5.4Special Cases: No Solution & Unbounded Regions
Not every system of inequalities has a neat, enclosed feasible region. Learn to recognize when a system has no solution at all, when the feasible region extends infinitely, and what each case means in real-world contexts.
Why This Matters
Recognizing when a system has no solution or an unbounded region is critical in optimization — it tells you when a problem has no feasible answer or infinite possibilities. This concept appears in linear programming and Precalculus.
Workbook
Lesson, vocabulary, worked examples, and practice problems.
Essential Question
How can you determine — from a graph or from the inequalities themselves — whether a system of inequalities has no solution, an unbounded solution, or a bounded solution, and what does each case mean in context?
Lesson Overview
In Chapters 1–3 every system you graphed had a visible, overlapping feasible region. But real-world constraints don't always cooperate. Sometimes two constraints directly contradict each other — no point can satisfy both at once, and the feasible region is empty (no solution). Other times the constraints only limit values in one direction, leaving the feasible region open and extending to infinity — an unbounded region. Recognizing these special cases is essential for interpreting mathematical models correctly.
Case 1: No Solution
Shaded regions don't overlap
Case 2: Unbounded
Overlap extends infinitely upward
Case 3: Bounded
Overlap is enclosed — finite region
Bounded vs. Unbounded vs. No Solution — Comparison
| Property | No Solution | Unbounded | Bounded |
|---|---|---|---|
| Feasible region exists? | No | Yes | Yes |
| Region extends to infinity? | N/A | Yes | No |
| Number of solutions | Zero | Infinitely many | Infinitely many |
| Enclosed by boundary lines? | N/A | No | Yes |
| Has corner vertices? | N/A | Some | Yes (finite set) |
| Real-world meaning | Constraints are contradictory | Some values are unlimited | All values are restricted |
| Example context | Impossible budget | No upper spending limit | Fixed budget & capacity |
Classification Flowchart
Graph the system of inequalities
Draw all boundary lines; shade each half-plane.
Is there any overlapping shaded region?
NO overlap
Shadings don't intersect
NO SOLUTION
Empty feasible region
YES overlap
Shadings intersect
Does the region extend infinitely?
YES → UNBOUNDED
NO → BOUNDED
Why Parallel Lines Cause No Solution
Parallel Boundary Lines → No Solution
Line 1: y > x + 3 (shade above)
Line 2: y < x − 1 (shade below)
Both lines have slope = 1. They are parallel and never intersect.
The region above y = x+3 and the region below y = x−1 are on opposite sides of two parallel lines — they can never overlap.
Step 1: Shade each separately
Red: above y=x+3 | Blue: below y=x−1
Step 2: Look for overlap
No overlap → No solution
Error Analysis
Error Analysis
Student's Claim ✗
'The system y > x+2 and y > x−1 has no solution because the lines are parallel.'
The Error
Parallel lines with shading in the SAME direction (both above) DO overlap — the region above the higher line satisfies both inequalities.
Correct Reasoning ✓
Both inequalities shade above their respective lines. The overlap is the region above y = x+2 (the higher line). The system has an UNBOUNDED solution.
Error Analysis
Student's Claim ✗
'An unbounded feasible region means the system has no solution.'
The Error
Unbounded means the feasible region extends infinitely — it still contains infinitely many solutions. No solution means the feasible region is completely empty.
Correct Reasoning ✓
Unbounded → infinitely many solutions with no upper limit. No solution → zero solutions (empty feasible region).
Worked Examples
No solution: y ≥ 2x + 4 and y ≤ 2x − 1 — parallel, opposite shading
Classify the system: y ≥ 2x + 4 and y ≤ 2x − 1. Graph and explain.
Both lines have slope 2 — they are parallel.
Line 1: y = 2x+4 (solid). Shade above (y ≥ 2x+4).
Line 2: y = 2x−1 (solid). Shade below (y ≤ 2x−1).
The region above y=2x+4 is always higher than the region below y=2x−1 (since 2x+4 > 2x−1 for all x).
The two shaded regions never overlap.
Classification: NO SOLUTION (empty feasible region).
Ex 1: y ≥ 2x+4 and y ≤ 2x−1
Parallel lines, opposite shading → No solution
Parallel Boundary Lines → No Solution
Line 1: y = 2x + 4 (slope = 2)
Line 2: y = 2x − 1 (slope = 2)
Same slope → parallel. Gap = 5 units.
Above y=2x+4 is always above below y=2x−1. No overlap possible.
Classify the system: y ≥ −x + 1 and y ≥ x − 3. Graph and explain.
Line 1: y = −x+1 (solid). Shade above.
Line 2: y = x−3 (solid). Shade above.
Find intersection: −x+1 = x−3 → 2x = 4 → x = 2, y = −1. Intersection: (2, −1).
Both shadings are above their respective lines. The overlap is the region above BOTH lines — a wedge opening upward.
The wedge extends infinitely upward.
Classification: UNBOUNDED feasible region.
Ex 2: y ≥ −x+1 and y ≥ x−3 (Unbounded)
Wedge opens upward — unbounded
Classify the system: y ≤ −x + 5, y ≥ x − 1, y ≥ 0. Graph and explain.
Line 1: y = −x+5 (solid). Shade below.
Line 2: y = x−1 (solid). Shade above.
Line 3: y = 0 (x-axis, solid). Shade above (y ≥ 0).
Find intersections: L1∩L2: −x+5=x−1 → x=3, y=2 → (3,2). L1∩L3: y=0 → x=5 → (5,0). L2∩L3: y=0 → x=1 → (1,0).
The feasible region is the triangle with vertices (1,0), (5,0), (3,2).
Classification: BOUNDED feasible region (triangle).
Ex 3: Bounded Triangle
Triangle with 3 vertices — bounded
Classify the system: y ≤ x + 4 and y ≤ x + 1. Graph and explain.
Both lines have slope 1 — they are parallel.
Line 1: y = x+4 (solid). Shade below.
Line 2: y = x+1 (solid). Shade below.
Both shadings are below their respective lines.
The region below y=x+1 is entirely contained within the region below y=x+4.
The overlap is the region below y=x+1 — which extends infinitely downward and to the right.
Classification: UNBOUNDED feasible region.
Ex 4: Parallel, same shading → Unbounded
Same slope, same shading direction → overlap is below y=x+1 (unbounded)
Budget: A company requires that monthly costs (c) are at most $5,000 AND at least $8,000. Write and classify the system.
Constraint 1: c ≤ 5000.
Constraint 2: c ≥ 8000.
These constraints say c must be both ≤ 5000 AND ≥ 8000 simultaneously.
No value of c can be both ≤ 5000 and ≥ 8000 at the same time.
Classification: NO SOLUTION.
Real-world meaning: The constraints are contradictory — it is impossible to have costs that are simultaneously at most $5,000 and at least $8,000.
Real-World Interpretation
Contradictory budget constraints
System
c ≤ 5,000 (costs at most $5,000)
c ≥ 8,000 (costs at least $8,000)
What This Means
No cost value can be simultaneously at most $5,000 and at least $8,000. The constraints are contradictory — the company's requirements are impossible to meet as stated. At least one constraint must be relaxed.
Guided Practice
Guided Practice Video: Special Cases: No Solution and Unbounded
Watch the guided practice walkthrough for special cases — no solution and unbounded systems — then complete the problems below.
Video by Sang Real Math
Watch on YouTube ↗Classify the system: y > 2x + 3 and y < 2x − 2. Graph and explain your reasoning.
Hint: Both lines have slope 2 (parallel). Shading: above y=2x+3 and below y=2x−2. Since 2x+3 > 2x−2 for all x, the 'above' region is always higher than the 'below' region — no overlap. Classification: NO SOLUTION.
Classify the system: y ≥ −2x + 1 and y ≥ 2x − 3. Graph and explain.
Hint: Find intersection: −2x+1 = 2x−3 → 4x=4 → x=1, y=−1. Both shade above → wedge opens upward from (1,−1). Classification: UNBOUNDED.
Classify the system: y ≤ 4, y ≥ −1, x ≤ 3, x ≥ 0. Graph and identify all vertices.
Hint: Four boundary lines form a rectangle. Vertices: (0,−1), (3,−1), (3,4), (0,4). The region is completely enclosed. Classification: BOUNDED (rectangle).
Real world: A store requires at least $500 in daily sales (s ≥ 500) and at most $400 in daily sales (s ≤ 400). Classify and interpret.
Hint: s ≥ 500 and s ≤ 400 are contradictory — no value of s can be both ≥ 500 and ≤ 400. Classification: NO SOLUTION. Real-world meaning: the requirements are impossible to meet simultaneously.
Add one constraint to make the system y ≥ x − 2 and y ≥ −x + 1 bounded. Explain your choice.
Hint: The system is currently unbounded (wedge opening upward). Adding y ≤ 4 (or any horizontal upper bound) encloses the region. New system: y ≥ x−2, y ≥ −x+1, y ≤ 4. Vertices: (−3,4), (5,4), (1.5,−0.5).
Key Vocabulary
No Solution (Empty Feasible Region)
A system of inequalities in which no ordered pair satisfies all inequalities simultaneously. The shaded regions do not overlap.
Unbounded Feasible Region
A feasible region that extends infinitely in at least one direction. The system has infinitely many solutions with no upper or lower limit in that direction.
Bounded Feasible Region
A feasible region that is completely enclosed by boundary lines. All solution values are restricted to a finite area.
Parallel Lines
Lines with the same slope but different y-intercepts. Parallel boundary lines with opposite shading directions produce no solution.
Contradictory Constraints
Two or more constraints that cannot be satisfied simultaneously, resulting in no feasible region.
Non-negativity Constraint
The implied constraints x ≥ 0 and y ≥ 0 that restrict the feasible region to the first quadrant. Adding these can convert an unbounded region to a bounded one.
Corner Point (Vertex)
A point where two boundary lines intersect at the edge of the feasible region. Bounded regions have a finite set of corner points.
Feasible Region Classification
The process of determining whether a system's solution set is empty (no solution), unbounded, or bounded based on the graph.
Interactive Practice — 5 Questions
A system has two parallel boundary lines with shading in opposite directions. What is the classification?
A system has two parallel boundary lines with shading in the same direction. What is the classification?
Which system has NO solution?
What is the minimum number of non-parallel boundary lines needed to create a BOUNDED feasible region?
Which statement about an unbounded feasible region is TRUE?
Independent Practice
Independent Practice
Classify each system without graphing. Explain your reasoning. (a) y ≥ x+4 and y ≤ x−1 (b) y ≤ −x+6 and y ≤ x+2 (c) y ≥ 2, y ≤ 5, x ≥ 0, x ≤ 4 (d) y > 3x+1 and y < 3x−4
Graph and classify: y ≥ −x + 2 and y ≥ x − 4. Find the vertex of the feasible region and describe the region.
Graph and classify: y ≤ x + 3, y ≥ −x − 1, y ≥ 0. Find all vertices of the feasible region.
Real world: A factory must produce at least 50 units of Product A (x ≥ 50) and at least 30 units of Product B (y ≥ 30). Classify the system and describe what the feasible region represents.
The system y ≥ x − 3 and y ≥ −2x + 1 is unbounded. (a) Find the vertex. (b) Add one constraint to make it bounded. (c) Find all vertices of the new bounded region.
Common Mistakes
Assuming parallel boundary lines always mean no solution — forgetting to check the shading direction.
Parallel lines with shading toward each other (opposite directions) give no solution. Parallel lines with shading in the SAME direction give an unbounded region.
Confusing 'unbounded' with 'no solution' — saying a region that extends infinitely has no solutions.
Unbounded means infinitely many solutions — the region exists but has no upper/lower limit. No solution means the shaded regions don't overlap at all.
Forgetting to check ALL inequalities when testing a point — accepting a point that satisfies only 2 out of 3 inequalities.
A point must satisfy every inequality in the system simultaneously to be in the feasible region.
Concluding 'no solution' from a small graph window — the region may extend beyond what's visible.
Always check algebraically if unsure. Parallel lines with opposite shading → no solution. Otherwise, look for overlap beyond the visible window.
Math Tips
Check slopes first: if two boundary lines have the same slope (parallel), check whether the shading directions are opposite — if so, the system has no solution.
Unbounded ≠ no solution: an unbounded region still has infinitely many solutions — it just has no upper (or lower) limit in some direction.
Non-negativity constraints (x ≥ 0, y ≥ 0) restrict the feasible region to the first quadrant — adding them can convert an unbounded system into a bounded one.
Test a point in the claimed feasible region: if no test point satisfies all inequalities, the region is empty (no solution).
A bounded region requires at least 3 non-parallel boundary lines to enclose an area — count your constraints.