An inequality is a statement comparing two expressions using <, >, ≤, or ≥. Linear inequalities involve linear expressions (no powers higher than 1). They arise naturally in real life — budgeting, resource allocation, and optimization problems all use inequalities.
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Basic Concepts
- Types of inequalities:
- Strict: a < b or a > b
- Non-strict (weak): a ≤ b or a ≥ b
- 1.Properties of Inequalities:
- 2.Adding or subtracting the same number from both sides preserves the inequality.
- 3.Multiplying or dividing both sides by a positive number preserves the inequality.
- 4.Multiplying or dividing both sides by a negative number reverses the inequality.
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Linear Inequalities in One Variable
To solve ax + b < c (a > 0): subtract b from both sides, then divide by a.
Solution set: The set of all real numbers satisfying the inequality.
Number line representation: Use open circles (o) for strict inequalities and closed circles (•) for non-strict.
- Interval notation:
- (a, b): a < x < b
- [a, b]: a ≤ x ≤ b
- (a, ∞): x > a
- (-∞, b]: x ≤ b
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Linear Inequalities in Two Variables
A linear inequality in two variables (e.g., 2x + 3y ≤ 6) has a half-plane as its solution.
- 1.Steps to graph:
- 2.Replace the inequality with = and draw the line (boundary).
- 3.Use a dashed line for strict inequalities; solid line for non-strict.
- 4.Test a point (usually origin) not on the line. If it satisfies the inequality, shade that half-plane; otherwise shade the other side.
System of Linear Inequalities: The solution is the intersection (overlapping region) of the individual half-planes.
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Worked Examples
Solve 3x − 5 < 7.
Add 5: 3x < 12. Divide by 3: x < 4. Solution: x ∈ (−∞, 4).
Solve −2x + 6 ≥ 2.
Subtract 6: −2x ≥ −4. Divide by −2 (inequality reverses): x ≤ 2. Solution: x ∈ (−∞, 2].
Solve 5 < 2x − 3 < 9.
Add 3 throughout: 8 < 2x < 12. Divide by 2: 4 < x < 6. Solution: x ∈ (4, 6).
Solve the system: x + y ≤ 5 and x − y ≥ 1 graphically (describe the process).
Draw x + y = 5 (solid line) and x − y = 1 (solid line). Shade below the first and to the right of the second. The feasible region is their intersection.
A shopkeeper sells two products. The profit per unit of product A is Rs 20 and product B is Rs 30. She must produce at least 5 units total and can make at most 20 units of A. Set up the inequalities.
Let x = units of A, y = units of B. x + y ≥ 5, x ≤ 20, x ≥ 0, y ≥ 0.
Solve |2x − 3| < 5.
−5 < 2x − 3 < 5. Add 3: −2 < 2x < 8. Divide by 2: −1 < x < 4. Solution: x ∈ (−1, 4).
Solve 2/(x − 1) > 0.
We need x − 1 > 0 (since the numerator 2 > 0). So x > 1. Solution: x ∈ (1, ∞).
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Common mistakes
- Forgetting to reverse the inequality when multiplying or dividing by a negative number.
- Using a solid line instead of a dashed line for strict inequalities when graphing.
- Not checking which half-plane to shade; always substitute a test point.
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Summary
Inequalities are solved by similar rules to equations, with one key difference: multiplying or dividing by a negative reverses the sign. The solution of a linear inequality in two variables is a half-plane. Systems of inequalities are solved by finding the intersection of all solution regions.