Chapter Notes: Inequalities (Systems & Graphs)

Table of Contents
1. Understanding Inequalities
2. Graphing Linear Inequalities
3. Systems of Linear Inequalities
4. Interpreting Solution Regions
5. Real-World Applications
6. Special Cases
7. Vertices and Boundary Points
8. Summary of Key Points
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In real life, we often face situations where we need to satisfy more than one condition at the same time. For example, you might need to save enough money for a concert ticket while also keeping enough to pay for lunch each day. In mathematics, we use systems of inequalities to model situations where multiple conditions must be met simultaneously. Graphing these systems helps us visualize all the possible solutions that work for every inequality at once. This chapter will teach you how to solve and graph systems of inequalities, interpret the solution regions, and apply these skills to real-world problems.

Understanding Inequalities

Before working with systems, we need to understand single inequalities clearly. An inequality is a mathematical statement that compares two expressions using symbols like <, >, ≤, or ≥. Unlike equations, which have specific solutions, inequalities usually have infinitely many solutions.

Inequality Symbols

The key difference between < and ≤ (or > and ≥) is whether the boundary value itself is included in the solution. The symbol ≤ means "less than or equal to," so the boundary value is part of the solution. The symbol < means "less than" only, so the boundary value is not included.

Linear Inequalities in Two Variables

A linear inequality in two variables looks similar to a linear equation but uses an inequality symbol instead of an equals sign. The general form is:

where \( a \), \( b \), and \( c \) are constants, and the inequality symbol can be <, >, ≤, or ≥. The solution to such an inequality is not a single point or line, but an entire region of the coordinate plane.

Graphing Linear Inequalities

Graphing a linear inequality involves two main steps: drawing the boundary line and shading the appropriate region.

Step 1: Graph the Boundary Line

First, treat the inequality symbol as an equals sign and graph the resulting line. This line is called the boundary line.

• If the inequality uses ≤ or ≥, draw a solid line. This means points on the line are included in the solution.
• If the inequality uses < or >, draw a dashed line. This means points on the line are not included in the solution.

Step 2: Shade the Solution Region

The boundary line divides the coordinate plane into two regions. One region contains all the solutions to the inequality, and the other does not. To determine which side to shade:

1. Choose a test point not on the boundary line. The point (0, 0) is usually the easiest choice unless the line passes through the origin.
2. Substitute the test point's coordinates into the original inequality.
3. If the inequality is true, shade the region containing the test point.
4. If the inequality is false, shade the region on the opposite side of the line.

Example:  Graph the inequality \( y < 2x + 1 \).

Solution:

Step 1: Graph the boundary line \( y = 2x + 1 \).

This line has a slope of 2 and a y-intercept of 1. Since the inequality uses < (not ≤), we draw a dashed line.

Step 2: Choose a test point. Let's use (0, 0).

Substitute into the inequality: \( 0 < 2(0) + 1 \)
\( 0 < 1 \) ✓

This is true, so we shade the region that contains (0, 0), which is below the dashed line.

The solution is the entire region below the line \( y = 2x + 1 \), not including the line itself.

Example:  Graph the inequality \( 3x + 2y ≥ 6 \).

Solution:

Step 1: Graph the boundary line \( 3x + 2y = 6 \).

Find the intercepts:
When \( x = 0 \): \( 2y = 6 \), so \( y = 3 \). Point: (0, 3)
When \( y = 0 \): \( 3x = 6 \), so \( x = 2 \). Point: (2, 0)

Since the inequality uses ≥, we draw a solid line through (0, 3) and (2, 0).

Step 2: Test the point (0, 0).

Substitute: \( 3(0) + 2(0) ≥ 6 \)
\( 0 ≥ 6 \) ✗

This is false, so we shade the region that does not contain (0, 0), which is the region above and to the right of the line.

The solution includes the line itself and all points above it.

Systems of Linear Inequalities

A system of linear inequalities consists of two or more inequalities considered together. The solution to the system is the set of all points that satisfy every inequality in the system simultaneously. Graphically, this solution is the region where all the shaded areas overlap.

Steps to Solve a System Graphically

1. Graph each inequality separately on the same coordinate plane.
2. Identify the boundary line for each inequality and determine whether it should be solid or dashed.
3. Shade the solution region for each inequality.
4. The solution to the system is the region where all shadings overlap.
5. If there is no overlapping region, the system has no solution.

Example:  Solve the system of inequalities graphically:
\( y < x + 3 \)
\( y ≥ -2x + 1 \)

What is the solution region?

Solution:

Graph the first inequality: \( y < x + 3 \)

Boundary line: \( y = x + 3 \) (slope = 1, y-intercept = 3)
Draw a dashed line.
Test (0, 0): \( 0 < 0 + 3 \) → \( 0 < 3 \) ✓
Shade below the dashed line.

Graph the second inequality: \( y ≥ -2x + 1 \)

Boundary line: \( y = -2x + 1 \) (slope = -2, y-intercept = 1)
Draw a solid line.
Test (0, 0): \( 0 ≥ -2(0) + 1 \) → \( 0 ≥ 1 \) ✗
Shade above the solid line.

Identify the solution region:

The solution is the region that is both below the dashed line and above the solid line. This is the overlapping shaded region. Points on the solid line \( y = -2x + 1 \) are included, but points on the dashed line \( y = x + 3 \) are not.

Systems with More Than Two Inequalities

Some systems involve three or more inequalities. The process is the same: graph each inequality and find where all the shaded regions overlap. The solution region may be a triangle, quadrilateral, or other polygon.

Example:  Graph the system:
\( x ≥ 0 \)
\( y ≥ 0 \)
\( x + y ≤ 4 \)

What shape is the solution region?

Solution:

Graph \( x ≥ 0 \):

This is a vertical solid line along the y-axis. Shade to the right (where \( x \) is positive).

Graph \( y ≥ 0 \):

This is a horizontal solid line along the x-axis. Shade above (where \( y \) is positive).

Graph \( x + y ≤ 4 \):

Boundary line: \( x + y = 4 \) passes through (4, 0) and (0, 4).
Draw a solid line.
Test (0, 0): \( 0 + 0 ≤ 4 \) ✓
Shade below and to the left of the line.

Solution region:

The overlapping region is a triangle with vertices at (0, 0), (4, 0), and (0, 4). All three boundary lines are solid, so the vertices and edges are included in the solution.

Interpreting Solution Regions

The solution region of a system of inequalities represents all the ordered pairs \( (x, y) \) that make every inequality true. Any point inside the shaded region (or on a solid boundary) is a solution. Any point outside the region is not a solution.

Testing Points

To verify whether a specific point is a solution to the system, substitute its coordinates into each inequality. The point is a solution only if it satisfies all inequalities.

Example:  Consider the system:
\( y < 2x + 1 \)
\( y ≥ -x + 3 \)
Is the point (1, 2) a solution?

Solution:

Test (1, 2) in the first inequality: \( 2 < 2(1) + 1 \)
\( 2 < 3 \) ✓

Test (1, 2) in the second inequality: \( 2 ≥ -(1) + 3 \)
\( 2 ≥ 2 \) ✓

Since (1, 2) satisfies both inequalities, (1, 2) is a solution to the system.

Real-World Applications

Systems of inequalities are powerful tools for solving problems with multiple constraints. They are used in business to maximize profit, in nutrition to meet dietary requirements, and in planning to allocate resources efficiently.

Example:  A student has $20 to spend on notebooks and pens.
Notebooks cost $4 each and pens cost $2 each.
She wants to buy at least 3 notebooks.

Write and graph a system of inequalities to represent this situation. What are some possible combinations she can buy?

Solution:

Let \( x \) = number of notebooks
Let \( y \) = number of pens

Constraints:

Budget constraint: \( 4x + 2y ≤ 20 \)
Notebook requirement: \( x ≥ 3 \)
Non-negative quantities: \( x ≥ 0 \) and \( y ≥ 0 \)

Graph the system:

Graph \( 4x + 2y ≤ 20 \): Boundary passes through (5, 0) and (0, 10). Solid line. Shade below.
Graph \( x ≥ 3 \): Vertical solid line at \( x = 3 \). Shade to the right.
Graph \( x ≥ 0 \) and \( y ≥ 0 \): First quadrant only.

Solution region:

The overlapping region shows all valid combinations. For example:
(3, 4): 3 notebooks and 4 pens → Cost = \( 4(3) + 2(4) = 12 + 8 = 20 \) ✓
(4, 2): 4 notebooks and 2 pens → Cost = \( 4(4) + 2(2) = 16 + 4 = 20 \) ✓
(3, 0): 3 notebooks and 0 pens → Cost = \( 4(3) = 12 \) ✓

Any point in the shaded region represents a valid purchase option.

Special Cases

No Solution

Sometimes the shaded regions of the inequalities do not overlap at all. In this case, the system has no solution. This means there are no points that satisfy all the inequalities simultaneously.

Example:  Consider the system:
\( y > x + 2 \)
\( y < x - 1 \)

Does this system have a solution?

Solution:

The first inequality requires \( y \) to be above the line \( y = x + 2 \).
The second inequality requires \( y \) to be below the line \( y = x - 1 \).

Since the line \( y = x + 2 \) is always above \( y = x - 1 \) (they are parallel with \( y = x + 2 \) shifted up), there is no region that is simultaneously above \( y = x + 2 \) and below \( y = x - 1 \).

This system has no solution.

Unbounded Solution Regions

Some systems have solution regions that extend infinitely in one or more directions. These are called unbounded regions. Even though the region is infinite, it is still a valid solution set.

Vertices and Boundary Points

When the solution region is a polygon, the corners of the polygon are called vertices. These vertices occur where boundary lines intersect. Vertices are important in optimization problems because maximum or minimum values often occur at these corner points.

To find a vertex, solve the system of equations formed by the two boundary lines that intersect at that point.

Example:  Find the vertices of the solution region for the system:
\( x ≥ 0 \)
\( y ≥ 0 \)
\( x + 2y ≤ 8 \)
\( 3x + y ≤ 9 \)

Solution:

Vertex 1: Intersection of \( x = 0 \) and \( y = 0 \) → (0, 0)

Vertex 2: Intersection of \( x = 0 \) and \( x + 2y = 8 \)
\( 0 + 2y = 8 \) → \( y = 4 \) → (0, 4)

Vertex 3: Intersection of \( y = 0 \) and \( 3x + y = 9 \)
\( 3x + 0 = 9 \) → \( x = 3 \) → (3, 0)

Vertex 4: Intersection of \( x + 2y = 8 \) and \( 3x + y = 9 \)

Solve the system:
From \( x + 2y = 8 \): \( x = 8 - 2y \)
Substitute into \( 3x + y = 9 \): \( 3(8 - 2y) + y = 9 \)
\( 24 - 6y + y = 9 \)
\( 24 - 5y = 9 \)
\( -5y = -15 \)
\( y = 3 \)

Then \( x = 8 - 2(3) = 8 - 6 = 2 \) → (2, 3)

The vertices of the solution region are (0, 0), (0, 4), (3, 0), and (2, 3).

Summary of Key Points

• A linear inequality in two variables has infinitely many solutions forming a region on the coordinate plane.
• The boundary line is solid for ≤ or ≥, and dashed for < or >.
• Use a test point to determine which side of the boundary line to shade.
• The solution to a system of inequalities is the overlapping region where all individual solution regions intersect.
• If there is no overlap, the system has no solution.
• Vertices occur where boundary lines intersect and are found by solving systems of equations.
• Real-world problems often involve multiple constraints that can be modeled with systems of inequalities.