2 (divide by –2, REVERSE inequality, Rule 2)

**Solution Set**: x ∈ (–2, ∞)

**Worked Example 3**: Solve (5 – 2x)/2 ≤ (x – 30)/5

**Solution**:

(5 – 2x)/2 ≤ (x – 30)/5

5(5 – 2x) ≤ 2(x – 30) (multiply by 10, the LCM, Rule 2 with positive number)

25 – 10x ≤ 2x – 60

25 + 60 ≤ 2x + 10x (Rule 1)

85 ≤ 12x

x ≥ 85/12

**Solution Set**: x ∈ [85/12, ∞)

Graphical Representation on Number Line

**For Strict Inequalities (< or >)**:

Use an **open circle (○)** at the boundary point

Draw a bold line extending in the direction of the inequality

**For Slack Inequalities (≤ or ≥)**:

Use a **closed circle (●)** at the boundary point

Draw a bold line extending in the direction of the inequality

**Worked Example**: Graph x < 3

Mark an open circle at 3

Draw a bold line to the left extending toward –∞

This represents all real numbers less than 3

**Worked Example**: Graph x ≥ 1

Mark a closed circle at 1

Draw a bold line to the right extending toward ∞

This represents all real numbers greater than or equal to 1

**Key Convention**:

**Open circle (○)** = excluded endpoint (strict inequality)

**Closed/filled circle (●)** = included endpoint (non-strict inequality)

Solutions for Different Sets of Numbers

**When x is a Natural Number**:

Solution set contains only positive integers satisfying the inequality

**Example**: Solve 30x < 200 when x is natural

30x < 200 → x < 20/3 ≈ 6.67

Solution: {1, 2, 3, 4, 5, 6}

**When x is an Integer**:

Solution set contains all integers (positive, negative, zero) satisfying the inequality

**Example**: Solve 30x < 200 when x is integer

x < 20/3 ≈ 6.67

Solution: {..., –3, –2, –1, 0, 1, 2, 3, 4, 5, 6}

**When x is a Real Number** (default unless stated):

Solution set is an interval on the real number line

**Example**: Solve 30x < 200 when x is real

x < 20/3

Solution: x ∈ (–∞, 20/3) — all real numbers less than 20/3

**Exam-Important**: Unless explicitly stated, assume solutions are required for real numbers and express as intervals.

Double Inequalities and Compound Statements

**Definition**: A double inequality involves three expressions with two inequality signs, e.g., a < x < b or a ≤ x ≤ b.

**Interpretation**: a < x < b means simultaneously x > a AND x < b.

**Solution Method**:

1. Treat as two separate inequalities

2. Solve each simultaneously

3. Find the intersection of both solution sets

**Worked Example**: Solve –8 ≤ 5x – 3 < 7

**Solution**:

This means: (–8 ≤ 5x – 3) AND (5x – 3 < 7)

From –8 ≤ 5x – 3: –5 ≤ 5x, so x ≥ –1

From 5x – 3 < 7: 5x < 10, so x < 2

**Combined**: –1 ≤ x < 2

**Solution Set**: x ∈ [–1, 2)

**Worked Example**: Solve –5 ≤ (5 – 3x)/2 ≤ 8

**Solution**:

From –5 ≤ (5 – 3x)/2: –10 ≤ 5 – 3x, so –15 ≤ –3x, thus x ≤ 5

From (5 – 3x)/2 ≤ 8: 5 – 3x ≤ 16, so –3x ≤ 11, thus x ≥ –11/3

**Combined**: –11/3 ≤ x ≤ 5

**Solution Set**: x ∈ [–11/3, 5]

Systems of Linear Inequalities in One Variable

**Definition**: Multiple inequalities that must all be satisfied simultaneously.

**Solution Method**:

1. Solve each inequality separately

2. Find the common region where all inequalities are true

3. Express the intersection as a single compound inequality

**Worked Example**: Solve the system:

3x – 7 < 5 + x ... (1)

11 – 5x ≤ 1 ... (2)

**Solution**:

From (1): 3x – 7 < 5 + x → 2x < 12 → x < 6

From (2): 11 – 5x ≤ 1 → –5x ≤ –10 → x ≥ 2

**Common solution**: 2 ≤ x < 6

**Solution Set**: x ∈ [2, 6)

**Graphical Representation**: Mark [2, 6) on number line with closed circle at 2, open circle at 6, and bold line between them.

**Exam-Important**: Always express the final answer as a single interval or inequality, not two separate statements.

Practical Applications and Word Problems

**Type 1 — Average/Weighted Problems**:

**Worked Example**: A student scored 62 and 48 marks in two terminal exams. Find minimum marks needed in the annual exam for an average of at least 60.

**Solution**:

Let x = marks in annual exam

(62 + 48 + x)/3 ≥ 60

110 + x ≥ 180

x ≥ 70

**Answer**: Student must score at least 70 marks

**Type 2 — Consecutive Number Problems**:

**Worked Example**: Find all pairs of consecutive odd natural numbers, both larger than 10, with sum less than 40.

**Solution**:

Let x and x + 2 be the consecutive odd numbers

Conditions: x > 10 AND (x + x + 2) < 40

From first: x > 10

From second: 2x + 2 < 40 → x < 19

**Combined**: 10 < x < 19

Since x must be odd: x ∈ {11, 13, 15, 17}

**Pairs**: (11, 13), (13, 15), (15, 17), (17, 19)

**Type 3 — Mixture Problems**:

**Worked Example**: A manufacturer has 600 litres of 12% acid solution. How many litres of 30% acid solution must be added so the resulting mixture has acid content between 15% and 18%?

**Solution**:

Let x = litres of 30% solution to add

Total mixture = x + 600 litres

Acid content conditions:

0.30x + 0.12(600) > 0.15(x + 600)

0.30x + 0.12(600) < 0.18(x + 600)

From first: 30x + 7200 > 15x + 9000 → 15x > 1800 → x > 120

From second: 30x + 7200 < 18x + 10800 → 12x < 3600 → x < 300

**Answer**: 120 < x < 300 litres

**Type 4 — Temperature Conversion Problems**:

**Worked Example**: A solution must be kept between 30°C and 35°C. What is the Fahrenheit range using C = (5/9)(F – 32)?

**Solution**:

Given: 30 < C < 35

Substitute C = (5/9)(F – 32):

30 < (5/9)(F – 32) < 35

Multiply by 9/5:

54 < F – 32 < 63

Add 32:

86 < F < 95

**Answer**: Temperature range is 86°F to 95°F

**Type 5 — Geometric/Physical Constraint Problems**:

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

**Solution**:

Let x = shortest side

Longest side = 3x

Third side = 3x – 2

Perimeter: x + 3x + (3x – 2) ≥ 61

7x – 2 ≥ 61

7x ≥ 63

x ≥ 9

**Answer**: Minimum length of shortest side is 9 cm

Common Mistakes and How to Avoid Them

**Mistake 1**: Forgetting to reverse the inequality when multiplying/dividing by negative numbers.

**Prevention**: Always identify whether the multiplier is positive or negative before applying Rule 2.

**Mistake 2**: Incorrectly handling fractions in inequalities.

**Correct Approach**: Multiply all terms by the LCM to clear fractions, keeping careful track of the inequality sign.

**Mistake 3**: Confusing open and closed circles on the number line.

**Prevention**: Use open circle (○) for < or >, closed circle (●) for ≤ or ≥.

**Mistake 4**: Treating compound inequalities as separate solutions instead of finding the intersection.

**Prevention**: Always find values that satisfy ALL conditions simultaneously.

**Mistake 5**: Forgetting to reverse inequality in double inequalities when multiplying by negatives.

**Correct Method**: Apply the reverse rule to ALL parts of the compound inequality.

**Worked Example of Mistake 5**:

Wrong: –5 ≤ (5 – 3x)/2 ≤ 8 → multiply by 2 → –10 ≤ 5 – 3x ≤ 16 ✓

But then: Subtract 5 → –15 ≤ –3x ≤ 11

Now divide by –3: must reverse BOTH inequalities → 5 ≥ x ≥ –11/3, written as –11/3 ≤ x ≤ 5 ✓

Summary of Key Formulas and Rules

**Rule 1**: a < b ⟹ a ± c < b ± c (holds for all operations with ±)

**Rule 2a**: a < b and c > 0 ⟹ ac < bc and a/c < b/c

**Rule 2b**: a < b and c < 0 ⟹ ac > bc and a/c > b/c (REVERSE)

**Double Inequality Solution**: a < f(x) < b is solved by finding intersection of a < f(x) AND f(x) < b

**System of Inequalities**: Solution is the intersection of all individual solution sets

**Interval Notation**:

(a, b) — open interval, neither endpoint included

[a, b] — closed interval, both endpoints included

[a, b) — half-open, includes a but not b

(–∞, a) — all real numbers less than a

[a, ∞) — all real numbers greater than or equal to a