Chapter 6: Linear Inequalities in Two Variables

Overview

Welcome to Chapter 6 of Form 4 Mathematics! This chapter introduces you to linear inequalities in two variables and how to represent and solve them graphically. You'll learn to graph boundary lines, determine solution regions, and solve systems of linear inequalities. These skills are essential for optimization problems and real-world constraint analysis.

What You'll Learn:

• Represent situations using linear inequalities
• Verify conjectures about points satisfying linear inequalities
• Determine and shade regions satisfying linear inequalities
• Solve systems of linear inequalities graphically

Learning Objectives

After completing this chapter, you will be able to:

• Represent situations in the form of linear inequalities
• Verify conjectures about points in a region that satisfy a linear inequality
• Determine and shade regions that satisfy a linear inequality
• Represent situations in the form of systems of linear inequalities
• Determine and shade regions that satisfy a system of linear inequalities

Key Concepts

Linear Inequalities

A linear inequality is a mathematical relationship between two linear expressions using inequality symbols (<, >, ≤, ≥).

General Forms:

• $y > mx + b$ (strict inequality)
• $y \geq mx + b$ (non-strict inequality)
• $ax + by < c$ (standard form)
• $ax + by \geq c$ (standard form)

Examples:

• $y > 2x + 1$
• $3x + 4y \leq 12$
• $x - 2y \geq 6$

Key Differences from Equations:

• Inequalities have a range of solutions rather than specific values
• The solution is a region rather than a line
• Special attention is needed for strict vs. non-strict inequalities
• Graphical representation involves shading entire regions

Boundary Line

The boundary line is a straight line that represents the linear equation (replacing the inequality symbol with =).

Mathematical Relationship: If we have an inequality $ax + by \leq c$, the boundary line is $ax + by = c$.

Examples:

• For $y > 2x + 1$, boundary line is $y = 2x + 1$
• For $3x + 4y \leq 12$, boundary line is $3x + 4y = 12$

Types of Boundary Lines

Solid Line (≤ or ≥):

• Used for inequalities with "less than or equal to" or "greater than or equal to"
• Mathematical notation: $\leq$, $\geq$
• Points on the line are included in the solution region
• Example: For $y \geq 2x + 1$, the line $y = 2x + 1$ is included

Dashed Line (< or >):

• Used for strict inequalities ("less than" or "greater than")
• Mathematical notation: $<$, $>$
• Points on the line are not included in the solution region
• Example: For $y > 2x + 1$, the line $y = 2x + 1$ is not included

Solution Region

The solution region is the area on the Cartesian plane that contains all points $(x, y)$ that satisfy the inequality. The solution region is either above or below the boundary line.

Mathematical Test: For a test point $(x_0, y_0)$, substitute into the inequality:

• If $ax_0 + by_0 \leq c$ is true, the point is in the solution region
• If false, the point is not in the solution region

How to Determine the Correct Region:

1. Choose a test point not on the line (usually $(0,0)$ if it's not on the line)
2. Substitute the point into the inequality: $ax_0 + by_0 \leq c$
3. If the inequality is true, shade that region
4. If the inequality is false, shade the opposite region

Special Cases:

• For vertical lines ($x = k$): left vs. right regions
• For horizontal lines ($y = k$): above vs. below regions

Important Formulas and Methods

Graphing Linear Inequalities

Step-by-Step Process:

1. Graph the boundary line:

   • For y = mx + b form, use slope-intercept method
   • For standard form ax + by = c, find x and y intercepts
   • Use solid line for ≤, ≥; dashed line for <, >

2. Choose a test point:

   • Use (0,0) if it's not on the line
   • Otherwise, choose another simple point like (1,0) or (0,1)

3. Test the inequality:

   • Substitute the test point coordinates
   • Determine if the inequality is true or false

4. Shade the correct region:

   • If true, shade the region containing the test point
   • If false, shade the opposite region

Systems of Linear Inequalities

Step-by-Step Process:

1. Graph each inequality separately:

   • Follow the steps above for each inequality
   • Use different shading patterns or colors for each inequality

2. Identify the common region:

   • The solution to the system is where all shaded regions overlap
   • This is the region that satisfies all inequalities simultaneously

3. Verify the solution:

   • Choose a test point in the common region
   • Verify it satisfies all inequalities

Step-by-Step Solved Examples

Example 1: Single Linear Inequality

Problem: Graph the inequality $y > 2x - 3$

Solution:

1. Graph boundary line: $y = 2x - 3$

   • y-intercept: $(0, -3)$
   • Slope: $m = 2$ (rise 2, run 1)
   • Use dashed line (strict inequality)

2. Choose test point: $(0,0)$ is not on the line

3. Test inequality: $0 > 2(0) - 3 \rightarrow 0 > -3$ ✓ True

4. Shade: Region containing $(0,0)$, which is above the line

Answer: Shade above the dashed line $y = 2x - 3$

Example 2: Inequality with Standard Form

Problem: Graph the inequality $2x + 3y \leq 6$

Solution:

1. Graph boundary line: $2x + 3y = 6$

   • x-intercept: $(3, 0)$ [when $y=0$]
   • y-intercept: $(0, 2)$ [when $x=0$]
   • Use solid line (non-strict inequality)

2. Choose test point: $(0,0)$ is not on the line

3. Test inequality: $2(0) + 3(0) \leq 6 \rightarrow 0 \leq 6$ ✓ True

4. Shade: Region containing $(0,0)$, which is below the line

Answer: Shade below the solid line connecting $(3,0)$ and $(0,2)$

Example 3: Inequality with Negative Slope

Problem: Graph the inequality $y \leq -x + 4$

Solution:

1. Graph boundary line: $y = -x + 4$

   • y-intercept: $(0, 4)$
   • Slope: $m = -1$ (rise -1, run 1)
   • Use solid line (non-strict inequality)

2. Choose test point: $(0,0)$ is not on the line

3. Test inequality: $0 \leq -(0) + 4 \rightarrow 0 \leq 4$ ✓ True

4. Shade: Region containing $(0,0)$, which is below the line

Answer: Shade below the solid line $y = -x + 4$

Example 4: System of Linear Inequalities

Problem: Graph and find the solution region for:

1. $y \geq x - 1$
2. $y \leq -x + 3$
3. $x \geq 0$
4. $y \geq 0$

Solution:

1. Graph each inequality:

   • $y \geq x - 1$: Solid line, slope 1, y-intercept -1, shade above
   • $y \leq -x + 3$: Solid line, slope -1, y-intercept 3, shade below
   • $x \geq 0$: Vertical line at x=0, shade right
   • $y \geq 0$: Horizontal line at y=0, shade above

2. Find common region: All shaded regions overlap in a triangular region

3. Identify vertices:

   • Intersection of $y = x - 1$ and $y = -x + 3$: $x - 1 = -x + 3 \rightarrow 2x = 4 \rightarrow x = 2, y = 1$
   • Intersection with axes: $(0,0)$, $(0,3)$, $(2,1)$

Example 5: Real-world Application

Problem: A farmer has 100 acres of land and wants to plant corn and wheat. Each acre of corn requires 2 units of fertilizer and 3 units of pesticide. Each acre of wheat requires 1 unit of fertilizer and 2 units of pesticide. The farmer has 160 units of fertilizer and 210 units of pesticide. Graph the constraints and find the feasible region.

Solution: Let $x$ = acres of corn, $y$ = acres of wheat

Constraints:

1. $x + y \leq 100$ (land constraint)
2. $2x + y \leq 160$ (fertilizer constraint)
3. $3x + 2y \leq 210$ (pesticide constraint)
4. $x \geq 0$, $y \geq 0$ (non-negativity)

Graphing:

1. $x + y = 100$: intercepts $(100,0)$, $(0,100)$
2. $2x + y = 160$: intercepts $(80,0)$, $(0,160)$
3. $3x + 2y = 210$: intercepts $(70,0)$, $(0,105)$

Vertices of Feasible Region:

• $(0,0)$
• $(0,100)$ - Intersection of $x=0$ and $x+y=100$
• $(50,50)$ - Intersection of $x+y=100$ and $2x+y=160$
• $(70,0)$ - Intersection of $y=0$ and $3x+2y=210$
• $(60,20)$ - Intersection of $2x+y=160$ and $3x+2y=210$

Answer: The feasible region is a polygon bounded by the lines $x + y = 100$, $2x + y = 160$, $3x + 2y = 210$, and the axes, with vertices at $(0,0)$, $(0,100)$, $(50,50)$, $(60,20)$, and $(70,0)$.

Real-world Applications

1. Business and Economics

Applications:

• Resource Allocation: Limited resources for multiple products
• Production Planning: Constraints on labor, materials, time
• Profit Maximization: Finding optimal production levels

2. Manufacturing

Applications:

• Production Scheduling: Limited machine time and labor
• Quality Control: Meeting quality standards
• Inventory Management: Storage constraints

3. Transportation

Applications:

• Route Planning: Distance and time constraints
• Vehicle Routing: Capacity and time limits
• Logistics: Multiple delivery constraints

4. Personal Finance

Applications:

• Budget Planning: Income and expense constraints
• Investment Portfolio: Risk and return constraints
• Loan Applications: Income and debt ratio limits

Important Terms

Summary Points

• Linear Inequality: Mathematical relationship with <, >, ≤, ≥
• Boundary Line: Replace inequality with = to find
• Solid Line: For ≤, ≥ (includes line)
• Dashed Line: For <, > (excludes line)
• Test Point: Determines which region to shade
• Solution Region: Area where inequality is true
• System Solution: Common region where all inequalities are true
• Applications: Resource allocation, optimization, constraint analysis

Practice Tips for SPM Students

1. Master Graphing Techniques

• Practice graphing linear equations quickly
• Learn to find intercepts efficiently
• Be comfortable with different line types (solid vs. dashed)

2. Inequality Testing

• Always choose an easy test point like (0,0)
• Practice substituting coordinates correctly
• Understand the logic of why regions are shaded

3. System Solutions

• Graph each inequality clearly
• Identify overlapping regions systematically
• Find vertices by solving systems of equations

4. Common Mistakes to Avoid

• Using wrong line types (solid vs. dashed)
• Testing points that are on the boundary line
• Shading the wrong region
• Missing constraints in real-world problems

SPM Exam Tips

Paper 1 (Multiple Choice)

• Look for inequality symbols and their meanings
• Remember the difference between solid and dashed lines
• Use test points to verify regions
• Practice quick graphing skills

Paper 2 (Structured)

• Always show your test point and substitution
• Label lines and regions clearly
• For systems, show all boundary lines
• Identify the solution region accurately

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Did You Know? Linear programming, which uses systems of linear inequalities, was developed during World War II to solve logistics and resource allocation problems for the military. It's now used in industries ranging from airlines to fast food chains for optimization!

Next Chapter: In Chapter 7, you'll explore graphs of motion and learn to interpret distance-time and speed-time graphs, which is essential for understanding kinematics and motion analysis.

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• Chapter 5: Graph Theory
• Chapter 7: Graphs of Functions