Systems of Linear Inequalities

[MathDefinition title="System of Linear Inequalities"] A system of linear inequalities consists of two or more linear inequalities in the same variables. The solution is the set of all points that satisfy all inequalities simultaneously. [/MathDefinition]

[MathDefinition title="Feasible Region"] The feasible region is the intersection of the solution regions of all inequalities in the system. It represents all possible solutions that satisfy every inequality. [/MathDefinition]

[MathKeyFormula title="Key Concepts"]

1. Each inequality represents a half-plane
2. The solution to a system is the intersection of all half-planes
3. The feasible region can be:
   • Bounded (finite area)
   • Unbounded (infinite area)
   • Empty (no solution)
4. Corner points (vertices) are important for optimization [/MathKeyFormula]

[MathNote title="Solving Strategy"]

1. Graph each inequality separately
2. Find the intersection of all regions
3. Identify the feasible region
4. Find corner points if needed
5. Check if the region is bounded or unbounded [/MathNote]

[MathExample title="Example 1: Two Inequalities"] Problem: Solve the system: x + y ≤ 5, x − y ≥ 1

Solution: Step 1: Graph x + y ≤ 5 Boundary: x + y = 5 Intercepts: (5, 0) and (0, 5) Solid line, shade below

Step 2: Graph x − y ≥ 1 Boundary: x − y = 1 Intercepts: (1, 0) and (0, −1) Solid line, shade above (test (0,0): 0 ≥ 1? No, shade opposite side)

Step 3: Find intersection The feasible region is the area that satisfies both inequalities.

Step 4: Find vertices Solving x + y = 5 and x − y = 1: Adding: 2x = 6, x = 3 y = 5 − 3 = 2 Vertex at (3, 2)

The feasible region is bounded by the lines x + y = 5 and x − y = 1. [/MathExample]

[MathExample title="Example 2: Three Inequalities"] Problem: Solve the system: x + y ≤ 4, x ≥ 0, y ≥ 0

Solution: Step 1: Graph x + y ≤ 4 Boundary: x + y = 4 Intercepts: (4, 0) and (0, 4) Solid line, shade below

Step 2: Graph x ≥ 0 Region to the right of y-axis (including y-axis)

Step 3: Graph y ≥ 0 Region above x-axis (including x-axis)

Step 4: Find intersection The feasible region is a triangle with vertices at:

• (0, 0) - intersection of x = 0 and y = 0
• (4, 0) - intersection of x + y = 4 and y = 0
• (0, 4) - intersection of x + y = 4 and x = 0

The feasible region is a bounded triangular region. [/MathExample]

[MathExample title="Example 3: Unbounded Region"] Problem: Solve the system: x + y ≥ 2, x ≥ 0, y ≥ 0

Solution: Step 1: Graph x + y ≥ 2 Boundary: x + y = 2 Intercepts: (2, 0) and (0, 2) Solid line, shade above

Step 2: Graph x ≥ 0 Region to the right of y-axis

Step 3: Graph y ≥ 0 Region above x-axis

Step 4: Find intersection The feasible region is unbounded, extending to infinity in the first quadrant above the line x + y = 2.

Vertices:

• (2, 0) - intersection of x + y = 2 and y = 0
• (0, 2) - intersection of x + y = 2 and x = 0

The region is unbounded with two vertices. [/MathExample]

[MathExample title="Example 4: No Solution"] Problem: Solve the system: x + y ≤ 1, x + y ≥ 3

Solution: Step 1: Graph x + y ≤ 1 Boundary: x + y = 1 Intercepts: (1, 0) and (0, 1) Solid line, shade below

Step 2: Graph x + y ≥ 3 Boundary: x + y = 3 Intercepts: (3, 0) and (0, 3) Solid line, shade above

Step 3: Find intersection The line x + y = 1 is below x + y = 3. The region x + y ≤ 1 is below the line x + y = 1. The region x + y ≥ 3 is above the line x + y = 3.

These two regions do not overlap.

Therefore, the system has no solution (empty feasible region). [/MathExample]