Chapter Overview
Linear Inequalities is Chapter 5 of CBSE Class 11 Mathematics and belongs to the Algebra unit (Unit II). It extends the idea of linear equations to inequalities — statements where one quantity is less than or greater than another. This chapter lays the groundwork for linear programming in Class 12 and optimisation problems in competitive exams.
This guide covers every NCERT topic with rules, tables, worked examples, graphical methods, common mistakes, and exam tips.
Key Definitions
| Term | Definition |
|---|---|
| Inequality | A statement that two expressions are not equal, using <, >, ≤, or ≥. |
| Strict inequality | Uses < or > — the boundary value is not included. Example: x > 3. |
| Slack inequality | Uses ≤ or ≥ — the boundary value is included. Example: x ≤ 5. |
| Linear inequality | An inequality involving a linear expression in one or two variables. |
| Solution set | The set of all values of the variable that satisfy the inequality. |
| Solution region | The region on a graph where all points satisfy the inequality. |
| Feasible region | The common region satisfying all inequalities in a system. |
Rules for Solving Inequalities
One-Variable Inequality Rules
| Rule | Effect on Inequality Sign | Example |
|---|---|---|
| Add the same number to both sides | No change | x − 3 > 5 → x > 8 |
| Subtract the same number from both sides | No change | x + 2 ≤ 7 → x ≤ 5 |
| Multiply both sides by a positive number | No change | 2x < 6 → x < 3 |
| Multiply both sides by a negative number | Sign reverses | −2x < 6 → x > −3 |
| Divide both sides by a positive number | No change | 3x ≥ 9 → x ≥ 3 |
| Divide both sides by a negative number | Sign reverses | −3x ≥ 9 → x ≤ −3 |
Critical Rule: Multiplying or Dividing by a Negative Number
When you multiply or divide both sides of an inequality by a negative number, the inequality sign reverses:
<becomes>>becomes<≤becomes≥≥becomes≤
This is the most important rule and the source of most errors.
Solving Linear Inequalities in One Variable
Procedure
1. Remove fractions by multiplying both sides by the LCM of denominators (watch the sign). 2. Expand brackets if present. 3. Collect all variable terms on one side and constants on the other. 4. Divide by the coefficient of the variable (reverse the sign if the coefficient is negative). 5. Express the solution as an interval or on a number line.
Example 1: Simple Inequality
Solve: 3x − 5 < 7
3x < 12 x < 4 Solution: x ∈ (−∞, 4)
Example 2: Inequality with Negative Coefficient
Solve: −4x + 7 ≥ 15
−4x ≥ 8 x ≤ −2 (sign reversed because we divided by −4) Solution: x ∈ (−∞, −2]
Example 3: Fractional Inequality
Solve: (x + 3)/(2) > (2x − 1)/3
Multiply both sides by 6 (LCM of 2 and 3): 3(x + 3) > 2(2x − 1) 3x + 9 > 4x − 2 9 + 2 > 4x − 3x 11 > x Solution: x ∈ (−∞, 11) or x < 11
Solving Linear Inequalities in Two Variables
A linear inequality in two variables has the form ax + by + c < 0, ax + by + c > 0, ax + by + c ≤ 0, or ax + by + c ≥ 0.
Graphical Solution Method
1. Replace the inequality sign with = to get the boundary line ax + by + c = 0. 2. Draw the line on the coordinate plane:
3. Pick a test point not on the line (usually the origin (0,0)):
- Solid line for ≤ or ≥ (boundary included)
- Dashed line for < or > (boundary excluded)
- If it satisfies the inequality, shade the region containing the test point.
- If it does not satisfy the inequality, shade the opposite region.
Example: Graph x + 2y > 4
1. Boundary line: x + 2y = 4 (dashed, since > is strict). 2. Test point (0,0): 0 + 0 = 0 > 4? No. 3. Shade the region NOT containing (0,0), i.e., the side away from the origin.
Solution Region Types
| Inequality | Boundary Line | Shading |
|---|---|---|
| ax + by + c < 0 | Dashed | Side where test point satisfies |
| ax + by + c > 0 | Dashed | Side where test point satisfies |
| ax + by + c ≤ 0 | Solid | Side where test point satisfies |
| ax + by + c ≥ 0 | Solid | Side where test point satisfies |
Systems of Linear Inequalities
A system of linear inequalities has two or more inequalities that must all be satisfied simultaneously.
Graphical Method
1. Graph each inequality on the same coordinate plane. 2. Shade the solution region for each. 3. The feasible region (common shaded area) is the solution — where all inequalities overlap.
Example: System of Two Inequalities
Solve graphically: x + y ≤ 5 and x − y ≥ 1
1. Line 1: x + y = 5 (solid). Shade below (test (0,0): 0 ≤ 5 ✓). 2. Line 2: x − y = 1 (solid). Shade above (test (0,0): 0 ≥ 1? No → shade opposite side). 3. Feasible region: the overlap of the two shaded regions.
Example: System of Three Inequalities
Solve: x ≥ 0, y ≥ 0, x + y ≤ 4
These three inequalities define a triangular feasible region in the first quadrant bounded by the axes and the line x + y = 4. Vertices: (0,0), (4,0), (0,4).
Word Problems Involving Inequalities
Example 1: Age Problem
Ravi is at least 5 years older than Priya. The sum of their ages is at most 30. If Priya is x years old, write the inequalities.
- Ravi's age ≥ x + 5 → Ravi ≥ x + 5
- x + Ravi ≤ 30
- Substituting: x + (x + 5) ≤ 30 → 2x + 5 ≤ 30 → 2x ≤ 25 → x ≤ 12.5
Example 2: Budget Problem
A student has at most Rs 120 to spend on books and notebooks. Each book costs Rs 25 and each notebook costs Rs 10. If she buys x books and y notebooks, write the inequalities.
- 25x + 10y ≤ 120
- x ≥ 0, y ≥ 0 (non-negative quantities)
This is a linear programming-style constraint.
Example 3: Score Problem
A student scores 60 marks in Maths and 75 marks in Science. She needs an average of at least 70 across three subjects. What minimum marks must she score in English?
Let English marks = x. (60 + 75 + x)/3 ≥ 70 135 + x ≥ 210 x ≥ 75 marks
Important Numericals
1. Compound Inequality
Solve: −3 ≤ 2x + 1 < 7
Split into two: −3 ≤ 2x + 1 AND 2x + 1 < 7 First: −3 ≤ 2x + 1 → −4 ≤ 2x → −2 ≤ x Second: 2x + 1 < 7 → 2x < 6 → x < 3 Combined: −2 ≤ x < 3, i.e., x ∈ [−2, 3)
2. System in One Variable
Solve: 3x − 1 ≥ 5 and 2x + 3 < 11.
First: 3x ≥ 6 → x ≥ 2 → [2, ∞) Second: 2x < 8 → x < 4 → (−∞, 4) Intersection: [2, 4)
3. Graphical Two-Variable Problem
Find the feasible region for: 2x + y ≥ 8, x + 2y ≥ 10, x ≥ 0, y ≥ 0.
- Line 1: 2x + y = 8. Test (0,0): 0 ≥ 8? No. Shade opposite.
- Line 2: x + 2y = 10. Test (0,0): 0 ≥ 10? No. Shade opposite.
- x ≥ 0, y ≥ 0: first quadrant.
- Feasible region: the unbounded region in Q-I above both lines.
- Corner point of intersection: Solve 2x + y = 8 and x + 2y = 10 → (2, 4).
Practice Problems
Problem 1
Solve: 24x < 100 when x is a natural number.
24x < 100 → x < 100/24 → x < 4.167 Since x is a natural number, x ∈ {1, 2, 3, 4} Solution set: {1, 2, 3, 4}
Problem 2
Solve the system: 2x + 3y ≤ 12, x + y ≥ 4, x ≥ 0, y ≥ 0 graphically.
- Line 1: 2x + 3y = 12 (solid). Test (0,0): 0 ≤ 12 ✓. Shade below.
- Line 2: x + y = 4 (solid). Test (0,0): 0 ≥ 4? No. Shade above.
- x ≥ 0, y ≥ 0: first quadrant.
- Corner points: (0, 4), (4, 0) from Line 2 with axes; intersection of 2x + 3y = 12 and x + y = 4: 2x + 3(4 − x) = 12 → 2x + 12 − 3x = 12 → x = 0, y = 4. So (0, 4) is a corner. Also x = 4, y = 0: check 2(4) + 3(0) = 8 ≤ 12 ✓. (4, 0) is a corner.
Problem 3
A manufacturer has 600 litres of a chemical solution. At least 120 litres must be used for Product A and at least 200 litres for Product B. If Product A uses 2 litres per unit and Product B uses 3 litres per unit, write the constraints.
Let x = units of Product A, y = units of Product B.
- 2x ≥ 120 (at least 120 litres for A) → x ≥ 60
- 3y ≥ 200 (at least 200 litres for B) → y ≥ 200/3 ≈ 66.67
- Total: 2x + 3y ≤ 600
- x ≥ 0, y ≥ 0
NCERT Exercise Mapping
| Exercise | Focus | Key Question Types |
|---|---|---|
| Exercise 6.1 | One-variable inequalities | Solve linear inequalities, express as intervals |
| Exercise 6.2 | Two-variable inequalities | Graph solution regions on coordinate plane |
| Exercise 6.3 | Systems of inequalities | Find feasible region for systems |
| Miscellaneous | Word problems | Real-world constraints, optimization setup |
Common Mistakes
1. Forgetting to reverse the inequality sign: When multiplying or dividing by a negative number, the sign MUST reverse. This is the number one mistake in the entire chapter. 2. Confusing < with ≤ on graphs: A dashed line means the boundary is NOT included. A solid line means it IS included. Drawing the wrong line type loses marks. 3. Wrong test point selection: Never pick a test point that lies on the boundary line. If (0,0) is on the line, pick (1,0) or (0,1) instead. 4. Ignoring the intersection in systems: The solution to a system is the OVERLAP of all regions, not the union. Only points in ALL shaded regions simultaneously satisfy the system. 5. Not stating the solution as an interval: Always express the final answer as a proper interval like [2, 5) or (−∞, 4), not just "x < 4". 6. Dividing by a variable that could be zero or negative: If you have ax > b and you do not know the sign of a, you cannot simply divide. Consider cases (a > 0 and a < 0) separately. 7. Shading the wrong half-plane: Always verify with a test point. Guessing which side to shade leads to errors.
Exam Tips
- 1-mark questions: Solve a simple one-variable inequality or identify strict vs slack.
- 2-mark questions: Solve an inequality with fractions, or solve a compound inequality like −5 ≤ 3x + 2 < 8.
- 4-mark questions: Graph a two-variable inequality and identify the solution region, or solve a system of two inequalities.
- 6-mark questions: Solve a system of three inequalities graphically and identify the feasible region with corner points, or solve a word problem with inequalities.
- Always write the rule you are applying (e.g., "Dividing both sides by −3, inequality reverses").
- For graphical problems, clearly label the boundary line, indicate solid/dashed, and mark the shaded region.
- For word problems, define variables, write all inequalities, and solve step by step.
- Check your answer by substituting a value from the solution region back into the original inequality.
Key Results Summary
| Result | Formula/Statement |
|---|---|
| Sign reversal rule | Multiply/divide by negative → flip inequality sign |
| Compound inequality | Split into two, solve each, take intersection |
| Solution region | Half-plane determined by boundary line and test point |
| Feasible region | Intersection of all individual solution regions |
| Boundary line | Solid (≤, ≥) or dashed (<, >) |
| Interval notation | Use brackets [ ] for included, parentheses ( ) for excluded |
| Word problem tip | Define variables, write all constraints, solve step by step |
Study Plan for This Chapter
| Day | Task | Duration |
|---|---|---|
| Day 1 | Read NCERT, learn inequality rules (sign reversal is critical) | 40 min |
| Day 2 | Practice one-variable inequalities with fractions and negatives | 45 min |
| Day 3 | Graph two-variable inequalities, learn test point method | 45 min |
| Day 4 | Solve systems of inequalities, find feasible regions | 50 min |
| Day 5 | NCERT exercise questions + word problems | 50 min |
| Day 6 | Revise with this guide, attempt mixed problems | 40 min |
Quick Revision Notes
- Inequalities use <, >, ≤, ≥. Strict: <, >. Slack: ≤, ≥.
- Adding or subtracting: no change to sign.
- Multiplying or dividing by positive: no change to sign.
- Multiplying or dividing by negative: REVERSE the sign.
- One-variable solution: express as interval on the number line.
- Two-variable solution: graph boundary line (solid for ≤/≥, dashed for </>), test a point, shade the correct half-plane.
- System of inequalities: the feasible region is the overlap (intersection) of all individual solution regions.
- Corner points of the feasible region are found by solving pairs of boundary line equations.
- Word problems: define variables, write all constraints as inequalities, solve.
- Always verify by substituting a test point into the original inequality.