Linear Inequalities: CBSE Class 11 Maths NCERT Guide

Key Definitions

Inequality

An inequality is a mathematical statement that compares two expressions using an inequality symbol: less than (<), greater than (>), less than or equal to (≤), or greater than or equal to (≥).

Linear Inequality

A linear inequality is an inequality where the variables involved have a maximum power of 1. Examples include ax + b < 0, ax + by ≥ c, etc. They can be in one or two variables.

Solution of an Inequality

A solution to an inequality is any value (or set of values for multiple variables) that, when substituted into the inequality, makes the statement true. Unlike equations which often have a finite number of solutions, inequalities typically have an infinite number of solutions.

Replacement Set

The replacement set is the set of values from which the variable(s) in an inequality can be chosen. For instance, it could be integers (Z), natural numbers (N), or real numbers (R).

Solution Set

The solution set is the subset of the replacement set containing all values that satisfy the given inequality. It is usually expressed using interval notation or set-builder notation.

Rules for Solving Linear Inequalities and Graphical Representation

Solving linear inequalities shares similarities with solving linear equations, but there's one critical difference that students often overlook. Let's first outline the fundamental rules that govern inequality manipulation:

1. Addition/Subtraction Rule: You can add or subtract the same number from both sides of an inequality without changing the direction of the inequality sign. For any real numbers a, b, and c:

• If a < b, then a + c < b + c and a - c < b - c.
• This rule applies similarly for >, ≤, and ≥.

2. Multiplication/Division Rule (The Crucial One):

• By a Positive Number: If you multiply or divide both sides of an inequality by the same positive number, the direction of the inequality sign remains unchanged. For a < b and c > 0, then ac < bc and a/c < b/c.
• By a Negative Number: If you multiply or divide both sides of an inequality by the same negative number, the direction of the inequality sign must be reversed. This is the most common source of error! For a < b and c < 0, then ac > bc and a/c > b/c.

Graphical Representation of Solutions:

A. In One Variable (on a Number Line):
When the solution set is a subset of real numbers, we represent it on a number line.

• For strict inequalities (< or >), use an open circle at the boundary point, indicating that the point is not included in the solution. The solution is then indicated by shading the appropriate part of the number line.
• For non-strict inequalities (≤ or ≥), use a closed (filled) circle at the boundary point, indicating that the point is included in the solution. Shade the corresponding region.
• Arrows are used to indicate that the solution extends infinitely in a direction.

B. In Two Variables (on a Cartesian Plane):
Solving linear inequalities in two variables (e.g., ax + by < c) involves identifying a region on the Cartesian plane, known as the feasible region or solution region.

1. Identify the Boundary Line: First, consider the corresponding linear equation by replacing the inequality sign with an equality sign (e.g., for ax + by < c, consider ax + by = c). This equation represents the boundary line of the solution region.
2. Draw the Boundary Line:

• If the inequality is strict (< or >), draw the boundary line as a dashed or broken line. This signifies that points on the line are not part of the solution.
• If the inequality is non-strict (≤ or ≥), draw the boundary line as a solid line. This means points on the line are part of the solution.

1. Choose a Test Point: Select a point that does not lie on the boundary line. The origin (0,0) is often the easiest choice if it doesn't lie on the line. Substitute the coordinates of this test point into the original inequality.
2. Determine the Solution Region:

• If the test point satisfies the inequality (makes it true), then the region containing the test point is the solution region. Shade this half-plane.
• If the test point does not satisfy the inequality (makes it false), then the region opposite to the test point is the solution region. Shade that half-plane.

1. For Systems of Inequalities: When dealing with multiple linear inequalities, you find the common region that satisfies all inequalities simultaneously. This common shaded region is the solution to the system.

Worked Examples

• Example 1: Solving a linear inequality in one variable and representing on a number line. Solve 5(2x - 7) - 3(2x + 3) ≤ 0 for real x, and represent the solution on a number line. Step 1: Simplify the inequality. Distribute the constants on both sides: 10x - 35 - 6x - 9 ≤ 0 Step 2: Combine like terms. Combine the x terms and constant terms: (10x - 6x) + (-35 - 9) ≤ 0 4x - 44 ≤ 0 Step 3: Isolate the variable term. Add 44 to both sides: 4x ≤ 44 Step 4: Solve for x. Divide both sides by 4 (a positive number, so the inequality sign does not change): x ≤ 11 Step 5: Represent the solution on a number line. Since x is real and x ≤ 11, we draw a closed circle at 11 and shade to the left, indicating all numbers less than or equal to 11. ---
• Example 2: Solving a linear inequality in two variables graphically. Solve 3x + 4y > 12 graphically. Step 1: Identify the boundary line. Replace the inequality sign with an equality sign: 3x + 4y = 12. Step 2: Find two points on the boundary line to draw it. If x = 0, then 4y = 12 => y = 3. So, (0, 3) is a point. If y = 0, then 3x = 12 => x = 4. So, (4, 0) is a point. Step 3: Draw the boundary line. Since the inequality is > (strict inequality), draw a dashed line connecting (0, 3) and (4, 0). This shows points on the line are not part of the solution. Step 4: Choose a test point. The origin (0,0) is a good choice as it does not lie on the line 3x + 4y = 12. Substitute (0,0) into the original inequality: 3(0) + 4(0) > 12 => 0 > 12. Step 5: Determine the solution region. Since 0 > 12 is false, the origin (0,0) does not satisfy the inequality. Therefore, the solution region is the half-plane not containing the origin. Shade the region above the dashed line 3x + 4y = 12. ---
• Example 3: Solving a system of linear inequalities graphically. Solve the system: x + y ≥ 5 and x - y ≤ 3 graphically. Step 1: Graph the first inequality: x + y ≥ 5 Boundary line: x + y = 5. Points: (0, 5) and (5, 0). Line type: Solid (due to ≥). Test point (0,0): 0 + 0 ≥ 5 => 0 ≥ 5 (False). So, shade the region not containing the origin (above the line). Step 2: Graph the second inequality: x - y ≤ 3 Boundary line: x - y = 3. Points: (0, -3) and (3, 0). Line type: Solid (due to ≤). Test point (0,0): 0 - 0 ≤ 3 => 0 ≤ 3 (True). So, shade the region containing the origin (above the line x-y=3 but visually below y=x-3). Step 3: Identify the common feasible region. The solution to the system is the region on the graph where the shaded areas for both inequalities overlap. This region is unbounded and lies above the line x+y=5 and above the line x-y=3 (or, graphically, the area where both individual solution regions intersect). Ensure to clearly mark this common region on your graph.

Exam Tips and Common Mistakes to Avoid

Mastering linear inequalities requires careful attention to detail. Here are some critical tips and common pitfalls to help you excel in your exams:

1. The Sign Reversal Rule is GOLD: This is, by far, the most frequent mistake students make. Always remember to reverse the inequality sign when you multiply or divide both sides of an inequality by a negative number. Forgetting this will lead to an incorrect solution set.

1. Careful with Parentheses: Always distribute constants correctly when they are multiplied by expressions within parentheses. Errors here can cascade throughout your solution.

1. Representing Solutions on a Number Line:

• Strict vs. Non-Strict: Use an open circle for > or < (meaning the endpoint is not included) and a closed (filled) circle for ≥ or ≤ (meaning the endpoint is included). This distinction is vital for accuracy.
• Shading Direction: Ensure you shade the correct side of the boundary point corresponding to the solution (e.g., for x > 5, shade to the right of 5).

1. Graphical Solutions in Two Variables:

• Dashed vs. Solid Lines: Draw a dashed line for strict inequalities (<, >) and a solid line for non-strict inequalities (≤, ≥). This correctly indicates whether the boundary points are part of the solution.
• Test Point Strategy: Always use a test point (like (0,0) if it's not on the boundary line) to correctly identify which half-plane represents the solution. Double-check your substitution and evaluation.
• Shading the Correct Region: For systems of inequalities, ensure you shade the common region that satisfies all given inequalities simultaneously. Use different colors or patterns if necessary to distinguish individual solutions before finding the overlap.

1. Domain of Variables: Pay close attention to whether the variable belongs to natural numbers (N), integers (Z), or real numbers (R), as this affects how you represent the solution set (e.g., individual points vs. continuous interval).

Practice Questions with Solutions

• Q: Solve (2x - 1)/3 ≥ (3x - 2)/4 - (2 - x)/5 for real x. Represent the solution on a number line. A: Step 1: Find the Least Common Multiple (LCM) of the denominators (3, 4, 5), which is 60. Multiply all terms by 60 to clear denominators: 60 (2x - 1)/3 ≥ 60 (3x - 2)/4 - 60 * (2 - x)/5 20(2x - 1) ≥ 15(3x - 2) - 12(2 - x) Step 2: Distribute and simplify. 40x - 20 ≥ 45x - 30 - 24 + 12x 40x - 20 ≥ (45x + 12x) - (30 + 24) 40x - 20 ≥ 57x - 54 Step 3: Collect x terms on one side and constants on the other. Subtract 57x from both sides: 40x - 57x - 20 ≥ -54 -17x - 20 ≥ -54 Add 20 to both sides: -17x ≥ -54 + 20 -17x ≥ -34 Step 4: Divide by -17. Remember to reverse the inequality sign! x ≤ (-34) / (-17) x ≤ 2 Step 5: Represent on a number line. Since x is real and x ≤ 2, draw a closed circle at 2 and shade to the left. Final answer: x ≤ 2 or (-∞, 2]
• Q: Find the solution set of 3x - 7 < 5x - 1 when x is an integer. A: Step 1: Solve the inequality algebraically. 3x - 7 < 5x - 1 Subtract 3x from both sides: -7 < 2x - 1 Add 1 to both sides: -7 + 1 < 2x -6 < 2x Divide by 2 (positive number, so no sign change): -3 < x Step 2: Identify integers that satisfy x > -3. The integers greater than -3 are -2, -1, 0, 1, 2, ... Final answer: Solution Set = {..., -2, -1, 0, 1, 2, ...}
• Q: Solve x + 2y < 8 graphically. A: Step 1: Draw the boundary line x + 2y = 8. Find intercepts: If x=0, 2y=8 => y=4 (Point (0,4)). If y=0, x=8 (Point (8,0)). Step 2: Determine line type. Since the inequality is < (strict), the line x + 2y = 8 will be a dashed line. Step 3: Choose a test point. Let's use (0,0). Substitute (0,0) into x + 2y < 8: 0 + 2(0) < 8 => 0 < 8. Step 4: Interpret the test point result. Since 0 < 8 is true, the region containing the origin is the solution region. Shade the half-plane that includes (0,0). Final answer: The region below the dashed line x + 2y = 8 is the solution.
• Q: Solve the following system of inequalities graphically: x + y ≤ 6 and x ≥ y. A: Step 1: Graph the first inequality x + y ≤ 6. Boundary line: x + y = 6. Points: (0,6), (6,0). Line type: Solid (due to ≤). Test point (0,0): 0 + 0 ≤ 6 => 0 ≤ 6 (True). Shade the region including the origin (below the line x+y=6). Step 2: Graph the second inequality x ≥ y (or y ≤ x). Boundary line: x = y. Points: (0,0), (1,1), (2,2). Line type: Solid (due to ≥). Test point (1,0) (not on the line): 1 ≥ 0 (True). Shade the region including (1,0) (below the line y=x). Step 3: Identify the common feasible region. The solution is the area where the shaded region for x + y ≤ 6 overlaps with the shaded region for x ≥ y. This forms a triangular region bounded by x+y=6, x=y, and the x-axis (implicitly, assuming positive values or just the intersection in 2D plane). The common region is below both lines, bounded by x+y=6 and y=x in the first and fourth quadrants. Final answer: The common shaded region below x+y=6 and below y=x.

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