Linear Inequalities
An inequality is a mathematical statement that compares two quantities using the symbols \(<, >, \le, \ge\). Unlike equations, inequalities do not yield a single value but a range of possible values.
When the comparison involves a linear expression in one variable, it is called a linear inequality. These form the foundation for solving real-world problems involving constraints such as profit limits, distance ranges, and optimization conditions.
Definition
A linear inequality in one variable can be expressed in any of the following standard forms:
\(ax + b < 0,\quad ax + b > 0,\quad ax + b \le 0,\quad ax + b \ge 0\)
where \(a, b \in \mathbb{R}\) and \(a \ne 0\).
The set of all real values of the variable that satisfy a given inequality is called its solution set.
Key Characteristics
- Solution of a linear inequality is usually an interval on the real number line.
- Unlike equations, inequalities can have infinitely many solutions.
- Graphical representation is done using open and closed circles.
- Direction of inequality reverses when multiplied or divided by a negative number.
Example
Consider the inequality:
\(2x + 3 > 7\)
Solving,
\(2x > 4 \Rightarrow x > 2\)
Hence, the solution set is all real numbers greater than 2.
Graphical Representation
The open circle indicates that 2 is not included, while the arrow shows all values greater than 2.
Importance in Board and Competitive Exams
- Frequently asked in CBSE Class 11 exams for conceptual clarity and graph-based questions.
- Forms the base for Linear Programming (Class 12).
- Essential for JEE, NEET, NDA in inequality manipulation and solution sets.
- Used in higher mathematics for interval analysis and optimization problems.
Real-Life Significance
Linear inequalities are widely used in practical situations such as budgeting, resource allocation, and decision-making where limits or constraints are involved. For example, determining minimum marks required to pass, maximum spending limits, or feasible production levels.
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Rules for Solving Linear Inequalities
Solving linear inequalities follows rules similar to solving linear equations, but with one critical difference: the inequality sign may change direction under certain operations. Understanding these rules is essential for accuracy in board exams and competitive tests like JEE and NEET.
Rule 1: Addition or Subtraction Rule
If the same real number is added to or subtracted from both sides of an inequality, the inequality sign remains unchanged.
If \(a < b\), then \(a + c < b + c\) for any real number \(c\).
Concept Insight: This operation shifts both numbers equally on the number line, preserving their relative order.
Example:
\(x - 5 > 2 \Rightarrow x > 7\)
Rule 2: Multiplication or Division by a Positive Number
If both sides are multiplied or divided by the same positive number, the inequality sign remains unchanged.
If \(a < b\) and \(c > 0\), then \(ac < bc\)
Concept Insight: Positive scaling stretches or compresses the number line without flipping it.
Example:
\(3x < 9 \Rightarrow x < 3\)
Rule 3: Multiplication or Division by a Negative Number
If both sides are multiplied or divided by the same negative number, the inequality sign reverses.
If \(a < b\) and \(c < 0\), then \(ac > bc\)
Concept Insight: Multiplication by a negative number reflects values across zero on the number line, reversing their order.
Example:
\(-2x > 6 \Rightarrow x < -3\)
Visual Insight: Sign Reversal
Representation on Number Line
Solutions of inequalities are represented graphically on a number line:
- Open Circle: Used for \(<\) or \(>\) (boundary excluded)
- Closed Circle: Used for \(\le\) or \(\ge\) (boundary included)
Linear Inequalities in Two Variables
A linear inequality in two variables is of the form:
\(ax + by < c\)
Its solution is a region in the Cartesian plane bounded by the line \(ax + by = c\). The boundary line is:
- Dashed for strict inequalities (\(<, >\))
- Solid for inclusive inequalities (\(\le, \ge\))
Exam-Oriented Insights
- Very common in CBSE Class 11 exams (2–4 mark questions)
- Core concept in JEE inequality manipulation
- Forms base for Linear Programming (Class 12)
- Sign reversal mistakes are one of the most frequent student errors
Common Mistakes to Avoid
- Forgetting to reverse sign when multiplying/dividing by a negative number
- Incorrect graph representation (open vs closed circle)
- Ignoring inequality direction in final answer
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Example 1: Solving Linear Inequality with Fractions
Solve \( \dfrac{5 - 2x}{3} \leq \dfrac{x}{6} - 5 \)
Solution
First, eliminate fractions by taking the LCM of denominators (which is 6). This simplifies calculations and reduces chances of error.
Multiply both sides by 6:
\( \dfrac{6(5 - 2x)}{3} \leq \dfrac{6x}{6} - 30 \)
\( 2(5 - 2x) \leq x - 30 \)
\( 10 - 4x \leq x - 30 \)
Bring like terms together:
\( 10 + 30 \leq x + 4x \)
\( 40 \leq 5x \)
\( 8 \leq x \)
\( x \geq 8 \)
Final Answer
Solution set:
\( x \in [8, \infty) \)
Graphical Representation
Concept Insight
- Always remove fractions early using LCM to simplify solving.
- No sign reversal occurs here because we multiplied by a positive number (6).
- Rewriting inequality as \(8 \le x\) helps avoid algebra mistakes.
Exam Tip
In CBSE and JEE exams, fractional inequalities are frequently asked. Students often make mistakes while clearing denominators or handling negative signs. Always check whether multiplication involves a negative number.
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Example 2: Solving a Simple Linear Inequality
Solve \(7x + 3 < 5x + 9\)
Solution
\(7x + 3 < 5x + 9\)
Bring like terms together:
\(7x - 5x < 9 - 3\)
\(2x < 6\)
Divide both sides by 2 (positive number, so inequality sign remains unchanged):
\(x < 3\)
Final Answer
\(x \in (-\infty, 3)\)
Graphical Representation
Concept Insight
- Always group variable terms on one side and constants on the other.
- Since we divide by a positive number, the inequality direction remains unchanged.
- The solution represents all values less than 3, not including 3.
Common Mistake
Students sometimes mistakenly reverse the inequality sign even when dividing by a positive number. This rule applies only when multiplying or dividing by a negative number.
Exam Relevance
- Very common 1–2 mark question in CBSE Class 11 exams
- Forms the base for solving compound inequalities in JEE/NEET
- Tests basic algebraic manipulation speed and accuracy
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Example 3: Solving Linear Inequality with Fractions
Solve \( \dfrac{3x - 4}{2} \geq \dfrac{x + 1}{4} - 1 \)
Solution
First simplify the right-hand side:
\( \dfrac{x + 1}{4} - 1 = \dfrac{x + 1 - 4}{4} = \dfrac{x - 3}{4} \)
Now eliminate fractions using LCM (4):
\( \dfrac{4(3x - 4)}{2} \geq \dfrac{4(x - 3)}{4} \)
\( 2(3x - 4) \geq x - 3 \)
\( 6x - 8 \geq x - 3 \)
Bring like terms together:
\( 6x - x \geq -3 + 8 \)
\( 5x \geq 5 \)
Divide both sides by 5 (positive number, so inequality sign remains unchanged):
\( x \geq 1 \)
Final Answer
\( x \in [1, \infty) \)
Graphical Representation
Concept Insight
- Simplify expressions before clearing denominators to reduce mistakes.
- LCM method is the most reliable approach for fractional inequalities.
- No sign reversal occurs because multiplication/division is done using positive numbers.
Common Mistakes
- Forgetting to distribute properly after removing brackets
- Incorrect simplification of fractions
- Missing sign reversal rule (though not applicable here)
Exam Relevance
- Frequently asked in CBSE Class 11 board exams
- Forms base for compound inequalities in JEE
- Tests algebraic simplification and accuracy under time pressure
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Example 4: Application of Linear Inequalities (Average Marks)
The marks obtained by a student in first and second terminal examinations are 62 and 48, respectively. Find the minimum marks he should obtain in the annual examination to have an average of at least 60 marks.
Solution
Let the marks obtained in the annual examination be \(x\).
The average of three exams is given by:
\( \dfrac{62 + 48 + x}{3} \geq 60 \)
Simplify the expression:
\( \dfrac{110 + x}{3} \geq 60 \)
Multiply both sides by 3:
\( 110 + x \geq 180 \)
\( x \geq 180 - 110 \)
\( x \geq 70 \)
Final Answer
The student must score at least:
\( x \geq 70 \)
Interpretation
Since the inequality is inclusive (\(\geq\)), a score of exactly 70 will also ensure the required average of 60.
Graphical Insight
Concept Insight
- Translate word problems into algebraic inequalities carefully.
- “At least” always corresponds to \(\geq\).
- Average-based problems are common in exams and require correct equation formation.
Exam Relevance
- Very common application-based question in CBSE Class 11 exams
- Strengthens mathematical modeling skills for JEE
- Helps in solving real-life constraint problems
Common Mistakes
- Using \(>\) instead of \(\geq\) for “at least”
- Incorrect average formula
- Arithmetic errors while simplifying constants
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Example 5: Consecutive Odd Numbers and Inequalities
Find all pairs of consecutive odd natural numbers, both greater than 10, such that their sum is less than 40.
Solution
Let the first odd number be \(x\). Then the next consecutive odd number will be \(x + 2\).
According to the question, we have two conditions:
- \(x > 10\)
- \(x + (x + 2) < 40\)
Solve the inequality:
\(2x + 2 < 40\)
\(2x < 38\)
\(x < 19\)
Combined Condition
\(10 < x < 19\)
Since \(x\) must be an odd natural number, the possible values are:
\(x = 11, 13, 15, 17\)
Final Answer
The required pairs are:
(11, 13), (13, 15), (15, 17), (17, 19)
Graphical Insight
Concept Insight
- Consecutive odd numbers differ by 2.
- Always combine multiple conditions into a single inequality range.
- Final answers must satisfy both algebraic and number-type constraints (odd natural numbers).
Exam Relevance
- Common in CBSE Class 11 exams as application-based questions
- Builds foundation for integer-based inequalities in JEE
- Tests logical filtering after solving inequalities
Common Mistakes
- Ignoring the condition “odd natural numbers”
- Including boundary values (10 or 19) incorrectly
- Missing one of the required constraints
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Compound Inequalities (AND / OR)
A compound inequality involves two or more inequalities combined using logical connectors such as AND or OR. These are crucial for representing ranges and multiple solution conditions.
Type 1: AND Inequality (Intersection of Solutions)
When two inequalities are joined using AND, the solution consists of values that satisfy both conditions simultaneously.
Example: \( 2 < x < 7 \)
This means \(x\) is greater than 2 and less than 7.
Interval Notation:
\( (2, 7) \)
Graphical Representation (AND)
Type 2: OR Inequality (Union of Solutions)
When inequalities are joined using OR, the solution includes values that satisfy at least one of the conditions.
Example: \( x < 2 \; \text{or} \; x > 7 \)
Interval Notation:
\( (-\infty, 2) \cup (7, \infty) \)
Graphical Representation (OR)
Key Differences
| Type | Meaning | Graph | Interval Form |
|---|---|---|---|
| AND | Common region | Middle portion | \((a, b)\) |
| OR | Combined regions | Two separate parts | \((-\infty, a) \cup (b, \infty)\) |
Concept Insight
- AND → Intersection (common values only)
- OR → Union (combine all valid values)
- Always visualize on number line for clarity
Exam Relevance
- Very important for JEE Main & Advanced
- Forms base for modulus inequalities
- Frequently asked in graph-based questions
Common Mistakes
- Confusing AND with OR conditions
- Incorrect interval notation
- Wrong shading in graphs
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Absolute Value Inequalities \(|x|\) Type
Absolute value inequalities involve expressions of the form \(|x|\), which represent the distance of a number from zero on the number line. These inequalities are crucial in transforming problems into compound inequalities (AND / OR).
Key Concept
For any real number \(x\):
\(|x| = \text{distance of } x \text{ from } 0\)
Case 1: \(|x| < a\) (AND Condition)
If \(a > 0\), then:
\(|x| < a \Rightarrow -a < x < a\)
This means \(x\) lies between \(-a\) and \(a\).
Interval Form:
\((-a, a)\)
Graphical Representation
Case 2: \(|x| > a\) (OR Condition)
If \(a > 0\), then:
\(|x| > a \Rightarrow x < -a \; \text{or} \; x > a\)
This means \(x\) lies outside the interval.
Interval Form:
\((-\infty, -a) \cup (a, \infty)\)
Graphical Representation
Important Variations
- \(|x| \le a \Rightarrow -a \le x \le a\)
- \(|x| \ge a \Rightarrow x \le -a \text{ or } x \ge a\)
Solved Example
Solve \( |2x - 1| < 5 \)
\(-5 < 2x - 1 < 5\)
\(-4 < 2x < 6\)
\(-2 < x < 3\)
\(x \in (-2, 3)\)
Concept Insight
- |x| < a → AND (inside region)
- |x| > a → OR (outside region)
- Always ensure \(a > 0\), otherwise solution behavior changes
Exam Relevance
- Very important for JEE Main & Advanced
- Used in modulus functions and graph-based problems
- Frequently appears in multi-step inequalities
Common Mistakes
- Confusing AND and OR cases
- Forgetting to split inequality correctly
- Ignoring condition \(a > 0\)
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Graph of Linear Inequalities in Two Variables (Shading & Half-Plane Concept)
A linear inequality in two variables is of the form:
\(ax + by < c,\; ax + by > c,\; ax + by \leq c,\; ax + by \geq c\)
The solution of such inequalities is represented as a region in the Cartesian plane, called a half-plane.
Step-by-Step Method
- Replace the inequality with an equation: \(ax + by = c\)
- Draw the corresponding straight line
- Use:
- Solid line for \(\leq\) or \(\geq\)
- Dashed line for \(<\) or \(>\)
- Choose a test point (usually \((0,0)\))
- Shade the region that satisfies the inequality
Half-Plane Concept
A straight line divides the plane into two regions called half-planes. Only one of these satisfies the given inequality.
The required solution is the shaded half-plane.
Example 1: \(x + y \leq 4\)
Boundary line: \(x + y = 4\)
- Draw a solid line (since ≤)
- Test point \((0,0)\): satisfies inequality
- Shade the region containing origin
Example 2: \(x - y > 2\)
Boundary line: \(x - y = 2\)
- Draw a dashed line (since >)
- Test point \((0,0)\): does NOT satisfy inequality
- Shade the opposite region
Key Observations
- Solution is always a region (not a single point)
- Boundary line inclusion depends on inequality type
- Test point method is the most reliable approach
Important Rules
- Use \((0,0)\) as test point unless line passes through origin
- If test point satisfies inequality → shade that region
- If not → shade the opposite region
Exam Relevance
- Core concept for Linear Programming (Class 12)
- Important for JEE graph-based questions
- Tests visualization and logical reasoning
Common Mistakes
- Using solid line instead of dashed (or vice versa)
- Incorrect shading region
- Skipping test point verification
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