Solved Exercises on Systems of Linear Inequalities in Integrated Math 1

Master systems of linear inequalities: solution regions, feasible areas, and real-world applications through these 5 detailed exercises with comprehensive solutions.

Solution: Exercises 1 to 3
1 Basic System
Exercise 1
Graph the system of inequalities:
\(y \leq x + 2\) and \(y > -x + 1\)
Identify the solution region and find at least one point that satisfies both inequalities.
Definition:

System of Linear Inequalities: A set of two or more linear inequalities with the same variables

Method for graphing systems:
  1. Graph each inequality separately
  2. Shade the solution region for each inequality
  3. Find the intersection of all shaded regions
  4. The overlapping area is the solution to the system
  5. Verify by testing a point in the solution region
Inequalities
y ≤ x + 2, y > -x + 1
Boundaries
y = x + 2 (solid), y = -x + 1 (dashed)
Solution
Intersection region
Step 1: Graph the first inequality \(y \leq x + 2\)

Boundary line: \(y = x + 2\) (solid line since ≤)

Y-intercept: (0, 2), Slope: 1

Test point (0, 0): \(0 \leq 0 + 2\) → \(0 \leq 2\) ✓

Shade below the line (including the line)

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

Boundary line: \(y = -x + 1\) (dashed line since >)

Y-intercept: (0, 1), Slope: -1

Test point (0, 0): \(0 > -0 + 1\) → \(0 > 1\) ✗

Shade above the line (excluding the line)

Step 3: Find the intersection of the shaded regions

The solution region is where both shadings overlap

This is the area above the dashed line and below the solid line

Step 4: Identify the solution region

The solution region is bounded by both lines

It extends infinitely in the direction where both inequalities are satisfied

Step 5: Verify with a test point in the solution region

Test point (1, 2):

First inequality: \(2 \leq 1 + 2\) → \(2 \leq 3\) ✓

Second inequality: \(2 > -1 + 1\) → \(2 > 0\) ✓

Solution: Region above y = -x + 1 AND below y = x + 2
Final answer:

The solution region is the area that lies above the dashed line \(y = -x + 1\) and below the solid line \(y = x + 2\). A point that satisfies both inequalities is (1, 2).

Applied rules:

• System Solution: Intersection of individual solution sets

• Shading Overlap: Points in overlapping regions satisfy all inequalities

• Verification: Test points must satisfy ALL inequalities in the system

2 Three Inequalities
Exercise 2
Graph the system of inequalities:
\(x \geq 0\), \(y \geq 0\), and \(2x + 3y \leq 12\)
Find the vertices of the solution region.
Definition:

Bounded Solution Region: A solution region that is enclosed and finite in area

System
x ≥ 0, y ≥ 0, 2x + 3y ≤ 12
Constraints
Quadrant I, bounded by line
Vertices
(0,0), (0,4), (6,0)
Step 1: Graph the constraint \(x \geq 0\)

This represents the region to the right of and including the y-axis

Boundary line: \(x = 0\) (the y-axis, solid line)

Step 2: Graph the constraint \(y \geq 0\)

This represents the region above and including the x-axis

Boundary line: \(y = 0\) (the x-axis, solid line)

Step 3: Graph the constraint \(2x + 3y \leq 12\)

Boundary line: \(2x + 3y = 12\) (solid line since ≤)

X-intercept: (6, 0), Y-intercept: (0, 4)

Test point (0, 0): \(2(0) + 3(0) = 0 \leq 12\) ✓

Shade below the line (including the line)

Step 4: Find the intersection of all three regions

The solution is in the first quadrant (due to x ≥ 0 and y ≥ 0)

And below the line 2x + 3y = 12

This creates a triangular region

Step 5: Find the vertices of the triangular region

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

Vertex 2: Intersection of x = 0 and 2x + 3y = 12 → (0, 4)

Vertex 3: Intersection of y = 0 and 2x + 3y = 12 → (6, 0)

Step 6: Verify each vertex satisfies all inequalities

Point (0, 0): \(0 \geq 0\), \(0 \geq 0\), \(2(0) + 3(0) = 0 \leq 12\) ✓

Point (0, 4): \(0 \geq 0\), \(4 \geq 0\), \(2(0) + 3(4) = 12 \leq 12\) ✓

Point (6, 0): \(6 \geq 0\), \(0 \geq 0\), \(2(6) + 3(0) = 12 \leq 12\) ✓

Vertices: (0,0), (0,4), (6,0)
Final answer:

The solution region is a triangle in the first quadrant with vertices at (0, 0), (0, 4), and (6, 0).

Applied rules:

• Non-Negativity Constraints: \(x \geq 0\) and \(y \geq 0\) restrict to Quadrant I

• Vertices of Polygon: Points where boundary lines intersect

• Bounded Region: Finite area solution region

3 Unbounded Region
Exercise 3
Graph the system: \(y \geq x - 1\) and \(y \leq 2x + 3\). Describe the shape and extent of the solution region.
Definition:

Unbounded Solution Region: A solution region that extends infinitely in at least one direction

System
y ≥ x - 1, y ≤ 2x + 3
Boundaries
y = x - 1 (solid), y = 2x + 3 (solid)
Region
Between the lines
Step 1: Graph the first inequality \(y \geq x - 1\)

Boundary line: \(y = x - 1\) (solid line since ≥)

Y-intercept: (0, -1), Slope: 1

Test point (0, 0): \(0 \geq 0 - 1\) → \(0 \geq -1\) ✓

Shade above the line (including the line)

Step 2: Graph the second inequality \(y \leq 2x + 3\)

Boundary line: \(y = 2x + 3\) (solid line since ≤)

Y-intercept: (0, 3), Slope: 2

Test point (0, 0): \(0 \leq 2(0) + 3\) → \(0 \leq 3\) ✓

Shade below the line (including the line)

Step 3: Identify the intersection of the regions

The solution region is between the two lines

It includes all points above \(y = x - 1\) and below \(y = 2x + 3\)

Step 4: Find the intersection point of the boundary lines

Solve the system: \(y = x - 1\) and \(y = 2x + 3\)

\(x - 1 = 2x + 3\)

\(-x = 4\)

\(x = -4\)

\(y = -4 - 1 = -5\)

Intersection point: (-4, -5)

Step 5: Describe the solution region

The region extends infinitely upward

It's bounded below by \(y = x - 1\) and above by \(y = 2x + 3\)

As x increases, the region between the lines expands

Step 6: Verify with a test point

Test point (0, 0):

First inequality: \(0 \geq 0 - 1\) → \(0 \geq -1\) ✓

Second inequality: \(0 \leq 2(0) + 3\) → \(0 \leq 3\) ✓

Region: x - 1 ≤ y ≤ 2x + 3
Final answer:

The solution region is the area between the lines \(y = x - 1\) and \(y = 2x + 3\), bounded below by the first line and above by the second. The region extends infinitely upward and is unbounded.

Applied rules:

• Unbounded Regions: Extend infinitely in one or more directions

• Intersection Point: Where boundary lines meet, forming the vertex of the region

• Between Lines: Solution region lies between two intersecting lines

Systems of Linear Inequalities Rules and Methods
\(\begin{cases} a_1x + b_1y \, \square_1 \, c_1 \\ a_2x + b_2y \, \square_2 \, c_2 \end{cases}\)
System of Linear Inequalities
System
Multiple constraints
Intersection of regions
Solution
Common region
Points satisfying all
Shape
Polygon or unbounded
Convex region
Key definitions:

System of Linear Inequalities: A set of two or more linear inequalities with the same variables

Solution Region: The set of all points that satisfy all inequalities in the system simultaneously

Feasible Region: The solution region in optimization problems

Corner Points/Vertices: Points where boundary lines intersect in the solution region

Complete methodology:
  1. Graph Each Inequality: Draw boundary lines and shade appropriate regions
  2. Identify Overlap: Find the area where all shadings intersect
  3. Mark Boundary: Distinguish between solid and dashed boundaries
  4. Find Vertices: Locate intersection points of boundary lines
  5. Verify Solution: Test points in the solution region
  6. Describe Region: State whether bounded/unbounded and shape
Tip 1: Always use different shading patterns or colors to distinguish regions.
Tip 2: The solution region is always convex (no indentations).
Tip 3: Test a point in each region to confirm correct shading.
Tip 4: Non-negativity constraints (x ≥ 0, y ≥ 0) restrict to Quadrant I.
Common errors: Shading the wrong side of boundary lines, confusing solid/dashed lines, not finding intersection correctly, missing corner points.
Exam preparation: Practice with different numbers of inequalities, work on identifying bounded vs. unbounded regions, memorize vertex-finding methods.
Formulas to know by heart:

• System Solution: Intersection of all individual solution sets

• Boundary Lines: Replace inequality with equals sign

• Line Type: ≤ or ≥ → solid, < or > → dashed

• Vertex Formula: Solve system of equations formed by intersecting boundary lines

Solution: Exercises 4 to 5
4 Cost and Revenue Problem
Exercise 4
A company produces two products. Product A requires 2 hours of labor and Product B requires 1 hour of labor. The company has at most 40 hours of labor available per day. Product A requires 1 unit of material and Product B requires 3 units of material. The company has at most 30 units of material available. Write a system of inequalities and graph the feasible region.
Definition:

Feasible Region: The solution region in optimization problems representing all possible solutions

Variables
x=A units, y=B units
System
2x + y ≤ 40, x + 3y ≤ 30, x ≥ 0, y ≥ 0
Vertices
(0,0), (0,10), (18,4), (20,0)
Step 1: Define variables

Let \(x\) = number of units of Product A produced per day

Let \(y\) = number of units of Product B produced per day

Step 2: Identify constraints

Labor constraint: \(2x + y \leq 40\) (at most 40 hours)

Material constraint: \(x + 3y \leq 30\) (at most 30 units)

Non-negativity: \(x \geq 0\), \(y \geq 0\) (can't produce negative amounts)

Step 3: Graph each inequality

Inequality 1: \(2x + y \leq 40\)

Boundary: \(2x + y = 40\), X-int: (20, 0), Y-int: (0, 40)

Inequality 2: \(x + 3y \leq 30\)

Boundary: \(x + 3y = 30\), X-int: (30, 0), Y-int: (0, 10)

Inequality 3: \(x \geq 0\) (right of y-axis)

Inequality 4: \(y \geq 0\) (above x-axis)

Step 4: Find intersection points

Intersection of \(2x + y = 40\) and \(x + 3y = 30\):

From first equation: \(y = 40 - 2x\)

Substitute into second: \(x + 3(40 - 2x) = 30\)

\(x + 120 - 6x = 30\)

\(-5x = -90\)

\(x = 18\), \(y = 40 - 2(18) = 4\)

Intersection point: (18, 4)

Step 5: Identify all vertices of the feasible region

Vertex 1: (0, 0) - intersection of \(x = 0\) and \(y = 0\)

Vertex 2: (0, 10) - intersection of \(x = 0\) and \(x + 3y = 30\)

Vertex 3: (18, 4) - intersection of \(2x + y = 40\) and \(x + 3y = 30\)

Vertex 4: (20, 0) - intersection of \(y = 0\) and \(2x + y = 40\)

Step 6: Verify the solution region

Test point inside region, such as (10, 5):

Labor: \(2(10) + 5 = 25 \leq 40\) ✓

Material: \(10 + 3(5) = 25 \leq 30\) ✓

Non-negativity: \(10 \geq 0\), \(5 \geq 0\) ✓

Feasible region with vertices (0,0), (0,10), (18,4), (20,0)
Final answer:

The feasible region is a quadrilateral with vertices at (0, 0), (0, 10), (18, 4), and (20, 0). Any point in this region represents a possible production plan that satisfies all constraints.

Applied rules:

• Resource Constraints: "at most" translates to ≤

• Non-negativity: Physical quantities must be non-negative

• Feasible Region: Intersection of all constraint regions

5 Investment Problem
Exercise 5
An investor wants to invest at most $10,000 in two stocks. Stock X costs $50 per share and Stock Y costs $30 per share. The investor wants to buy at least 100 shares total but no more than 150 shares of Stock X. Write a system of inequalities and describe the feasible region.
Definition:

Investment Constraints: Limitations on financial resources and purchasing decisions

Variables
x=X shares, y=Y shares
System
50x + 30y ≤ 10000, x + y ≥ 100, x ≤ 150, x ≥ 0, y ≥ 0
Region
Bounded polygon
Step 1: Define variables

Let \(x\) = number of shares of Stock X

Let \(y\) = number of shares of Stock Y

Step 2: Identify all constraints

Budget constraint: \(50x + 30y \leq 10000\) (at most $10,000)

Total shares: \(x + y \geq 100\) (at least 100 shares)

Stock X limit: \(x \leq 150\) (no more than 150 shares of X)

Non-negativity: \(x \geq 0\), \(y \geq 0\)

Step 3: Simplify constraints if possible

Budget constraint: \(5x + 3y \leq 1000\) (divide by 10)

Total shares: \(x + y \geq 100\)

Stock X: \(x \leq 150\)

Non-negativity: \(x \geq 0\), \(y \geq 0\)

Step 4: Find boundary intersections

Key intersections involve all five constraints

Vertices occur where constraint boundaries intersect

Step 5: Describe the feasible region

The region is bounded by five constraints

It's a pentagon-shaped region in the first quadrant

Any point in this region represents a valid investment strategy

Step 6: Example verification

Test point (100, 50):

Budget: \(50(100) + 30(50) = 5000 + 1500 = 6500 \leq 10000\) ✓

Total: \(100 + 50 = 150 \geq 100\) ✓

Stock X: \(100 \leq 150\) ✓

System: 5x + 3y ≤ 1000, x + y ≥ 100, x ≤ 150, x ≥ 0, y ≥ 0
Final answer:

The system of inequalities is: \(5x + 3y \leq 1000\), \(x + y \geq 100\), \(x \leq 150\), \(x \geq 0\), \(y \geq 0\). The feasible region is a bounded polygon representing all valid investment combinations.

Applied rules:

• Investment Modeling: Translate financial constraints into inequalities

• Multiple Constraints: Each limitation adds a boundary to the region

• Feasibility: All constraints must be satisfied simultaneously

Systems of Linear Inequalities Fundamentals
\(\begin{cases} a_1x + b_1y \leq c_1 \\ a_2x + b_2y \geq c_2 \end{cases}\)
System of Linear Inequalities
Key definitions:

System of Linear Inequalities: A collection of linear inequalities that must all be satisfied simultaneously

Solution Region: The area of the coordinate plane containing all points that satisfy every inequality in the system

Feasible Region: The solution region in optimization contexts

Corner Points: The vertices of the solution region where boundary lines intersect

Complete methodology:
  1. Read Problem Carefully: Identify variables and constraints
  2. Translate to Inequalities: Convert verbal constraints to mathematical form
  3. Graph Each Inequality: Draw boundary lines and shade solution regions
  4. Find Intersection: Identify the overlapping solution region
  5. Locate Vertices: Find intersection points of boundary lines
  6. Verify Solution: Check that the region satisfies all constraints
  7. Interpret Results: Connect mathematical solution to real-world context
Tip 1: Look for keywords like "at most", "at least", "no more than", "no less than" to identify inequality types.
Tip 2: Always include non-negativity constraints (x ≥ 0, y ≥ 0) for physical quantities.
Tip 3: The solution region is always convex (no inward curves).
Tip 4: Corner points often represent critical solutions in optimization problems.
Applications: Business (production planning), economics (budget constraints), engineering (design specifications), agriculture (crop planning), manufacturing (resource allocation).
Properties: Solution region is always convex; bounded regions have finite area; unbounded regions extend infinitely in some direction.
Essential formulas:

• System Representation: \(\begin{cases} a_1x + b_1y \, \square_1 \, c_1 \\ a_2x + b_2y \, \square_2 \, c_2 \end{cases}\)

• Boundary Lines: Replace inequality with equals sign

• Line Type: ≤ or ≥ → solid line, < or > → dashed line

• Intersection Points: Solve system of boundary equations

Systems of Inequalities Visualization
Exercise 6: Different System Types
Compare these systems:
Bounded: x ≥ 0, y ≥ 0, x + y ≤ 5
Unbounded: y ≥ x, y ≤ 2x + 1
Empty: x + y ≤ 2, x + y ≥ 5

Analysis: The chart shows how different systems create different solution regions.

  • Bounded systems: Finite solution region (triangle, rectangle, etc.)
  • Unbounded systems: Solution region extends infinitely
  • Empty systems: No intersection of all constraints

Questions & Answers

Question: How do I know if a system of inequalities has no solution?

Answer: A system of inequalities has no solution when the solution regions of the individual inequalities do not overlap:

  • Graphically: The shaded regions from each inequality do not intersect
  • Algebraically: You'll reach a contradiction when trying to satisfy all constraints
  • Common scenario: Opposing constraints like \(x > 5\) and \(x < 3\)

For example, the system \(x + y > 5\) and \(x + y < 2\) has no solution because no point can simultaneously satisfy both inequalities.

In word problems, no solution often indicates contradictory constraints or impossible scenarios.

Question: What's the difference between a bounded and unbounded solution region?

Answer: The difference relates to the extent of the solution region:

Bounded Region:

  • Finite area that can be enclosed in a circle
  • Has vertices (corner points) at intersections of boundary lines
  • Often occurs with constraints like x ≥ 0, y ≥ 0, and upper limits
  • Example: Triangle formed by x ≥ 0, y ≥ 0, x + y ≤ 5

Unbounded Region:

  • Extends infinitely in at least one direction
  • Cannot be enclosed in a finite circle
  • Often occurs with only lower limits or directional constraints
  • Example: Region above y = x and below y = 2x + 3

In optimization problems, bounded regions guarantee that optimal solutions exist at corner points.

Question: How do I handle word problems with more than two constraints?

Answer: For systems with more than two constraints, follow the same process but graph each constraint:

  1. Define variables for unknown quantities
  2. Write an inequality for each constraint
  3. Graph each inequality on the same coordinate plane
  4. Identify the region where ALL shadings overlap
  5. Find vertices at intersections of boundary lines
  6. Verify the solution region satisfies all constraints

For example, with three constraints, you'll have three boundary lines and the solution region is where all three shaded areas overlap. With more constraints, the solution region becomes more restricted, possibly creating a polygon with more sides.

The key is to be systematic: graph each constraint carefully and verify that your final region satisfies all original inequalities.

Question: How can I check if my solution region is correct?

Answer: Use these verification methods:

  • Test Interior Point: Choose a point clearly inside the solution region and verify it satisfies all inequalities
  • Test Exterior Point: Choose a point outside the region and verify at least one inequality is violated
  • Test Boundary Points: Points on solid boundaries should satisfy the corresponding equality
  • Visual Check: Ensure all boundary lines are correctly drawn (solid/dashed) and shaded appropriately

For example, if your solution region is defined by \(y \leq x + 2\) and \(y \geq -x + 1\), test a point like (1, 2):

  • \(2 \leq 1 + 2 = 3\) ✓
  • \(2 \geq -1 + 1 = 0\) ✓

This confirms that (1, 2) is in the solution region. Testing multiple points increases confidence in your solution.