5.2Graphing Systems of Inequalities
Graph two or more linear inequalities on the same coordinate plane and identify the overlapping feasible region — the set of all points that satisfy every inequality simultaneously.
Why This Matters
Graphing inequality systems makes the solution region visible — a powerful tool for understanding constraints. This skill directly connects to linear programming in business and optimization problems in Precalculus and AP Calculus.
Workbook
Lesson, vocabulary, worked examples, and practice problems.
Essential Question
How do you graph a system of linear inequalities and identify the region that satisfies all constraints simultaneously?
Lesson Overview
In Chapter 1 you learned what a system of inequalities is and how to verify solutions. In this chapter you will graph systems of inequalities step by step. Each inequality divides the coordinate plane into two half-planes. You graph the boundary line for each inequality, shade the correct half-plane, and then identify the feasible region — the area where all shadings overlap. Every point in the feasible region is a solution to the system.
y ≤ x − 1: solid boundary, shade below
y > −x + 3: dashed boundary, shade above
Feasible region: overlap of y ≤ x − 1 and y > −x + 3
Steps to Graph a System of Inequalities
- Graph the boundary line for Inequality 1. Use a solid line for ≤ or ≥; use a dashed line for < or >.
- Choose a test point (usually (0, 0) unless the line passes through the origin — then use (1, 0) or (0, 1)).
- Substitute the test point into Inequality 1. If it satisfies the inequality, shade the side containing the test point. Otherwise, shade the opposite side.
- Repeat Steps 1–3 for every remaining inequality.
- Identify the feasible region: the area where ALL shaded regions overlap is the solution set.
- Verify: pick a point in the feasible region and confirm it satisfies every inequality.
Step-by-Step Visual: y ≤ x + 2 and y > −x + 1
Step 1 — Graph y ≤ x + 2 (solid, shade below)
Step 2 — Graph y > −x + 1 (dashed, shade above)
Step 3 — Feasible region = overlap of both shadings
Solid vs. Dashed Boundary Lines
Solid — ≤ or ≥
Dashed — < or >
Solid vs. dashed boundary line comparison
Shading Direction: Above vs. Below
y > x → shade above
y ≤ x → shade below
Shading direction: above vs. below the boundary line
The Test Point Method
Test point (0,0) determines which side to shade
Test Point Rule
- Use (0, 0) when the boundary line does NOT pass through the origin.
- If the line passes through (0, 0), use (1, 0) or (0, 1).
- Test point satisfies inequality → shade that side.
- Test point fails → shade the opposite side.
Error Analysis
Error Analysis — Wrong Shading Side
Shaded below — INCORRECT
Shaded above — CORRECT
y > x + 1 means y is greater than x + 1 → shade the region ABOVE the line
Error Analysis — Solid vs. Dashed Line
Solid line — INCORRECT for <
Dashed line — CORRECT for <
Strict inequalities (< and >) always use a dashed boundary line
Before and After: Building the Feasible Region
Before — each inequality separately
y ≤ x+2
y > −x+1
After — combined feasible region
Overlap = solution
Before-and-after: individual shadings → combined feasible region
Solid (≤ or ≥) vs. dashed (< or >) boundary lines
Worked Examples
Graph the system: y ≤ 2x − 1 and y > −x + 2.
Inequality 1: y ≤ 2x − 1. Draw a SOLID line (≤ includes the boundary). Shade BELOW.
Test (0,0): 0 ≤ 2(0)−1 = −1? No → shade the side away from the origin (below the line).
Inequality 2: y > −x + 2. Draw a DASHED line (> is strict). Shade ABOVE.
Test (0,0): 0 > −0+2 = 2? No → shade the opposite side (above the line).
Feasible region: below the solid line AND above the dashed line. Verify (3,1): 1≤2(3)−1=5 ✓; 1>−3+2=−1 ✓.
y ≤ 2x − 1 and y > −x + 2
Graph the system: y ≥ x + 1 and y ≤ −x + 3. Identify the feasible region and one solution.
Inequality 1: y ≥ x + 1. SOLID line (≥). Shade ABOVE. Test (0,0): 0 ≥ 0+1? No → shade above.
Inequality 2: y ≤ −x + 3. SOLID line (≤). Shade BELOW. Test (0,0): 0 ≤ 0+3? Yes → shade below (origin side).
Find intersection: x+1 = −x+3 → 2x=2 → x=1, y=2. Vertex: (1, 2).
Feasible region: above y=x+1 AND below y=−x+3 — a bounded wedge with vertex at (1,2).
Verify (0,2): 2≥0+1=1 ✓; 2≤−0+3=3 ✓.
y ≥ x + 1 and y ≤ −x + 3 — bounded region
Graph the system: y > x + 3 and y < x − 1. Classify the system.
Inequality 1: y > x + 3. DASHED line. Shade ABOVE.
Inequality 2: y < x − 1. DASHED line. Shade BELOW.
Both lines have slope m = 1 — they are PARALLEL.
Shading for Ineq 1 is above y=x+3; shading for Ineq 2 is below y=x−1. These regions never overlap.
Parallel lines — no solution (empty feasible region)
Graph the system: x ≥ 0, y ≥ 0, and x + y ≤ 4. Describe the feasible region.
x ≥ 0: SOLID vertical line x = 0 (y-axis). Shade to the RIGHT.
y ≥ 0: SOLID horizontal line y = 0 (x-axis). Shade ABOVE.
x + y ≤ 4: rewrite as y ≤ −x + 4. SOLID line. Test (0,0): 0+0=0 ≤ 4 ✓ → shade origin side.
Feasible region: triangle with vertices (0,0), (4,0), and (0,4).
Bounded region — x ≥ 0, y ≥ 0, x + y ≤ 4
Graph the system: y ≥ x − 1 and y ≤ −x + 3. Find the intersection point of the boundary lines.
Inequality 1: y ≥ x − 1. SOLID line. Shade ABOVE.
Inequality 2: y ≤ −x + 3. SOLID line. Shade BELOW.
Find intersection: x−1 = −x+3 → 2x=4 → x=2, y=1. Intersection: (2,1).
Feasible region: above y=x−1 AND below y=−x+3. Boundary lines are perpendicular (slopes 1 and −1).
Verify (1,2): 2≥1−1=0 ✓; 2≤−1+3=2 ✓.
Perpendicular lines — bounded feasible region
Guided Practice
Guided Practice Video: Graphing Systems of Inequalities
Watch the guided practice walkthrough for graphing systems of inequalities, then complete the problems below.
Video by Sang Real Math
Watch on YouTube ↗Graph the system: y ≤ x + 3 and y > −2x + 1. Identify the boundary line type (solid/dashed) and shading direction for each inequality.
Hint: y ≤ x+3: solid line, shade below. y > −2x+1: dashed line, shade above. The feasible region is where both shadings overlap.
Graph the system: y ≥ 2x − 4 and y < x + 1. Find the intersection point of the two boundary lines.
Hint: Set 2x−4 = x+1 → x = 5, y = 6. The boundary lines intersect at (5, 6). Feasible region: above the solid line and below the dashed line.
Is (2, 5) a solution to the system y ≥ x + 2 and y ≤ −x + 8? Show your work.
Hint: Check Ineq 1: 5 ≥ 2+2=4 ✓. Check Ineq 2: 5 ≤ −2+8=6 ✓. Both satisfied — yes, (2,5) is a solution.
Graph the system: x ≥ 0, y ≥ 0, y ≤ 3. Describe the shape of the feasible region.
Hint: x ≥ 0: right of y-axis. y ≥ 0: above x-axis. y ≤ 3: below the horizontal line y = 3. The feasible region is an infinite strip in the first quadrant bounded above by y = 3.
Graph the system: y > 2x + 1 and y > −x − 1. Is the feasible region bounded or unbounded?
Hint: Both inequalities shade above their respective lines. The feasible region is the area above both lines — it extends to infinity upward, so it is unbounded.
Key Vocabulary
Boundary Line
The line that forms the edge of a half-plane. Solid (≤ or ≥) if the line is part of the solution; dashed (< or >) if not.
Half-Plane
The region on one side of a boundary line. Each inequality divides the plane into two half-planes.
Feasible Region
The overlapping shaded area where all inequalities in the system are satisfied simultaneously.
Test Point
A point substituted into an inequality to determine which half-plane to shade. The origin (0, 0) is the most common choice.
Bounded Region
A feasible region that is enclosed — it does not extend to infinity in any direction.
Unbounded Region
A feasible region that extends infinitely in at least one direction.
Empty Solution Set
When no point satisfies all inequalities simultaneously — the shaded regions do not overlap.
Constraint
A condition expressed as an inequality that limits the possible values of the variables.
Interactive Practice — 5 Questions
For the inequality y ≤ 3x − 2, which boundary line type and shading direction are correct?
Which ordered pair is a solution to the system y ≥ x + 1 and y ≤ −x + 5?
The feasible region of a system of inequalities is:
Two parallel boundary lines have shading in opposite directions. The system has:
A student graphs y > 2x − 1 with a solid line. What error did the student make?
Independent Practice
Independent Practice
Graph the system: y ≤ x + 2 and y ≥ −x − 1. Identify the feasible region and verify with a test point.
Graph the system: y > 2x − 3 and y < −x + 4. Is the feasible region bounded or unbounded?
Is (3, 2) a solution to the system y ≤ x + 1 and y ≥ −x + 3? Verify algebraically.
Graph the system: x ≥ 0, y ≥ 0, and y ≤ −x + 5. Name all vertices of the feasible region.
Graph the system: y > x + 4 and y < x − 2. Does this system have a solution? Explain.
Common Mistakes
Using a solid line for a strict inequality (< or >) when graphing the boundary.
Dashed line for < or >. Solid line for ≤ or ≥. The boundary line type signals whether it is included.
Shading each inequality independently without finding the overlap region.
Graph both boundaries, shade each inequality lightly, then identify and darken the overlap.
Testing a point on the boundary line to determine which side to shade.
Test a point NOT on the boundary. The origin (0, 0) is usually the easiest choice unless it is on the line.
Forgetting that the feasible region can be empty if the shadings do not overlap.
If no region satisfies all inequalities simultaneously, the system has no solution — this is valid, not an error.
Math Tips
Solid vs. dashed: ≤ and ≥ use a solid boundary line; < and > use a dashed line — the line is NOT part of the solution.
Shading shortcut: for y > mx + b or y ≥ mx + b shade above; for y < mx + b or y ≤ mx + b shade below.
Test point: use (0, 0) unless the boundary line passes through the origin — then use (1, 0) or (0, 1).
The feasible region is the INTERSECTION (overlap) of all shaded regions — not the union.
No overlap between shaded regions means the system has no solution — an empty feasible region.